﻿ Theory - Practical Electronics for Inventors, Fourth Edition - Paul Scherz, Simon Monk ﻿

## Practical Electronics for Inventors, Fourth Edition - Paul Scherz, Simon Monk (2016)

### Chapter 2. Theory

2.1 Theory of Electronics

This chapter covers the basic concepts of electronics, such as current, voltage, resistance, electrical power, capacitance, and inductance. After going through these concepts, this chapter illustrates how to mathematically model currents and voltage through and across basic electrical elements such as resistors, capacitors, and inductors. By using some fundamental laws and theorems, such as Ohm’s law, Kirchhoff’s laws, and Thevenin’s theorem, the chapter presents methods for analyzing complex networks containing resistors, capacitors, and inductors that are driven by a power source. The kinds of power sources used to drive these networks, as we will see, include direct current (dc) sources, alternating current (ac) sources (including sinusoidal and nonsinusoidal periodic sources), and nonsinusoidal nonperiodic sources. We will also discuss transient circuits, where sudden changes in state (such as flipping a switch within a circuit) are encountered. At the end of the chapter, the approach needed to analyze circuits that contain nonlinear elements (diodes, transistors, integrated circuits, etc.) is discussed.

We recommend using a circuit simulator program if you’re just starting out in electronics. The web-based simulator CircuitLab (www.circuitlab.com) is extremely easy to use and has a nice graphical interface. There are also online calculators that can help you with many of the calculations in this chapter. Using a simulator program as you go through this chapter will help crystallize your knowledge, while providing an intuitive understanding of circuit behavior. Be careful—simulators can lie, or at least they can appear to lie when you don’t understand all the necessary parameters the simulator needs to make a realistic simulation. It is always important to get your hands dirty—get out the breadboards, wires, resistors, power supplies, and so on, and construct. It is during this stage that you gain the greatest practical knowledge that is necessary for an inventor.

It is important to realize that components mentioned in this chapter are only “theoretically” explained. For example, in regard to capacitors, you’ll learn how a capacitor works, what characteristic equations are used to describe a capacitor under certain conditions, and various other basic tricks related to predicting basic behavior. To get important practical insight into capacitors, however, such as real-life capacitor applications (filtering, snubbing, oscillator design, etc.), what type of real capacitors exist, how these real capacitors differ in terms of nonideal characteristics, which capacitors work best for a particular application, and, more important, how to read a capacitor label, requires that you jump to Chap. 3Sec. 3.6, which is dedicated to these issues. This applies to other components mentioned in this theory portion of the book.

The theoretical and practical information regarding transformers and nonlinear devices, such as diodes, transistors, and analog and digital integrated circuits (ICs), is not treated within this chapter. Transformers are discussed in full in Chap. 3Sec. 3.8, while the various nonlinear devices are treated separately in the remaining chapters of this book.

A word of advice: if the math in a particular section of this chapter starts looking scary, don’t worry. As it turns out, most of the nasty math in this chapter is used to prove, say, a theorem or law or to give you an idea of how hard things can get if you do not use some mathematical tricks. The actual amount of math you will need to know to design most circuits is surprisingly small; in fact, basic algebra may be all you need to know. Therefore, when the math in a particular section in this chapter starts looking ugly, skim through the section until you locate the useful, nonugly formulas, rules, and so on, that do not have weird mathematical expressions in them. You don’t have to be a mathematical whiz to be able to design decent circuits.

2.2 Electric Current

Electric current is the total charge that passes through some cross-sectional area A per unit time. This cross-sectional area could represent a disk placed in a gas, plasma, or liquid, but in electronics, this cross-sectional area is most frequently a slice through a solid material, such as a conductor.

If ΔQ is the amount of charge passing through an area in a time interval Δt, then the average current Iave is defined as:

FIGURE 2.1

If the current changes with time, we define the instantaneous current I by taking the limit as Δt → 0, so that the current is the instantaneous rate at which charge passes through an area:

 (2.1)

The unit of current is coulombs per second, but this unit is also called the ampere (A), named after Andre-Marie Ampere:

1 A = 1 C/s

To sound less nerdy, the term amp can be used in place of ampere. Because the ampere is a rather large unit, current is also expressed in milliamps (1 mA = 1 × 10−3 A), microamps (1 µA = 1 × 10−6 A), and nanoamps (1 nA = 1 × 10−9 A).

Within conductors such as copper, electrical current is made up of free electrons moving through a lattice of copper ions. Copper has one free electron per copper atom. The charge on a single electron is given by:

 Qelectron = (− e) = −1.602 × 10−19 C (2.2.a)

This is equal to, but opposite in sign of, the charge of a single copper ion. (The positive charge is a result of the atom donating one electron to the “sea” of free electrons randomly moving about the lattice. The loss of the electron means there is one more proton per atom than electrons.) The charge of a proton is:

 Qproton = (+ e) = +1.602 × 10−19 C (2.2.b)

The conductor, as a whole, is neutral, since there are equal numbers of electrons and protons. Using Eq. 2.2, we see that if a current of 1 A flows through a copper wire, the number of electrons flowing by a cross section of the wire in 1 s is equal to:

Now, there is a problem! How do we get a negative number of electrons flowing per second, as our result indicates? The only two possibilities for this would be to say that either electrons must be flowing in the opposite direction as the defined current, or positive charges must be moving in our wire instead of electrons to account for the sign. The last choice is an incorrect one, since experimental evidence exists to prove electrons are free to move, not positive charges, which are fixed in the lattice network of the conductor. (Note, however, there are media in which positive charge flow is possible, such as positive ion flow in liquids, gases, and plasmas.) It turns out that the first choice—namely, electrons flowing in the opposite direction as the defined current flow—is the correct answer.

Long ago, when Benjamin Franklin (often considered the father of electronics) was doing his pioneering work in early electronics, he had a convention of assigning positive charge signs to the mysterious (at that time) things that were moving and doing work. Sometime later, a physicist by the name of Joseph Thomson performed an experiment that isolated the mysterious moving charges. However, to measure and record his experiments, as well as to do his calculations, Thomson had to stick with using the only laws available to him—those formulated using Franklin’s positive currents. But these moving charges that Thomson found (which he called electrons) were moving in the opposite direction of the conventional current I used in the equations, or moving against convention. See Fig. 2.2.

FIGURE 2.2 Thomson changed the notion that positive charges were what were moving in conductors, contrary to Franklin’s notion. However, negative electrons going one way is equivalent to positive charges going the opposite direction, so the old formulas still work. Since you deal with the old formulas, it’s practical to adopt Franklin’s conventional current—though realize that what’s actually moving in conductors is electrons.

What does this mean to us, to those of us not so interested in the detailed physics and such? Well, not too much. We could pretend that there were positive charges moving in the wires and various electrical devices, and everything would work out fine: negative electrons going one way are equivalent to positive charges going in the opposite direction. In fact, all the formulas used in electronics, such as Ohm’s law (V = IR), “pretend” that the current I is made up of positive charge carriers. We will always be stuck with this convention. In a nutshell, it’s convenient to pretend that positive charges are moving. So when you see the term electron flow, make sure you realize that the conventional current flow I is moving in the opposite direction. In a minute, we’ll discuss the microscopic goings-on within a conductor that will clarify things a bit better.

Example 1: How many electrons pass a given point in 3 s if a conductor is carrying a 2-A current?

FIGURE 2.3

Answer: The charge that passes a given point in 3 s is:

ΔQ = I × Δt = (2 A)(3 s) = 6 C

One electron has a charge of 1.6 × 10−19 C, so 6 C worth of electrons is:

# Electrons = 6 C/1.602 × 10−19 C = 3.74 × 1019

Example 2: Charge is changing in a circuit with time according to Q(t) = (0.001 C) sin [(1000/s) t]. Calculate the instantaneous current flow.

Answer: If we plug in a specific time within this equation, we get an instantaneous current for that time. For example, if t = 1, the current would be 0.174 A. At t = 3 s, the current would be − 0.5 A, the negative sign indicating that the current is in the opposite direction—a result of the sinusoidal nature.

Note: The last example involved using calculus—you can read about the basics of calculus in App. C if you’re unfamiliar with it. Fortunately, as we’ll see, rarely do you actually need to work in units of charge when doing electronics. Usually you worry only about current, which can be directly measured using an ammeter, or calculated by applying formulas that usually require no calculus whatsoever.

2.2.1 Currents in Perspective

What’s considered a lot or a little amount of current? It’s a good idea to have a gauge of comparison when you start tinkering with electronic devices. Here are some examples: a 100-W lightbulb draws about 1 A; a microwave draws 8 to 13 A; a laptop computer, 2 to 3 A; an electric fan, 1 A; a television, 1 to 3 A; a toaster, 7 to 10 A; a fluorescent light, 1 to 2 A; a radio/stereo, 1 to 4 A; a typical LED, 20 mA; a mobile (smart) phone accessing the web uses around 200 mA; an advanced low-power microchip (individual), a few µA to perhaps even several pA; an automobile starter, around 200 A; a lightning strike, around 1000 A; a sufficient amount of current to induce cardiac/respiratory arrest, around 100 mA to 1 A.

2.3 Voltage

To get electrical current to flow from one point to another, a voltage must exist between the two points. A voltage placed across a conductor gives rise to an electromotive force (EMF) that is responsible for giving all free electrons within the conductor a push.

As a technical note, before we begin, voltage is also referred to as a potential difference or just potential—they all mean the same thing. We’ll avoid using these terms, however, because it is easy to confuse them with the term potential energy, which is not the same thing.

A simple flashlight circuit, consisting of a battery connected to a lamp, through two conductors and a switch, is shown in Fig. 2.4. When the switch is open (“off”), no current will flow. The moment the switch is closed, however, the resistance of the switch falls to almost zero, and current will flow. This voltage then drives all free electrons, everywhere within the circuit, in a direction that points from negative to positive; conventional current flow, of course, points in the opposite direction (see Benjamin Franklin).

FIGURE 2.4

It is important to note that the battery needs the rest of the circuit, just as the rest of the circuit needs the battery. Without the linkage between its terminals, the chemical reactions within the battery cannot be carried out. These chemical reactions involve the transfer of electrons, which by intended design can only occur through a link between the battery’s terminals (e.g., where the circuit goes). Figure 2.5 shows this process using an alkaline dry cell battery. Notice that the flow of current is conserved through the circuit, even though the nature of the current throughout the circuit varies—ionic current within sections of the battery, electron current elsewhere.

FIGURE 2.5

As free electrons within the lamp filament experience an EMF due to the applied voltage, the extra energy they gain is transferred to the filament lattice atoms, which result in heat (filament atomic vibrations) and emitted light (when a valence electron of a lattice atom is excited by a free electron and the bound electron returns to a lower energy configuration, thus releasing a photon).

A device that maintains a constant voltage across it terminals is called a direct current voltage source (or dc voltage source). A battery is an example of a dc voltage source. The schematic symbol for a battery is .

2.3.1 The Mechanisms of Voltage

To get a mental image of how a battery generates an EMF through a circuit, we envision that chemical reactions inside yield free electrons that quickly build in number within the negative terminal region (anode material), causing an electron concentration. This concentration is full of repulsive force (electrons repel) that can be viewed as a kind of “electrical pressure.” With a load (e.g., our flashlight lamp, conductors, switch) placed between the battery’s terminals, electrons from the battery’s negative terminal attempt to alleviate this pressure by dispersing into the circuit. These electrons increase the concentration of free electrons within the end of the conductor attached to the negative terminal. Even a small percentage difference in free electron concentration in one region gives rise to great repulsive forces between free electrons. The repulsive force is expressed as a seemingly instantaneous (close to the speed of light) pulse that travels throughout the circuit. Those free electrons nearest to the pumped-in electrons are quickly repulsed in the opposite direction; the next neighboring electrons get shoved, and so on down the line, causing a chain reaction, or pulse. This pulse travels down the conductor near the speed of light. See Fig. 2.6.

FIGURE 2.6

The actual physical movement of electrons is, on average, much slower. In fact, the drift velocity (average net velocity of electrons toward the positive terminal) is usually fractions of a millimeter per second—say, 0.002 mm/s for a 0.1-A current through a 12-gauge wire. We associate this drift movement of free electrons with current flow or, more precisely, conventional current flow I moving in the opposite direction. (As it turns out, the actual motion of electrons is quite complex, involving thermal effects, too—we’ll go over this in the next section.)

It is likely that those electrons farther “down in” the circuit will not feel the same level of repulsive force, since there may be quite a bit of material in the way which absorbs some of the repulsive energy flow emanating from the negative terminal (absorbing via electron-electron collisions, free electron–bond electron interactions, etc.). And, as you probably know, circuits can contain large numbers of components, some of which are buried deep within a complex network of pathways. It is possible to imagine that through some of these pathways the repulsive effects are reduced to a weak nudge. We associate these regions of “weak nudge” with regions of low “electrical pressure,” or voltage. Electrons in these regions have little potential to do work—they have low potential energy relative to those closer to the source of pumped-in electrons.

Voltage represents the difference in potential energy. A unit charge has been at one location relative to another within a region of “electrical pressure”—the pressure attributed to new free electrons being pumped into the system. The relationship between the voltage and the difference in potential energy is expressed as:

FIGURE 2.7

Implicit in the definition of voltage is the notion that voltage is always a measurement between two points, say point A and point B. That is the reason for the subscript “AB” in VAB. The symbol ΔV means the same. Both infer that there is an absolute scale on which to measure and give individual points a specific voltage value. In electronics, we can create such a scale by picking a point, often the point where there is the lowest electrical pressure, and defining this point as the zero point, or 0-V reference. In many dc circuits, people choose the negative terminal of the battery as the 0-V reference, and let everyone know by inserting a ground symbol  (more on this later). In practice, you rarely see voltages expressed using subscripts (VAB) or deltas (ΔV), but instead you simply see the symbol V, or you may see a symbol like VR. The “blank symbol” V, however, is always modified with a phrase stating the two points across which the voltage is present. In the second case, VR, the subscript means that the voltage is measured across the component R—in this case, a resistor. In light of this, we can write a cleaner expression for the voltage/potential energy expression:

Just make sure you remember that the voltage and potential energy variables represent the difference in relation to two points. As we’ll discover, all the big electronics laws usually assume that variables of voltage or energy are of this “clean form.”

In our flashlight example, we can calculate the difference in potential energy between an electron emanating from the negative terminal of the 1.5-V battery and one entering the positive terminal.

ΔU = ΔVq = (1.5 V)(1.602 × 10−19 C) = 2.4 × 10−19 J

Notice that this result gives us the potential energy difference between the two electrons, not the actual potential energy of either the electron emanating from the negative terminal (U1) or the electron entering the positive terminal (U0). However, if we make the assumption that the electron entering the positive terminal is at zero potential energy, we can figure that the electron emanating from the negative terminal has a relative potential energy of:

U1 = ΔU + U0 = ΔU + 0 = 2.4 × 10−19 J

Note: Increasing positive potential energy can be associated with similar charges getting closer together. Decreasing energy can be associated with similar charges getting farther apart. We avoided the use of a negative sign in front of the charge of the electron, because voltages are defined by a positive test charge. We are in a pickle similar to the one we saw with Benjamin Franklin’s positive charges. As long as we treat the potential relative to the pumped-in electron concentration, things work out.

In a real circuit, where the number of electrons pumped out by the battery will be quite large—hundreds to thousands of trillions of electrons, depending on the resistance to electron flow—we must multiply our previous calculation by the total number of entering electrons. For example, if our flashlight draws 0.1 A, there will be 6.24 × 1017 electrons pumped into it by the battery per second, so you calculate the potential energy of all the new electrons together to be about 0.15 J/s.

What about the potential energies of free electrons at other locations throughout the circuit, such as those found in the lamp filament, those in the positive wire, those in the negative wire, and so on? We can say that somewhere in the filament of the lamp, there is an electron that has half the potential energy of a fresh pumped-in electron emanating from the negative terminal of the battery. We attribute this lower energy to the fact that other free electrons up the line have lost energy due to collision mechanisms, which in turn yields a weaker electrical repulsive pressure (shoving action) that our electron in question experiences. In fact, in our flashlight circuit, we attribute all loss in electrical pressure to be through the lamp filament as free-electron energy is converted into heat and light.

In regard to potential energies of free electrons within the conductors leading to and from the battery, we assume all electrons within the same conductor have the same potential energy. This assumes that there is no voltage difference between points in the same conductor. For example, if you take a voltmeter and place it between any two points of a single conductor, it will measure 0 V. (See Fig. 2.8.) For practical purposes, we accept this as true. However, in reality it isn’t. There is a slight voltage drop through a conductor, and if we had a voltmeter that was extremely accurate we might measure a voltage drop of 0.00001 V or so, depending on the length of the conductor, current flow, and conductor material type. This is attributed to internal resistance within conductors—a topic we’ll cover in a moment.

FIGURE 2.8

2.3.2 Definition of Volt and Generalized Power Law

We come now to a formal definition of the volt—the unit of measure of voltage. Using the relationship between voltage and potential energy difference V = U/q, we define a volt to be:

(Be aware that the use of “V” for both an algebraic quantity and a unit of voltage is a potential source of confusion in an expression like V = 1.5 V. The algebraic quantity is in italic.)

Two points with a voltage of 1 V between them have enough “pressure” to perform 1 J worth of work while moving 1 C worth of charge between the points. For example, an ideal 1.5-V battery is capable of moving 1 C of charge through a circuit while performing 1.5 J worth of work.

Another way to define a volt is in terms of power, which happens to be more useful in electronics. Power represents how much energy per second goes into powering a circuit. According to the conservation of energy, we can say the power used to drive a circuit must equal the power used by the circuit to do useful work plus the power wasted, as in the case of heat. Assuming that a single electron loses all its potential energy from going through a circuit from negative to positive terminal, we say, for the sake of argument, that all this energy must have been converted to work—useful and wasted (heat). By definition, power is mathematically expressed as dW/dt. If we substitute the potential energy expression U = Vq for W, assuming the voltage is constant (e.g., battery voltage), we get the following:

Since we know that current I = dq/dt, we can substitute this into the preceding expression to get:

 P = VI (2.3)

This is referred to as the generalized power law. This law is incredibly powerful, and it provides a general result, one that is independent of type of material and of the nature of the charge movement. The unit of this electrical power is watts (W), with 1 W = 1 J/s, or in terms of volts and amps, 1 W = 1 VA.

In terms of power, then, the volt is defined as:

The generalized power law can be used to determine the power loss of any circuit, given only the voltage applied across it and the current drawn, both of which can easily be measured using a voltmeter and an ammeter. However, it doesn’t tell you specifically how this power is used up—more on this when we get to resistance. See Fig. 2.9.

FIGURE 2.9

Example 1: Our 1.5-V flashlight circuit draws 0.1 A. How much power does the circuit consume?

P = VI = (1.5 V)(0.1 A) = 0.15 W

Example 2: A 12-V electrical device is specified as consuming 100 W of power. How much current does it draw?

2.3.3 Combining Batteries

To get a larger voltage capable of supplying more power, we can place two batteries in series (end to end), as shown in Fig. 2.10. The voltage across the combination is equal to the individual battery voltages added together. In essence, we have placed two charge pumps in series, increasing the effective electrical pressure. Chemically speaking, if the batteries are of the same voltage, we double the number of chemical reactions, doubling the number of electrons that can be pumped out into the circuit.

FIGURE 2.10

In Fig. 2.10, we use the notion of a ground reference, or 0-V reference, symbolized . Though this symbol is used to represent an earth ground (which we define a bit later), it can also be used to indicate the point where all voltage measurements are to be referenced within a circuit. Logically, whenever you create a scale of measure, you pick the lowest point in the scale to be zero—0 V here. For most dc circuits, the ground reference point is usually placed at the negative terminal of the voltage source. With the notion of ground reference point, we also get the notion of a point voltage, which is the voltage measured between the ground reference and a specific point of interest within the circuit. For example, the single battery shown in Fig. 2.10 has a voltage of 1.5 V. We place a ground reference at the negative terminal and give this a 0-V point voltage, and place a 1.5-V point voltage marker at the positive terminal.

In the center of Fig. 2.10, we have two 1.5-V batteries in series, giving a combined voltage of 3.0 V. A ground placed at the negative terminal of the lower battery gives us point voltages of 1.5 V between the batteries, and 3.0 V at the positive terminal of the top battery. A load placed between ground and 3.0 V will result in a load current that returns to the lower battery’s negative terminal.

Finally, it is possible to create a split supply by simply repositioning the 0-V ground reference, placing it between the batteries. This creates +1.5 V and −1.5 V leads relative to the 0-V reference. Many circuits require both positive and negative voltage relative to a 0-V ground reference. In this case, the 0-V ground reference acts as a common return. This is often necessary, say, in an audio circuit, where signals are sinusoidal and alternate between positive and negative voltage relative to a 0-V reference.

2.3.4 Other Voltage Sources

There are other mechanisms besides the chemical reactions within batteries that give rise to an electromotive force that pushes electrons through circuits. Some examples include magnetic induction, photovoltaic action, thermoelectric effect, piezoelectric effect, and static electric effect. Magnetic induction (used in electrical generators) and photovoltaic action (used in photocells), along with chemical reactions, are, however, the only mechanisms of those listed that provide enough power to drive most circuits. The thermoelectric and piezoelectric effects are usually so small (mV range, typically) that they are limited to sensor-type applications. Static electric effect is based on giving objects, such as conductors and insulators, a surplus of charge. Though voltages can become very high between charged objects, if a circuit were connected between the objects, a dangerous discharge of current could flow, possibly damaging sensitive circuits. Also, once the discharge is complete—a matter of milliseconds—there is no more current to power the circuit. Static electricity is considered a nuisance in electronics, not a source of useful power. We’ll discuss all these different mechanisms in more detail throughout the book.

2.3.5 Water Analogies

It is often helpful to use a water analogy to explain voltage. In Fig. 2.11, we treat a dc voltage source as a water pump, wires like pipes, Benjamin Franklin’s positive charges as water, and conventional current flow like water flow. A load (resistor) is treated as a network of stationary force-absorbing particles that limit water flow. We’ll leave it to you to compare the similarities and differences.

FIGURE 2.11

Here’s another water analogy that relies on gravity to provide the pressure. Though this analogy falls short of being accurate in many regards, it at least demonstrates how a larger voltage (greater water pressure) can result in greater current flow.

FIGURE 2.12

It’s not wise to focus too much attention on these water analogies—they fall short of being truly analogous to electric circuits. Take them with a grain of salt. The next section will prove how true this is.

Example 1: Find the voltage between the various points indicated in the following figures. For example, the voltage between points A and B in Fig. 2.13a is 12 V.

FIGURE 2.13

Answer: a. VAC = 0, VBD = 0, VAD = 0, VBC = 0. b. VAC = 3 V, VBD = 0 V, VAD = 12 V, VBC = 9 V. c. VAC = 12 V, VBD = 9V. VAD = 21V, VBC = 0 V. d. VAC = 3 V, VAB = 6 V, VCD = 1.5 V, VAD = 1.5 V, VBD = 4.5 V.

Example 2: Find the point voltages (referenced to ground) at the various locations indicated in the following figures.

FIGURE 2.14

Answer: a. A = 3 V, B = −3 V, C = 3 V, D = 3 V, E = 3 V, F = 3 V, G = 6 V, H = 9 V. b. A = 1.5 V, B = 0 V, C = 1.5 V, D = 1.5 V, E = −1.5 V, F = −3.0 V, G = 1.5 V, H = −1.5 V.

2.4 A Microscopic View of Conduction (for Those Who Are Interested)

At a microscopic level, a copper conductor resembles a lattice of copper balls packed together in what’s called a face-centered-cubic lattice structure, as shown in Fig. 2.15. For copper, as well as other metals, the bonding mechanism that holds everything together is referred to as metallic bonding, where outermost valence electrons from the metal atoms form a “cloud of free electrons” which fill the space between the metal ions (positively charged atoms missing an electron that became “free”—see the planetary model in Fig. 2.15b). This cloud of free electrons acts as a glue, holding the lattice metal ions together.

FIGURE 2.15 (a) Copper nucleus composed of protons and neutrons held together by nuclear forces that is roughly 137 times stronger than the electromagnetic force. (b) Copper atoms, as viewed by the classic planetary model, consisting of valance electrons held in orbit by electric forces. Quantum mechanics is required to explain why electrons exist in discrete energy levels, and why they don’t fall into the nucleus or radiate electromagnetic energy as they orbit. (c) Copper lattice has a face-centered cubic packing arrangement. (d) Scanning tunneling electron microscope (STM) image of copper 100, courtesy of Institut für Allgemeine Physik, TU Wien. (e) Ball packing model of lattice, showing irregularities in lattice geometry, partly caused by impurities (other kinds of atoms). (f) Lattice view showing that lattice atoms vibrate due to external thermal interactions as well as interactions with free electrons. Free electrons move about randomly, at varying speeds and directions, colliding with other electrons and lattice ions. Under normal conditions, they do not leave the surface of the metal.

Each free electron within the cloud of free electrons moves about in random directions and speeds, colliding and rebounding “off” metal ions and other imperfections (impurities in lattice and grain boundary transitions, etc.). It is important to realize that this is occurring in a chunk of copper, at room temperature, without any applied voltage.

At room temperature, no free electrons ever leave the surface of the metal. A free electron cannot escape the coulomb (electric) attractive forces presented by the positive metal ions in the lattice. (We’ll see later that under special conditions, using unique mechanisms, it is possible for electrons to escape.)

According to what’s called the free-electron model—a classical model that treats free electrons as a gas of noninteracting charges—there is approximately one free electron per copper atom, giving a copper conductor a free electron concentration of ρn = 8.5 × 1028 electrons/m3. This model predicts that, under normal conditions (a piece of copper just sitting there at room temperature), the thermal velocity v of electrons (or root-mean-square speed) within copper is about 120 km/s (1.2 × 105 m/s), but depends on temperature. The average distance an electron travels before it collides with something, called the mean free path λ, is about 0.000003 mm (2.9 × 10−9 m), with the average time between collisions τ of roughly 0.000000000000024 s (2.4 × 10−14 s). The free-electron model is qualitatively correct in many respects, but isn’t as accurate as models based on quantum mechanics. (The speed, path, and time are related by v = λ/τ.)

In quantum mechanics, electrons obey velocity-distribution laws based on quantum physics, and the movement of electrons depends on these quantum ideas. It requires that we treat electrons as though they were waves scattering from the lattice structure of the copper. The quantum view shows the thermal speed (now called Fermi velocity vF) of a free electron to be faster than that predicted by the free-electron model, now around 1.57 × 106 m/s, and contrarily, it is essentially independent of temperature. In addition, the quantum model predicts a larger mean free path, now around 3.9 × 10−8 m, which is independent of temperature. The quantum view happens to be the accepted view, since it gives answers that match more precisely with experimental data. Table 2.1 shows the Fermi velocities of electrons for various metals.

TABLE 2.1 Condensed Matter Properties of Various Metals

Also, the surface binding energy (caused by electrostatic attraction) that prevents electrons from exiting the surface of the metal, referred to as the work function, is about 4.7 eV for copper (1 eV = 1.6022 × 10−19 J). The only way to eject electrons is through special processes, such as thermionic emission, field emission, secondary emission, and photoelectric emission.

(Thermionic emission: increase in temperature provides free electrons enough energy to overcome work function of the material. The emitted electron is referred to as a thermoelectron. Field emission: additional energy from an electric field generated by a high-voltage conductor provides an attractive enough positive field to free electrons from the surface. This requires huge voltages [MV per cm between emitting surface and positive conductor]. Secondary emission: electrons are emitted from a metallic surface by the bombardment of high-speed electrons or other particles. Photoelectric emission: electron in material absorbs energy from incoming photon of particular frequency, giving it enough energy to overcome work function. A photon must be of the correct frequency, governed by W = hf0, for this to occur [Planck’s constant h = 6.63 × 10−34 J-s or 4.14 × 10−14 eV; f0 is in hertz]).

2.4.1 Applying a Voltage

Next, we wish to see what happens when we apply a voltage across the conductor—say, by attaching a thick copper wire across a battery. When we do this, our randomly moving free electrons all experience a force pointing toward the positive end of the wire due to the electric field set up within the wire. (The field is due to the negative concentration of pumped-in electrons at one end relative to the neutral [positive relative to negative] concentration at the other end.) The actual influence this force has on the motion of the random free electrons is small—the thermal velocity is so large that it is difficult to change the momentum of the electrons. What you get is a slightly parabolic deviation in path, as shown in Fig. 2.16.

FIGURE 2.16 (a) Simplistic view of an electron randomly moving through a copper lattice, rebounding off lattice atoms and impurities. (b) An electron collides frequently with the ions and impurities in a metal and scatters randomly. In an electric field, the electron picks up a small component of velocity opposite the field. The differences in the paths are exaggerated. The electron’s path in an electric field is slightly parabolic. (c) Model illustrating current density, drift velocity, charge density, thermal velocity, and current.

Normally, the field present in the wire would create a net acceleration component in the direction of the force; however, the constant collisions electrons experience create a drag force, similar to the drag experienced by a parachute. The net effect is an average group velocity referred to as the drift velocity vd. Remarkably, this velocity is surprisingly small. For example, the voltage applied to a 12-gauge copper wire to yield a 0.100-A current will result in a drift velocity of about 0.002 mm per second! The drift velocity is related, determined by

vd = J/(ρee)

where J is the current density—the current flowing through an area (J = I/A), ρe is the free-electron density in the material, and e is the charge of an electron. Table 2.1 shows free-electron densities for various materials. As you can see, the drift velocity varies with current and diameter of the conductor.

The drift velocity is so slow, only fractions of a millimeter per second, that it is worth pondering how a measurable current can even flow. For example, what happens when you flip the switch on a flashlight? Of course, we don’t have to wait hours for electrons to move down the conductors from the battery. When we throw the switch, the electric field of the incoming electron has a repulsive effect on its neighbor within the wire. This neighbor then moves away toward another neighbor, creating a chain of interactions that propagates through the material at near the speed of light. (See Fig. 2.17.) This reaction, however, really isn’t the speed of light, but a fraction less, depending on the medium. The free electrons spreading throughout the conductor all start moving at once in response—those nearest the switch, as well as those nearest the light filament or LED. A similar effect occurs in fluid flow, as when you turn on a garden hose. Because if the hose is already full of water, the outflow starts immediately. The force of the water at the faucet end is quickly transmitted all along the hose, and the water at the open end of the hose flows almost at the moment the faucet is opened.

FIGURE 2.17 Illustration of how the electric field propagates down a wire as electrons are pumped into one end.

In the case of alternating current, the field reverses directions in a sinusoidal fashion, causing the drift velocity component of electrons to swish back and forth. If the alternating current has a frequency of 60 Hz, the velocity component would be vibrating back and forth 60 times a second. If our maximum drift velocity during an ac cycle is 0.002 mm/s, we could roughly determine that the distance between maximum swings in the drift distance would be about 0.00045 mm. Of course, this doesn’t mean that electrons are fixed in an oscillatory position. It means only that the drift displacement component of electrons is—if there is such a notion. Recall that an electron’s overall motion is quite random and its actual displacement quite large, due to the thermal effects.

2.5 Resistance, Resistivity, Conductivity

As was explained in the last section, free electrons in a copper wire at room temperature undergo frequent collisions with other electrons, lattice ions, and impurities within the lattice that limit their forward motion. We associate these microscopic mechanisms that impede electron flow with electrical resistance. In 1826, Georg Simon Ohm published experimental results regarding the resistance of different materials, using a qualitative approach that wasn’t concerned with the hidden mechanisms, but rather considered only the net observable effects. He found a linear relationship between how much current flowed through a material when a given voltage was applied across it. He defined the resistance to be the ratio of the applied voltage divided by the resultant current flow, given by:

 (2.4)

This statement is called Ohm’s law, where R is the resistance, given in volts per ampere or ohms (abbreviated with the Greek symbol omega, Ω). One ohm is the resistance through which a current of 1 A flows when a voltage of 1 V is applied:

1 Ω = 1 V/1 A

The symbol  is used to designate a resistor.

Now, Ohm’s law isn’t really a law, but rather an empirical statement about the behavior of materials. In fact, there are some materials for which Ohm’s law actually doesn’t work.

FIGURE 2.18

Ohm’s law can be applied only to ohmic materials—materials whose resistance remains constant over a range of voltages. Nonohmic materials, on the other hand, do not follow this pattern; they do not obey Ohm’s law. For example, a diode is a device that allows current to pass easily when the voltage is positive, but prevents current flow (creates a high resistance) when the voltage is negative.

Usually you see Ohm’s law written in the following form:

V = I × R

However, in this form it is tempting to define voltage in terms of resistance and current. It is important to realize that R is the resistance of an ohmic material and is independent of V in Ohm’s law. In fact, Ohm’s law does not say anything about voltage; rather, it defines resistance in terms of it and cannot be applied to other areas of physics such as static electricity, because there is no current flow. In other words, you don’t define voltage in terms of current and resistance; you define resistance in terms of voltage and current. That’s not to say that you can’t apply Ohm’s law to, say, predict what voltage must exist across a known resistance, given a measured current. In fact, this is done all the time in circuit analysis.

2.5.1 How the Shape of a Conductor Affects Resistance

The resistance of a conducting wire of a given material varies with its shape. Doubling the length of a wire doubles the resistance, allowing half the current to flow, assuming similar applied voltages. Conversely, doubling the cross-sectional area A has the opposite effect—the resistance is cut in half, and twice as much current will flow, again assuming similar applied voltages.

Increasing resistance with length can be explained by the fact that down the wire, there are more lattice ions and imperfections present for which an applied field (electric field instigated by added electrons pumped in by the source) must shove against. This field is less effective at moving electrons because as you go down the line, there are more electrons pushing back—there are more collisions occurring on average.

Decreasing resistance with cross-sectional area can be explained by the fact that a larger-volume conductor (greater cross-sectional area) can support a larger current flow. If you have a thin wire passing 0.100 A and a thick wire passing 0.100 A, the thinner wire must concentrate the 0.100 A through a small volume, while the thick wire can distribute this current over a greater volume. Electrons confined to a smaller volume tend to undergo a greater number of collisions with other electrons, lattice ions, and imperfections than a wire with a larger volume. The concentration of free-electron flow, according to Benjamin Franklin’s convention, represents a concentration of conventional current flow in the opposite direction. This concentration of current flow is called the current density J—the rate of current flows per unit area. For a wire: J = I/A. Figure 2.19 demonstrates how the current density is greater in a thin 12-gauge wire than in a thicker 4-gauge wire. It also shows that the drift velocity in the thick wire is less than the drift velocity in the thin wire—a result of a “decrease in electron field pressure” lowering the average “push” in the current flow direction.

FIGURE 2.19 Effects of wire diameter on resistance. A thinner wire has more resistance per unit length than a thicker wire.

2.5.2 Resistivity and Conductivity

We have left out the most important aspect of resistance that has nothing to do with the physical length or diameter of the material. How does the “chemistry” of the material affect the resistance? For example, if you have a copper wire with the same dimensions as a brass wire, which metal has the greater overall resistance? To answer this question, as well as provide a way to categorize materials, we adopt the concept of resistivity. Unlike resistance, resistivity is entirely independent of the dimensions of the material. Resistivity is a property unique to the material. The resistivity ρ is defined as follows:

 (2.5)

where A is the cross-sectional area, L is the length, and R is the overall resistance of the material, as measured across its length. The units of resistivity are ohm-meters (Ωm).

For some, resistivity is too negative a concept—it tells you how “bad” something is at passing current. Optimists prefer the concept of conductivity—how “good” something is at passing current. Conductivity, symbolized σ, is simply the mathematical inverse of resistivity:

 (2.6)

The units of conductivity are siemens, S = (Ωm)−1. (Mathematical note: [Ωm]−1 = 1/[Ωm]). Both conductivity and resistivity contain the same important underlining information. Some prefer to play with equations that use the optimistic notion of conductivity (“glass half full”); others prefer the pessimistic notion of resistivity (“glass half empty”).

In terms of resistivity and conductivity, we can rewrite Ohm’s law as follows:

 (2.7)

Table 2.2 shows the conductivities (resistivities, for the pessimists) of various materials. (Consult a technical handbook, such as The Handbook of Chemistry and Physics, for a more complete list.) The conductivity of metals, such as copper and silver, is a factor of 1021 higher than that of a good insulator, such as Teflon. Though both copper and silver are great conductors, silver is simply too expensive for practical use. Though aluminum is a fairly good conductor and was actually used at one time in home circuitry, it quickly became apparent that it oxidized badly, inhibiting good electrical contacts and limiting current flow to channels of limited size. The result produced fire hazards.

TABLE 2.2 Conductivity of Various Materials

An important feature of resistivity (or conductivity) is its temperature dependency. Generally, within a certain temperature range, the resistivity for a large number of metals obeys the following equation:

 ρ = ρ0[1 + α(T − T0)] (2.8)

where ρ is the calculated resistivity based on a set reference resistivity ρ0 and temperature T0. Alpha α is called the temperature coefficient of resistivity, given in units of 1/°C or (°C)−1. The resistivity for most metals increases with temperature because lattice atom vibrations caused by thermal energy (increased temperature) impede the drift velocity of conducting electrons.

Are Water, Air, and Vacuums Considered Insulators or Conductors?

These media require special mention. See comments presented in Table 2.3.

TABLE 2.3 Resistivity of Special Materials

2.6 Insulators, Conductors, and Semiconductors

As we have seen, the electrical resistivities of materials vary greatly between conductors and insulators. A decent conductor has around 10−8 Ωm; a good insulator has around 1014 Ωm; a typical semiconductor from 10−5 to 103 Ωm—depending on temperature. What is the microscopic explanation for these differences?

The answer to this question rests upon the quantum nature of electrons. In classical physics the energy of an electron in a metal can take on any value—it is said the energy values form a continuum. (Here, electron energy is considered zero at infinite distance from the nucleus and becomes more negative in energy closer to the nucleus, relative to the zero reference state. Negative energy infers that there is electric attraction between the positive nucleus and the electron—it is electric potential energy.) However, a quantum description of electrons in metals shows that the energy values of electrons are quantized, taking on discrete values. This comes from the wavelike nature of electrons—analogous to standing waves on strings existing only at discrete frequencies. Figure 2.20 shows an energy diagram illustrating the possible energy levels of an electron (ignoring lattice influences). The diagram illustrates only the possible energy levels—electrons are not necessarily in each level.

FIGURE 2.20 (Left) Energy diagram showing all possible energy levels of an electron in a solid; but it takes no account of the influence of an atomic crystalline lattice structure. (Right) Energy diagram that shows possible energy levels of an electron within a material made of a regular lattice of atoms. Electron energies are restricted to lie within allowed bands, and there is a large energy gap where no electrons are allowed. Even within the allowed bands, the possible electron energies are closely spaced discrete levels.

When a set of atoms forms a regular background lattice, the possible energy values of the electrons are altered even further. We still have discrete energy regions, called allowed bands, but we now get what are called energy gaps. Energy gaps are forbidden regions to the electron, and represent regions where no traveling wave (electron) can exist when placed in the periodic electric potential of the metal’s positive lattice ions. These gaps are quite large in the scale of atomic physics—within electron-volts range. Again, energy levels presented in the band diagram specify only possible values of electron energies—they may or may not be occupied.

Now, quantum physics has an interesting property, called the Pauli exclusion principle, which has a critical role in determining the properties of materials. The Pauli exclusion principle says that no two electrons in an atom can be in the same quantum state. The lowest common divisor of quantum states is the spin quantum number ms, which states that no more than two electrons with opposing spin (up or down) can be located in the same energy level. Now, if we consider a solid that has many free electrons, which is in an equilibrium state, electrons fill the lowest energy levels available in the allowed band, up to two in each level. Those electrons further down in energy are more tightly bound and are called innermost electrons. When all the electrons are placed in the lowest energy state, we are left with two possible outcomes. In the first case, the highest level to be filled is somewhere in the middle of a band. In the second case, the electrons just fill one or more bands completely. We assume that the material is at low enough temperature to prevent electrons from jumping to higher energy levels due to thermal effects.

Now if we add some energy to the free electrons by applying an electric field (attach a voltage source), for example, the electrons in the lower energy levels cannot accept that energy, because they cannot move into an already filled higher energy level. The only electrons that can accept energy are those that lie in the top levels, and then only if there are nearby empty levels into which they can move. Materials with electrons in only partly filled bands are conductors. When the top layer of their electrons moves freely into the empty energy levels immediately above, there is a current. The electrons that jump from a lower level to a higher level are said to be excited. The valance band is an allowed occupied band. The conduction band is an allowed empty band. The energy-band structure for conductors is shown in Fig. 2.21a and c.

FIGURE 2.21 Four possible band structures for a solid: (a) Conductor—allowed band is only partially full, so electrons can be excited to nearby states. (b) Insulator—forbidden band with large energy gap between the filled band and the next allowed band. (c) Conductor—allowed bands overlap. (d) Semiconductor—energy gap between the filled band and the next allowed band is very small, so some electrons are excited to the conduction band at normal temperatures, leaving holes in the valence band.

If the highest-energy electrons of a material fill a band completely, then a small electric field will not give these electrons enough energy to jump the large energy gap to the bottom of the next (empty) band. We then have an insulator (see Fig. 2.21b). An example of a good insulator is diamond, whose energy gap is 6 eV.

In semiconductors, the highest-energy electrons fill a band (the valance band) at T = 0, as in insulators. However, unlike insulators, semiconductors have a small energy gap between that band and the next, the conduction band. Because the energy gap is so small, a modest electric field (or finite temperature) will allow electrons to jump the gap and thereby conduct electricity. Thus, there is a minimum electric field under the influence of which a material changes from insulator to conductor. Silicon and germanium have energy gaps of 1.1 eV and 0.7 eV, respectively, and are semiconductors. For semiconductors, an increase in temperature will give a fraction of the electrons enough thermal energy to jump the gap. For an ordinary conductor, a rise in temperature increases the resistivity, because the atoms, which are obstacles to electron flow, vibrate more vigorously. A temperature increase in a semiconductor allows more electrons into the empty band and thus lowers the resistivity.

When an electron in the valance band of a semiconductor crosses the energy gap and conducts electricity, it leaves behind what is known as a hole. Other electrons in the valence band near the top of the stack of energy levels can move into this hole, leaving behind their own holes, in which still other electrons can move, and so forth. The hole behaves like a positive charge that conducts electricity on its own as a positive charge carrier. An electron excited from the valence band to the conduction band is thus doubly effective at conducting electricity in semiconductors.

Besides the intrinsic elemental semiconductors, such as silicon and germanium, there are hybrid compounds—compounds such as gallium arsenide. Other semiconductors are made by introducing impurities into a silicon lattice. For example, an atom in the chemical group of phosphorous, arsenic, and antimony can replace one of the silicon atoms in a lattice without affecting the lattice itself too much. However, each of these impurities has one more electron in its valence level than the silicon atom has; this extra electron, for which there is no room in the valence band, takes a place in the conduction band and can conduct electricity. A semiconductor with impurities of this sort is called an n-type semiconductor, and the extra electrons are called donor electrons.

Atoms of elements in the same chemical group as boron, aluminum, and gallium have one less valence electron than silicon has. If an atom is added to a lattice of silicon as an impurity, there is one less electron than is needed to form a bond that holds the lattice together. This electron must be provided by the electrons of the valence band of the lattice material, and holes are created in this band. These holes act as positive charge carriers. The impurity atoms are called acceptors. A semiconductor with such impurities is called a p-type semiconductor.

We will see later on how n-type and p-type semiconductors are used for making one-way gates for current flow (diodes) and voltage-controlled current switches (transistors).

2.7 Heat and Power

In Sec. 2.3, we discovered the generalized power law, which tells us that if we can measure the current entering a device, as well as the voltage across the device, we know the power that is used by the device:

 P = VI (2.9)

The generalized power law tells us how much power is pumped into a circuit but doesn’t say anything about how this power is used up. Let’s consider a two-lead black box—an unknown circuit that may contain all sorts of devices, such as resistors, lamps, motors, or transistors. If all we can do is measure the current entering the black box and the voltage across it—say, using an ammeter and voltmeter (or singularly, a wattmeter)—we could apply the generalized power law, multiply the measured current and voltage readings together, and find the power pumped into the black box. For example, in Fig. 2.22, we measure 0.1 A entering when a voltage of 10 V is applied, giving us a total consumed power of 1 W.

FIGURE 2.22

Knowing how much power is pumped into the black box is incredibly useful—it allows for quick power consumption measurements and often simplifies circuit analysis—as we’ll see later. But let’s say that we are interested in figuring out how much power is lost to heating (energy that goes into lattice vibrations, emissive radiation, etc.). We really can’t say, assuming we aren’t allowed to look inside the black box. There could be devices inside that take some of the initial energy and use it to do useful work, such as generating magnetic fields in the armature (rotor) and stator sections of a motor, causing the stator to rotate; or generating a magnetic field in a voice coil attached to a paper speaker cone that compresses air; or generating light energy, radio waves, and so on. There may be power converted into other weird forms not really coined as heat, such as driving chemical reactions, generating hysteresis effects, or eddy current in transformers.

The only time we can say for certain that power is totally converted into heat energy is if we assume our black box is a perfect resistor (100 percent ohmic in nature). Only then can we substitute Ohm’s law into the generalized power equation:

 P = VI = V(V/R) = V2/R (2.10)

or

P = VI = (IR)I = I2R

In this form, the power lost due to heating is often called Ohmic heating, Joule heating, or I2R loss. Be careful how you interpret this law. For example, let’s consider our black box that drew 1 W of power. Given the power and the current—which we measure—it would be easy to assume that the resistance of the black box is:

Accordingly, we would say the black box is a 100-Ω resistor generating 1 W of heat. As you can see, this is an erroneous assumption, since we have disregarded the internal workings of the black box—we didn’t account for devices that perform useful work. You’ll often see people treat any load (black box) as a resistor when doing circuit analysis and such. This will get you the right answer when solving for a particular variable, but it is an analysis trick and shouldn’t be used to determine how much heat is being generated, unless, of course, the black box is actually a resistor.

The following example provides some insight into where power is being used and how much of it is being converted into heat.

FIGURE 2.23

The total power pumped into this circuit gets converted into useful work and heat. The total “pump-in” power is:

Ptot = IV = (0.757 A)(12 V) = 9.08 W

(This is based on an open terminal battery measurement, without the rest of the circuit attached.) From there, we notice that some of the total power is wasted within the internal resistance of the battery, within the internal resistance of the wires, and within the resistance of the current-limiting resistor used by the LED. The power used to create light from the lamp and LED can be considered useful power used. However, since we can’t really separate the heat from the light power for these devices, we must apply the generalized power law to them and be content with that. According to the conservation of energy (or power), all individual powers within the circuit add up to the total power.

Example: With an ammeter and voltmeter, you measure the current drawn by a computer to be 1.5 A and the voltage entering to be 117 V. How much power does the computer consume? Can we say how much power is lost to heat?

Answer: P = IV = (1.5 A)(117 V) = 176 W. Knowing how much of the power is lost to heating is practically impossible to measure without taking the computer apart.

Example: Determine the resistance of the following four round rods of material, each 1 m long and 2 mm in diameter: copper, brass, stainless steel, and graphite. Also figure out how much power is lost to heating if a current of 0.2 A flows through each one.

Using Table 2.2:

ρcopper = 1.72 × 10−8 Ωm, ρbrass = 7.0 × 10−8 Ωm, ρsteel = 7.2 × 10−7 Ωm, ρgraphite = 3.5 × 10−5 Ωm

Substituting this into the resistance expression, we get:

Rcopper = 5.48 × 10−3 Ω, Rbrass = 2.23 × 10−2 Ω, Rsteel = 2.31 × 10−1 Ω, Rgraphite = 11.1 Ω

The power loss we get by using Eq. 2.10P = I2R = (0.2 A)2R = (0.04 A2)R:

Pcopper = 2.2 × 10−4 W, Pbrass = 8.9 × 10−4 W, Psteel = 9.2 × 10−3 W, Pgraphite = 0.44 W

2.8 Thermal Heat Conduction and Thermal Resistance

How is the energy transferred in heating? Within a gas, heat transfer represents the transfer of energy between colliding gas molecules. Gas molecules at a hotter temperature move around more quickly—they have a high kinetic energy. When they are introduced into another, colder-temperature gas, the “hotter,” fast-moving molecules impart their energy to the slower-moving molecules. Gases tend to be the worst thermal conductors, due to a low density of molecules.

In nonmetals, heat transfer is a result of the transfer of energy due to lattice vibrations—energetically vibrating atoms in one region of a solid (e.g., the region near a flame) transfer their energy to other regions of a solid that have less energetically vibrating atoms. The transfer of heat can be enhanced by cooperative motion in the form of propagating lattice waves, which in the quantum limit are quantized as phonons. The thermal conductivity of nonmetals varies, depending greatly on the lattice structure.

In terms of metals, heat transfer is a result of both lattice vibration effects (as seen in nonmetals) as well as kinetic energy transfer due to mobile free electrons. Recall that free electrons within a metal are moving quite fast (∼106m/s for most metals) at room temperature. Though quantum mechanics is required, it is possible to treat these electrons like a dense gas, capable of increasing its overall energy as heat is added, and likewise, capable of transporting this energy to regions of the metal that are lower in temperature. Notice, however, that an increase in metal temperature, as a whole, also increases the electrical resistance—the drift velocity component of the free electron goes down, due to both increased lattice vibrations and an increase in thermal velocity component. It becomes harder to influence the free electrons with an applied field. Metals are the best thermal conductors, due in part to the additional free electrons.

The energy an object has at temperature T is correlated to the internal energy—a result of its internal atomic/molecular/electron motions. It is not correct, however, to use the word heat, such as “object possesses heat.” Heat is reserved to describe the process of energy transfer from a high-temperature object to a lower-temperature object. According to the first law of thermodynamics, which is a statement about the conservation of energy, the change in internal energy of a system ΔU is equal to the heat QH added to the system and the work W done by the system: ΔU = QH − W. If we assume that no work is done (energy transferred to, say, move a piston, in the case of a gas), we say ΔU = QH. With this assumption, we can take heat to be not a measure of internal energy of the system, but a change in internal energy. The main reason for this concept is that it is very difficult to determine the actual internal energy of a system; changes in internal energy are more meaningful and measurable.

In practice, what is most useful is the rate of heat transfer—the power loss due to heating. With the help of experimental data, the following formula can be used to determine how well certain materials transfer heat:

 (2.11)

Here, k is called the thermal conductivity (measured in W/m °C) of the material in question, and ∇T is the temperature gradient:

Now, the gradient is probably scary to a lot of you—it is simply a way to represent temperature distributions in 3-D, with time. To keep things simple, we will stick to 2-D, and represent the gradient as acting through an area A through a thickness L, and assume steady-state conditions:

 (2.12)

where ΔT = Thot − Tcold, measured at points across the length L of the material. The material may be steel, silicon, copper, PC board material, and so on. Figure 2.24 shows a picture of the situation.

FIGURE 2.24

When one end of a block of material is placed at a hot temperature, heat will be conducted through the material to the colder end. The rate of heat transfer, or power due to heating, depends on the thermal resistance of the material, which in turn is dependent on the geometry of the material and the material’s thermal resistivity. A weird-looking resistor-like symbol is used here to represent the thermal resistance. Table 2.4 shows the thermal resistivities of various materials.

TABLE 2.4 Typical Thermal Resistivities (λ) in Units of °C·in/W

 Divide λ by 39 to get value in units of °C·m/W MATERIAL λ MATERIAL λ MATERIAL λ Diamond 0.06 Lead 1.14 Quartz 27.6 Silver 0.10 Indium 2.1 Glass (774) 34.8 Copper 0.11 Boron nitride 1.24 Silicon grease 46 Gold 0.13 Alumina ceramic 2.13 Water 63 Aluminum 0.23 Kovar 2.34 Mica 80 Beryllia ceramic 0.24 Silicon carbide 2.3 Polyethylene 120 Molybdenum 0.27 Steel (300) 2.4 Nylon 190 Brass 0.34 Nichrome 3.00 Silicon rubber 190 Silicon 0.47 Carbon 5.7 Teflon 190 Platinum 0.54 Ferrite 6.3 P.P.O. 205 Tin 0.60 Pyroceram 11.7 Polystyrene 380 Nickel 0.61 Epoxy (high conductivity) 24 Mylar 1040 Tin solder 0.78 Air 2280

Thermal conductivity k, like electrical conductivity, has an inverse—namely, thermal resistivity λ. Again, one tells you “how good” a material is at transferring heat, the other tells you “how bad” it is at doing it. The two are related by k = 1/λ.

If we consider the geometry of the material, we can create a notion of thermal resistance ℜtherm (analogous to electrical resistance), which depends on the cross-sectional area A, the length of the block of material L, and the thermal conductivity k, or resistivity λ:

 (2.13)

Thermal resistance has units of °C/W.

Thus, putting everything together, the power transfer of heat across a block of material, from one point at one temperature to another point at a different temperature, can be expressed as:

(2.14)

(k = thermal conductance) (λ = thermal resistivity) (ℜtherm = thermal resistance)

A very useful property of Eq. 2.14 is that it is exactly analogous to Ohm’s law, and therefore the same principles and methods apply to heat flow problems as to circuit problems. For example, the following correspondences hold:

 Thermal conductivity k [W/(m °C)] Electrical conductivity σ [S/m or (Ωm)−1] Thermal resistivity λ [m°C/W] Electrical resistivity ρ [Ωm] Thermal resistance ℜtherm [°C/W] Electrical resistance R [Ω] Thermal current (heat flow) Pheat [W] Electrical current I [A] Thermal potential difference ΔT [°C] Electrical potential difference or voltage V [V] Heat source Current source

Example: Calculate the temperature of a 4-in (0.1 m) piece of #12 copper wire at the end that is being heated by a 25-W (input power) soldering iron, and whose other end is clamped to a large metal vise (assume an infinite heat sink), if the ambient temperature is 25°C (77°F).

Answer: First, calculate the thermal resistance of the copper wire (diameter #12 wire is 2.053 mm, cross-sectional area is 3.31 × 10−6 m2):

Then, rearranging the heat flow equation and making a realistic assumption that only around 10 W of heat actually is transferred to the wire from the 25-watt iron, we get:

ΔT = Pheattherm = (10 W)(77.4°C/W) = 774°C

So the wire temperature at the hot end is estimated at:

25°C + ΔT = 799°C (or 1470°F)

It’s important to realize that in the preceding example, we assume steady-state conditions, where the soldering iron has been held in place for a long time. Also, the assumption that only 10 W were transferred is an important point, since there is a lot of heat being radiated off as heat to the air and the iron handle, and so on. In any case, things can get very hot, even when moderate power levels are in question.

2.8.1 Importance of Heat Production

Heat production has its place in electronics (toasters, hair dryers, water heaters, etc.), but most of the time heat represents power loss that is to be minimized whenever possible, or at least taken into consideration when selecting components. All real circuit components—not just resistors but things like capacitors, transformers, transistors, and motors—contain inherent internal resistances. Though these internal resistances can often be neglected, in some situations they cannot be ignored.

Major problems arise when unintended heat generation increases the temperature of a circuit component to a critical point, causing component failure by explosion, melting, or some other catastrophic event. Less severe problems may surface as a component becomes thermally damaged, resulting in a change in characteristic properties, such as a shift in resistance that may cause undesirable effects in circuit behavior.

To avoid problems associated with heat production, it’s important to use components that are rated to handle two to three or more times the maximum power they are expected to dissipate. In cases where heat presents a shift in component parameter performance, selecting a component with a lower temperature coefficient (TC) will help.

Heat dissipation (more correctly, the efficient removal of generated heat) becomes very important in medium- to high-power circuits—power supplies, amplifier stages, transmitting circuits, and power-hungry circuits with power transistors. There are various techniques to remove heat from a circuit in order to lower the operating temperature of components to below critical levels. Passive methods include heat sinks, careful component layout, and ventilation. Heat sinks are special devices that are used to draw heat away from temperature-sensitive devices by increasing the radiating surface in air—which acts like a cooling fluid for conduction. Active methods include forced air (fans) or some sort of liquid cooling. We’ll discuss these methods throughout the book.

Example: Figure 2.25 shows a thin-film resistor in an integrated circuit. How hot will it get with 2 W dissipated over its 0.1 × 0.2-in surface (100 W/in2)? Assume the ground plane is at 80°C.

FIGURE 2.25

Answer: Since we have three different media through which this heat will be transferred, we must take into consideration thermal conduction within each. With the help of Eq. 2.14 and Table 2.4, we get the individual heat transfers through each region:

ΔT1−4 = 5.3°C + 9.2°C + 2.9°C = 17.4°C

Adding this to the 80°C for the aluminum ground plane leads to an estimated 100°C for maximum resistor temperature. This is a conservative estimate, since it neglects transverse heat spreading.

2.9 Wire Gauges

In Sec. 2.5, we saw that the current density within a copper wire increased as the diameter of the wire decreased. As it turns out, a higher current density translates into a hotter wire; there are more collisions occurring between electrons and copper lattice ions. There is a point where the current density can become so large that the vibrating effect can overpower the copper-lattice binding energy, resulting in wire meltdown (also referred to as the fusing point). To prevent this from occurring, it is important to select the appropriate wire size for anticipated current levels. Wire size is expressed in gauge number—the common standard being the American Wire Gauge (AWG)—whereby a smaller gauge number corresponds to a larger-diameter wire (high current capacity). Table 2.5 shows a short list of common AWG wires. Section 3.1, on wires and cables, provides a more in-depth list.

FIGURE 2.26

TABLE 2.5 Copper Wire Specifications (Bare and Enamel-Coated Wire)

 WIRE SIZE (AWG) DIAMETER (MILS)* AREA (CM)† FEET PER POUND BARE OHMS PER 1000 FT, 25°C CURRENT CAPACITY (AMPS) 4 204.3 41738.49 7.918 0.2485 59.626 8 128.5 16512.25 25.24 0.7925 18.696 10 101.9 10383.61 31.82 0.9987 14.834 12 80.8 6528.64 50.61 1.5880 9.327 14 64.1 4108.81 80.39 2.5240 5.870 18 40.3 1624.09 203.5 6.3860 2.320 20 32 1024.00 222.7 10.1280 1.463 22 25.3 640.09 516.3 16.2000 0.914 24 20.1 404.01 817.7 25.6700 0.577 28 12.6 158.76 2081 65.3100 0.227 32 8.0 64.00 5163 162.0000 0.091 40 3.1 9.61 34364 1079.0000 0.014 * 1 mil = 0.001 in or 0.0254 mm. † A circular mil (CM) is a unit of area equal to that of a 1-mil-diameter circle. The CM area of a wire is the square of the mil diameter. Diameters of wires in Fig. 2.26 are relative and not to scale.

Example: A load device that is known to vary in output power from 0.1 mW to 5 W is to be connected to a 12-V source that is 10 ft away from the load. Determine the minimum wire gauge, provided by Table 2.5, that can safely support any anticipated current drawn by the load.

Answer: We only care about the maximum power level, so using the generalized power law:

Given only the selection of wire gauges provided by Table 2.5, a 22-gauge wire with a 0.914-A rating would work, though we could be conservative and select an 18-gauge wire with a 2.32-A rating. Since the length is so short, there is no appreciable drop in voltage through the wire, so we can ignore the length.

Example: A 10-Ω heating device is powered by a 120-VAC source. How much current does it draw, and what size conductors should be used to connect to the device?

Answer: 120 VAC is an RMS value of a sinusoidal voltage—in this case, household line voltage. Though we’ll discuss this later when we cover ac, it can be treated like a dc voltage in terms of power dissipated through a resistor. So,

A 10-gauge wire would support this kind of current, though a larger 8-gauge wire would be safer.

Example: Why shouldn’t you connect a wire across a voltage source? For example, if you connect a 12-gauge wire directly across a 120-V source (120-V mains outlet), what do you think will happen? What will happen when you do this to a 12-V dc supply, or to a 1.5-V battery?

Answer: In the 120-V mains case, you will likely cause a huge spark, possibly melting the wire and perhaps in the process receiving a nasty shock (if the wire isn’t insulated). But more likely, your circuit breaker in the home will trip, since the wire will draw a huge current due to its low resistance—breakers trip when they sense a large level of current flowing into one of their runs. Some are rated at 10 A, others at 15 A, depending on setup. In a good dc supply, you will probably trip an internal breaker or blow a fuse, or in a bad supply, ruin the inner circuitry. In the case of a battery, there is internal resistance in the battery, which will result in heating of the battery. There will be less severe levels of current due to the internal resistance of the battery, but the battery will soon drain, possibly even destroying the battery, or in an extreme case causing the battery to rupture.

2.10 Grounds

As we left off in Sec. 2.3, we saw that understanding voltage is a relativity game. For example, to say that a point in a circuit has a voltage of 10 V is meaningless unless you have another point in the circuit with which to compare it. Often you define a point in the circuit to be a kind of 0-V reference point on which to base all other voltage measurements. This point is often called the ground, and is frequently represented by the symbol shown in Fig. 2.27:

FIGURE 2.27

For example, Fig. 2.28 shows various ways in which to define voltages by selecting a ground—which in this case is simply a 0-V reference marker. The single battery provides a 1.5-V potential difference or voltage between its terminals. We can simply place a 0-V reference ground at the negative terminal and then state that the positive terminal sits at 1.5 V relative to the 0-V reference ground. The 0-V reference, or the negative terminal of the battery, is called the return. If a load, such as a lamp or resistor, is placed between the terminals, a load current will return to the negative terminal.

FIGURE 2.28

In the center diagram in Fig. 2.28, we have two 1.5-V batteries placed end to end. When batteries are linked this way, their voltages add, creating a combined total voltage of 3.0 V. With the 0-V reference ground at the bottom, we get 1.5-V and 3.0-V readings at the locations shown in the figure. A load placed across the two batteries (3.0-V difference) will result in a load current that returns to the lower battery’s negative terminal. In this case, the return is through the 0-V reference—the lower battery’s negative terminal.

Finally, it is possible to create a split supply by simply repositioning the 0-V ground reference, placing it between the batteries. This creates +1.5 V and −1.5 V leads relative to the 0-V reference. Many circuits require both positive and negative voltage relative to a 0-V ground reference. In this case, the 0-V ground reference acts as a common return. This is often necessary—say, in an audio circuit—where signals are sinusoidal and alternate between positive and negative voltage relative to a 0-V reference.

Now, the ground symbol shown in Fig. 2.27 as a 0-V reference, or as a return, is used all the time by various people. As it turns out, however, it really is supposed to represent a true earth ground—a physical connection to the earth through a conductive material buried in the earth. For whatever reasons, the symbol’s dual meaning has survived, and this can often be a source of confusion to beginners.

2.10.1 Earth Ground

The correct definition of an earth ground is usually a connection terminated at a rod driven into the earth to a depth of 8 ft or more. This earth ground rod is wired directly to a mains breaker box’s ground bar and sent to the various ac outlets in one’s home via a green-coated or bare copper wire that is housed within the same mains cable as hot and neutral wires. The ground can then be accessed at the outlet at the ground socket. Metal piping buried in the earth is often considered an earth ground. See Fig. 2.29.

FIGURE 2.29 Earth ground.

A physical link to the earth is important because the earth provides an electrically neutral body; equal numbers of positive and negative charges are distributed through its entirety. Due to the earth’s practically infinite charge neutrality, attempts at changing the earth potential, via electrical generators, batteries, static electrical mechanisms, or the like, will have essentially no measurable effect. Any introduction of new charge into the earth is quickly absorbed (the earth’s moist soil is usually rather conductive). Such charge interactions occur constantly throughout the planet, and the exchanges average out to zero net charge.

For practical purposes, then, the earth is defined to be at a zero potential (relative to other things)—a potential that is practically immune to wavering. This makes the earth a convenient and useful potential on which to reference other signals. By connecting various pieces of electronics equipment to the earth ground, they can all share the earth’s ground reference potential, and thus all devices share a common reference.

The actual physical connection to earth ground at a particular piece of equipment is usually through the power cord’s ground wire that links to the mains ground wire network when the device is plugged in. The ground wire from the power cord is typically connected internally to the equipment chassis (frame) and, more important for our discussion, to the return portion of a channel that emanates from the interior circuitry. This is then brought out as a ground lead terminal. For example, in Fig. 2.30, an oscilloscope, function generator, and generic audiovisual device use BNC and UHF connectors for input and output channels. Internally, the outer connector body of the BNC or UHF jack is wired to the return (or source) portion of the channel, while a central conductor wire (insulated from the outer body) is wired to the source (or return) portion of the channel. The important part, now, is that the return, or outer, connector is also internally wired to the mains ground wire through the power cord cable. This sets the return to an earth ground reference. In the case of the dc power supply, a separate earth ground terminal is presented at the face in the form of a banana jack terminal. In order to ground the dc supply, a jumper wire must be connected between the negative supply terminal and the ground terminal. If no jumper is used, the supply is said to be floating.

FIGURE 2.30 Illustration showing how various test instruments and an audiovisual device share a common ground connection through the mains ground wiring.

All the grounded pieces of equipment share a common ground. To prove this to yourself, try measuring the resistance between the ground terminals of any two separate pieces of test equipment in your lab. If the devices are properly grounded, you will get a measurement of 0 Ω (but a bit more for internal resistances).

Besides acting simply as a reference point, grounding reduces the possibility of electrical shock if something within a piece of equipment should fail and the chassis or cabinet becomes “hot.” If the chassis is connected to a properly grounded outlet via a three-wire electrical system ground, the path of current flow from the hot chassis will be toward ground, not through your body (which is a more resistive path). A ground system to prevent shock is generally referred to as a dc ground. We’ll discuss shock hazards and grounding later, when we cover ac.

Grounding also helps eliminate electrostatic discharge (ESD) when a statically charged body comes in contact with sensitive equipment. The charged body could be you, after a stroll you took across the carpet. Some ICs are highly vulnerable to damage from ESD. By providing a grounded work mat or using a grounded wrist strap while working with sensitive ICs, you can avoid destroying your chips by ensuring that charge is drained from your body before you touch anything.

Another big job the ground system does is provide a low-impedance path to ground for any stray RF current caused by stray radiofrequency-producing devices, such as electrical equipment, radio waves, and so on. Stray RF can cause equipment to malfunction and contributes to RFI problems. This low-impedance path is usually called RF ground. In most cases, the dc ground and the RF ground are provided by the same system.

Common Grounding Error

As previously mentioned, the ground symbol, in many cases, has been used as a generic symbol in circuit diagrams to represent the current return path, even though no physical earth ground is used. This can be confusing for beginners when they approach a three-terminal dc power supply that has a positive (+), negative (−), and ground terminal. As we have learned, the ground terminal of the supply is tied to the case of the instrument, which in turn is wired to the mains earth ground system. A common mistake for a novice to make is to attempt to power a load, such as a lamp, using the positive and ground terminals of the supply, as shown in Fig. 2.31a. This, however, doesn’t complete a current return path to the energy source (supply), so no current will flow from the source; hence, the load current will be zero. The correct procedure, of course, is to either connect the load between the positive and negative terminals directly, thus creating a floating load, or, using a jumper wire between the ground and negative supply, create a grounded load. Obviously, many dc circuits don’t need to be grounded—it will generally neither help nor hinder performance (e.g., battery-powered devices need no such connection).

FIGURE 2.31

Circuits that require both positive and negative voltage require a power supply to provide each polarity. The supply for the positive voltage will have the negative terminal as a return, and the negative supply will have the positive terminal as the return. These two terminals are connected together, forming a common return path for load current, as shown in Fig. 2.32. The connection between the negative and the positive terminals of the supplies results in a common or floating return. The floating common may be connected to the earth ground terminal of a supply, if a particular circuit requests this. Generally, it will neither help nor hinder circuit performance.

FIGURE 2.32

Unfortunately, the earth ground symbol is used a bit too loosely in electronics, often meaning different things to different people. It is used as a 0-V reference, even though no actual connection is made to Earth. Sometimes it actually means to connect a point in a circuit to earth ground. Sometimes it is used as a generic return—to eliminate the need to draw a return wire. It could be used as an actual earth ground return (using the mains ground copper wire), though this is unwise. See Fig. 2.33. To avoid complications, alternative symbols are used, which we discuss next.

FIGURE 2.33

2.10.2 Different Types of Ground Symbols

To avoid misinterpretations regarding earth grounding, voltage references, and current return paths, less ambiguous symbols have been adopted. Figure 2.34 shows an earth ground (could be Earth or reference), a frame or chassis ground, and a digital and analog reference ground. Unfortunately, the common return for digital and analog is a bit ambiguous, too, but usually a circuit diagram will specify what the symbol is referring to.

FIGURE 2.34

Table 2.6 provides a rundown on meanings behind these symbols.

TABLE 2.6 Types of Ground Symbols

2.10.3 Loose Ends on Grounding

There are a few loose ends on grounding that need mentioning. They are discussed here, with reference to Fig. 2.35.

FIGURE 2.35

Shock Hazard

In instances where high voltages are required and chassis grounds or metal frames are used as return paths, shock hazardous conditions can be created if the earth grounds are neglected. For example, in Fig. 2.35a when a load circuit uses a metal enclosure as a chassis ground, resistive leakage paths (unwanted resistive paths) can exist, which result in high voltages between the enclosure and earth ground. If, inadvertently, an earth-grounded object, such as a grounded metal pipe, and the circuit chassis are simultaneously touched, a serious shock will result. To avoid this situation, the chassis is simply wired to an earth ground connection, as shown in Fig. 2.35b. This places the metal pipe and the enclosure at the same potential, eliminating the shock hazard. Similar hazardous conditions can develop in household appliances. Electrical codes require that appliance frames, such as washers and dryers, be connected to earth ground.

Grounding and Noise

The most common cause of noise in large-scale electronic systems is lack of good grounding practices. Grounding is a major issue for practicing design and system engineers. Though it is not within the scope of this book to get into the gory details, we’ll mention some basic practices to avoid grounding problems in your circuits.

If several points are used for ground connections, differences in potential between points caused by inherent impedance in the ground line can cause troublesome ground loops, which will cause errors in voltage readings. This is illustrated in Fig. 2.35c, where two separated chassis grounds are used. VG represents a voltage existing between signal ground and the load ground. If voltage measurements are made between the load ground and the input signal, VS, an erroneous voltage, (VS + VG) is measured. A way to circumvent this problem is to use a single-point ground, as shown in Fig. 2.35d.

The single-point ground concept ensures that no ground loops are created. As the name implies, all circuit grounds are returned to a common point. While this approach looks good on paper, it is usually not practical to implement. Even the simplest circuits can have 10 or more grounds. Connecting all of them to a single point becomes a nightmare. An alternative is to use a ground bus.

Analog and Digital Grounds

Devices that combine analog and digital circuitry should, in general, have their analog and digital grounds kept separate, and eventually connected together at one single point. This is to prevent noise from being generated within the circuits due to a ground current. Digital circuits are notorious for generating spikes of current when signals change state. Analog circuits can generate current spikes when load currents change or during slewing. In either case, the changing currents are impressed across the ground-return impedance, causing voltage variations (use Ohm’s law) at the local ground plane with respect to the system reference ground, often located near the power supply. The ground return impedance consists of resistive, capacitive, and inductive elements, though resistance and inductance are predominant. If a constant current is impressed across the ground return, resistance is primary, and a dc offset voltage will exist. If the current is alternating, resistance, inductance, and capacitance all play a role, and a resulting high-frequency ac voltage will exist. In either case, these voltage variations get injected into the local circuits and are considered noise—capable of screwing up sensitive signal levels used within the local circuits. There are a number of tricks to reduce noise (such as adding capacitors to counterbalance the inductance), but a good trick is to keep the digital and analog ground separated, then attach them together at one single point.

Example: What do the following symbols represent?

FIGURE 2.36

Answer: (a) Probably an analog circuit ground that is terminated at the supply to an actual earth ground connection. (b) A chassis ground that is connected to earth ground, probably to prevent shock hazards. (c) Appears like an analog ground return that is linked to both the chassis and the earth ground. (d) A floating chassis that is connected to circuit return ground—a potential shock hazard. (e) Separate analog and digital grounds that are linked to a common ground point near the supply, which in turn is grounded to earth.

2.11 Electric Circuits

Though we have already shown circuits, let’s define circuits in basic terms. An electric circuit is any arrangement of resistors, wires, or other electrical components (capacitors, inductors, transistors, lamps, motors, etc.) connected together that has some level of current flowing through it. Typically, a circuit consists of a voltage source and a number of components connected together by means of wires or other conductive means. Electric circuits can be categorized as series circuits, parallel circuits, or a combination of series and parallel parts. See Fig. 2.37.

Basic Circuit

A simple lightbulb acts as a load (the part of the circuit on which work must be done to move current through it). Attaching the bulb to the battery’s terminals, as shown, will initiate current flow from the positive terminal to the negative terminal. In the process, the current will power the filament of the bulb, and light will be emitted. (Remember that the term current here refers to conventional current—electrons are actually flowing in the opposite direction.)

Series Circuit

Connecting load elements (lightbulbs) one after the other forms a series circuit. The current through all loads in series will be the same. In this series circuit, the voltage drops by a third each time current passes through one of the bulbs (all bulbs are exactly the same). With the same battery used in the basic circuit, each light will be one-third as bright as the bulb in the basic circuit. The effective resistance of this combination will be three times that of a single resistive element (one bulb).

Parallel Circuit

A parallel circuit contains load elements that have their leads attached in such a way that the voltage across each element is the same. If all three bulbs have the same resistance values, current from the battery will be divided equally into each of the three branches. In this arrangement, lightbulbs will not have the dimming effect, as was seen in the series circuit, but three times the amount of current will flow from the battery, hence draining it three times as fast. The effective resistance of this combination will be one-third that of a single resistive element (one bulb).

Combination of Series and Parallel

A circuit with load elements placed both in series and in parallel will have the effects of both lowering the voltage and dividing the current. The effective resistance of this combination will be three-halves that of a single resistive element (one bulb).

FIGURE 2.37

Circuit Analysis

Following are some important laws, theorems, and techniques used to help predict what the voltage and currents will be within a purely resistive circuit powered by a direct current (dc) source, such as a battery.

2.12 Ohm’s Law and Resistors

Resistors are devices used in circuits to limit current flow or to set voltage levels within circuits. Figure 2.38 shows the schematic symbol for a resistor; two different forms are commonly used. Schematic symbols for variable resistors—resistors that have a manually adjustable resistance, as well as a model of a real-life resistor, are also shown. (The real-life model becomes important later on when we deal with high-frequency ac applications. For now, ignore the model.)

FIGURE 2.38

If a dc voltage is applied across a resistor, the amount of current that will flow through the resistor can be found using Ohm’s law. To find the power dissipated as heat by the resistor, the generalized power (with Ohm’s law substitution) can be used.

 V = I × R (2.15) Ohm’s law
 P = IV = V2/R = I2R (2.16) Ohm’s power law

R is the resistance or the resistor expressed in ohms (Ω), P is the power loss in watts (W), V is the voltage in volts (V), and I is the current in amperes (A).

UNIT PREFIXES

Resistors typically come with resistance values from 1 Ω to 10,000,000 Ω. Most of the time, the resistance is large enough to adopt a unit prefix convention to simplify the bookkeeping. For example, a 100,000-Ω resistance can be simplified by writing 100 kΩ (or simply 100k, for short). Here k = ×1000. A 2,000,000-Ω resistance can be shortened to 2 MΩ (or 2M, for short). Here M = ×1,000,000.

Conversely, voltages, currents, and power levels are usually small fractions of a unit, in which case it is often easier to use unit prefixes such as m (milli or ×10−3), µ (micro or × 10−6), n (nano or ×10−9), or even p (pico or ×10−12). For example, a current of 0.0000594 A (5.94 × 10−5 A) can be written in unit prefix form as 59.4 µA. A voltage of 0.0035 (3.5 × 10−3 V) can be written in unit prefix form as 3.5 mV. A power of 0.166 W can be written in unit prefix form as 166 mW.

Example: In Fig. 2.39, a 100-Ω resistor is placed across a 12-V battery. How much current flows through the resistor? How much power does the resistor dissipate?

FIGURE 2.39

2.12.1 Resistor Power Ratings

Determining how much power a resistor dissipates is very important when designing circuits. All real resistors have maximum allowable power ratings that must not be exceeded. If you exceed the power rating, you’ll probably end up frying your resistor, destroying the internal structure, and thus altering the resistance. Typical general-purpose resistors come in ⅛-, ¼-, ½-, and 1-W power ratings, while high-power resistors can range from 2 to several hundred watts.

So, in the preceding example, where our resistor was dissipating 1.44 W, we should have made sure that our resistor’s power rating exceeded 1.44 W; otherwise, there could be smoke. As a rule of thumb, always select a resistor that has a power rating at least twice the maximum value anticipated. Though a 2-W resistor would work in our example, a 3-W resistor would be safer.

To illustrate how important power ratings are, we examine the circuit shown in Fig. 2.40. The resistance is variable, while the supply voltage is fixed at 5 V. As the resistance increases, the current decreases, and according to the power law, the power decreases, as shown in the graphs. As the resistance decreases, the current and power increase. The far right graph shows that as you decrease the resistance, the power rating of the resistor must increase; otherwise, you’ll burn up the resistor.

FIGURE 2.40

Example 1: Using an ammeter, you measure a current of 1.0 mA through a 4.7-kΩ resistor. What voltage must exist across the resistor? How much power does the resistor dissipate?

Answer: V = IR = (0.001 A)(4700 Ω) = 4.7 V; P = I2 × R = (0.001 A)2 × (4700 Ω) = 0.0047 W = 4.7 mW

Example 2: Using a voltmeter, you measure 24 V across an unmarked resistor. With an ammeter, you measure a current of 50 mA. Determine the resistance and power dissipated in the resistor.

Answer: R = V/I = (24 V)/(0.05 A) = 480 Ω; P = I × V = (0.05 A) × (24 V) = 1.2 W

Example 3: You apply 3 V to a 1-MΩ resistor. Find the current through the resistor and the power dissipated in the process.

Answer: I = V/R = (3 V)/(1,000,000 Ω) = 0.000003 A = 3 µA; P = V2/R = (3 V)2/(1,000,000 Ω) = 0.000009 W = 9 µW

Example 4: You are given 2-Ω, 100-Ω, 3-kΩ, 68-kΩ, and 1-MΩ resistors, all with 1-W power ratings. What’s the maximum voltage that can be applied across each of them without exceeding their power ratings?

Answer: P = V2/R ⇒ V = ; voltages must not exceed 1.4 V (2 Ω), 10.0 V (100 Ω), 54.7 V (3 kΩ), 260.7 V (68 kΩ), 1000 V (1 MΩ)

2.12.2 Resistors in Parallel

Rarely do you see circuits that use a single resistor alone. Usually, resistors are found connected in a variety of ways. The two fundamental ways of connecting resistors are in series and in parallel.

When two or more resistors are placed in parallel, the voltage across each resistor is the same, but the current through each resistor will vary with resistance. Also, the total resistance of the combination will be lower than that of the lowest resistance value present. The formula for finding the total resistance of resistors in parallel is:

 (2.17)
 (2.18) Two resistors in parallel

The dots in the equation indicate that any number of resistors can be combined. For only two resistances in parallel (a very common case), the formula reduces to Eq. 2.18.

(You can derive the resistor-in-parallel formula by noting that the sum of the individual branch currents is equal to the total current: Itotal = I1 + I2 + I3 + … IN. This is referred to as Kirchhoff’s current law. Then, applying Ohm’s law, we get: Itotal = V1/R1 + V2/R2 + V3/R3 + … VN/RN. Because all resistor voltages are equal to Vtotal since they share the same voltage across them, we get: Itotal = Vtotal/R1 + Vtotal/R2 + Vtotal/R3 + … Vtotal/RN. Factoring out Vtotal, we get: Itotal = Vtotal (1/R1 + 1/R2 + 1/R3 … 1/R4). We call the term in brackets Rtotal.

Note that there is a shorthand for saying that two resistors are in parallel. The shorthand is to use double bars | | to indicate resistors in parallel. So to say R1 is in parallel with R2, you would write R1 | | R2. Thus, you can express two resistors in parallel in the following ways:

In terms of arithmetic order of operation, the | | can be treated similar to multiplication or division. For example, in the equation Zin = R1 + R2 | | Rload, you calculate R2 and Rload in parallel first, and then you add R1.

Example 1: If a 1000-Ω resistor is connected in parallel with a 3000-Ω resistor, what is the total or equivalent resistance? Also calculate total current and individual currents, as well as the total and individual dissipated powers.

To find how much current flows through each resistor, apply Ohm’s law:

These individual currents add up to the total input current:

Iin = I1 + I2 = 12 mA + 4 mA = 16 mA

This statement is referred to as Kirchhoff’s current law. With this law, and Ohm’s law, you come up with the current divider equations, shown at the bottom of Fig. 2.41. These equations come in handy when you know the input current but not the input voltage.

FIGURE 2.41

We could have just as easily found the total current using:

Iin = Vin/Rtotal = (12 V)/(750 Ω) = 0.016 A = 16 mA

To find how much power resistors in parallel dissipate, apply the power law:

Ptot = IinVin = (0.0016 A)(12 V) = 0.192 W = 192 mW

P1 = I1Vin = (0.012 A)(12 V) = 0.144 W = 144 mW

P2 = I2Vin = (0.004 A)(12 V) = 0.048 W = 48 mW

Example 2: Three resistors R1 = 1 kΩ, R2 = 2 kΩ, R3 = 4 kΩ are in parallel. Find the equivalent resistance. Also, if a 24-V battery is attached to the parallel circuit to complete a circuit, find the total current, individual currents through each of the resistors, total power loss, and individual resistor power losses.

The total resistance for resistors in parallel:

The current through each of the resistors:

I1 = V1/R1 = 24 V/1000 Ω = 0.024 A = 24 mA

I2 = V2/R2 = 24 V/2000 Ω = 0.012 A = 12 mA

I3 = V3/R3 = 24 V/4000 Ω = 0.006 A = 6 mA

The total current, according to Kirchhoff’s current law:

Itotal = I1 + I2 + I3 = 24 mA + 12 mA + 6 mA = 42 mA

The total power dissipated by the parallel resistors:

Ptotal = Itotal × Vtotal = 0.042 A × 24 V = 1.0 W

The power dissipated by individual resistors is shown in Fig. 2.42.

FIGURE 2.42

2.12.3 Resistors in Series

When a circuit has a number of resistors connected in series, the total resistance of the circuit is the sum of the individual resistances. Also, the amount of current flowing through each resistor in series is the same, while the voltage across each resistor varies with resistance. The formula for finding the total resistance of resistors in series is:

 Rtotal = R1 + R2 + R3 + R4 + … (2.19)

The dots indicate that as many resistors as necessary may be added.

You can derive this formula by noting that the sum of all the voltage drops across each series resistor will equal the applied voltage across the combination Vtotal = V1 + V2 + V3 + … + VN. This is referred to as Kirchhoff’s voltage law. Applying Ohm’s law, and noting that the same current I flows through each resistor, we get: IRtotal = IR1 + IR2 + IR3 + … + IRN. Canceling the I’s you get: Rtotal = R1 + R2 + R3 + R4 + …

Example 1: If a 1.0-kΩ resistor is placed in series with a 2.0-kΩ resistor, the total resistance becomes:

Rtot = R1 + R2 = 1000 Ω + 2000 Ω = 3000 Ω = 3 kΩ

When these series resistors are placed in series with a battery, as shown in Fig. 2.43, the total current flow I is simply equal to the applied voltage Vin, divided by the total resistance:

FIGURE 2.43

Since the circuit is a series circuit, the currents through each resistor are equal to the total current I:

I1 = 3 mA, I2 = 3 mA

To find the voltage drop across each resistor, apply Ohm’s law:

V1 = I1 × R1 = 0.003 A × 1000 Ω = 3 V

V2 = I2 × R2 = 0.003 A × 2000 Ω = 6 V

Now, we didn’t really have to calculate the current. We could have just plugged I = Vin/Rtot into I1 and I2 in the preceding equations and got:

(Voltage divider
equations)

These equations are called voltage divider equations and are so useful in electronics that it is worth memorizing them. (See Fig. 2.43.) Often V2 is called the output voltage Vout.

The voltage drop across each resistor is directly proportional to the resistance. The voltage drop across the 2000-Ω resistor is twice as large as that of the 1000-Ω resistor. Adding both voltage drops together gives you the applied voltage of 9 V:

Vin = V1 + V2 9 V = 3 V + 6 V

The total power loss and individual resistor power losses are:

Ptot = IVin = (0.003 A)(9 V) = 0.027 W = 27 mW

(Ptot = I2Rtot = (0.003 A)2(3000 Ω) = 0.027 W = 27 mW)

P1 = I2R1 = (0.003 A)2(1000 Ω) = 0.009 W = 9 mW

P2 = I2R2 = (0.003 A)2(2000 Ω) = 0.018 W = 18 mW

The larger resistor dissipates twice as much power.

Example 2: The input of an IC requires a constant 5 V, but the supply voltage is 9 V. Use the voltage divider equations to create a voltage divider with an output of 5 V. Assume the IC has such a high input resistance (10 MΩ) that it practically draws no current from the divider.

FIGURE 2.44

Answer: Since we assume the IC draws no current, we can apply the voltage divider directly:

We must choose voltage divider resistors, making sure our choice doesn’t draw too much current, causing unnecessary power loss. To keep things simple for now, let’s choose R2 to be 10 kΩ. Rearranging the voltage divider and solving for R1:

Example 3: You have a 10-V supply, but a device that is to be connected to the supply is rated at 3 V and draws 9.1 mA. Create a voltage divider for the load device.

Answer: In this case, the load draws current and can be considered a resistor in parallel with R2. Therefore, using the voltage divider relation without taking the load into consideration will not work. We must apply what is called the 10 percent rule.

FIGURE 2.45

The 10 Percent Rule: This rule is a standard method for selecting R1 and R2 that takes into account the load and minimizes unnecessary power losses in the divider.

The first thing you do is select R2 so that I2 is 10 percent of the desired load current. This resistance and current are called the bleeder resistance and bleeder current. The bleeder current in our example is:

Ibleed = I2 = (0.10)(9.1 mA) = 0.91 mA

Using Ohm’s law, next we calculate the bleeder resistance:

Rbleed = R2 = 3 V/0.00091 A = 3297 Ω

Considering resistor tolerances and standard resistance values, we select a resistor in close vicinity—3300 Ω.

Next, we need to select R1, so that the output is maintained at 3 V.

To do this, simply calculate the total current through the resistor and use Ohm’s law:

I1 = I2 + Iload = 0.91 mA + 9.1 mA = 10.0 mA = 0.01 A

In terms of power ratings:

PR1 = V12/R1 = (7 V)2/(700 Ω) = 0.07 W = 70 mW

PR2 = V22/R2 = (3 V)2/(3300 Ω) = 0.003 W = 3 mW

Low-power ¼-W resistors will suffice.

In actual practice, the computed value of the bleeder resistor does not always come out to an even value. Since the rule of thumb for bleeder current is only an estimated value, the bleeder resistor can be of a value close to the computed value. (If the computed value of the resistance were 500Ω, a 5 percent 510Ω resistor could be used. See the standard resistor values table in the front matter of the book.) Once the actual value of the bleeder resistor is selected, the bleeder current must be recomputed. The voltage developed by the bleeder resistor must be equal to the voltage requirement of the load in parallel with the bleeder resistor. We’ll discuss voltage dividers, as well as more complex divider arrangements, in the section on resistors in Chap. 3.

Example: Find the equivalent resistance of series resistors R1 = 3.3k, R2 = 4.7k, R3 = 10k. If a 24-V battery is attached to the series resistors to complete a circuit, find the total current flow, the individual voltage drops across the resistors, the total power loss, and the individual power loss of each resistor.

FIGURE 2.46

The equivalent resistance of the three resistors:

Rtotal = R1 + R2 + R3

Rtotal = 3.3 kΩ + 4.7 kΩ + 10 kΩ = 18 kΩ

The total current flow through the resistors is:

Using Ohm’s law, the voltage drops across the resistors are:

V1 = Itotal × R1 = 1.33 mA × 3.3 kΩ = 4.39 V

V2 = Itotal × R2 = 1.33 mA × 4.7 kΩ = 6.25 V

V3 = Itotal × R3 = 1.33 mA × 10 kΩ = 13.30 V

The total power dissipated is:

Ptotal = Itotal × Vtotal = 1.33 mA × 24 V = 32 mW

Power dissipated by individual resistors is shown in Fig. 2.46.

2.12.4 Reducing a Complex Resistor Network

To find the equivalent resistance for a complex network of resistors, the network is broken down into series and parallel combinations. A single equivalent resistance for these combinations is then found, and a new and simpler network is formed. This new network is then broken down and simplified. The process continues over and over again until a single equivalent resistance is found. Here’s an example of how reduction works.

Example 1: Find the equivalent resistance of the network attached to the battery by using circuit reduction. Afterward, find the total current flow supplied by the battery to the network, the voltage drops across all resistors, and the individual current through each resistor.

FIGURE 2.47

R2 and R3 form a parallel branch that can be reduced to:

This equivalent resistance and R1 are in series, so their combined resistance is:

Req(2) = R1 + Req(1) = 5.0 kΩ + 4.4 kΩ = 9.4 kΩ

The total current flow through the circuit and through R1 is:

The voltage across R2 and R3 is the same as that across Req(1):

VReq(1) = Itotal × Req(1) = 26.6 mA × 4.4 kΩ = 117 V

V2 = V3 = 117 V

The current through R2 and R3 are found using Ohm’s law:

The voltage across R1 is, by Kirchhoff’s voltage law:

V1 = 250 V − 117 V = 133 V

Alternatively we could have used Ohm’s law:

V1 = I1 × R1 = (26.6 mA)(5.0 kΩ) = 133 V

Example 2: Find the equivalent resistance of the following network, along with the total current flow, the individual current flows, and individual voltage drops across the resistors.

FIGURE 2.48

R3 and R4 can be reduced to resistors in series:

Req(1) = R3 + R4 = 3.3 kΩ + 10.0 kΩ = 13.3 kΩ

This equivalent resistance is in parallel with R2:

This new equivalent resistance is in series with R1:

Req(3) = R1 + Req(2) = 1.0 kΩ + 4.3 kΩ = 5.3 kΩ

The total current flow is:

The voltage across Req(2) or point voltage at b is:

VReq(2) = Itotal × Req(2) = 2.26 mA × 4.3 kΩ = 9.7 V

The voltage across R1 is:

VR1 = Itotal × R1 = 2.26 mA × 1.0 kΩ = 2.3 V

You can confirm this using Kirchhoff’s voltage law:

12 V − 9.7 V = 2.3 V

The current through R2:

The current through Req(1), also through R3 and R4:

You can confirm this using Kirchhoff’s current law:

2.26 mA − 1.43 mA = 0.73 mA.

The voltage across R3:

VR3 = I3 × R3 = 0.73 mA × 3.3 kΩ = 2.4 V

The voltage across R4:

VR4 = I4 × R4 = 0.73 mA × 10.0 kΩ = 7.3 V

You can confirm this using Kirchhoff’s voltage law:

9.7 V − 2.4 V = 7.3 V

2.12.5 Multiple Voltage Dividers

Example 1: You wish to create a multiple voltage divider that powers three loads: load 1 (75 V, 30 mA), load 2 (50 V, 10 mA), and load 3 (25 V, 10 mA). Use the 10 percent rule and Fig. 2.49 to construct the voltage divider.

FIGURE 2.49

An important point in determining the resistor when applying the 10 percent rule of thumb is that to calculate the bleeder current, you must take 10 percent of the total load current. The steps are as follows:

Find the bleeder current, which is 10 percent (0.1) of the total current:

IR4 = 0.1 × (10 mA + 10 mA + 30 mA) = 5 mA

To find R4 (bleeder resistor), use Ohm’s law:

R4 = (25 V − 0 V)/(0.005 A) = 5000 Ω

The current through R3 is equal to the current through R4 plus the current through load 3:

IR3 = IR4 + Iload3 = 5 mA + 10 mA = 15 mA

To find R3, use Ohm’s law, using the voltage difference between load 2 and load 3: R3 = (50 V − 25 V)/(0.015 A) = 1667 Ω or 1.68 kΩ, considering tolerances and standard resistance values.

The current through R2 is:

IR2 = IR3 + Iload2 = 15 mA + 10 mA = 25 mA

Using Ohm’s law, R2 = (75 V − 50 V)/(0.025 A) = 1000 Ω

The current through R1 is:

IR1 = IR2 + Iload1 = 25 mA + 30 mA = 55 mA

Using Ohm’s law: R1 = (100 V − 75 V)/(0.055 A) = 455 Ω

To determine resistor power ratings and total power losses in loads use P = IV. See results in Fig. 2.49.

Example 2: In many cases, the load for a voltage divider requires both positive and negative voltages. Positive and negative voltages can be supplied from a single-source voltage by connecting the ground return between two of the divider resistors. The exact point in the circuit at which the ground is placed depends upon the voltages required by the loads.

For example, the voltage divider in Fig. 2.50 is designed to provide the voltage and current to three loads from a given source voltage:

FIGURE 2.50

Load 1: +50 V @ 50 mA

Load 2: +25 V @ 10 mA

Load 3: −25 V @ 100 mA

The values of R4R2, and R1 are computed exactly as was done in the preceding example. IR4 is the bleeder current and can be calculated as follows:

Calculating R4:

R4 = 25 V/0.016 A = 1562 Ω

To calculate current through R3, use Kirchhoff’s current law (at point A):

IR3 + 10 mA + 50 mA − 16 mA − 100 mA = 0

IR3 = 56 mA

Calculating R3R3 = 25 V/(0.056 A) = 446 Ω.

Calculating IR2IR2 = IR3 + Iload2 = 56 mA + 10 mA = 66 mA.

Then R2 = 25 V/(0.066 A) = 379 Ω.

Calculating IR1 = IR2 + Iload1 = 66 mA + 50 mA = 116 mA.

Then R1 = 25 V/0.116 A = 216 Ω.

Though voltage dividers are simple to apply, they are not regulated in any way. If a load’s resistance changes, or if there is variation in the supply voltage, all loads will experience a change in voltage. Therefore, voltage dividers should not be applied to circuits where the divider will be weighed down by changes in load. For applications that require steady voltage and that draw considerable current, it is best to use an active device, such as a voltage regulator IC—more on this later.

2.13 Voltage and Current Sources

An ideal voltage source is a two-terminal device that maintains a fixed voltage across its terminals. If a variable load is connected to an ideal voltage source, the source will maintain its terminal voltage regardless of changes in the load resistance. This means that an ideal voltage source will supply as much current as needed to the load in order to keep the terminal voltage fixed (in I = V/R, I changes with R, but V is fixed). Now a fishy thing with an ideal voltage source is that if the resistance goes to zero, the current must go to infinity. Well, in the real world, there is no device that can supply an infinite amount of current. If you placed a real wire between the terminals of an ideal voltage source, the calculated current would be so large it would melt the wire. To avoid this theoretical dilemma, we must define a real voltage source (a battery, plug-in dc power supply, etc.) that can supply only a maximum finite amount of current. A real voltage source resembles an ideal voltage source with a small series internal resistance or source resistance rs, which is a result of the imperfect conducting nature of the source (resistance in battery electrolyte and lead, etc.). This internal resistance tends to reduce the terminal voltage, the magnitude of which depends on its value and the amount of current that is drawn from the source (or the size of the load resistance).

FIGURE 2.51

In Fig. 2.52, when a real voltage source is open-circuited (no load connected between its terminals), the terminal voltage (VT) equals the ideal source voltage (VS)—there is no voltage drop across the resistor, since current can’t pass through it due to an incomplete circuit condition.

FIGURE 2.52

However, when a load Rload is attached across the source terminals, Rload and rs are connected in series. The voltage at the terminal can be determined by using the voltage divider relation:

From the equation, you can see that when Rload is very large compared to rs, (1000 times greater or more), the effect of rs is so small that it may be ignored. However, when Rload is small or closer to rs in size, you must take rs into account when doing your calculations and designing circuits. See the graph in Fig. 2.52.

In general, the source resistance for dc power supplies is usually small; however, it can be as high as 600 Ω in some cases. For this reason, it’s important to always adjust the power supply voltage with the load connected. In addition, it is a good idea to recheck the power supply voltage as you add or remove components to or from a circuit.

Another symbol used in electronics pertains to dc current sources—see Fig. 2.52. An ideal current source provides the same amount of source current IS at all times to a load, regardless of load resistance changes. This means that the terminal voltage will change as much as needed as the load resistance changes in order to keep the source current constant.

In the real world, current sources have a large internal shunt (parallel) resistance rs, as shown in Fig. 2.52. This internal resistance, which is usually very large, tends to reduce the terminal current IT, the magnitude of which depends on its value and the amount of current that is drawn from the source (or the size of load resistance).

When the source terminals are open-circuited, IT must obviously be zero. However, when we connect a load resistance Rload across the source terminals, Rload and rs form a parallel resistor circuit. Using the current divider expression, the terminal current becomes:

From this equation, you can see that when Rload is small compared to rs, the dip in current becomes so small it usually can be ignored. However, when Rload is large or closer to rs in size, you must take rs into account when doing your calculations.

A source may be represented either as a current source or a voltage source. In essence, they are duals of each other. To translate between the voltage source and current source representation, the resistance value is kept the same, while the voltage of the source is translated into the current of the source using Ohm’s law. See Fig. 2.53 for details.

FIGURE 2.53

One way to look at an ideal current source is to say it has an internal resistance that is infinite, which enables it to support any kind of externally imposed potential difference across its terminals (e.g., a load’s resistance changes). An approximation to an ideal current source is a voltage source of very high voltage V in series with a very large resistance R. This approximation would supply a current V/R into any load that has a resistance much smaller than R. For example, the simple resistive current source circuit shown in Fig. 2.54 uses a 1-kV voltage source in series with a 1-MΩ resistor. It will maintain the set current of 1 mA to an accuracy within 1 percent over a 0- to 10-V range (0 < Rload < 10 kΩ). The current is practically constant even though the load resistance varies, since the source resistance is much greater than the load resistance, and thus the current remains practically constant. (I = 1000 V/(1,000,000 Ω + 10,0000 Ω. Since 1,000,000 Ω is so big, it overshadows Rload.)

FIGURE 2.54

A practical current source is usually an active circuit made with transistors, as shown in Fig. 2.54cVin drives current through R1 into the base of the second transistor, so current flows into the transistor’s collector, through it, and out its emitter. This current must flow through R2. If the current gets too high, the first transistor turns on and robs the second transistor of base current, so its collector current can never exceed the value shown. This is an excellent way of either making a current source or limiting the available current to a defined maximum value.

2.14 Measuring Voltage, Current, and Resistance

Voltmeters, ammeters, and ohmmeters, used to measure voltage, current, and resistance, ideally should never introduce any effects within the circuit under test. In theory, an ideal voltmeter should draw no current as it measures a voltage between two points in a circuit; it has infinite input resistance Rin. Likewise, an ideal ammeter should introduce no voltage drop when it is placed in series within the circuit; it has zero input resistance. An ideal ohmmeter should provide no additional resistance when making a resistance measurement.

Real meters, on the other hand, have limitations that prevent them from making truly accurate measurements. This stems from the fact that the circuit of a meter requires a sample current from the circuit under test in order to make a display measurement. Figure 2.55 shows symbols for an ideal voltmeter, ammeter, and ohmmeter, along with the more accurate real-life equivalent circuits.

FIGURE 2.55

An ideal voltmeter has infinite input resistance and draws no current; a real voltmeter’s input resistance is several hundred MΩs. An ideal ammeter has zero input resistance and provides no voltage drop; a real ammeter’s input resistance is fractions of ohms. An ideal ohmmeter has zero internal resistance; a real ohmmeter’s internal resistance is fractions of ohms. It is important to read your instrument manuals to determine the internal resistances.

The effects of meter internal resistance are shown in Fig. 2.56. In each case, the internal resistance becomes part of the circuit. The percentages of error in measurements due to the internal resistances become more pronounced when the circuit resistances approach the meters’ internal resistance.

FIGURE 2.56

Figure 2.57 shows the actual makeup of a basic analog multimeter. Note that this device is overly simplistic; real meters are much more sophisticated, with provisions for range selection, ac measurements, and so on. At any rate, it shows you that the meters are nonideal due to the sample current required to drive the galvanometer’s needle.

FIGURE 2.57 The heart of this simple analog multimeter is a galvanometer which measures current flow IG. Current sent through the galvanometer’s wire leads will generate a magnetic field from the coil windings in the center rotor. Since the coil is tilted relative to the N-S alignment of the permanent magnet, the rotor will rotate in accordance with the amount of current flow. Normally you get full deflection of the needle at extremely low current levels, so you can add a parallel resistor to divert current flow away from the galvanometer. The galvanometer can also be used as a voltmeter by placing its lead at the point voltages within the circuit. If there is a voltage difference at the leads, a current will flow into the galvanometer and move the needle in proportion to the voltage magnitude. A series resistor is used to limit current flow and needle deflection. A galvanometer can be used to make an ohmmeter, too, provided you place a battery in series with it. Again, to calibrate needle deflection, a series resistor RO is used.

2.15 Combining Batteries

The two battery networks in Fig. 2.58 show how to increase the supply voltage and/or increase the supply current capacity. To increase the supply voltage, batteries are placed end to end or in series; the terminal voltages of each battery add together to give a final supply voltage equal to the sum from the batteries. To create a supply with added current capacity (increased operating time), batteries can be placed in parallel—positive terminals are joined together, as are negative terminals, as shown in Fig. 2.58. The power delivered to the load can be found using Ohm’s power law: P = V2/R, where V is the final supply voltage generated by all batteries within the network. Note that the ground symbol shown here acts simply as a 0-V reference, not as an actual physical connection to ground. Rarely would you ever connect a physical ground to a battery-operated device.

FIGURE 2.58

Note: When placing batteries in parallel, it’s important that the voltages and chemistry be the same—choose all similar batteries, all fresh. If the voltages are different, you can run into problems.

2.16 Open and Short Circuits

The most common problems (faults) in circuits are open circuits and shorts. A short circuit in all or part of a circuit causes excessive current flow. This may blow a fuse or burn out a component, which may result in an open-circuit condition. An open circuit represents a break in the circuit, preventing current from flowing. Short circuits may be caused by a number of things—from wire crossing, insulation failure, or solder splatter inadvertently linking two separate conductors within a circuit board. An open circuit may result from wire or component lead separation from the circuit, or from a component that has simply burned out, resulting in a huge resistance. Figure 2.59 shows cases of open- and short-circuit conditions. A fuse, symbolized , is used in the circuit and will blow when the current through it exceeds its current rating, given in amps.

FIGURE 2.59

A weird thing about a short circuit, if you consider all components to be ideal in nature, is that an infinite current will flow if an ideal voltage source is short-circuited, while the voltage across the short goes to zero. Real voltage sources have internal resistance, as do the conductors, so the maximum current level is reduced. However, there is usually still plenty of current available to do damage.

You can often diagnose a short circuit by noting a burning smell, or by placing your hand nearby to sense any components overheating. To prevent short circuits from destroying circuits, various protection devices can be used, such as fuses, transient voltage suppressors, and circuit breakers. These devices sense when too much current is flowing, and they will blow or trip, thus creating an open-circuit condition to limit excessive current damage.

Example 1: In the series circuit in Fig. 2.60, determine how much current flows through the normal circuit (a), then determine how much current flows when there is a partial short (b), and determine how much current would flow during a full short if the 1-A fuse weren’t in place. Assume an internal 3-Ω circuit resistance at the moment of a full-short condition.

FIGURE 2.60

Answer: a. 11 mA, b. 109 mA, c. 4 A—fuse blows.

Example 2: In the parallel circuit in Fig. 2.61, determine the total current flow in the normal circuit (a), the open circuit (b), and the current flow in the short circuit (c). Assume the internal resistance of the battery is 0.2 Ω below 3 A, but goes to 2 Ω during a short-circuit condition.

FIGURE 2.61

Answer: a. 3.4 A, b. 2.3 A, c. 6 A—fuse blows.

Example 3: In the parallel circuit in Fig. 2.62, loads B, C, and D receive no power when S2 is closed (assume all other switches are already closed); however, load A receives power. You notice the fuse is blown. After replacing the fuse, you close S2 with S3 and S4 open, and the fuse doesn’t blow. Closing S3 power on loads B and C has no effect. Closing S4 causes B and C to turn off—load D receives no power, either. The fuse is blown again. What is the problem?

FIGURE 2.62

2.17 Kirchhoff’s Laws

Often, you will run across a circuit that cannot be analyzed with simple resistor circuit reduction techniques alone. Even if you could find the equivalent resistance by using circuit reduction, it might not be possible for you to find the individual currents and voltage through and across the components within the network. Likewise, if there are multiple sources, or complex networks of resistors, using Ohm’s law, as well as the current and voltage divider equations, may not cut it. For this reason, we turn to Kirchhoff’s laws.

Kirchhoff’s laws provide the most general method for analyzing circuits. These laws work for either linear (resistor, capacitors, and inductors) or nonlinear elements (diodes, transistors, etc.), no matter how complex the circuit gets. Kirchhoff’s two laws are stated as follows.

Kirchhoff’s Voltage Law (or Loop Rule): The algebraic sum of the voltages around any loop of a circuit is zero:

 (2.20)

In essence, Kirchhoff’s voltage law is a statement about the conservation of energy. If an electric charge starts anywhere in a circuit and is made to go through any loop in a circuit and back to its starting point, the net change in its potential is zero.

To show how Kirchhoff’s voltage law works, we consider the circuit in Fig. 2.63. We start anywhere we like along the circuit path—say, at the negative terminal of the 5-V battery. Then we start making a loop trace, which in this case we choose to go clockwise, though it doesn’t really matter which direction you choose. Each time we encounter a circuit element, we add it to what we’ll call our ongoing loop equation. To determine the sign of the voltage difference, we apply the loop trace rules shown in the shaded section of the figure. We continue adding elements until we make it back to the start of the loop, at which point we terminate the loop equation with an “= 0.”

FIGURE 2.63

As noted, Kirchhoff’s voltage law applies to any circuit elements, both linear and nonlinear. For example, the rather fictitious circuit shown here illustrates Kirchhoff’s voltage law being applied to a circuit that has other elements besides resistors and dc sources—namely, a capacitor, an inductor, and a nonlinear diode, along with a sinusoidal voltage source. We can apply the loop trick, as we did in the preceding example, and come up with an equation. As you can see, the expressions used to describe the voltage changes across the capacitor, inductor, and diode are rather complex, not to mention the solution to the resulting differential equation. You don’t do electronics this way (there are tricks), but it nevertheless demonstrates the universality of Kirchhoff’s law.

FIGURE 2.64

Kirchhoff’s Current Law (or Junction Rule): The sum of the currents that enter a junction equals the sum of the currents that leave the junction:

 (2.21)

Kirchhoff’s current law is a statement about the conservation of charge flow through a circuit: at no time are charges created or destroyed.

The following example shows both Kirchhoff’s current and voltage laws in action.

Example: By applying Kirchhoff’s laws to the following circuit, find all the unknown currents, I1I2I3I4I5I6, assuming that R1R2R3R4R5R6, and V0 are known. After that, the voltage drops across the resistors V1V2V3V4V5, and V6 can be found using Ohm’s law: Vn = InRn.

FIGURE 2.65

Answer: To solve this problem, you apply Kirchhoff’s voltage law to enough closed loops and apply Kirchhoff’s current law to enough junctions that you end up with enough equations to counterbalance the unknowns. After that, it is simply a matter of doing some algebra. Figure 2.66 illustrates how to apply the laws in order to set up the final equations.

In Fig. 2.66, there are six equations and six unknowns. According to the rules of algebra, as long as you have an equal number of equations and unknowns, you can usually figure out what the unknowns will be. There are three ways we can think of to solve for the unknowns in this case. First, you could apply the old “plug and chug” method, better known as the substitution method, where you combine all the equations together and try to find a single unknown, and then substitute it back into another equation, and so forth. A second method, which is a lot cleaner and perhaps easier, involves using matrices. A book on linear algebra will tell you all you need to know about using matrices to solve for the unknown.

FIGURE 2.66

A third method that we think is useful—practically speaking—involves using a trick with determinants and Cramer’s rule. The neat thing about this trick is that you do not have to know any math—that is, if you have a mathematical computer program or calculator that can do determinants. The only requirement is that you be able to plug numbers into a grid (determinant) and press “equals.” We do not want to spend too much time on this technique, so we will simply provide you with the equations and use the equations to find one of the solutions to the resistor circuit problem. See Fig. 2.67a.

FIGURE 2.67a

For example, you can find Δ for the system of equations from the resistor problem by plugging all the coefficients into the determinant and pressing the “evaluate” button on the calculator or computer. See Fig. 2.67b.

FIGURE 2.67b

Now, to find, say, the current through R5 and the voltage across it, you find ΔI5, then use I5 = ΔI5/Δ to find the current. Then you use Ohm’s law to find the voltage. Figure 2.67c shows how it is done.

FIGURE 2.67c

To solve for the other currents (and voltages), simply find the other ΔI’s and divide by Δ.

The last approach, as you can see, requires a huge mathematical effort to get a single current value. For simplicity, you can find out everything that’s going on in the circuit by running it through a circuit simulator program. For example, using MultiSim, we get the results in Fig. 2.68:

FIGURE 2.68

Doing long calculations is good theoretical exercise, but in practice it’s a waste of time. A problem such as this only takes a few minutes to solve using a simulator. The results from simulation:

V1 = 2.027 V   I1 = 2.027 A

V2 = 2.351 V   I2 = 1.175 A

V3 = 2.555 V   I3 = 0.852 A

V4 = 0.204 V   I4 = 0.051 A

V5 = 5.622 V   I5 = 1.124 A

V6 = 5.417 V   I6 = 0.903 A

Plugging the results back into the diagram, you can see Kirchhoff’s voltage and current laws in action, as shown in Fig. 2.69. Take any loop, sum the changes in voltage across components, and you get 0 (note that the voltages indicated in black shadow represent point voltages relative to 0-V reference ground). Also, the current that enters any junction will equal the sum of the currents exiting the junction, and vice versa—Kirchhoff’s current law.

FIGURE 2.69

Now, before you get too gung ho about playing with equations or become lazy by grabbing a simulator, you should check out a special theorem known as Thevenin’s theorem. This theorem uses some very interesting tricks to analyze circuits, and it may help you avoid dealing with systems of equations or having to resort to simulation. Thevenin’s theorem utilizes something called the superposition theorem, which we must consider first.

2.18 Superposition Theorem

The superposition theorem is an important concept in electronics that is useful whenever a linear circuit contains more than one source. It can be stated as follows:

·    Superposition theorem: The current in a branch of a linear circuit is equal to the sum of the currents produced by each source, with the other sources set equal to zero.

The proof of the superposition theorem follows directly from the fact that Kirchhoff’s laws applied to linear circuits always result in a set of linear equations, which can be reduced to a single linear equation in a single unknown. This means that an unknown branch current can thus be written as a linear superposition of each of the source terms with an appropriate coefficient. Be aware that the superposition should not be applied to nonlinear circuits.

It is important to make clear what it means to set sources equal to zero when interpreting the superposition theorem. A source may be a voltage source or a current source. If the source is a voltage source, to set it to zero means that the points in the circuit where its terminals were connected must be kept at the same potential. The only way to do this is to replace the voltage source with a conductor, thus creating a short circuit. If the source is a current source, to set it to zero means to simply remove it and leave the terminals open, thus creating an open circuit. A short circuit causes the voltage to be zero; an open circuit causes the current to be zero.

In Fig. 2.70, we will analyze the circuit using the superposition theorem. The circuit contains two resistors, a voltage source, and a current source.

FIGURE 2.70 The circuit in (a) can be analyzed using the superposition theorem by considering the simpler circuits in (b) and (c).

First we remove the current source (create an open circuit at its terminals), as shown in Fig. 2.70b. The new current through R2 due to the voltage source alone is just VA divided by the equivalent resistance:

This current is called the partial current in branch 2 due to source 1. Next, the voltage source is removed and set to zero by replacing it with a conductor (we short it); see Fig. 2.70c. The resulting circuit is a current divider, and the resulting partial current is given by:

Applying the superposition, we add the partial current to get the total current:

The current through R1 could have been determined in a similar manner, with the following result:

The superposition theorem is an important tool whose theory makes possible complex impedance analysis in linear, sinusoidally driven circuits—a subject we will cover a bit later. The superposition is also an underlying mechanism that makes possible two important circuit theorems: Thevenin’s theorem and Norton’s theorem. These two theorems, which use some fairly ingenious tricks, are much more practical to use than the superposition. Though you will seldom use the superposition directly, it is important to know that it is the base upon which many other circuit analysis tools rest.

2.19 Thevenin’s and Norton’s Theorems

2.19.1 Thevenin’s Theorem

Say that you are given a complex circuit such as that shown in Fig. 2.71. Pretend that you are only interested in figuring out what the voltage will be across terminals A and F (or any other set of terminals, for that matter) and what amount of current will flow through a load resistor attached between these terminals. If you were to apply Kirchhoff’s laws to this problem, you would be in trouble—the amount of work required to set up the equations would be a nightmare, and then after that you would be left with a nasty system of equations to solve.

FIGURE 2.71 The essence of Thevenin’s theorem.

Luckily, a man by the name of Thevenin came up with a theorem, or trick, to simplify the problem and produce an answer—one that does not involve “hairy” mathematics. Using what Thevenin discovered, if only two terminals are of interest, these two terminals can be extracted from the complex circuit, and the rest of the circuit can be considered a black box. Now the only things left to work with are these two terminals. By applying Thevenin’s tricks (which you will see in a moment), you will discover that this black box, or any linear two-terminal dc network, can be represented by a voltage source in series with a resistor. (This statement is referred to as Thevenin’s theorem.) The voltage source is called the Thevenin voltage VTHEV, and the resistance is called the Thevenin resistance RTHEV; together, the two form what is called the Thevenin equivalent circuit. From this simple equivalent circuit you can easily calculate the current flow through a load placed across its terminals by using Ohm’s law: I = VTHEV/(RTHEV + RLOAD).

Note that circuit terminals (black box terminals) might actually not be present in a circuit. For example, instead, you may want to find the current and voltage across a resistor (RLOAD) that is already within a complex network. In this case, you must remove the resistor and create two terminals (making a black box) and then find the Thevenin equivalent circuit. After the Thevenin equivalent circuit is found, you simply replace the resistor (or place it across the terminals of the Thevenin equivalent circuit), calculate the voltage across it, and calculate the current through it by applying Ohm’s law again: I = VTHEV/(RTHEV + RLOAD). However, two important questions remain: What are the tricks? And what are VTHEV and RTHEV?

First, VTHEV is simply the voltage across the terminals of the black box, which can be either measured or calculated. RTHEV is the resistance across the terminals of the black box when all the dc sources (e.g., batteries) are replaced with shorts, and it, too, can be measured or calculated.

As for the tricks, we can generalize and say that the superposition theorem is involved. However, Thevenin figured out, using the tricks presented in the following example, that the labor involved with applying the superposition theorem (removing sources one at a time, calculating partial currents, and adding them, etc.) could be reduced by removing all sources at once, and finding the Thevenin resistance. An example is the best cure. See Fig. 2.72.

FIGURE 2.72 Here’s an example of how Thevenin’s theorem can be used in a voltage divider circuit to easily calculate the voltage across and the current flow through an attached load.

First remove the load R3 and open up the terminals of interest (A and B). Then, determine the Thevenin voltage VTHEV using Ohm’s law or the voltage divider equation—it’s the open-circuit voltage across A and B.

Next, calculate the Thevenin resistance RTHEV across the terminals A and B by replacing the dc source (VBAT) with a short and calculating or measuring the resistance across A and B. RTHEV is simply R1 and R2 in parallel.

The final Thevenin equivalent circuit is then simply VTHEV in series with RTHEV. The voltage across the load and the current through the load (R3) are:

2.19.2 Norton’s Theorem

Norton’s theorem is another tool for analyzing complex networks. Like Thevenin’s theorem, it takes a complex two-terminal network and replaces it with a simple equivalent circuit. However, instead of a Thevenin voltage source in series with a Thevenin resistance, the Norton equivalent circuit consists of a current source in parallel with a resistance—which happens to be the same as the Thevenin resistance. The only new trick is finding the value of the current source, which is referred to as the Norton current INORTON. In essence, Norton’s theorem is to current sources as Thevenin’s theorem is to voltage sources. Its underlying mechanism, like that of Thevenin’s, is the superposition theorem.

Figure 2.73 shows how the circuit just analyzed by means of Thevenin’s theorem can be analyzed using Norton’s theorem. The Norton current INORTON represents the short-circuit current through terminals A and B.

FIGURE 2.73 To find INORTON, first remove load R3 and replace it with a short. Using Ohm’s law, and noticing that no current will ideally flow through R2 since it’s shorted, you find the short-circuit current, or Norton current, to be:

Next, find the Thevenin resistance—simply use the result from the previous example:

RTHEV = 800 Ω.

The final Norton equivalent circuit can then be constructed. Replacing R3, you can now determine how much current will flow through R3, using Ohm’s law, or simply applying the current divider equation.

A Norton equivalent circuit can be transformed into a Thevenin equivalent circuit and vice versa. The equivalent resistor stays the same in both cases; it is placed in series with the voltage source in the case of a Thevenin equivalent circuit and in parallel with the current source in the case of the Norton equivalent circuit. The voltage for a Thevenin equivalent source is equal to the nonload voltage appearing across the resistor in the Norton equivalent circuit. The current for a Norton equivalent source is equal to the short-circuit current provided by the Thevenin source.

Examples: Find the Thevenin and Norton equivalent circuits for everything between points A and B for each of the four circuits in Fig. 2.74.

FIGURE 2.74

Answers: a: VTHEV = 2 V, RTHEV = 100 Ω, INORT = 0.02 A, b: VTHEV = 6 V, RTHEV = 300 Ω, INORT = 0.02 A, c: VTHEV = 3 V, RTHEV = 60 Ω, INORT = 0.05 A, d: VTHEV = 0.5 V, RTHEV = 67 Ω, INORT = 0.007 A

Example: Here’s an example where Thevenin’s theorem can be applied a number of times to simplify a complex circuit that has more than one source. In essence, you create Thevenin subcircuits and combine them. Often this is easier than trying to find the whole Thevenin equivalent circuit in one grand step. Refer to Fig. 2.75.

FIGURE 2.75

Here we’re interested in finding the current that will flow through a load resistor Rload when attached to terminals c and d. To simplify matters, first find the Thevenin equivalent circuit for everything to the left of a and b. Using the voltage divider and resistors-in-parallel formulas:

(Recall that we replaced the 5-V source with a short when finding RTHEV.)

Incorporating this equivalent circuit back into the main circuit, as shown in the second circuit down, we then determine the Thevenin equivalent circuit for everything to the left of c and d. Using Kirchhoff’s voltage law and resistors-in-series formulas:

VTHEV (c,d) = 2.5 V − 3.5 V = −1.0 V

(In terms of the diagram, this represents a polarity reversal, or simply flipping the battery.)

RTHEV (c,d) = 500 Ω + 1000 Ω = 1500 Ω

(Both sources were shorted to find RTHEV.)

Now we attach our load of 500 Ω and get the current:

Example: To increase the current capacity, batteries are placed in parallel. If the internal resistance is 0.2 Ω for each 1.5-V battery, find the equivalent Thevenin circuit. Refer to Fig. 2.76.

FIGURE 2.76

Answer: RTHEV = 0.04 Ω, VTHEV = 1.5 V. (Apply Thevenin’s theorem in steps.) As you can see, the net internal resistance of the Thevenin circuit is much less—a result of placing batteries in parallel.

2.20 AC Circuits

A circuit is a complete conductive path through which electrons flow from source to load and back to source. As we’ve seen, if the source is dc, electrons will flow in only one direction, resulting in a direct current (dc). Another type of source that is frequently used in electronics is an alternating source that causes current to periodically change direction, resulting in an alternating current (ac). In an ac circuit, not only does the current change directions periodically, the voltage also periodically reverses.

Figure 2.77 shows a dc circuit and an ac circuit. The ac circuit is powered by a sinusoidal source, which generates a repetitive sine wave that may vary in frequency from a few cycles per second to billions of cycles per second, depending on the application.

FIGURE 2.77

The positive and negative swings in voltage/current relative to a zero volt/amp reference line simply imply that the electromotive force of the source has switched directions, causing the polarity of the voltage source to flip, and forcing current to change directions. The actual voltage across the source terminals at a given instant in time is the voltage measured from the 0-V reference line to the point on the sinusoidal waveform at the specified time.

2.20.1 Generating AC

The most common way to generate sinusoidal waveforms is by electromagnetic induction, by means of an ac generator (or alternator). For example, the simple ac generator in Fig. 2.78 consists of a loop of wire that is mechanically rotated about an axis while positioned between the north and south poles of a magnet. As the loop rotates in the magnetic field, the magnetic flux through it changes, and charges are forced through the wire, giving rise to an effective voltage or induced voltage. According to Fig. 2.78, the magnetic flux through the loop is a function of the angle of the loop relative to the direction of the magnetic field. The resultant induced voltage is sinusoidal, with angular frequency ω (radians per second).

FIGURE 2.78 Simple ac generator.

Real ac generators are, of course, more complex than this, but they operate under the same laws of induction, nevertheless. Other ways of generating ac include using a transducer (e.g., a microphone) or even using a dc-powered oscillator circuit that uses special inductive and capacitive effects to cause current to resonate back and forth between an inductor and a capacitor.

Why Is AC Important?

There are several reasons why sinusoidal waveforms are important in electronics. The first obvious reason has to do with the ease of converting circular mechanical motion into induced current via an ac generator. However, another very important reason for using sinusoidal waveforms is that if you differentiate or integrate a sinusoid, you get a sinusoid. Applying sinusoidal voltage to capacitors and inductors leads to sinusoidal current. It also avoids problems on systems, a subject that we’ll cover later. But one of the most important benefits of ac involves the ability to increase voltage or decrease voltage (at the expense of current) by using a transformer. In dc, a transformer is useless, and increasing or decreasing a voltage is a bit tricky, usually involving some resistive power losses. Transformers are very efficient, on the other hand, and little power is lost in the voltage conversion.

2.20.2 Water Analogy of AC

Figure 2.79 shows a water analogy of an ac source. The analogy uses an oscillating piston pump that moves up and down by means of a cam mechanism, driven by a hand crank.

FIGURE 2.79

In the water analogy, water particles, on average, appear to simply swish back and forth as the crank is turned. In an ac electrical circuit, a similar effect occurs, though things are a bit more complex. One way to envision what’s going on is that within a conductor, the drift velocity of the sea of electrons is being swished back and forth in a sinusoidal manner. The actual drift velocity and distance over which the average drift occurs are really quite small (fractions of millimeter-per-second range, depending on conductor and applied voltage). In theory, this means that there is no net change in position of an “average” electron over one complete cycle. (This is not to be confused with an individual electron’s thermal velocity, which is mostly random, and at high velocity.) Also, things get even more complex when you start applying high frequencies, where the skin effect enters the picture—more on this later.

2.20.3 Pulsating DC

If current and voltage never change direction within a circuit, then from one perspective, we have a dc current, even if the level of the dc constantly changes. For example, in Fig. 2.80, the current is always positive with respect to 0, though it varies periodically in amplitude. Whatever the shape of the variations, the current can be referred to as “pulsating dc.” If the current periodically reaches 0, it is referred to as “intermittent dc.”

FIGURE 2.80

From another perspective, we may look at intermittent and pulsating dc as a combination of an ac and a dc current. Special circuits can separate the two currents into ac and dc components for separate analysis or use. There are also circuits that combine ac and dc currents and voltages.

2.20.4 Combining Sinusoidal Sources

Besides combining ac and dc voltages and currents, we can also combine separate ac voltages and currents. Such combinations will result in complex waveforms. Figure 2.81 shows two ac waveforms fairly close in frequency, and their resultant combination. The figure also shows two ac waveforms dissimilar in both frequency and wavelength, along with the resultant combined waveform.

FIGURE 2.81 (Left) Two ac waveforms of similar magnitude and close in frequency form a composite wave. Note the points where the positive peaks of the two waves combine to create high composite peaks: this is the phenomenon of beats. The beat note frequency is f2  f1 = 500 Hz. (Right) Two ac waveforms of widely different frequencies and amplitudes form a composite wave in which one wave appears to ride upon the other.

Later we will discover that by combining sinusoidal waveforms of the same frequency—even though their amplitudes and phases may be different—you always get a resultant sine wave. This fact becomes very important in ac circuit analysis.

2.20.5 AC Waveforms

Alternating current can take on many other useful wave shapes besides sinusoidal. Figure 2.82 shows a few common waveforms used in electronics. The squarewave is vital to digital electronics, where states are either true (on) or false (off). Triangular and ramp waveforms—sometimes called sawtooth waves—are especially useful in timing circuits. As we’ll see later in the book, using Fourier analysis, you can create any desired shape of periodic waveform by adding a collection of sine waves together.

FIGURE 2.82

An ideal sinusoidal voltage source will maintain its voltage across its terminals regardless of load—it will supply as much current as necessary to keep the voltage the same. An ideal sinusoidal current source, on the other hand, will maintain its output current, regardless of the load resistance. It will supply as much voltage as necessary to keep the current the same. You can also create ideal sources of other waveforms. Figure 2.83 shows schematic symbols for an ac voltage source, an ac current source, and a clock source used to generate squarewaves.

FIGURE 2.83

In the laboratory, a function generator is a handy device that can be used to generate a wide variety of waveforms with varying amplitudes and frequencies.

2.20.6 Describing an AC Waveform

A complete description of an ac voltage or current involves reference to three properties: amplitude (or magnitude), frequency, and phase.

Amplitude

Figure 2.84 shows the curve of a sinusoidal waveform, or sine wave. It demonstrates the relationship of the voltage (or current) to relative positions of a circular rotation through one complete revolution of 360°. The magnitude of the voltage (or current) varies with the sine of the angle made by the circular movement with respect to the zero point. The sine of 90° is 1, which is the point of maximum current (or voltage); the sine of 270° is −1, which is the point of maximum reverse current (or voltage); the sine of 45° is 0.707, and the value of current (or voltage) at the 45° point of rotation is 0.707 times the maximum current (or voltage).

FIGURE 2.84

2.20.7 Frequency and Period

A sinusoidal waveform generated by a continuously rotating generator will generate alternating current (or voltage) that will pass through many cycles over time. You can choose an arbitrary point on any one cycle and use it as a marker—say, for example, the positive peak. The number of times per second that the current (or voltage) reaches this positive peak in any one second is called the frequency of the ac. In other words, frequency expresses the rate at which current (or voltage) cycles occur. The unit of frequency is cycles per second, or hertz—abbreviated Hz (after Heinrich Hertz).

The length of any cycle in units of time is the period of the cycle, as measured from two equivalent points on succeeding cycles. Mathematically, the period is simply the inverse of the frequency:

 (2.22)

and

 (2.23)

Example: What is the period of a 60-Hz ac current?

Example: What is the frequency of an ac voltage that has period of 2 ns?

The frequency of alternating current (or voltage) in electronics varies over a wide range, from a few cycles per second to billions of cycles per second. To make life easier, prefixes are used to express large frequencies and small periods. For example: 1000 Hz = 1 kHz (kilohertz), 1 million hertz = 1 MHz (megahertz), 1 billion hertz = 1 GHz (gigahertz), 1 trillion hertz = 1 THz (terahertz). For units smaller than 1, as in the measurements of period, the basic unit of a second can become millisecond (1 thousandth of a second, or ms), microsecond (1 millionth of a second, or µs), nanosecond (1 billionth of a second, or ns), and picosecond (1 trillionth of a second, or ps).

2.20.8 Phase

When graphing a sine wave of voltage or current, the horizontal axis represents time. Events to the right on the graph take place later, while events to the left occur earlier. Although time can be measured in seconds, it actually becomes more convenient to treat each cycle of a waveform as a complete time unit, divisible by 360°. A conventional starting point for counting in degrees is 0°—the zero point as the voltage or current begins a positive half cycle. See Fig. 2.85a.

FIGURE 2.85 (a) An ac cycle is divided into 360° that are used as a measure of time or phase. (b) When two waves of the same frequency start their cycles at slightly different times, the time difference or phase difference is measured in degrees. In this drawing, wave B starts 45° (one-eighth cycle) later than wave A, and so lags 45° behind A. (c and d) Two special cases of phase difference: In (c) the phase difference between A and B is 90°; in (d) the phase difference is 180°.

By measuring the ac cycle this way, it is possible to do calculations and record measurements in a way that is independent of frequency. The positive peak voltage or current occurs at 90° during a cycle. In other words, 90° represents the phase of the ac peak relative to the 0° starting point.

Phase relationships are also used to compare two ac voltage or current waveforms at the same frequency, as shown in Fig. 2.85b. Since waveform B crosses the zero point in the positive direction after A has already done so, there is a phase difference between the two waveforms. In this case, B lags A by 45°; alternatively, we can say that A leads B by 45°. If A and B occur in the same circuit, they add together, producing a composite sinusoidal waveform at an intermediate phase angle relative to the individual waveforms. Interestingly, adding any number of sine waves of the same frequency will always produce a sine wave of the same frequency—though the magnitude and phase may be unique.

In Fig. 2.85c we have a special case where B lags A by 90°. B’s cycle begins exactly one-quarter cycle after A’s. As one waveform passes through zero, the other just reaches its maximum value.

Another special case occurs in Fig. 2.85d, where waveforms A and B are 180° out of phase. Here it doesn’t matter which waveform is considered the leading or lagging waveform. Waveform B is always positive when waveform A is negative, and vice versa. If you combine these two equal voltage or current waveforms together within the same circuit, they completely cancel each other out.

2.21 AC and Resistors, RMS Voltage, and Current

Alternating voltages applied across a resistor will result in alternating current through the resistor that is in phase with the voltage, as seen in Fig. 2.86. Given the ac voltage and resistance, you can apply Ohm’s law and find the ac current. For example, a sinusoidal waveform generated by a function generator can be mathematically described by:

 V(t) = VP sin (2π × f × t) (2.24)

where VP is the peak amplitude of the sinusoidal voltage waveform, f is the frequency, and t is the time. Using Ohm’s law and the power law, you get the following:

 (2.25)

If you graph both V(t) and I(t) together, as shown in Fig. 2.86, you notice that the current and the voltage are in phase with each other. As the voltage increases in one direction, the current also increases in the same direction. Thus, when an ac source is connected to a purely resistive load, the current and voltage are in phase. If the load isn’t purely resistive (e.g., has capacitance and inductance), it’s a whole different story—more on that later.

FIGURE 2.86

To find the power dissipated by the resistor under sinusoidal conditions, we can simply plug the sinusoidal voltage expression into Ohm’s power law to get an instantaneous power expression:

 (2.26)

Expressing voltage, current, and power in an instantaneous fashion is fine, mathematically speaking; however, in order to get a useful result, you need to plug in a specific time—say, t = 1.3 s. But how often do you need to know that the voltage, current, or power are at exactly t = 1.3 s? Better yet, when do you start counting? These instantaneous values are typically inconvenient for any practical use. Instead, it is more important to come up with a kind of averaging scheme that can be used to calculate effective power dissipation without dealing with sinusoidal functions.

Now, you might be clever and consider averaging the sinusoidal voltage or current over a complete cycle to get some meaningful value. However, the average turns out to be zero—positive and negative sides of waveforms cancel. This may be a bit confusing, in terms of power, since the positive-going part of the wave still delivers energy, as does the negative-going part. If you’ve ever received a shock from 120-V line voltage, you’ll be able to attest to that.

The measurement that is used instead of the average value is the RMS or root mean square value, which is found by squaring the instantaneous values of the ac voltage or current, then calculating their mean (i.e., their average), and finally taking the square root of this—which gives the effective value of the ac voltage or current. These effective, or RMS, values don’t average out to zero and are essentially the ac equivalents of dc voltages and currents. The RMS values of ac voltage and current are based upon equating the values of ac and dc power required to heat a resistive element to exactly the same degree. The peak ac power required for this condition is twice the dc power needed. Therefore, the average ac power equivalent to a corresponding average dc power is half the peak ac power.

 (Average dc power equivalent of ac waveform) (2.27)

Mathematically, we can determine the RMS voltage and current values for sinusoidal waveforms V(t) = VP sin (2π × f × t), and I(t) = IP sin (2π × f × t):

 (2.28) RMS voltage
 (2.29) RMS current

Notice that the RMS voltage and current depend only on the peak voltage or current—they are independent of time or frequency. Here are the important relations, without the scary calculus:

For example, a U.S. electric utility provides your home with 60 Hz, 120 VAC (in Europe and many other countries it’s 50 Hz, 240V AC). The “VAC” unit tells you that the supplied voltage is given in RMS. If you were to attach an oscilloscope to the outlet, the displayed waveform on the screen would resemble the following function: V(t) = 170 V sin (2π × 60 Hz × t), where 170 V is the peak voltage.

With our new RMS values for voltage and current, we can now substitute them into Ohm’s law to get what’s called ac Ohm’s law:

 VRMS = IRMS × R (2.30) ac Ohm’s law

Likewise, we can use the RMS voltage and current and substitute them into Ohm’s power law to get what’s called the ac power law, which gives the effective power dissipated (energy lost per second):

 (2.31) ac power law

Again, these equations apply only to circuits that are purely resistive, meaning there is virtually no capacitance and/or inductance. Doing power calculation on circuits with inductance and capacitance is a bit more complicated, as we’ll see a bit later.

Figure 2.87 shows the relationships between RMS, peak, peak-to-peak and half-wave average values of voltage and current. Being able to convert from one type to another is important, especially when dealing with component maximum voltage and current ratings—sometimes you’ll be given the peak value, other times the RMS value. Understanding the differences also becomes crucial when making test measurements—see the note on pages 91–92 about making RMS test measurements. Most of the time, when dealing with ac voltage you can assume that voltage is expressed as an RMS value unless otherwise stated.

FIGURE 2.87

Conversion Factors for AC Voltage and Current

 FROM TO MULTIPLY BY Peak Peak-to-peak 2 Peak-to-peak Peak 0.5 Peak RMS 1/  or 0.7071 RMS Peak or 1.4142 Peak-to-peak RMS 1/(2) or 0.35355 RMS Peak-to-peak (2 × ) or 2.828 Peak Average* 2/π or 0.6366 Average* Peak π/2 or 1.5708 RMS Average* (2 × )/π or 0.9003 Average* RMS π/(2 × ) or 1.1107 * Represents average over half a cycle.

Example 1: How much current would flow through a 100-Ω resistor connected across the hot and neutral sockets of a 120-VAC outlet? How much power would be dissipated? What would the results be using 1000-Ω, 10,000-Ω, and 100,000-Ω resistors?

FIGURE 2.88

First, don’t try this with any ordinary resistor; you’d need a power resistor or special heating element with a power rating of greater than 144 W! (Also, simply don’t attempt attaching a resistor with the outlets powered.) A 1000-Ω resistor plugged into the same outlet would dissipate 14.4 W; a 10,000-Ω resistor would dissipate 1.44 W; a 100,000-Ω resistor would dissipate 0.14 W.

Example 2: What is the peak voltage on a capacitor if the RMS voltage of a sinusoidal waveform signal across it is 10.00 VAC?

Answer: VAC means RMS, so VP =  × VRMS = 1.414 × 10 V = 14.14 V.

Example 3: A sinusoidal voltage displayed on an oscilloscope has a peak amplitude of 3.15 V. What is the RMS value of the waveform?

Example 4: A 200-W resistive element in a heater is connected to a 120-VAC outlet. How much current is flowing through the resistive element? What’s the resistance of the element, assuming it’s an ideal resistor?

Answer: IRMS = PAVE/VRMS = 200 W/120 VAC = 1.7 A. R = VRMS/IRMS = 120 V/1.7 A = 72 Ω.

Example 5: A sinusoidal voltage supplied by a function generator is specified as 20 V peak to peak at 1000 Hz. What is the minimum resistance value of a ⅛-W resistor you can place across the generator’s output before exceeding the resistor’s power rating?

Answer: VP = 1/2 VPP = 10 V; VRMS = 0.707 × VP = 7.1 VAC; R = VRMS2/PAVE = (7.1)2/(1/8 W) = 400 Ω.

Example 6: The output of an oscillator circuit is specified as 680 mVAC. If you feed this into an op amp with an input resistance of 10 MΩ, how much current enters the IC?

Answer: IRMS = VRMS/R = 0.68 V/(10,000,000 Ω) = 0.000000068 = 68 nA.

MEASURING RMS VOLTAGES AND CURRENTS

There are many digital multimeters that do not measure the RMS value of an ac voltage directly. Often the meter will simply measure the peak value and calculate the equivalent RMS value—assuming the measured waveform is sinusoidal—and then display this value. Analog meters usually measure the half-wave average value, but are made to indicate the equivalent RMS.

True RMS multimeters, on the other hand, measure the true RMS value of voltages and current, and are especially handy for nonsinusoidal voltage and currents. Though relatively expensive, these meters are worth the price. It’s important to note that true RMS meters will also include the contribution of any dc voltages or current components present along with the ac.

Fortunately, you can still get a fairly accurate idea of the RMS value of a sine waveform, knowing one of the other measurements such as the half-wave average, peak, or peak-to-peak value. This can be done by calculation, or using the table in Fig. 2.89. As you can see, it’s also possible to work out the RMS value of a few other symmetrical and regular waveforms, such as square and triangular waves, knowing their peak, average, or peak-to-peak values.

An important thing to note when using the table is that you need to know the exact basis on which your meter’s measurement is made. For example, if your meter measures the peak value, and then calculates and displays the equivalent sine wave RMS figure, this means you’ll need to use the table differently when compared to the situation where the meter really measures the half-wave average and calculates the sine wave RMS figure from that. So take care, especially if you’re not sure exactly how your meter works.

FIGURE 2.89

2.22 Mains Power

In the United States, three wires run from the pole transformers (or underground or surface enclosed transformer) to the main service panel at one’s home. One wire is the A-phase wire (usually black in color), another is the B-phase wire (usually black in color), and the third is the neutral wire (white in color). Figure 2.90 shows where these three wires originate from the pole transformer. The voltage between the A-phase and the B-phase wires, or the hot-to-hot voltage, is 240 V, while the voltage between the neutral wire and either the A-phase or the B-phase wire, or the neutral-to-hot voltage, is 120 V. (These voltages are nominal and may vary from region to region, say 117 V instead of 120 V.)

FIGURE 2.90

At the home, the three wires from the pole/green box transformer are connected through a wattmeter and then enter a main service panel that is grounded to a long copper rod driven into the ground or to the steel in a home’s foundation. The A-phase and B-phase wires that enter the main panel are connected through a main disconnect breaker, while the neutral wire is connected to a terminal referred to as the neutral bar or neutral bus. A ground bar also may be present within the main service panel. The ground bar is connected to the grounding rod or to the foundation’s steel supports.

Within the main service panel, the neutral bar and the ground bar are connected together (they act as one). However, within subpanels (service panels that get their power from the main service panel but which are located some distance from the main service panel), the neutral and ground bars are not joined together. Instead, the subpanel’s ground bar receives a ground wire from the main services panel. Often the metal conduit that is used to transport the wires from the main service panel to the subpanel is used as the ground wire. However, for certain critical applications (e.g., computer and life-support systems), the ground wire probably will be included within the conduit. Also, if a subpanel is not located in the same building as the main panel, a new ground rod typically is used to ground the subpanel. Note that different regions within the United States may use different wiring protocols. Therefore, do not assume that what we are telling you is standard practice where you live. Contact your local electrical inspector.

Now, if you want to supply power to 240-V appliances (ovens, washers, etc.), you insert a double-pole breaker between the A-phase and B-phase bus bars in the main (or subpanel). Next, you take a 240-V three-wire cable and attach one of its hot wires to the A-phase terminal of the breaker and attach its other hot wire to the B-phase terminal of the breaker. The ground wire (green or bare) is connected to the ground bar. You then run the cable to where the 240-V loads are located and attach the wires to the corresponding terminals of the load (typically within a 240-V outlet). Also, 120-V/240-V appliances are wired in a similar manner, except you use a four-wire cable that contains an additional neutral (white) wire that is joined at the neutral bar within the main panel (or subpanel).

In addition, in many places, modifications to mains wiring must be carried out or checked by a certified electrician. As a note of caution, do not attempt home wiring unless you are sure of your abilities. If you feel that you are capable, just make sure to flip the main breaker off before you start work within the main service panel. When working on light fixtures, switches, and outlets that are connected to an individual breaker, tag the breaker with tape so that you do not mistakenly flip the wrong breaker when you go back to test your connections.

2.23 Capacitors

If you take two oppositely charged parallel conducting plates separated a small distance apart by an insulator—such as air or a dielectric such as ceramic—you have created what’s called a capacitor. Now, if you apply a voltage across the plates of the capacitor using a battery, as shown in Fig. 2.91, an interesting thing occurs. Electrons are pumped out the negative battery terminal and collect on the lower plate, while electrons are drawn away from the upper plate into the positive battery terminal. The top plate becomes deficient in electrons, while the lower plate becomes abundant in electrons.

FIGURE 2.91

Very quickly, the top plate reaches a positive charge +Q and the negative plate reaches a negative charge −Q. Accompanying the charge is a resultant electric field between the plates and a voltage equal to the battery voltage.

The important thing to notice with our capacitor is that when we remove the voltage source (battery), the charge, electric field, and corresponding voltage (presently equal to the battery voltage) remain. Ideally, this state of charge will be maintained indefinitely. Even attaching an earth ground connection to one of the plates—doesn’t matter which one—will not discharge the system. For example, attaching an earth ground to the negative terminal doesn’t cause the electrons within that plate to escape to the earth ground where neutral charge is assumed. (See Fig. 2.92.)

FIGURE 2.92

It might appear that the abundance of electrons would like to escape to the earth ground, since it is at a lower potential (neutral). However, the electric field that exists within the capacitor acts like a glue; the positive charge on the upper plate “holds” onto the abundance of electrons on the negative plate. In other words, the positive plate induces a negative charge in the grounded plate.

In reality, a real-life charged capacitor that is charged and removed from the voltage source would eventually lose its charge. The reason for this has to do with the imperfect insulating nature of the gas or dielectric that is placed between the plates. This is referred to as leakage current and, depending on the construction of the capacitor, can discharge a capacitor within as little as a few seconds to several hours, if the source voltage is removed.

To quickly discharge a capacitor you can join the two plates together with a wire, which creates a conductive path for electrons from the negative plate to flow to the positive plate, thus neutralizing the system. This form of discharge occurs almost instantaneously.

The ratio of charge on one of the plates of a capacitor to the voltage that exists between the plates is called capacitance (symbolized C):

 (2.32) Capacitance relatedto charge and voltage

C is always taken to be positive, and has units of farads (abbreviated F). One farad is equal to one coulomb per volt:

1 F = 1 C/1 V

Devices that are specifically designed to hold charge (electrical energy in the form of an electric field) are called capacitors. Figure 2.93 shows various symbols used to represent capacitors, along with a real capacitor model that we’ll discuss a bit later.

FIGURE 2.93

The equation C = Q/V is a general one; it really doesn’t tell you why one capacitor has a larger or smaller capacitance than another. However, in practice, when you buy a capacitor all you’ll be interested in is the capacitance value labeled on the device. (A voltage rating and other parameters are important, too, but we’ll talk about them later.) Most commercial capacitors are limited to a range from 1 pF (1 × 10−12 F) to 4700 µF (1 × 10−6 F), with typical values for the first two digits of the capacitance of 10, 12, 15, 18, 22, 27, 33, 39, 47, 56, 68, 82, 100. (Examples: 27 pF, 100 pF, 0.01 µF, 4.7 µF, 680 µF).

Having a wide range of capacitances allows you to store different amounts of charge for a given potential difference, as well as maintaining different potential differences for a given charge. With the appropriate capacitor, you can therefore control the storage and delivery of charge, or control potential differences.

Example 1: Five volts are applied across a 1000-µF capacitor until the capacitor is fully charged. How much charge exists on the positive and negative plates?

Answer: Q = CV = (1000 × 10−6 F)(5 V) = 5 × 10−3 C. This is the charge on the positive plate. The charge on the negative plate is the same, but opposite in sign.

Example 2: A 1000-µF capacitor and a 470-µF capacitor are arranged in the circuit shown in Fig. 2.94, with a 10-V dc supply. Initially, the switch is at position B then thrown to position A, and then thrown to position B, and then to position A, and finally to position B. Assuming the capacitors have enough time to fully charge or discharge during the interval between switches, what is the final voltage across each capacitor after the last switch takes place?

FIGURE 2.94

Answer: When the switch is thrown from B to A the first time, C1 charges to:

Q1 = C1V = (1000 × 10−6 F)(10 V) = 0.01 C

When the switch is then thrown to B, the circuit becomes essentially one big capacitor equal to C1 + C2 or 1470 µF. Charge will flow from C1 to C2, since the system wants to go to the lowest energy configuration. The charge on each capacitor is the percentage of capacitance to the total capacitance for each capacitor multiplied by the initial charge on C1 before the switch was thrown to position B:

The voltage at the new equilibrium is:

V1 = Q1/C1 = 0.0068/1000 µF = 6.8 V

V2 = Q2/C2 = 0.0032/470 µF = 6.8 V

The rest of the results are obtained in using similar calculations—the final result yields 9.0 V, as shown in the graph to the right.

We could be content with this limited knowledge. However, if you want to build your own capacitors, as well as understand time-dependent behavior, such as displacement current and capacitive reactance, a deeper understanding of capacitance is needed.

2.23.1 Determining Capacitance

The capacitance of a capacitor is determined by plate area A, plate separation d, and insulating material or dielectric. If a voltage V is applied between two parallel plates, an electric field equal to E = V/d will be produced. From Gauss’s law, each plate must contain an equal and opposite charge given by:

 (2.33)

where ε is the permittivity of the dielectric. Free space (vacuum) has a permittivity given by:

 ε0 = 8.85 × 10−12 C2/N ⋅ m2 (2.34)

The constant εA/d term in the equation is the capacitance,

 (2.35)

The relative permittivity of a material referenced to the permittivity in vacuum is referred to as the dielectric constant, which is given by:

Plugging this into the previous expression, we get the capacitance in terms of dielectric constant:

 (2.36)

where C is in farads, A is in meters squared, and d is in meters.

The dielectric constant varies from 1.00059 for air (1 atm) to over 105 for some types of ceramic. Table 2.6 shows relative dielectric constants for various materials often used in constructing capacitors.

Capacitors often have more than two plates, the alternate plates being connected to form two sets, as shown in the lower drawing in Fig. 2.95. This makes it possible to obtain a fairly large capacitance in a small space. For a multiple-plate capacitor, we use the following expression to find the capacitance:

FIGURE 2.95

 (2.37)

where the area A is in meters squared, the separation d is in meters, and the number of plates n is an integer.

Example: What is the capacitance of a multiple-plate capacitor containing two plates, each with an area of 4 cm2, a separation of 0.15 mm, and a paper dielectric?

2.23.2 Commercial Capacitors

Commercial capacitors, like those shown in Fig. 2.95, are constructed from plates of foil with a thin solid or liquid dielectric sandwiched between, so relatively large capacitance can be obtained in a small unit. The solid dielectrics commonly used are mica, paper, polypropylene, and special ceramics.

Electrolytic capacitors use aluminum-foil plates with a semiliquid conducting chemical compound between them. The actual dielectric is a very thin film of insulating material that forms on one set of the plates through electrochemical action when a dc voltage is applied to the capacitor. The capacitance obtained with a given area in an electrolytic capacitor is very large compared to capacitors having other dielectrics, because the film is so thin—much less than any thickness practical with a solid dielectric. Electrolytic capacitors, due to the electrochemical action, require that one lead be placed at a lower potential than the other. The negative lead (−) is indicated on the package, and some surface mount electrolytics mark the positive end. This polarity adherence means that, with the exception of special nonpolarized electrolytics, electrolytic capacitors shouldn’t be used in ac applications. It is okay to apply a superimposed ac signal riding upon a dc voltage, provided the peak voltage doesn’t exceed the maximum dc voltage rating of the electrolytic capacitor.

2.23.3 Voltage Rating and Dielectric Breakdown

The dielectric within a capacitor acts as an insulator—its electrons do not become detached from atoms the way they do in conductors. However, if a high enough voltage is applied across the plates of the capacitor, the electric field can supply enough force on electrons and nuclei within the dielectric to detach them, resulting in a breakdown in the dielectric. Failed dielectrics often puncture and offer a low-resistance current path between the two plates.

The breakdown voltage of a dielectric depends on the dielectric’s chemical composition and thickness. A gas dielectric capacitor breakdown is displayed as a spark or arc between the plates. Spark voltages are typically given in units of kilovolts per centimeter. For air, the spark voltage may range from 100 kV/cm for gaps as narrow as 0.005 cm to 30 kV/mm for gaps as wide as 10 cm. Other things that contribute to the exact breakdown voltage level are electrode shape, gap distance, air pressure or density, the voltage, impurities within the dielectric, and the nature of the external circuit (air humidity, temperature, etc.).

Dielectric breakdown can occur at a lower voltage between points or sharp-edged surfaces than between rounded and polished surfaces, since electric fields are more concentrated at sharp projections. This means that the breakdown voltage between metal plates can be increased by buffing the edges to remove any sharp irregularities. If a capacitor with a gas dielectric, such as air, experiences breakdown, once the arc is extinguished, the capacitor can be used again. However, if the plates become damaged due to the spark, they may require polishing, or the capacitor may need to be replaced. Capacitors with solid dielectrics are usually permanently damaged if there is dielectric breakdown, often resulting in a short or even an explosion.

Manufacturers provide what’s called a dielectric withstanding voltage (dwv), expressed in voltage per mil (0.001 in) at a specified temperature. They also provide a dc working voltage (dcwv) that takes into account other factors such as temperature and safety margin, which gives you a guideline to the maximum safe limits of dc voltage that can be applied before dielectric breakdown. The dcwv rating is most useful in practice.

As a rule of thumb, it is not safe to connect a capacitor across an ac power line unless it is designed for it. Most capacitors with dc ratings may short the line. Special ac-rated capacitors are available for such tasks. When used with other ac signals, the peak value of ac voltage should not exceed the dc working voltage.

2.23.4 Maxwell’s Displacement Current

An interesting thing to notice with our parallel-plate capacitor is that current appears to flow through the capacitor as it is charging and discharging, but doesn’t flow under steady dc conditions. You may ask: How is it possible for current to flow through a capacitor, ever, if there is a gap between the plates of the capacitor? Do electrons jump the gap? As it turns out, no actual current (or electron flow) makes it across the gap, at least in an ideal capacitor.

As we calculated a moment ago using Gauss’s law, the charge on an air-filled capacitor plate can be expressed in terms of the electric field, area, and permittivity of free space:

 (2.38)

Some time ago, Scottish physicist James Clerk Maxwell (1831–1879) noted that even if no real current passed from one capacitor plate to the other, there was nevertheless a changing electric flux through the gap of the capacitor that increased and decreased with the magnitude and direction of the electric flux. (Electric flux for a parallel-plate capacitor is approximated by ΦE = EA, while a changing electric flux is expressed as dΦE/dt). Maxwell believed the electric flux permeated the empty space between the capacitor plates and induced a current in the other plate. Given the state of knowledge of electrodynamics at the time, he envisioned a displacement current (which he coined) crossing the empty gap, which he associated with a kind of stress within the ether (accepted at the time)—the “stress” being electric and magnetic fields. (The displacement current helped supply Maxwell with the missing component to complete a set of electromagnetic formulas known as Maxwell’s equations.) He associated the displacement current with displacements of the ether. With a bit of theoretical reckoning, as well as some help from some experimental data, he came up with the following equation, known as the displacement current, to explain how current could appear to enter one end of a capacitor and come out the other end.

 (2.39)

Maxwell’s displacement expression appears to provide the correct answer, even though his notion of the ether has lost favor in the realm of physics. Modern physics provides a different model for displacement current than that envisioned by Maxwell and his ether. Nevertheless, the results obtained using Maxwell’s equation closely correlate with experiment.

As a side note, there also exists a magnetic field due to the displacement current, as shown in the drawing on the right in Fig. 2.96. You can calculate the magnetic field using what’s called Maxwell’s generalized form of Ampere’s law; however, the size of the magnetic field turns out to be so small that it essentially has no practical influence when compared to the electric field.

FIGURE 2.96

However deep you go when trying to explain the physical phenomena within a capacitor, such as using Maxwell’s equations or even modern physics, the practical equations that are useful in electronics really don’t require such detail. Instead, you can simply stick with using the following charge-based model.

2.23.5 Charge-Based Model of Current Through a Capacitor

Though Maxwell’s displacement current provides a model to explain the apparent current flow through a capacitor in terms of changing electric fields, it really isn’t needed to define capacitor performance. Instead, we can treat the capacitor as a black box with two leads, and define some rules relating the current entering and exiting the capacitor as the applied voltage across the capacitor changes. We don’t need to worry about the complex physical behavior within.

Now, the question remains: how do we determine the rules if we don’t understand the complex behavior within? Simple—we use the general definition of capacitance and the general definition of current, and combine the two. Though the mathematics for doing this is simple, understanding why this makes logical sense is not entirely obvious. The following parallel-plate example provides an explanation. Refer to Fig. 2.97.

FIGURE 2.97

If we work in differentials (small changes), we can rewrite the general expression for capacitance as dQ = CdV, where C remains constant (with charge, voltage, or time). The general expression for current is I = dQ/dt, which, when combined with the last differential expression of capacitance, provides the expression:

 (2.40) Apparent current“through” capacitor

Looking at Fig. 2.97, a “small” chunk dQ, which is equal to CdVc, enters the right plate during dt, while an equal-sized chunk exits the left plate. So a current equal to dQ/dt = CdVc/dt enters the left plate while an equal-sized current exits the right plate. (Negative electrons flow in the opposite direction.) Even though no actual current (or electrons) passes across the gap, Eq. 2.40 makes it appear that it does. After our little exercise using differentials, however, we can see that there is really no need to assume that a current must flow across the gap to get an apparent current flow “through” the capacitor.

Moving on, we can take the capacitor current expression just derived, rearrange things, and solve for the voltage across the capacitor:

 (2.41) Voltage across capacitor

It’s important to note that these equations are representative of what’s defined as an ideal capacitor. Ideal capacitors, as the equation suggests, have several curious properties that are misleading if you are dealing with real capacitors. First, if we apply a constant voltage across an ideal capacitor, the capacitor current would be zero, since the voltage isn’t changing (dV/dt = 0). In a dc circuit, a capacitor thus acts like an open circuit. On the other hand, if we try to change the voltage abruptly, from 0 to 9 V, the quantity dV/dt = 9 V/0 V = infinity and the capacitor current would have to be infinite (see Fig. 2.98). Real circuits cannot have infinite currents, due to resistivity, available free electrons, inductance, capacitance, and so on, so the voltage across the capacitor cannot change abruptly. A more accurate representation of a real capacitor, considering construction and materials, looks like the model shown in Fig. 2.93.

FIGURE 2.98

Under steady-state dc, a capacitor cannot pass current. It can only store or discharge charge (which it collects from current) when the voltage across it changes. Here, when the “charge switch” is closed (shorted), the 9-V battery voltage is applied instantly across the capacitor. In a real capacitor, the capacitor charges up to its maximum value practically instantly. But upon closer examination, the charge takes time to build up, due to internal resistance, meaning the displacement current cannot reach infinity. Instead, the current jumps to Vbattery/Rinternal and quickly drops exponentially as the capacitor reaches full charge, during which the voltage rises exponentially until it levels off at the applied voltage. The graph to the left shows voltage and current curves as the capacitor charges up. Note the impossible behavior that an ideal treatment of a capacitor introduces.

When the discharge switch is closed, a conductive path from positive to negative plate is made, and charge electrons will flow to the plate with a deficiency of electrons. The current that results is in the opposite direction as the charging current, but resembles it in that it initially peaks to a value of Vbattery/Rinternal and decays as the charge neutralizes. The voltage drops exponentially in the process.

If the equations for an ideal capacitor are screwy, what do we do for real-life calculations? For the most part, you don’t have to worry, because the capacitor will be substituted within a circuit that has resistance, which eliminates the possibility of infinite currents. The circuit resistance is also usually much greater than the internal resistance of the capacitor, so that the internal resistance of the capacitor can typically be ignored. In a minute, we’ll see a few resistor-capacitor circuits that will demonstrate this.

2.23.6 Capacitor Water Analogy

For those of you who are having problems with the previous explanations of apparent or displacement current, perhaps the following water analogy will help. Take it with a grain of salt, however, since what’s going on in a real capacitor isn’t analogous in all regards. Refer to Fig. 2.99.

FIGURE 2.99

The water capacitor in Fig. 2.99 resembles a tube with a rubber membrane in the middle. The rubber membrane is somewhat analogous to the insulator or dielectric of a capacitor, while each separate compartment is analogous to each plate of a capacitor. If there is no pressure (analogous to voltage) across the water capacitor, each compartment contains the same amount of water (analogous to the number of free electrons). However, if pressure is suddenly applied across the water capacitor, the pressure within the top chamber increases, causing the membrane to expand downward, displacing water out from the lower chamber. Even though no water from the top makes it through the membrane, it appears as though current flows through the water capacitor, since the membrane is pushing the water within the lower chamber out. This is analogous to displacement current. Increasing the chamber size and altering the membrane strength are analogous to changing the capacitance and dielectric strength.

Example 1: A 10-µF capacitor is connected to a 50-mA constant current source. Determine the voltage across the capacitor after 10 µs, 10 ms, and 1 s.

Answer: Since IC is constant, it can be moved in front of the integral:

obviously typical capacitor won’t survive.

Example 2: A 47-µF capacitor is charged by the voltage sources having the following waveform. Determine the charging current. Refer to Fig. 2.100. Assume that the voltage source is ideal and has no resistance.

FIGURE 2.100

Answer: Since dV/dt represents the slope of the waveform, it is simply 10 V/10 ms, and the current becomes:

Example 3: If the voltage across a 100-µF capacitor is 5 V et, what is the capacitor current?

(Remember, these examples assume ideal capacitors. If you used real capacitors, the results would follow the same trends, but would be limited in current.)

2.23.7 Energy in a Capacitor

Finally, energy cannot be dissipated in an ideal capacitor. (This is not the case for real capacitors because of internal resistance, but since the internal resistance is so small, that energy lost to heating is often ignored.) Energy can only be stored in the electric field (or potential that exists between the plates) for later recovery. The energy stored in a capacitor is found by substituting the capacitor current into the generalized power law (P = IV), then inserting the resulting power into the definition of power (P = dE/dt), and solving for E by integration:

 (2.42)

Example: How much energy is stored in a 1000-µF capacitor with an applied voltage of 5 V?

2.23.8 RC Time Constant

When a capacitor is connected to a dc voltage source, it will charge up almost instantaneously. (In reality, a capacitor has internal resistance, as well as inductance, therefore the term “almost”—see Sec. 3.6, on capacitors, for real-life characteristics.) Likewise, a charged capacitor that is shorted with a wire will discharge almost instantaneously. However, with some resistance added, the rate of charge or discharge follows an exponential pattern, as shown in Fig. 2.101. There are numerous applications that use controlled charge and discharge rates, such as timing ICs, oscillators, waveform shapers, and low-discharge power backup circuits.

FIGURE 2.101

Current and voltage equations for RC charging circuit

(2.43)

where I is the current in amps, VS is the source voltage in volts, R is the resistance in ohms, C is the capacitance in farads, t is the time in seconds after the source voltage is applied, e = 2.718, VR is the resistor voltage in volts and VC is the capacitor voltage in volts. Graph shown to the left is for circuit with R = 10 kΩ, and C = 100 µF. Decreasing the resistance means the capacitor charges up more quickly and the voltage across the capacitor rises more quickly.

For a charging capacitor, the following equations are used.

(You can derive the preceding expressions using Kirchhoff’s law, by summing the voltage around the closed loop: VS = RI + (1/CIdt. Differentiating and solving the differential equation, given an initial condition of current equal to V/R, the voltage across the resistor = VS, and the voltage across the capacitor VC = 0, you get the solution: I = (V/R)et/RC. The voltages across the resistor and capacitor are simply found by plugging the current into the expression for the voltage across a resistor VR = IR and the voltage across a capacitor: VC = (1/CIdt. We’ll discuss solving such circuits in detail in the section on transients in dc circuits.)

Theoretically, the charging process never really finishes, but eventually the charging current drops to an immeasurable value. A convention often used is to let t = RC, which makes V(t) = 0.632 VS. The RC term is called the time constant of the circuit and is the time in seconds required to charge the capacitor to 63.2 percent of the supply voltage. The lowercase tau τ is often used to represent RC: τ = RC. After two time constants (t = 2RC = 2τ) the capacitor charges another 63.2 percent of the difference between the capacitor voltage at one time constant and the supply voltage, for a total change of 86.5 percent. After three time constants, the capacitor reaches 95 percent of the supply voltage, and so on, as illustrated in the graph in Fig. 2.101. After five time constants, a capacitor is considered fully charged, having reached 99.24 percent of the source voltage.

Example: An IC uses an external RC charging network to control its timing. The IC requires 3.4 V at VIN to trigger its output to switch from high to low, at which time an internal transistor (switch) is activated, allowing the capacitor to be discharged to ground. If a 5-s timing period before the trigger point is required, what value of R is required if C = 10 µF?

and solve for R:

FIGURE 2.102

Current and voltage equations for discharging an RC circuit

 (2.44)

where I is the current in amps, VS is the source voltage in volts, R is the resistance in ohms, C is the capacitance in farads, t is the time in seconds after the source voltage is removed, e = 2.718, VR is the resistor voltage in volts, and VC is the capacitor voltage in volts. Graph shown to the left in Fig. 2.103 is for circuit with R = 3 kΩ and C = 0.1 µF.

For a discharging capacitor, the following equations are used.

FIGURE 2.103

(You can derive this expression using Kirchhoff’s law, by summing the voltage around the closed loop: 0 = RI + (1/CIdt. Differentiating and solving the differential equation, given an initial condition of current equal to 0, the voltage across the resistor = 0, and the voltage across the capacitor VC = VS, you get the solution: I = (V/R)et/RC. The voltage across the resistor and capacitor is found simply by plugging the current into the expression for the voltage across a resistor VR = IR and the voltage across a capacitor: VC = (1/CIdt. We’ll discuss solving such circuits in detail in the section on transients in dc circuits.)

The expression for a discharging capacitor is essentially the inverse of that for a charging capacitor. After one time constant, the capacitor voltage will have dropped by 63.2 percent from the supply voltage, so it will have reached 37.8 percent of the supply voltage. After five time constants, the capacitor is considered fully discharged, it will have dropped 99.24 percent, or down to 0.76 percent of the supply voltage.

Example: If a 100-µF capacitor in a high-voltage power supply is shunted by a 100k resistor, what is the minimum time before the capacitor is considered fully discharged when power is turned off?

Answer: After five time constants, a capacitor is considered discharged:

t = 5τ = 5RC = (5)(100 × 103 Ω)(100 × 10−6 F) = 50 s

2.23.9 Stray Capacitance

Capacitance doesn’t exist only within capacitors. In fact, any two surfaces at different electrical potential, and that are close enough together to generate an electric field, have capacitance, and thus act like a capacitor. Such effects are often present within circuits (e.g., between conductive runs or component leads), even though they are not intended. This unintended capacitance is referred to as stray capacitance, and it can result in a disruption of normal current flow within a circuit. Designers of electric circuits figure out ways to minimize stray capacitance as much as possible, such as keeping capacitor leads short and grouping components in such a way as to eliminate capacitive coupling. In high-impedance circuits, stray capacitance may have a greater influence, since capacitive reactance (which we will discuss shortly) may be a greater portion of the circuit impedance. In addition, since stray capacitance usually appears in parallel with a circuit, it may bypass more of the desired signal at higher frequencies. Stray capacitance usually affects sensitive circuits more profoundly.

2.23.10 Capacitors in Parallel

When capacitors are placed in parallel, their capacitances add, just like resistors in series:

 Ctot = C1 + C2 + … Cn (2.45)

(You derive this formula by applying Kirchhoff’s current law at the top junction, which gives Itot = I1 + I2 + I3 + … IN. Making use of the fact that the voltage V is the same across C1 and C2, you can substitute the displacement currents for each capacitor into Kirchhoff’s current expression as follows:

The term in brackets is the equivalent capacitance.)

Intuitively, you can think of capacitors in parallel representing one single capacitor with increased plate surface area. It’s important to note that the largest voltage that can be applied safely to a group of capacitors in parallel is limited to the voltage rating of the capacitor with the lowest voltage rating. Both the capacitance and the voltage rating are usually included next to the capacitor symbol in schematics, but often the voltage rating is missing; you must figure out the rating based on the expected voltages present at that point in the circuit.

FIGURE 2.104

2.23.11 Capacitors in Series

When two or more capacitors are connected in series, the total capacitance is less than that of the smallest capacitor in the group. The equivalent capacitance is similar to resistors in parallel:

 (2.46)

(You derive this by applying Kirchhoff’s voltage law. Since the current I must flow through each capacitor, Kirchhoff’s voltage expression becomes:

The term in parentheses is called the equivalent capacitance for capacitors in series.)

Capacitors may be connected in series to enable the group to withstand a larger voltage than any individual capacitor is rated to withstand (the maximum voltage ratings add). The trade-off is a decrease in total capacitance—though that could be what you intend to do, if you can’t find a capacitor or create a parallel arrangement that gives you the desired capacitance value. Notice in Fig. 2.104 that the voltage does not divide equally between capacitors. The voltage across a single capacitor—say, C2—is a fraction of the total, expressed as (Ctotal/C2)VIN. There are various circuits that tap the voltage between series capacitors.

Use care to ensure that the voltage rating of any capacitor in the group is not exceeded. If you use capacitors in series to withstand larger voltages, it’s a good idea to also connect an equalizing resistor across each capacitor. Use resistors with about 100 Ω per volt of supply voltage, and be sure they have sufficient power-handling capability. With real capacitors, the leakage resistance of the capacitor may have more effect on the voltage division than does the capacitance. A capacitor with a high parallel resistance will have the highest voltage across it. Adding equalizing resistors reduces this effect.

Example: (a) Find the total capacitance and maximum working voltage (WV) for the capacitor in the parallel network. (b) Find the total capacitance, WV, V1, and V2. (c) Find the total capacitance and WV for the network of capacitors. (d) Find the value of C that gives a total capacitance of 592 pF with a total WV of 200 V. Individual capacitor WV values are in parentheses. Refer to Fig. 2.105.

FIGURE 2.105

·        (a) 157 µF (35 V)

·        (b) 0.9 µF (200 V), V1 = 136 V, V2 = 14 V

·        (c) Ctot = 3.2 µF (20 V)

·        (d) C = 470 pF (WV > 100 V)

2.23.12 Alternating Current in a Capacitor

Everything that was discussed about capacitors in dc circuits also applies in ac circuits, with one major exception. While a capacitor in a dc circuit will block current flow (except during brief periods of charging and discharging), a capacitor in an ac circuit will either pass or limit current flow, depending on frequency. Unlike a resistor that converts current energy into heat to reduce current flow, a capacitor stores electrical energy and returns it to the circuit.

The graph in Fig. 2.106 shows the relationship between current and voltage when an ac signal is applied to a capacitor. The ac sine wave voltage has a maximum value of 100. In interval 0 to A, the applied voltage increases from 0 to 38 and the capacitor charges up to that voltage. In interval AB, the voltage increases to 71, so the capacitor has gained 33 V more. During this interval, a smaller amount of charge has been added than 0A, because the voltage rise during AB is smaller than 0A. In interval BC, the voltage rises by 21, from 71 to 92. The increase in current in this interval is still smaller. In interval CD, the voltage increases only 8, and therefore the increase in current is also smaller.

FIGURE 2.106

If you were to divide the first quarter cycle into a very large number of intervals, you’d see that the current charging the capacitor has the shape of a sine wave, just like the applied voltage. The current is largest at the start of the cycle and goes to zero at the maximum value of voltage, so there is a phase difference of 90° between the voltage and the current.

In the second quarter cycle, from time D to H, the applied voltage decreases and the capacitor loses its charge. Using the same reasoning as before, it’s apparent that the current is small during period DE and continues to increase during the other periods. The current is flowing against the applied voltage; however, since the capacitor is discharging into the circuit, the current flows in the negative direction during this quarter cycle.

The third and fourth quarter cycles repeat the events of the first and second, respectively, with one difference: the polarity of the applied voltage has reversed, and current changes to correspond. In other words, an alternating current flows in the circuit because of the alternate charging and discharging of the capacitor. As shown in the graph in Fig. 2.106, the current starts its cycle 90° before the voltage, so the current in a capacitor leads the applied voltage by 90°.

2.23.13 Capacitive Reactance

The amount of charge on a capacitor is equal to the voltage drop across the capacitor times the capacitance (Q = CV). Within an ac circuit, the amount of charge moves back and forth in the circuit every cycle, so the rate of movement of charge (current) is proportional to voltage, capacitance, and frequency. When the effect of capacitance and frequency are considered together, they form a quantity similar to resistance. However, since no actual heat is being generated, the effect is termed capacitive reactance. The unit for reactance is the ohm, just as for resistors, and the formula for calculating the reactance of a capacitor at a particular frequency is given by:

 (2.47) Capacitive reactance

where XC is the capacitive reactance in ohms, f is the frequency in Hz, C is the capacitance in farads, and π = 3.1416. Often, omega ω is used in place of 2πf, where ω is called the angular frequency.

(You can derive this by noting that a sinusoidal voltage source placed across a capacitor will allow displaced current to flow through it because the voltage across it is changing (recall that I = CdV/dt for a capacitor). For example, if the voltage source is given by V0 cos (ωt), you plug this voltage into V in the expression for the displacement current for a capacitor, which gives:

Maximum current or peak current I0 occurs when sin (ωt) = −1, at which point I0 = ωCV0. The ratio of peak voltage to peak current V0/I0 resembles a resistance in light of Ohm’s law, and is given in units of ohms. However, because the physical phenomenon for “resisting” is different from that of a traditional resistor (heating), the effect is given the name capacitive reactance.)

As the frequency goes to infinity, XC goes to 0, and the capacitor acts like a short (wire) at high frequencies; capacitors like to pass current at high frequencies. As the frequency goes to 0, XC goes to infinity, and the capacitor acts like an open circuit; capacitors do not like to pass low frequencies.

It’s important to note that even though the unit of reactance is the ohm, there is no power dissipated in reactance. The energy stored in the capacitor during one portion of the cycle is simply returned to the circuit in the next. In other words, over a complete cycle, the average power is zero. See the graph in Fig. 2.106.

Example 1: Find the reactance of a 220-pF capacitor at an applied frequency of 10 MHz.

(Note: 1 MHz = 1 × 106 Hz, 1 µF = 1 × 10−6, 1 nF = 1 × 10−9 F, 1 pF = 1 × 10−12 F.)

Example 2: What is the reactance of a 470-pF capacitor at 7.5 MHz and 15.0 MHz?

Answer: XC @ 7.5 MHz = 45.2 Ω, XC @ 15 MHz = 22.5 Ω

As you can see, the reactance decreases with an increase in frequency and/or an increase in capacitance. The left graph in Fig. 2.107 shows the reactance versus frequency of a capacitor. Real capacitors don’t follow the graph and equation so precisely, a result of parasitic effects—see “Real Capacitor Models” in Fig. 2.93.

FIGURE 2.107 (a) Graph showing how reactance decreases with frequency for various sizes of capacitors—all capacitors are considered ideal in nature. (b) Graph showing the frequency response for real-life capacitors, which takes into account the parasitic resistances and inductances present within a real capacitor package. The pointy dips within the graph represent self-resonance, where the capacitive and inductive reactances cancel and only internal resistance within the capacitor package is left. The frequency at which point this occurs is called the self-resonant frequency.

2.23.14 Capacitive Divider

Capacitive dividers can be used with ac input signals or even dc, since the capacitors will rapidly reach a steady state. The formula for determining the ac output voltage of a capacitive divider is different from the resistive divider, because the series element, C1 is in the numerator, not C2, the shunt element. See Fig. 2.108.

FIGURE 2.108

Note that the output voltage is independent of the input frequency. However, if the reactance of the capacitors is not large at the frequency of interest (i.e., the capacitance is not large enough), the output current capability will be very low.

2.23.15 Quality Factor

Components that store energy, like a capacitor (and as we’ll see, an inductor), may be compared in terms of quality factor Q, also known as the merit. The Q of any such component is the ratio of its ability to store energy to the sum total of all energy losses within the component. Since reactance is associated with stored energy and resistance is associated with energy loss, we can express the quality factor as:

 (2.48)

Q has no units. When considering a capacitor, the reactance (in ohms) is simply the capacitive reactance X = XC. (For an inductor, we’ll see that X = XL, where XL is the inductive reactance.) R is the sum of all resistances associated with the energy losses in the component (in ohms). The Q of capacitors is ordinarily high. Good-quality ceramic capacitors and mica capacitors may have Q values of 1200 or more. Small ceramic trimmer capacitors may have Q values too small to ignore in some applications. Microwave capacitors can have poor Q values, 10 or less at 10 GHz or higher.

2.24 Inductors

In the preceding section we saw how a capacitor stored electrical energy in the form of an electric field. Another way to store electrical energy is in a magnetic field. Circular radiating magnetic fields can be generated about a wire any time current passes through it. Increasing or decreasing current flow through the wire increases and decreases the magnetic field strength, respectively. During such changes in magnetic field strength, we encounter a phenomenon known as inductance. Inductance is a property of circuits somewhat analogous to resistance and capacitance; however, it is not attributed to heat production or charge storage (electric field), but rather it is associated with magnetic fields—more specifically, how changing magnetic fields influence the free electrons (current) within a circuit. Theoretically, any device capable of generating a magnetic field has inductance. Any device that has inductance is referred to as an inductor. To understand inductance requires a basic understanding of electromagnetic properties.

FIGURE 2.109 The three cornerstones of electronics are resistance, capacitance, and inductance. Inductance, unlike the other two, involves alternations in a circuit’s current and voltage characteristics as a result of forces acting upon free electrons resulting in the creation and collapse of magnetic fields, usually concentrated in a discrete inductor device. Like a capacitor, however, inductive effects occur only during times of change, when the applied voltage/current increases or decreases with time. Resistance doesn’t have a time dependency. Can you guess what will happen to the brightness of the lamp in each of the circuits in the figure when the switch is closed? What do you think will occur when the switch is later opened? We’ll discuss this a bit later.

2.24.1 Electromagnetism

According to the laws of electromagnetism, the field of a charge at rest can be represented by a uniform, radial distribution of electric field lines or lines of force (see Fig. 2.110a). For a charge moving at a constant velocity, the field lines are still radial and straight, but they are no longer uniformly distributed (see Fig. 2.110b). At the same time, the electron generates a circular magnetic field (see Fig. 2.110c). If the charge accelerates, things get a bit more complex, and a “kink” is created in the electromagnetic field, giving rise to an electromagnetic wave that radiates out (see Fig. 2.110d and e).

FIGURE 2.110 The electric and magnetic fields are part of the same phenomenon called electromagnetism. Magnetic fields appear whenever a charge is in motion. Interestingly, if you move along with a moving charge, the observable magnetic field would disappear—thanks to Einstein’s relativity.

As depicted in Fig. 2.110c, the electric field (denoted E) of a moving electron—or any charge for that matter—is, in effect, partially transformed into a magnetic field (denoted B). Hence, it is apparent that the electric and magnetic fields are part of the same phenomenon. In fact, physics today groups electric and magnetic fields together into one fundamental field theory, referred to as electromagnetism. (The work of Maxwell and Einstein helped prove that the two phenomena are linked. Today, certain fields in physics paint a unique picture of field interactions using virtual photons being emitted and absorbed by charges to explain electromagnetic forces. Fortunately, in electronics you don’t need to get that detailed.)

The simplest way to generate a magnetic field is to pass a current through a conductor. Microscopically, each electron within a wire should be generating a magnetic field perpendicular to its motion. However, without any potential applied across the wire, the sheer randomness of the electrons due to thermal effects, collision, and so on, cause the individual magnetic fields of all electrons to be pointing in random directions. Averaged over the whole, the magnetic field about the conductor is zero (see Fig. 2.111a.1). Now, when a voltage is applied across the conductor, free electrons gain a drift component pointing from negative to positive—conventional current in the opposite direction. In terms of electron speed, this influence is very slight, but it’s enough to generate a net magnetic field (see Fig. 2.111a.2). The direction of this field is perpendicular to the direction of conventional current flow and curls in a direction described by the right-hand rule: your right thumb points in the direction of the conventional current flow; your finger curls in the direction of the magnetic field. See Fig. 2.111b. (When following electron flow instead of conventional current flow, you’d use your left hand.)

FIGURE 2.111 (a) Magnetic field generated free electrons moving in unison when voltage is applied. (b) Right-hand rule showing direction of magnetic field in relation to conventional current flow. (c) Permanent magnet. (d) Magnetic dipole radiation pattern created by current flowing through single loop of wire. (e) Energized solenoid with a magnetic field similar to permanent magnet. (f) Electromagnet, using ferromagnetic core for increased field strength.

The magnetic field created by sending current through a conductor is similar in nature to the magnetic field of a permanent magnet. (In reality, the magnetic field pattern of a permanent bar magnet, in Fig. 2.111c, is more accurately mimicked when the wire is coiled into a tight solenoid, as shown in Fig. 2.111e.) The fact that both a current-carrying wire and a permanent magnet produce magnetic fields is no coincidence. Permanent magnets made from ferromagnetic materials exhibit magnetic closed-loop fields, mainly as a result of the motion of unpaired electrons orbiting about the nucleus of an atom, as shown in Fig. 2.112, generating a dipole magnetic field. The lattice structure of the ferromagnetic material has an important role of fixing a large portion of the atomic magnetic dipoles in a fixed direction, so as to set up a net magnetic dipole pointing from north to south. This microscopic motion of the unpaired electron about the nucleus of the atom resembles the flow of current through a loop of wire, as depicted in Fig. 2.111d. (Electron spin is another source of magnetic fields, but is far weaker than that due to the electron orbital.)

FIGURE 2.112 Microscopically, the magnetic field of a permanent magnet is a result of unpaired valence electrons fixed in a common direction, generating a magnetic dipole. The fixing in orientation is a result of atomic bonding in the crystalline lattice structure of the magnet.

By coiling a wire into a series of loops, a solenoid is formed, as shown in Fig. 2.111e. Every loop of wire contributes constructively to make the interior field strong. In other words, the fields inside the solenoid add together to form a large field component that points to the right along the axis, as depicted in Fig. 2.111e. By placing a ferromagnetic material (one that isn’t initially magnetized) within a solenoid, as shown in Fig. 2.111f, a much stronger magnetic field than would be present with the solenoid alone is created. The reason for such an increase in field strength has to do with the solenoid’s field rotating a large portion of the core’s atomic magnetic dipoles in the direction of the field. Thus, the total magnetic field becomes the sum of the solenoid’s magnetic field and the core’s temporarily induced magnetic field. Depending on material and construction, a core can magnify the total field strength by a factor of 1000.

2.24.2 Magnetic Fields and Their Influence

Magnetic fields, unlike electric fields, only act upon charges that are moving in a direction that is perpendicular (or has a perpendicular component) to the direction of the applied field. A magnetic field has no influence on a stationary charge, unless the field itself is moving. Figure 2.113a shows the force exerted upon a moving charge placed within a magnetic field. When considering a positive charge, we use our right hand to determine the direction of force upon the moving charge—the back of the hand points in the direction of the initial charge velocity, the fingers curl in the direction of the external magnetic field, and the thumb points in the direction of the force that is exerted upon the moving charge. For a negative charge, like an electron, we can use the left hand, as shown in Fig. 2.113b. If a charge moves parallel to the applied field, it experiences no force due to the magnetic field—see Fig. 2.113c.

FIGURE 2.113 Illustration showing the direction of force upon a moving charge in the presence of a fixed magnetic field.

In terms of a group of moving charges, such as current through a wire, the net magnetic field of one wire will exert a force on the other wire and vice versa (provided the current is fairly large), as shown in Fig. 2.114. (The force on the wire is possible, since the electrostatic forces at the surface of the lattice structure of the wire prevent electrons from escaping from the surface.)

FIGURE 2.114 Forces exerted between two current-carrying wires.

Likewise, the magnetic field of a fixed magnet can exert a force on a current-carrying wire, as shown in Fig. 2.115.

FIGURE 2.115 (a) The force a current-carrying wire experiences in the presence of a magnet’s field. (b) Illustration showing how bar magnets attract and repel.

Externally, a magnet is given a north-seeking (or “north,” for short) and south-seeking (or “south,” for short) pole. The north pole of one magnet attracts the south pole of another, while like poles repel—see Fig. 2.115b. You may ask how two stationary magnets exert forces on each other. Isn’t the requirement that a charge or field must be moving for a force to be observed? We associate the macroscopic (observed) force with the forces on the moving charges that comprise the microscopic internal magnetic dipoles which are, at the heart, electrons in motion around atoms. These orbitals tend to be fixed in a general direction called domains—a result of the lattice binding forces.

Another aspect of magnetic fields is their ability to force electrons within conductors to move in a certain direction, thus inducing current flow. The induced force is an electromotive force (EMF) being set up within the circuit. However, unlike, say, a battery’s EMF, an induced EMF depends on time and also on geometry. According to Faraday’s law, the EMF induced in a circuit is directly proportional to the time rate of change of the magnetic flux through the circuit:

 (for a coiled wire of N loops) (2.49)

where ΦM is the magnetic flux threading through a closed-loop circuit (which is equal to the magnetic field B dotted with the direction surface area—both of which are vectors, and summed over the entire surface area—as the integral indicates). According to the law, the EMF can be induced in the circuit in several ways: (1) the magnitude of B can vary with time; (2) the area of the circuit can change with time; (3) the angle between B and the normal of A can change with time; (4) any combination of these can occur. See Fig. 2.116.

FIGURE 2.116 Illustration of Faraday’s law of induction.

The simple ac generator in Fig. 2.117 shows Faraday’s law in action. A simple rotating loop of wire in a constant magnetic field generates an EMF that can be used to power a circuit. As the loop rotates, the magnetic flux through it changes with time, inducing an EMF and a current in an external circuit. The ends of the loop are connected to slip rings that rotate with the loop, while the external circuit is linked to the generator by stationary brushes in contact with the slip rings.

FIGURE 2.117 Basic ac generator.

A simple dc generator is essentially the same as the ac generator, except that the contacts to the rotating loop are made using a split ring or commutator. The result is a pulsating direct current, resembling the absolute value of a sine wave—there are no polarity reversals.

A motor is essentially a generator operating in reverse. Instead of generating a current by rotating a loop, a current is supplied to the loop by a battery, and the torque acting on the current-carrying loop causes it to rotate. Real ac generators and motors are much more complex than the simple ones demonstrated here. However, they still operate under the same fundamental principles of electromagnetic induction.

The circuit in Fig. 2.118 shows how it is possible to induce current within a secondary coil of wire by suddenly changing the current flow through a primary coil of wire. As the magnetic field of the primary expands, an increasing magnetic flux permeates the secondary. This induces an EMF that causes current to flow in the secondary circuit. This is the basic principle behind how transformers work; however, a real transformer’s primary and secondary coils are typically wound around a common ferromagnetic core to increase magnetic coupling.

FIGURE 2.118 An induced voltage, or EMF, is generated in the secondary circuit whenever there is a sudden change in current in the primary.

2.24.3 Self-Inductance

In the previous section, we saw how an EMF could be induced in a closed-loop circuit whenever the magnetic flux through the circuit changed with time. This phenomenon of electromagnetic induction is used in a number of mechanisms, such as motors, generators, and transformers, as was pointed out. However, in each of these cases, the induced EMF was a result of an external magnetic field, such as the primary coil in relation to a secondary coil. Now, however, we will discuss a phenomenon called self-induction. As the name suggests, self-induction typically involves a looped wire inflicting itself with an induced EMF that is generated by the varying current that passes through it. According to Faraday’s law of induction, the only time that our loop can self-inflict is when the magnetic field grows or shrinks in strength (as a result of an increase or decrease in current). Self-induction is the basis for the inductor, an important device used to store energy and release energy as current levels fluctuate in time-dependent circuits.

Consider an isolated circuit consisting of a switch, a resistor, and a voltage source, as shown in Fig. 2.119. If you close the switch, you might predict that the current flow through the circuit would jump immediately from zero to V/R, according to Ohm’s law. However, according to Faraday’s law of electromagnetic induction, this isn’t entirely accurate. Instead, when the switch is initially closed, the current increases rapidly. As the current increases with time, the magnetic flux through the loop rises rapidly. This increasing magnetic flux then induces an EMF in the circuit that opposes the current flow, giving rise to an exponentially delayed rise in current. We call the induced EMF a self-induced EMF.

FIGURE 2.119 (a) Circuit is open and thus no current or magnetic field is generated. (b) The moment the circuit is closed, current begins to flow, but at the same time, an increasing magnetic flux is generated through the circuit loop. This increasing flux induces a back EMF that opposes the applied or external EMF. After some time, the current levels off, the magnetic flux reaches a constant value, and the induced EMF disappears. (d) If the switch is suddenly opened, the current attempts to go to zero; however, during this transition, as the current goes to zero, the flux decreases through the loop, thus generating a forward induced voltage of the same polarity as the applied or external EMF. As we’ll see later, when we incorporate large solenoid and toroidal inductors within a circuit, opening a switch such as this can yield a spark—current attempting to keep going due to a very large forward EMF.

Self-induction within a simple circuit like that shown in Fig. 2.119 is usually so small that the induced voltage has no measurable effect. However, when we start incorporating special devices that concentrate magnetic fields—namely, discrete inductors—time-varying signals can generate significant induced EMFs. For the most part, unless otherwise noted, we shall assume the self-inductance of a circuit is negligible compared with that of a discrete inductor.

2.24.4 Inductors

Inductors are discrete devices especially designed to take full advantage of the effects of electromagnetic induction. They are capable of generating large concentrations of magnetic flux, and they are likewise capable of experiencing a large amount of self-induction during times of great change in current. (Note that self-induction also exists within straight wire, but it is usually so small that it is ignored, except in special cases, e.g., VHF and above, where inductive reactance can become significant.)

The common characteristic of inductors is a looplike geometry, such as a solenoid, toroid, or even a spiral shape, as shown in Fig. 2.120. A solenoid is easily constructed by wrapping a wire around a hollow plastic form a number of times in a tight-wound fashion.

FIGURE 2.120 Various coil configurations of an inductor—solenoid, toroid, and spiral.

The basic schematic symbol of an air core inductor is given by . Magnetic core inductors (core is either iron, iron powder, or a ferrite-type ceramic), adjustable core inductors, and a ferrite bead, along with their respective schematic symbols are shown in Fig. 2.121.

FIGURE 2.121

Magnetic core inductors are capable of generating much higher magnetic field densities than air core inductors as a result of the internal magnetization that occurs at the atomic level within the core material due to the surrounding wire coil’s magnetic field. As a result, these inductors experience much greater levels of self-induction when compared to air core inductors. Likewise, it is possible to use fewer turns when using a magnetic core to achieve a desired inductance. The magnetic core material is often iron, iron powder, or a metallic oxide material (also called a ferrite, which is ceramic in nature). The choice of core material is a complex process that we will cover in a moment.

Air core inductors range from a single loop in length of wire (used at ultrahigh frequencies), through spirals in copper coating of an etched circuit board (used at very high frequencies), to large coils of insulated wire wound onto a nonmagnetic former. For radio use, inductors often have air cores to avoid losses caused by magnetic hysteresis and eddy current that occur within magnetic core–type inductors.

Adjustable inductors can be made by physically altering the effective coil length—say, by using a slider contact along the uncoated coil of wire, or more commonly by using a ferrite, powdered iron, or brass slug screwed into the center of the coil. The idea behind the slug-tuned inductor is that the inductance depends on the permeability of the material within the coil. Most materials have a relative permeability close to 1 (close to that of vacuum), while ferrite materials have a large relative permeability. Since the inductance depends on the average permeability of the volume inside the coil, the inductance will change as the slug is turned. Sometimes the slug of an adjustable inductor is made of a conducting material such as brass, which has a relative permeability near 1, in which case eddy currents flow on the outside of the slug and eliminate magnetic flux from the center of the coil, reducing its effective area.

Inductor Basics

An inductor acts like a time-varying current-sensitive resistance. It only “resists” during changes in current; otherwise (under steady-state dc conditions), it passes current as if it were a wire. When the applied voltage increases, it acts like a time-dependent resistor whose resistance is greatest during times of rapid increase in current. On the other hand, when the applied voltage decreases, the inductor acts like a time-dependent voltage source (or negative resistance) attempting to keep current flowing. Maximal sourcing is greatest during times of rapid decreases in current.

In Fig. 2.122a, when an increasing voltage is applied across an inductor, resulting in an increasing current flow, the inductor acts to resist this increase by generating a reverse force on free electrons as a result of an increasing magnetic flux cutting across the coils of the solenoid (or crossing through the coil loops). This reverse force on free electrons can be viewed as an induced EMF that points in the opposite direction of the applied voltage. We deem this induced EMF a reverse EMF, also referred to as a back EMF. The result is that the inductor strongly resists during a sudden increase in current flow, but quickly loses its resistance once the current flow levels off to a constant value.

FIGURE 2.122

In Fig. 2.122b, when a decreasing voltage is applied across an inductor, resulting in a decreasing current flow, the inductor acts to resist this decrease by generating a forward force on free electrons as a result of a decreasing magnetic flux cutting across the coils of the solenoid (or crossing through the coil loops). This forward force on free electrons can be viewed as an induced EMF that points in the same direction as that of the applied voltage (the applied voltage that existed before the sudden change). We deem this induced EMF a forward EMF. The result is that the inductor acts like a voltage source during a sudden decrease in current flow, but quickly disappears once the current flow levels off to a constant value.

Another view of how inductors work is to consider energy transfer. The transfer of energy to the magnetic field of an inductor requires that work be performed by a voltage source connected across it. If we consider a perfect inductor where there is no resistance, the energy that goes into magnetic field energy is equal to the work performed by the voltage source. Or, in terms of power, the power is the rate at which energy is stored (P = dW/dt). Using the generalized power law P = IV, we can equate the powers and see that there must be a voltage drop across an inductor while energy is being stored in the magnetic field. This voltage drop, exclusive of any voltage drop due to resistance in a circuit, is the result of an opposing voltage induced in the circuit while the field is building up to its final value. Once the field becomes constant, the energy stored in the magnetic field is equal to the work performed by the voltage source.

Figure 2.123 illustrates what occurs when an inductor is energized suddenly as a switch is closed.

FIGURE 2.123

When an inductor is being energized, we say that electrical energy from an externally applied source is transformed into a magnetic field about the inductor. Only during times of change—say, when the switch is thrown, and the magnetic field suddenly grows in size—do we see inductive behavior that affects circuit dynamics.

In the circuit to the left, when the switch is thrown from position B to position A, a sudden change in voltage is applied across the inductor, resulting in a sudden increase in current flow. Where there was no magnetic field, there now is a rapidly growing magnetic field about the inductor. The inductor is said to be energizing. The expanding magnetic field cuts across its own inductor coils, which, by Faraday’s law, exerts a force on free electrons within the coil. The force on these electrons is such as to be pointing in the opposite direction as the applied voltage. This effective reverse force and the effect it has on the free electrons is deemed a reverse EMF—it’s analogous to a little imaginary battery placed in the reverse direction of the applied voltage. (See Fig. 2.123b.) The result is that the inductor resists during increases in current. After a short while, the current flow through the circuit levels off, and the magnetic field stops growing and assumes a constant value. With no change in magnetic field strength, there is no increase in magnet flux through our fixed coils, so there is no more reverse EMF. The inductor acts as a simple conductor. Figure 2.123c shows the induced voltage and voltage across the resistor as a function of time.

Figure 2.123d shows the resultant current flow through the inductor as a result of the resultant voltage. Mathematically, the current flow is expressed by the following equation:

(You might question what happens if we assume an ideal inductor with zero internal resistance R. This is a very good question—one that we’ll discuss when we get to defining inductance mathematically.)

In terms of energy, this can be viewed as the electrical energy being transformed into magnetic field energy. In terms of power, we see that a voltage drop occurs as energy is pumped into the magnetic field. (We associate the drop with the back EMF.) Once the current levels off, no more energy goes into the field; hence, no reverse voltage (or voltage drop) is present.

Figure 2.124 illustrates what occurs as an inductor is deenergized suddenly by opening a switch.

FIGURE 2.124

When an inductor is being deenergized, we say that the magnetic field energy of the inductor is “transformed” back into electrical energy. Again, only during times of change, which in our case involves a decreasing current flow, do we notice effects of induction.

In the circuit to the left, when the switch is thrown from position A to B, a sudden change voltage occurs across the inductor. The inductor initially opposes this decrease by generating a collapsing magnetic field that cuts across the inductor coils. According to Faraday’s law, a decreasing magnetic flux passes through the loops of the inductor, thus imparting a force upon free electrons within the coil in the same direction the applied voltage was pointing right before the switch occurred. Since the force is in the same direction, we deem the effect a forward EMF. Hence, the inductor sources current when attempts are made to decrease current flow. The energy for it to do so comes from the magnetic field, whose energy drops in proportion to the electrical energy delivered to the circuit. See 2.124(b).

Part (c) shows the resultant voltage across an inductor adding the applied voltage to the inducted voltage.

Part (d) shows the resultant current flow through the inductor as a result of the resultant voltage.

Mathematically, the current flow is expressed by the following equation:

(Again, you might question what happens if we assume an ideal inductor with zero internal resistance R. This is a very good question—one that we’ll discuss when we get to defining inductance mathematically.)

Note: We must assume that when the switch is thrown from A to B, the transition occurs instantaneously. We’ll see in a moment that when a physical break occurs, cutting current flow through an inductive circuit, the collapsing magnetic field can be large enough to generate an EMF capable of causing a spark to jump between the break points (i.e., switch contacts).

2.24.5 Inductor Water Analogy

The property of inductance in electric circuits is closely analogous to mass inertia in mechanical systems. For example, the property of an inductor resisting any sudden changes in current flow (increasing or decreasing) is similar to a mass on a spinning wheel resisting any change in motion (increasing or decreasing in speed). In the following water analogy, we take this mass analogy to heart, incorporating a water turbine/flywheel device to represent a “water inductor.”

To start, we consider a basic electrical inductor circuit, as shown to the left in Fig. 2.125. The field generated by a suddenly applied voltage creates a reverse induced voltage of opposite polarity that initially “resists” current flow. Quickly, depending on the inductance value, the reverse voltage disappears as the magnetic field becomes constant, at which point it has reached a maximum strength and energy. A collapsing field generated when the applied voltage is removed creates a forward induced voltage that attempts to keep current flowing. Quickly, depending on the inductance value, the forward voltage disappears and the magnetic field goes to zero.

FIGURE 2.125

In the water analogy, the turbine with attached flywheel resists any sudden changes in current flow. If water pressure is suddenly applied, the turbine initially resists water flow due to its mass and that of the attached flywheel. However, the pressure exerted over the turbine blades quickly gives rise to mechanical motion. Depending on the mass of the flywheel, the time it takes for the flywheel to reach a steady angular velocity will vary—a heavier flywheel will require more time (analogous to a high-value inductor requiring more time to reach a constant current after a sudden increase in applied voltage). When the flywheel reaches this constant angular rotation, the water inductor has maximum rotational momentum and energy. This is analogous to the magnetic field strength reaching a constant maximum strength and energy when the reverse voltage disappears. If there is any sudden interruption in applied pressure—say, by turning the tap to position B-S, as shown in Fig. 2.125—the flywheel’s angular momentum will attempt to keep current flowing. This is analogous to the collapsing magnetic field in an inductor inducing a forward voltage.

Example: What will happen when the switches are closed in the following three circuits? What happens when the switches are later opened? How does the size of the inductance and capacitance influence behavior? Refer to Fig. 2.126.

FIGURE 2.126

Answer: (a) When the switch is closed in the inductor circuit, the lamp will suddenly light up brightly, but then quickly dims out. This is because the moment the switch is closed, the inductor has very high impedance to current flow, but quickly loses its impedance as the current becomes constant (the magnetic field is no longer expanding). Once constant, the inductor acts like a short, and all current is diverted away from the lamp through the inductor. (We assume here that there is sufficient dc internal resistance in the inductor to prevent excessive current flow. Also, we assume the internal dc resistance of the inductor is much smaller than the internal resistance of the lamp.) A larger inductance value increases the time it takes for the lamp to dim out completely.

(b) When the switch is closed in the capacitor circuit, the opposite effect occurs; the lamp slowly builds up in brightness until it reaches maximum illumination. This is because when the switch is closed, the capacitor initially has very low impedance to current flow during rapid changes in applied voltage. However, as the capacitor charges up, the capacitor’s impedance rises toward infinity, and consequently resembles an open circuit—hence, all current is diverted through the lamp. A larger capacitance value increases the time it takes for the lamp to reach full brightness.

(c) When the switch is closed in the resistor circuit, free-electron flow throughout the system is essentially instantaneous. Aside from small inherent inductance and capacitance built into the circuit, there are no time-dependent effects on current flow caused by discrete inductance or capacitance. Note that since the voltage source is ideal, no matter what value the parallel resistor has, there will always be 12 V across the lamp, and hence the brightness of the lamp does not change over time.

2.24.6 Inductor Equations

Conceptually, you should now understand that the amplitude of the induced voltage—be it reverse or forward induced—is proportional to the rate at which the current changes, or the rate at which the magnetic flux changes. Quantitatively, we can express this relationship by using the following equation:

 (2.50) Voltage across aninductor = induced EMF

If we integrate and solve for IL, we get:

 (2.51) Current throughan inductor

In Eq. 2.50, the proportionality constant L is called the inductance. This constant depends on a number of physical inductor parameters, such as coil shape, number of turns, and core makeup. A coil with many turns will have a higher L value than one with few turns, if both coils are otherwise physically similar. Furthermore, if an inductor is coiled around a magnetic core, such as iron or ferrite, its L value will increase in proportion to the permeability of that core (provided circuit current is below the point at which the core saturates).

FIGURE 2.127 The voltage measured across an ideal inductor is the induced voltage, or EMF, symbolized VL. When there is a steady-state dc current flow, there is no induced voltage (VL is zero) and the inductor resembles a short. As we’ll see, the inductor equations can yield surprisingly unrealistic values, if we don’t consider internal resistance and capacitance inherently present within a real inductor.

The basic unit of inductance L is the henry, abbreviated H. One henry equals an induced voltage of 1 V when the current is varying at a rate of 1 A/s:

(Definition of a henry)

Making inductors from scratch is common in electronics, unlike the construction of capacitors, which is left almost exclusively to the manufacturers. Though we’ll examine how to make inductors in a moment, it’s worth taking a look at some commercial inductors. Note the inductance range, core type, current, and frequency limits listed in Table 2.7.

TABLE 2.7 Typical Characteristics of Commercial Inductors

 CORE TYPE MINIMUM H MAXIMUM H ADJUSTABLE? HIGH CURRENT? FREQUENCY LIMIT Air core, self-supporting 20 nH 1 mH Yes Yes 1 GHz Air core, on former 20 nH 100 mH No Yes 500 MHz Slug tuned open winding 100 nH 1 mH Yes No 500 MHz Ferrite ring 10 mH 20 mH No No 500 MHz RM Ferrite Core 20 mH 0.3 H Yes No 1 MHz EC or ETD Ferrite Core 50 mH 1 H No Yes 1 MHz Iron 1 H 50 H No Yes 10 kHz

Typical values for commercial inductors range from fractions of a nanohenry to about 50 H. Inductances are most commonly expressed in terms of the following unit prefixes:

 nanohenry (nH): 1 nH = 1 × 10−9 H = 0.000000001 H microhenry (µH): 1 µH = 1 × 10−6 H = 0.000001 H millihenry (mH): 1 mH = 1 × 10−3 H = 0.001 H

Example 1: Rewrite 0.000034 H, 1800 mH, 0.003 mH, 2000 µH, and 0.09 µH in a more suitable unit prefix format (1 ≤ numeric value < 1000).

Answer: 34 µH, 1.8 H, 3 µH, 2 mH, and 90 nH.

From another standpoint, inductance can be determined from basic physics principles. Theoretically, you can determine the inductance any time by stating that the inductance is always the ratio of the magnetic flux linkage (NΦM) to the current:

 (2.52)

For an air-filled solenoid, as shown in Fig. 1.128, if a current I flows through the coil, Ampere’s law allows us to calculate the magnetic flux:

where nunit is the turns per unit length:

 nunit = N/ℓ (2.53)

where N is the total turns and ℓ is the length. The variable A is the cross-sectional area of the coil, and µ is the permeability of the material on which the coil is wound. For most materials (excluding iron and ferrite materials), the permeability is close to the permeability of free space:

µ0 = 4π × 10−7 T ⋅ m/A

FIGURE 2.128

According to Faraday’s law, an induced voltage is in each loop of the solenoid, resulting in a net induced voltage across the solenoid equal to n times the change in magnetic flux:

The term in front of dI/dt in the equation we call the inductance of the solenoid:

 (2.54)

Inductance varies as the square of the turns. If the number of turns is doubled, the inductance is quadrupled. This relationship is inherent in the equation but is often overlooked. For example, if you want to double the inductance of a coil, you don’t double the turns; rather you make the number of turns (or 1.41) times the original number of turns, or 40 percent more turns.

Example 2: Find the inductance of a cylindrical coil of length 10 cm, radius 0.5 cm, having 1000 turns of wire wrapped around a hollow plastic form.

L = µ0N2A/ℓ = (4π × 10−7) 106(π × 0.0052)/0.1

= 1 × 10−3H = 1 mH

Luckily, there are some simple formulas, shown in Fig. 2.129, for air core inductors, as well as for a multiple-wound and a spiral-wound inductor. Note that the formulas’ answers will not be in standard form—they assume the results are in units of microhenrys.

FIGURE 2.129 Practical air core inductor equations.

Example 3: What is the inductance of a coil wrapped around a 0.5-in-diameter plastic form if the coil has 38 turns wound at 22 turns per inch?

Answer: First the total length is determined:

Next, using the equation for an air core inductor shown in Fig. 2.129, noting the result will be in units of microhenrys, we get:

Example 4: Design a solenoid inductor with an inductance of 8 µH if the form on which the coil is wound has a diameter of 1 in and a length of 0.75 in.

Answer: Rearranging the equation in the previous example:

A 20-turn coil would be close enough in practical work. Since the coil will be 0.75 in long, the number of turns per inch will be 19.6/0.75 = 26.1. A #17 enameled wire (or anything smaller) can be used. In practice, you wind the required number of turns on the form and then adjust the spacing between the turns to make a uniformly spaced coil of 0.75 in long.

On the Internet you’ll find a number of free web-based inductor calculators. Some are quite advanced, allowing you to input inductance, diameter, and length values, and afterward providing you with the required number of turns, possible number of layers required, dc resistance of the wire, wire gauge to use, and so on. Check these tools out—you can save yourself the trouble of doing calculations and looking up wire dimensions and such.

2.24.7 Energy Within an Inductor

An ideal inductor, like an ideal capacitor, doesn’t dissipate energy, but rather stores it in the magnetic field and later returns it to the circuit when the magnetic field collapses. The energy EL stored in the inductor is found by using the generalized power law P = IV, along with the definition of power P = dW/dt, and the inductor equation V = L dI/dt. By equating the work W with EL we get:

 (2.55)

where EL = energy in joules, I = current in amps, and L = inductance in henrys. Note that in a real inductor a small portion of energy is lost to resistive heating through the inductor’s internal resistance.

2.24.8 Inductor Cores

To conserve space and material, inductors are often wound on a magnetic core material, such as laminated iron or a special molded mix made from iron powder or ferrite material (iron oxide mixed with manganese, zinc, nickel, and other ingredients). A magnetic core increases the magnetic flux density of a coil greatly, and thus increases the inductance. This is further intensified if the magnetic core is formed into a doughnut-shaped toroid. See Fig. 2.130.

FIGURE 2.130 Magnetic core inductors.

The reason a magnetic core is so influential has to do with the magnetization that occurs within it as the outer coil passes current. When current is sent through the wire coil, a relatively weak magnetic field is set up at its center. This magnetic field, which we’ll call the external magnetic field, causes the atomic rearrangement of magnetic dipoles (refer to Fig. 2.112) within the core material. This realignment is such as to rotate the dipole moments in a common direction. As more current is passed through the coil, more and more dipoles line up. The core itself is now generating a magnetic field as a result of the dipole alignment. The net magnetic field Btotal generated by the whole inductor (coil and core) then becomes the sum of the external field (coil) and a term that is proportional to the magnetization M present in the core itself, as the external field is applied by the coil through the core. Mathematically, this is expressed:

(1)

Btotal = Bext + µ0M

where µ0 is the permeability of free space. The magnetic intensity H due to real current in the coil, as opposed to the magnetic intensity generated by atomic magnetization of the core, is expressed:

(2)

This can be further reduced, using the notions of susceptibility and permeability (consult a physics book), to:

(3)

Btotal = µH

where µ is the permeability of the core material.

The ratio of magnetic flux density produced by a given core material compared to the flux density produced by an air core is called the relative permeability of the material µR = µ/µ0. For example, an air core generating a flux density of 50 lines of force per square inch can be made to generate 40,000 lines of force per square inch with an iron core inserted. The ratio of these flux densities, iron core to air core, or the relative permeability is 40,000/50, or 800. Table 2.8 shows permeabilities of some popular high-permeability materials.

TABLE 2.8 Permeability of Various Materials

 MATERIAL APPROXIMATE MAXIMUM PERMEABILITY (H/M) APPROXIMATE MAXIMUM RELATIVE PERMEABILITY APPLICATIONS Air 1.257 × 10−6 1 RF Ferrite U60 1.00 × 10−5 8 UHF chokes Ferrite M33 9.42 × 10−4 750 Resonant circuits Ferrite N41 3.77 × 10−3 3000 Power circuits Iron (99.8% pure) 6.28 × 10−3 5000 Ferrite T38 1.26 × 10−2 10,000 Broadband transformers 45 Permalloy 3.14 × 10−2 25,000 Silicon 60 steel 5.03 × 10−2 40,000 Dynamos, transformers 78 Permalloy 0.126 100,000 Supermalloy 1.26 1,000,000 Recording heads

Problems with Magnetic Cores

When a magnetic core material is conductive (e.g., steel), there is a phenomenon known as eddy currents that arise within the core material itself when there is an applied magnetic field that is changing. For example, in Fig. 2.131a, when an increasing current is supplied through the outer coil, a changing magnetic flux passes through the core. This in turn induces a circular current flow within the core material. Eddy currents that are induced in the core represent loss in the form of resistive heating and can be a significant disadvantage in certain applications (e.g., power transformers). Eddy current losses are higher in materials with low resistivity.

FIGURE 2.131 (a) Large eddy currents within core. (b) Eddy currents are reduced within laminated core.

To avoid eddy currents, the conductive core (steel, in this case) can be laminated together with insulating sheets of varnish or shellac. Current will still be induced in the sheets of the core, as shown in Fig. 2.131b, but because the area of the sheets is limited, so is the flux change, and therefore so are these currents.

Ferrite materials have quite high intrinsic resistivity compared to, say, steel (10 to 1,000 Ω-cm for Mn-Zn ferrites; 105 to 107 Ω-cm for Ni-Zn ferrites). Eddy current losses are therefore much less of a problem in ferrites, and this is the fundamental reason they are used in higher-frequency applications. Powdered iron cores, with their insulating compound present between iron particles, also reduce eddy currents, since the path for the eddy current is limited to the powder particle size.

A second difficulty with iron is that its permeability is not constant, but varies with the strength of the magnetic field and, hence, with the current in the windings. (It also varies with temperature.) In fact, at sufficiently high magnetic fields, the core will saturate and its relative permeability will drop to a value near unity. Not only that, but the magnetic field in the iron depends on the past history of the current in the winding. This property of remanence is essential in a permanent magnet, but in an inductor it gives rise to additional losses, called hysteresis losses. See Fig. 2.132.

FIGURE 2.132 Hysteresis curve showing how the magnetization of the core material does not represent a reversible process. At point a, no current is applied through the coil. Along path a-b-c, coil current, and thus H (applied magnetic field or magnetizing force), increases, causing the core’s magnetic dipole moments (in this case localized in domains) to rotate in proportion. As we approach point c, the core reaches saturation—increasing H causes no appreciable increase in M (magnetization or magnetic dipole density of core)—the domain’s dipole moments are aligned as much as possible parallel to H. Saturation causes a rapid decrease in permeability. The saturation point of a magnetic core varies with material makeup; air and other nonmagnetic materials do not saturate—they have a permeability of 1. As H is decreased, the magnetization M does not follow the same path back down, but follows path c-d-e. Notice that when H goes to zero, the core remains magnetized. In essence, the core has become a permanent magnet. The term retentivity is used to describe this effect, and it presents another set of losses caused by hysteresis. In order to demagnetize the core, a reverse force is necessary to overcome the residual magnetism retained by the core. In other words, H must go negative in the opposite direction, driving the magnetic domains back into a random orientation. At point e, the core has again reached saturation; however, the magnetic dipoles (domains) are now pointing in the opposite direction. To reach saturation in the opposite direction again, H must be applied, as shown in path e-f-c. Air cores and other materials with a permeability of 1, such as brass and aluminum, are immune to hysteresis effects and losses.

To avoid loses associated with hysteresis, it is important to not run the core inductor into saturation. This can be accomplished by running the inductor at lower current, using a larger core, altering the number of turns, using a core with lower permeability, or using a core with an air gap.

It is possible for the eddy current and hysteresis losses to be so large that the inductor behaves more like a resistor. Furthermore, there is always some capacitance between the turns of the inductor, and under some circumstances an inductor may act like a capacitor. (We’ll discuss this a bit later.)

Table 2.9 shows a comparison between the various core inductors.

TABLE 2.9 Comparison of Inductor Cores

 Air core Relative permeability is equal to 1. The inductance is independent of current-flow because air does not saturate. Limited to low inductance values, but can be operated at very high frequencies (e.g., RF applications ∼1 GHz). Iron core Permeability is a factor of 1000 greater than for air core, but inductance is highly dependent on the current flowing through the coil because the core saturates. Chiefly used in power-supply equipment. Highly prone to eddy currents (power loss) due to the core’s high conductivity. Losses caused by eddy currents can be reduced by laminating the core (cutting it into thin strips separated by insulation, such as varnish or shellac. Experience significant power losses due to hysteresis. Eddy currents and hysteresis losses in iron increase rapidly as the frequency of ac increases. Limited to power-line and audio frequencies up to about 15,000 Hz. Laminated iron cores are useless at radio frequencies. Powdered iron Ground-up iron (powder) mixed with a binder or insulating material reduces eddy currents significantly, since the iron particles are insulated from one another. Permeability is low compared to iron core due to insulator concentration. Slug-tuned inductors, where the slug is powered iron can be made adjustable and are useful in RF work, even up to the VHF range. Manufacturers offer a wide variety of core materials, or mixes, to provide units that will perform over a desired frequency range with a reasonable permeability. Toroidal Cores are considered self-shielding. Manufacturers provide an inductance index AL for their toroidal cores. For powdered-iron toroids, AL provides the inductance in µH per 100 turns of wire on the core, arranged in a single layer. Formulas found in Fig. 2.133, and example problems 5 and 6 illustrate how to calculate inductance of a powdered-iron toroidal inductor. Ferrite Composed of nickel-zinc ferrites for lower permeability ranges and of manganese-zinc ferrites for higher permeabilities, these cores span the permeability range from 20 to above 10,000. They are used for RF chokes and wideband transformers. Ferrites are often used, since they are nonconductors and are immune to eddy currents. Like powdered-iron toroids, ferrite toroids also have an AL value provided by the manufacturers. However, unlike powdered-iron toroids, ferrite AL values are given in mH per 1000 turns. Formulas found in Fig. 2.133 and example problems 7 and 8 illustrate how to calculate inductance of a ferrite toroidal inductor.

FIGURE 2.133

Example 5: What is the inductance of a 100-turn coil on a powdered iron toroidal core with an index of inductance of 20?

Answer: To solve this you’d have to consult the manufacturer’s data sheets, but, here, use a T-12-2 from the table in Fig. 2.133. Using the equation for powdered iron toroids in Fig. 2.133, and inserting AL = 20 for the T-12-2, and N = 100:

Example 6: Calculate the number of turns needed for a 19.0-µH coil if the index of inductance for the powdered iron toroid is 36.

Example 7: What is the inductance of a 50-turn coil on a ferrite toroid with an index of inductance of 68?

Answer: To solve this you’d have to consult the manufacturer’s data sheets, but here we’ll use a FT-50 with 61-Mix, as shown in the table in Fig. 2.133. Using the equation for ferrite toroids in Fig. 2.133, and inserting AL = 68 for the FT-50-61, and N = 50:

Example 8: How many turns are needed for a 2.2-mH coil if the index of inductance for the ferrite toroid is 188?

2.24.9 Understanding the Inductor Equations

The preceding inductor equation (which we derived earlier) has some curious properties that may leave you scratching your head. For starters, let’s consider the dIL/dt term. It represents the rate of change in current through the inductor with time. If there is no change in current flow through the inductor, there is no measured voltage across the inductor. For example, if we assume that a constant dc current has been flowing through an inductor for some time, then dIL/dt is zero, making VL zero as well. Thus, under dc conditions the inductor acts like a short—a simple wire.

However, if the current IL is changing with time (increasing or decreasing), dIL/dt is no longer zero, and an induced voltage VL is present across the inductor. For example, consider the current waveform shown in Fig. 2.134. From time interval 0 to 1s, the rate of current change dIL/dt is 1 A/s—the slope of the line. If the inductance L is 0.1 H, the induced voltage is simply (1 A/s)(0.1 H) = 0.1 V during this time interval—see the lower waveform. During interval 1s to 2s, the current is constant, making the dIL/dt zero and, hence, the induced voltage zero. During interval 2s to 3s, dIL/dt is −1 A/s, making the induced voltage equal to (−1 A/s)(0.1 H) = −0.1 V. The induced voltage waveform paints the rest of the picture.

FIGURE 2.134

Example 9: If the current through a 1-mH inductor is given by the function 2t, what is the voltage across it?

Example 10: The current through a 4-mH inductor is given by IL = 3 − 2e−10t A. What is the voltage across the inductor?

Example 11: Suppose the current flow through a 1-H inductor decreases from 0.60 A to 0.20 A during a 1-s period. Find the average voltage across the inductor during this period. See how this compared to the average induced voltage if the period is 100 ms, 10 ms, and 1 ms.

Answer: Here we ignore changes between the 1-s intervals, and take averages:

From Example 11, notice how the induced voltage grows considerably larger as the change in current flow is more abrupt. When the change in current flow is instantaneous, the inductor equation predicts an infinite induced voltage. How can this be?

The answer to this dilemma is explained by the following example. Say you have an ideal inductor attached to a 10-V battery via a switch, as shown in Fig. 2.135. The moment the switch is closed, according to the inductor equation, dI/dt should be infinite (assuming ideal battery, wires, and coil wire). This means that the induced voltage should rise in proportion with the applied voltage, forever, and no current should flow. In other words, the reverse voltage should jump to infinity. (See Fig. 135b.) Likewise, if the switch is opened, as shown in Fig. 135c, an infinite forward voltage is predicted. The answer to this problem is rather subtle, but fundamentally important. As it turns out, these infinities we predict are never observed in the real world. There is always internal resistance within a real inductor as well as internal capacitance (as well as internal resistance and capacitance in the rest of the circuit). A realistic model of an inductor is shown in Fig. 135d, incorporating internal resistance and capacitance. It is these “imperfections” that explain why the predictions are not observed.

FIGURE 2.135 Real inductors never generate infinite induced voltages.

As we can see from the last example, ignoring internal resistance can cause conceptual headaches. You may be wondering why someone didn’t simply create an inductor equation that automatically included an internal resistance term. The fact of the matter is that we should, especially when considering a simple circuit like that shown in Fig. 2.135. However, it is important to define inductance as a unique quantity and associate its effects with changing magnetic field energy alone, not with resistive heating within the coil or magnetic core losses, or with distributed capacitance between the coil loops. As it turns out, when you start analyzing more complex circuits (say RL circuits and RLC circuits), the discrete resistance present within the circuit will keep the inductor equation from freaking out. In more precise circuits, knowing the internal resistance of the inductor is critical. A standard practice is to represent an inductor as an ideal inductor in series with a resistor RDC, where RDC is called the dc resistance of the inductor. More accurate models throw in parallel capacitance (interloop capacitance) and parallel resistance (representing magnetic core losses), which become important in high-frequency applications.

Before we move on, it should be noted that even though we must assume there is internal resistance within an inductor, it is still possible for induced voltages to reach surprisingly high values during transient conditions. For example, turning off inductive circuits can lead to some dangerously high voltages that can cause arcing and other problems requiring special handling.

Example 12: Suppose you apply a linearly increasing voltage across an ideal 1-H inductor. The initial current through the inductor is 0.5 A, and the voltage ramps from 5 to 10 V over a 10 ms interval. Calculate the current through the inductor as a function of time.

Integrating this gives:

or:

Carrying out the integration gives:

At t = 0.01 s (the end of the ramp), the current is:

2.24.10 Energizing RL Circuit

When a resistor is placed in series with an inductor, the resistance controls the rate at which energy is pumped into the magnetic field of the inductor (or pumped back into the circuit when the field collapses). When we consider an RL circuit with a dc supply and a switch, as shown in Fig. 2.136, the energizing response that begins the moment the switch is closed (t = 0) is illustrated by the voltage and current response curves in Fig. 2.136, as well as the equations to the right.

You can derive the expressions for the energizing response of an RL circuit by applying Kirchhoff’s law, summing the voltages around the closed loop:

FIGURE 2.136

Current and voltage equations for RL energizing circuit:

where I is the current in amps, VS is the source voltage in volts, R is the resistance in ohms, L is the inductance in henrys, t is the time in seconds after the source voltage is applied, e = 2.718, VR is the resistor voltage in volts, and VL is the inductor voltage in volts. The graph shown to the left is for a circuit with R = 100 Ω and L = 20 mH.

Rewriting in the standard form gives:

After solving this linear, first-order nonhomogeneous differential equation, using the initial condition that the current before the switch was closed was zero I (0) = 0, the solution for the current becomes:

We plug this into Ohm’s law to find the voltage across the resistor:

VR = IR = VS (1 − et/(L/R))

and plug it into the expression for the inductor voltage:

To understand what’s going on within the energizing RL circuit, we first pretend the resistor value is zero. With no resistance, when the switch is closed, the current would increase forever (according to Ohm’s law and assuming an ideal voltage source), always growing just fast enough to keep the self-induced voltage equal to the applied voltage.

But when there is resistance in the circuit, the current is limited—Ohm’s law defines the value the current can reach. The reverse voltage generated in L must only equal the difference between the applied voltage and the drop across R. This difference becomes smaller as the current approaches the final Ohm’s law value. Theoretically, the reverse voltage never quite disappears, and so the current never quite reaches the no-inductor value. In practical terms, the differences become immeasurable after a short duration.

The time in seconds required for the current to build up to 63.2 percent of the maximum value is called the time constant, and is equal to L/R. After each time interval equal to this constant, the circuit conducts an additional 63.2 percent of the remaining current. This behavior is graphed in Fig. 2.137. As is the case with capacitors, after five time constants the current is considered to have reached its maximum value.

FIGURE 2.137

Example 13: If an RL circuit has an inductor of 10 mH and a series resistor of 10 Ω, how long will it take for the current in the circuit to reach full value after power is applied?

Answer: Since reaching maximum current takes approximately five time constants, t = 5τ = 5 (L/R) = 5 (10 × 10−3 H)/10 Ω = 5.0 × 10−3 s or 5.0 ms.

Note that if the inductance is increased to 1.0 H, the required time increases to 0.5 s. Since the circuit resistance didn’t change, the final current is the same for both cases in this example. Increasing inductance increases the time required to reach full current. Figure 2.137 shows current response curves for various inductors with the same resistance.

2.24.11 Deenergizing RL Circuit

Unlike a capacitor that can store energy in the form of an electric field when an applied voltage is cut (say, by means of a switch), an inductor does not remain “charged,” or energized, since its magnetic field collapses as soon as current ceases. The energy stored in the magnetic field returns to the circuit. Now, when current flow is cut (say, by means of a switch), predicting the current flow and voltage drops within an RL circuit is a bit tricky. We can say that the instant the switch is opened, a rapid collapse of the magnetic field induces a voltage that is usually many times larger than the applied voltage, since the induced voltage is proportional to the rate at which the field changes. The common result of opening the switch in such a circuit is that a spark or arc forms at the switch contacts during the instant the switch opens—see Fig. 2.138a. When the inductance is large and the current in the circuit is high, large amounts of energy are released in a very short time. It is not unusual for the switch contacts to burn or melt under such circumstances. The spark or arc at the opened switch can be reduced or suppressed by connecting a suitable capacitor and resistor in series across the contacts. Such an RC combination is called a snubber network. Transistor switches connected to large inductive loads, such as relays and solenoids, also require protection. In most cases, a small power diode connected in reverse across the relay coil will prevent field-collapse currents from harming the transistor—more on this in a moment.

If the excitation is removed without breaking the circuit, as theoretically diagrammed in Fig. 2.138b, the current will decay according to the following waveforms and equations.

You can derive the expressions for a deenergizing RL circuit by applying Kirchhoff’s law, summing the voltage around the closed loop:

Vs is 0 because the battery is no longer in the circuit. Rewriting in the standard form gives:

After solving this linear, first-order nonhomogeneous differential equation, using the initial condition that the current before the switch was closed was simply:

the solution for the current becomes:

We plug this into Ohm’s law to find the voltage across the resistor:

VR = IR = VSet/(L/R)

and plug it into the expression for the inductor voltage:

As with the energizing RL circuit, the deenergizing RL circuit current response can be modeled in terms of time constants. After five time constants, the inductor is considered fully deenergized. Increasing the inductance increases this time, as shown in Fig. 2.139.

FIGURE 2.138

In A, a break in current flow caused by opening a switch will cause a large inductive load’s magnetic field to collapse, thus generating a large forward voltage. This voltage can be so big that the “electron pressure” between the switch contacts becomes so large that electrons escape the metal surface of one switch contact and jump toward the other contact. As the liberated electrons make the jump, they collide with airborne molecules, causing ionization reactions, which lead to a spark discharge across the switch contacts. The current and voltage response curves under such conditions are rather complex.

In B, if we remove the applied voltage but prevent a break in the circuit by moving the discharge to the ground (switch thrown to position B), we get predictable current and voltage expressions:

where I is the current in amps, VS is the source voltage in volts, R is the resistance in ohms, L is the in-ductance in henrys, t is the time in seconds after the source voltage is applied, e = 2.718, VR is the resistor voltage in volts, and VL is the inductor voltage in volts.

The graph shown in the figure is for a circuit with R = 100 Ω and L = 20 mH.

FIGURE 2.139

Inductance, be it intended or not, can have a major influence on signals. For example, the output signals across the inductor and resistor in the RL circuit to the left become increasingly distorted as the inductance increases. With a constant 1.0-kHz, 0–5-V squarewave source, and fixed 10-Ω resistance, we increase the inductance and note the changes in the waveforms. First we note that the period of the squarewave is:

Now let’s see how the waveforms change as we choose increasing inductance values.

Graph A: Inductance: L = 0.1 mH

Time constant: τ = 0.0001 H/10 Ω = 0.01 ms

Here the RL time constant is 1 percent of the period, so the induced voltage spikes are narrow during squarewave high-to-low and low-to-high transitions. Considering that an inductor is fully energized or deenergized after five time constants or, in this case, 0.05 ms, the inductor easily completely energizes and deenergizes during a half period of 0.5 ms. The voltage across the resistor is slightly rounded at the edges as a result.

Graph B: Inductance: L = 1 mH

Time constant: τ = 0.001 H/10 Ω = 0.1 ms

The RL time constant is 10 percent of the period, so induced voltage and the effects of exponential rise and fall during source voltage transitions are clearly visible. To fully energize or deenergize requires five time constants, or 0.5 ms, which is exactly equal to the half period of the signal. Hence, the magnetic field about the inductor is capable of absorbing and giving up all its magnetic field energy during each consecutive half cycle.

Graph C: Inductance: L = 10 mH

Time constant: τ = 0.01 H/10 Ω = 1 ms

The RL time constant is equal to the period of the squarewave. However, since it requires five time constants, or 5 ms, to be fully energized or deenergized, the resulting voltages appear linear—you get to see only a small portion of the exponential rise or fall. The magnetic field about the inductor isn’t able to absorb or give up all its energy during a half cycle.

Graph D: Inductance: L = 1 H

Time constant: τ = 1 H/10 Ω = 0.1 s

The RL time constant is 100 times as large as the period of the squarewave. Again, to become fully energized or deenergized requires five time constants, or 500 times the period. This means that there is practically no time for the inductor to fully energize or deenergize. Though there is exponential rise and decay, practically, you see only the first 1500 portion, which appears linear.

FIGURE 2.140

2.24.12 Voltage Spikes Due to Switching

Inductive voltage spikes are common in circuits where large inductive loads, such as relays, solenoids, and motors, are turned on and off by a mechanical or transistor-like switch. Spikes as high as a couple hundred volts are possible, even when the supply voltages are relatively small. Depending on the circuit design, these voltage spikes can cause arcing, leading to switch contact degradation, or damage to transistor or other integrated switching devices. Figure 2.141 shows a diode (a device that acts as a one-way gate to current flow) placed across the coil of a relay to provide a “pressure release” path for the inductive spike, should the supply voltage circuit be broken.

FIGURE 2.141

2.24.13 Straight-Wire Inductance

Every conductor passing current has a magnetic field associated with it and, therefore, inductance—the conductor need not be formed into a coil. For example, a straight wire has an inductance associated with it. This is attributed to the average alignment of magnetic fields of individual free electrons as they obtain a drift component under an applied EMF. The inductance of a straight nonmagnetic wire or rod in free space is given by:

 (2.56)

where L is the inductance in µH, a is the wire radius in inches, b is the length in inches, and ln is the natural logarithm.

Example 14: Find the inductance of a #18 wire (diameter = 0.0403 in) that is 4 in long.

Answer: Here a = 0.0201 and b = 4, so:

Skin effects change Eq. 2.56 slightly at VHF (30–300 MHz) and above. As the frequency approaches infinity, the constant 0.75 in the preceding equation approaches 1.

Straight wire inductance is quite small, and is typically referred to as parasitic inductance. We can apply the concept of reactance to inductors, as we did with capacitive reactance earlier in this chapter. Parasitic inductive reactance at low frequencies (AF to LF) is practically zero. For instance, in our example, at 10 MHz the 0.106-µH inductance provides only 6.6 Ω of reactance. However, if we consider a frequency of 300 MHz, the inductive reactance goes to 200 Ω—a potential problem. For this reason, when designing circuits for VHF and above it is important to keep component leads as short as possible (capacitor leads, resistor leads, etc.). Parasitic inductance in a component is modeled by adding an inductor of appropriate value in series with the component (since wire lengths are in series with the component).

FIGURE 2.142

2.24.14 Mutual Inductance and Magnetic Coupling

When two inductor coils are placed near each other with their axes aligned, the current applied through coil 1 creates a magnetic field that propagates through coil 2. See Fig. 2.143.

FIGURE 2.143

As a result, a voltage is induced in coil 2 whenever the field strength of coil 1 changes. The voltage induced in coil 2 is similar to the voltage of self-induction, but since it acts upon the external coil 2, it is called mutual induction—the two coils are said to be inductively coupled. The closer the coils are together, the greater the mutual inductance. If the coils are farther apart or are aligned off axis, the mutual inductance is relatively small—the coils are said to be loosely coupled. The ratio of actual mutual inductance to the maximum possible mutual inductance is called the coefficient of coupling, which is usually given as a percentage. The coefficient with air core coils may run as high as 0.6 to 0.7 if one coil is wound over the other, but much less if the two coils are separated. It is possible to achieve near 100 percent coupling only when the coils are wound on a closed magnetic core—a scenario used in transformer design. Mutual inductance also has an undesirable consequence within circuit design, where unwanted induced voltages are injected into circuits by neighboring components or by external magnetic field fluctuations generated by inductive loads or high-current alternating cables.

2.24.15 Unwanted Coupling: Spikes, Lightning, and Other Pulses

There are many phenomena, both natural and man-made, that generate sufficiently large magnetic fields capable of inducing voltages in wires leading into and out of electrical equipment. A mutual inductance exists between the external source and the affected circuit in question. Parallel-wire cables linking elements of electronic equipment consist of long wires in close proximity to each other. Signal pulses can couple both magnetically and capacitively from one wire to another. Since the magnetic field of a changing current decreases as the square of distance, separating the signal-carrying lines diminishes inductive coupling. Unless they are well shielded and filtered, however, the lines are still susceptible to the inductive coupling of pulses from other sources. This is often experienced when using a long ground lead for a scope probe—external magnetic interference can couple to the ground lead of the probe and appear within the displayed signal as unwanted noise. Such coupling between external sources and electrical equipment can be quite problematic—even more so when the source generates a “burst” in magnetic field strength. Sudden bursts have a tendency to induce high-level voltage spikes onto ac and dc power lines, which can migrate into the inner circuitry where sensitive components lay prone to damage. For example, lightning in the vicinity of the equipment can induce voltages on power lines and other conductive paths (even ground conductors) that lead to the equipment location. Lightning that may appear some distance away can still induce large spikes on power lines that ultimately lead to equipment. Heavy equipment with electrical motors can induce significant spikes into power lines within the equipment location. Even though the power lines are straight, the powerful magnetic fields of a spike source can induce damaging voltages on equipment left plugged in during electrical storms or during the operation of heavy equipment that inadequately filters its spikes.

2.24.16 Inductors in Series and Parallel

FIGURE 2.144

When two or more inductors are connected in series, the total inductance is equal to the sum of the individual inductances, provided the coils are sufficiently separated, so that coils are not in the magnetic field of one another:

 Ltot = L1 + L2 + L3+ … + LN (2.57) Inductors in series

You can derive the inductors-in-series formula by applying Kirchhoff’s voltage law. Taking the voltage drop across L1 to be L1dI/dt and the voltage drop across L2 to be L2dI/dt, and across L3 to be L3dI/dt, you get the following expression:

L1 + L2 + L3 is called the equivalent inductance for three inductors in series.

If inductors are connected in parallel, and if the coils are separated sufficiently, the total inductance is given by:

 (2.58) Inductors in parallel

When only two inductors are in parallel, the formula simplifies to Ltot = (L1 × L2)/(L1 + L2).

You derive this by applying Kirchhoff’s current law to the junction, which gives I = I1 + I2 + I3, making use of the fact that the voltage V is the same across L1L2, and L3. Thus I1 becomes 1/L1I2 becomes 1/L2 ∫ Vdt, and I3becomes 1/L3 ∫ Vdt. The final expression for I is:

1/L1 + 1/L2 + 1/L3 is called the reciprocal equivalent inductance of three inductors in parallel.

Example 15: What value of L2 is required to make the total equivalent inductance of the circuit shown in Fig. 2.145 equal to 70 mH?

FIGURE 2.145

2.24.17 Alternating Current and Inductors

When an alternating voltage is applied to an ideal inductance, the current that flows through the inductor is said to lag the applied voltage by 90°. Or if you like, the applied voltage leads the current by 90°. (This is exactly the opposite of what we saw with capacitors under ac.) The primary cause for the current lag in an inductor is due to the reverse voltage generated in the inductance. The amplitude of the reverse voltage is proportional to the rate at which the current changes. This can be demonstrated by the graph in Fig. 2.146. If we start at time segment 0A, when the applied voltage is at its positive maximum, the reverse or induced voltage is also at a maximum, allowing the least current to flow. The rate at which the current is changing is the highest, a 38 percent change in the time period 0A. In the segment AB, the current changes by only 33 percent, yielding a reduced level of induced voltage, which is in step with the decrease in the applied voltage. The process continues in time segments BC and CD, the latter producing only an 8 percent rise in current as the applied and induced voltage approach zero.

FIGURE 2.146

In interval DE, the applied voltage changes direction. The induced voltage also changes direction, returning current to the circuit as the magnetic field collapses. The direction of this current is now opposite to the applied voltage, which sustains the current in the positive direction. As the applied voltage continues to increase negatively, the current—although positive—decreases in value, reaching zero as the applied voltage reaches its negative maximum. The negative half cycle continues just as did the positive half cycle. Thus, we say that within a pure inductive ac circuit, the current lags the voltage by 90°.

2.24.18 Inductive Reactance

The amplitude of alternating current in an inductor is inversely proportional to the applied frequency. Since the reverse voltage is directly proportional to inductance for a given rate of current change, the current is inversely proportional to inductance for a given applied voltage and frequency. The combined effect of inductance and frequency is called inductive reactance. Like capacitive reactance, inductive reactance is expressed in ohms. The inductive reactance is given by the following formula:

 XL = 2πfL (2.59) Inductive reactance

where XL is inductive reactance, π = 3.1416, f is frequency in Hz, and L is inductance in henrys. Inductive reactance, in angular form (where ω = 2πf), is equal to:

XL = ωL.

You derive the expression for inductive reactance by connecting an inductor to a sinusoidal voltage source. To make the calculations more straightforward, we’ll use a cosine function instead of a sine function—there is really no difference here. For example, if the source voltage is given by V0 cos (ωt), the current through the inductor becomes

the maximum current or peak current through an inductor occurs when sin (ωt) = 1, at which point it is equal to:

The ratio of peak voltage to peak current resembles a resistance and has units of ohms. However, because the physical phenomenon doing the “resisting” (e.g., reverse induced voltage working against forward voltage) is different from a resistor (heating), the effect is given a new name, inductive reactance:

As ω goes to infinity, XL goes to infinity, and the inductor acts like an open circuit (inductors do not like to pass high-frequency signals). However, as ω goes to 0, the XL goes to zero (inductors have an easier time passing low-frequency signals and ideally present no “resistance” to dc signals).

Figure 2.147 shows a graph of inductive reactance versus frequency for 1-µH, 10-µH, and 100-µH inductors. Notice the response is linear—increasing the frequency increases the reactance in proportion. However, in real inductors, the reactive response is a bit more complicated, since real inductors have parasitic resistance and capacitance built in. Figure 2.147 shows a real-life impedance versus frequency graph. Notice that when the frequency approaches what is called the resonant frequency, the impedance is no longer linear in appearance, but peaks and falls. (This will make more sense when we cover resonant circuits a bit later.)

FIGURE 2.147

Example 16: What is the reactance of an ideal 100-µH coil with applied frequencies of 120 Hz and 15 MHz?

·    120 Hz: XL = 2πfL = 2π(120 Hz)(100 × 10−6 H) = 0.075 Ω

·    15 MHz: XL = 2πfL = 2π(15 × 106 Hz)(100 × 10−6 H) = 9425 Ω

Example 17: What inductance provides 100 Ω of reactance with an applied frequency of 100 MHz?

Example 18: At what frequency will the reactance of a 1-µH inductor reach 2000 Ω?

By the way, inductive reactance has an inverse called inductive susceptance, given by the following expression:

 (2.60)

Inductive susceptance uses units of siemens, abbreviated S (S = 1/Ω). It simply tells you “how well” an inductor passes current, as opposed to “how bad” it does—as reactance implies.

2.24.19 Nonideal Inductor Model

Though the ideal inductor model and associated ideal equations are fundamentally important in circuit analysis, using them blindly without consideration of real inductor imperfections, such as internal resistance and capacitance, will yield inaccurate results. When designing critical devices, such as high-frequency filters used in radiofrequency receivers, you must use a more accurate real-life inductor model.

In practice, a real inductor can be modeled by four passive ideal elements: a series inductor (L), a series resistor RDC, a parallel capacitor CP, and a parallel resistor RP. RDC represents the dc resistance, or the measured resistance drop when a dc current passes through the inductor. Manufacturers provide the dc resistance of their inductors on their specification sheets (e.g., 1900 series 100-µH inductor with an RDC of 0.0065 Ω). RP represents magnetic core losses and is derived from the self-resonant frequency f0—the point at which the reactance of the inductor is zero (i.e., the impedance is purely resistive). It can be calculated from the quality factor Q, as we’ll see in a moment. The parallel resistor limits the simulated self-resonance from rising to infinity. CP represents the distributed capacitance that exists between coils and leads within the inductor, as shown in Fig. 2.148. When a voltage changes due to ac current passing through a coil, the effect is that of many small capacitors acting in parallel with the inductance of the coil. The graph in Fig. 2.148 shows how this distributed capacitance resonates with the inductance. Below resonance, the reactance is inductive, but it increases as the frequency increases. Above resonance, the reactance is capacitive and decreases with frequency.

FIGURE 2.148 Inductors exhibit distributed capacitance, as explained in the text. The graph shows how this distributed capacitance resonates with the inductance. Below resonance, the reactance is inductive, but it decreases as frequency increases. Above resonance, the reactance is capacitive and increases with frequency.

Inductors are subject to many types of electrical energy losses, however—wire resistance, core losses, and skin effect. All electrical conductors have some resistance through which electrical energy is lost as heat. Moreover, inductor wire must be sized to handle the anticipated current through the coil. Wire conductors suffer additional ac losses because alternating current tends to flow on the conductors’ surface. As the frequency increases, the current is confined to a thinner layer on the conductor surface. This property is called the skin effect. If the inductor’s core is a conductive material, such as iron, ferrite, or brass, the core will introduce additional losses of energy.

2.24.20 Quality Factor

Components that store energy, such as capacitors and inductors, can be described in terms of a quality factor Q. The Q of such a component is a ratio of its ability to store energy to the total of all energy losses within the component. In essence, the ratio reduces to: Q = X/R, where Q is the quality factor (no units), X is the reactance (inductive or capacitive), and R is the sum of all resistance associated with the real energy losses within the component, given in ohms.

For a capacitor, Q is ordinarily high, with quality ceramic capacitors obtaining values of 1200 or more. Small ceramic trimmer capacitors may have Q values that are so small that they shouldn’t be ignored in certain applications.

The quality factor for an inductor is given by Q = 2πfL/RDC. The Q value for an inductor rarely if ever approaches capacitor Q in a circuit where both components work together. Although many circuits call for the highest Q inductor obtainable, other circuits call for a specific Q, which may, in fact, be very low.

Inductive Divider

Inductive dividers can be used with ac input signals. A dc input voltage would split according to the relative resistances of the two inductors by using the resistive voltage divider. The formula for determining the ac output voltage of an inductive divider (provided the inductors are separated—that is, not wound on the same core—and have no mutual inductance) is shown in Fig. 2.149.

FIGURE 2.149

Note that the output voltage is independent of the input frequency. However, if the reactance of the inductors is not high at the frequency of operation, (i.e., the inductance is not large enough), there will be a very large current drawn by the shunt element (L2).

2.24.21 Inductor Applications

Inductors’ basic function in electronics is to store electrical energy in a magnetic field. Inductors are used extensively in analog circuits and signal processing, including radio reception and broadcasting. Inductors, in conjunction with capacitors and other components, can form electric filters used to filter out specific signal frequencies. Two (or more) coupled inductors form a transformer that is used to step up or step down an ac voltage. An inductor can be used as the energy storage device in a switching regulator power supply—the inductor is charged for a specific fraction of the regulator’s switching frequency and discharged for the remainder of the cycle. This charge/discharge ratio determines the output-to-input voltage ratio. Inductors are also employed in electrical transmission systems, where they are used to intentionally depress system voltages or limit fault current. In this field, they are more commonly referred to as reactors.

2.25 Modeling Complex Circuits

As a note, let us tell you that this section is designed to scare you. You may have a hard time understanding some parts of this section if you don’t have a decent math background. However, this section is worth reading for theoretical footing and, more important, because it stresses the need to come up with alternative tricks to avoid the nasty math.

Theoretically, given enough parameters, any complex electric circuit can be modeled in terms of equations. In other words, Kirchhoff’s laws always hold, whether our circuit consists of linear or nonlinear elements. Linear devices have responses that are proportional to the applied signal. For example, doubling the voltage across a resistor doubles the current through it. With a capacitor, doubling the frequency of the applied voltage across it doubles the current through it. With an inductor, doubling the frequency of voltage across it halves the current through it. We found that we could apply the following equations to model resistors, capacitors, and inductors:

In terms of voltage and current sources, we have, up until now, mainly discussed dc and sinusoidal sources, which we can express mathematically as:

VS = Constant, IS = Constant, VS = V0 sin (ωt), IS = I0 sin (ωt)

If a circuit contains only resistors, capacitors, inductors, and one of these sources, we simply apply Kirchhoff’s laws and come up with an equation or set of equations that accurately describes how the voltages and currents within our circuit will behave with time. Linear dc circuits are described by linear algebraic equations. On the other hand, linear time-dependent circuits are described by linear differential equations. The time dependence may be a result of a sinusoidal source, or it can simply be a dc source that is turned on or off abruptly—this is referred to as a transient.

As an example, if we have a series RLC circuit (see Fig. 2.150) with a dc source VS, we can write Kirchhoff’s voltage loop equations as:

FIGURE 2.150

This equation isn’t of any practical use at this stage. Mathematically, we need to simplify it in order to get rid of the integral. You’d start out by first differentiating everything with respect to time, then rearrange things to look like this:

This is an example of a linear second-order homogeneous differential equation. To solve it requires some mathematical tricks and defining initial conditions—at which point the switch is turned on or off.

Now, let’s take the same RLC circuit, but remove the switch and dc supply and insert a sinusoidal supply (see Fig. 2.151). The supply is expressed mathematically as V0 cos (ωt).

FIGURE 2.151

Applying Kirchhoff’s voltage equation:

or,

Again, we must simplify it to get rid of the integral:

This expression is a linear second-order nonhomogeneous differential equation. To find the solution to this equation, you could apply, say, the technique of variation of parameters or the method of undetermined coefficients. After the solution for the current is found, finding the voltages across the resistor, capacitor, and inductor is a simple matter of plugging the current into the characteristic voltage/current equation for that particular component. However, coming up with the solution for the current in this case is not easy because it requires advanced math.

As you can see, things don’t look promising, mathematically speaking. Things get even worse when we start incorporating sources that are nonsinusoidal, such as a squarewave source or a triangle wave source. For example, how do we mathematically express a squarewave voltage source? As it turns out, the simplest way is to use the following Fourier series:

where V0 is the peak voltage of the squarewave. If we take our RLC circuit and attach this squarewave voltage source, then apply Kirchhoff’s voltage law around the loop, we get:

As you might guess, the solution to this equation isn’t trivial.

There are other sources we haven’t considered yet, such as nonsinusoidal nonrepetitive sources (impulse, etc.). And things, of course, get worse when you consider circuits with more than three linear elements. Let’s not forget nonlinear devices, like diodes and transistors, which we haven’t even discussed yet.

When circuits get complex and the voltage and current sources start looking weird, setting up Kirchhoff’s equations and solving them can require fairly sophisticated mathematics. There are a number of tricks used in electronic analysis to prevent the math from getting out of hand, but there are situations where avoiding the nasty math is impossible. The comments in Fig. 2.152 should give you a feeling for the difficulties ahead.

FIGURE 2.152

In the next section, we’ll discuss complex numbers. Complex numbers, along with a concept known as complex impedances, are tricks that we’ll use to avoid setting up complex differential equations, at least under special circumstances.

2.26 Complex Numbers

Before we touch upon the techniques used to analyze sinusoidally driven circuits, a quick review of complex numbers is helpful. As you will see in a moment, a sinusoidal circuit shares a unique trait with a complex number. By applying some tricks, you will be able to model and solve sinusoidal circuit problems using complex numbers and the arithmetic that goes with it, and—this is the important part—you will be able to avoid differential equations in the process.

A complex number consists of two parts: a real part and an imaginary part. (See Fig. 2.153.)

FIGURE 2.153

Both a and b are real numbers, whereas  is an imaginary unit, thereby making the term ib an imaginary number or the imaginary part of a complex number. In practice, to avoid confusing i (imaginary unit) with the symbol i (current), the imaginary unit i is replaced with a j.

A complex number can be expressed graphically on a complex plane (argand or gaussian plane), with the horizontal axis representing the real axis and the vertical axis representing the imaginary axis. (See Fig. 2.154.)

FIGURE 2.154

In terms of the drawing, a complex number can be interpreted as the vector from 0 to P having a magnitude of length of:

 (2.61)

that makes an angle relative to the positive real axis of:

 (2.62)

Now let’s go a bit further—for the complex number to be useful in circuit analysis, it must be altered slightly. If you replace a with r cos θ and replace b with r sin θ, the complex number takes on what is called the polar trigonometric form of a complex number. (See Fig. 2.155.)

FIGURE 2.155

Okay, now you are getting there; just one more thing to cover. Long ago, a man by the name of Euler noticed that the cos θ + j sin θ part of the trigonometric form of the complex number was related to ejθ by the following expression:

 ejθ = cos θ + j sin θ (2.63)

You can prove this by taking the individual power series for ejθ, cos θ, j sin θ. When the power series for cos θ and j sin θ are added, the result equals the power series for ejθ. This means that the complex number can be expressed as follows:

 z = rejθ (2.64)

This represents the polar exponential form of a complex number. A shorthand version of this form can be written as:

 z = r ∠ θ (2.65)

Though this form goes by the name of polar coordinate form and has its basis in adding vectors and angles, and so on, it helps to think of it as a shorthand version of the polar exponential form, since it is really no different from the exponential form (same arithmetic rules apply), but it turns out to be a bit more intuitive, as well as easier to deal with in calculations.

Now we have basically four ways to express a complex number:

z = a + jb, z = r cos θ + jr sin θ, z = rejθz = r ∠ θ

Each form is useful in its own right. Sometimes it is easier to use z = a + jb, and sometimes it is easier to use z = rejθ (or z = r ∠ θ)—it all depends on the situation, as we’ll see in a moment.

The model shown in Fig. 2.156 is designed to help you get a feeling for how the various forms of a complex number are related. For what follows, you will need to know the arithmetic rules for complex numbers, too, which are summarized in Table 2.10.

FIGURE 2.156

TABLE 2.10 Arithmetic Rules for Complex Numbers

 FORM OF COMPLEX NUMBER ADDITION/SUBTRACTION MULTIPLICATION DIVISION Rectangular form z1 = a + jb z2 = c + jd Z1 ± Z2 = (a ± c) + j(b ± d) Example: Z1 = 3 + j4, Z2 = 5 − j7 Z1 + Z2 = (3 + 5) + j(4 − 7) = 8 − j3 Z1 × Z2 = (ac − bd) + j(ad + bc) Example: Z1 = 5 + j2, Z2 = −4 + j3 Z1 × Z2 = [5(−4) − 2(3)] + j[5(3) + 2(−4)]             = −26 + j7 Example: Z1 = 1 + j, Z2 = 3 + j2: Trigonometric form z1 = r1 cos θ1 + jr1 sin θ1 z2 = r2 cos θ2 + jr2 sin θ2 Can be done but involves using trigonometric identities; it is easier to convert this form into rectangular form and then add or subtract. z1 × z2 = r1r2 [cos (θ1 + θ2) + j sin (θ1 + θ2)] Exponential form z1 = r1ejθ1 z2 = r2ejθ2 Does not make much sense to add or subtract in this form because the result will not be in simplified form, except if r1 and r2 are equal; it is better to first convert to rectangular form and then add or subtract. Z1 × Z2 = r1r2ej(θ1 + θ2) Example:         Z1 = 5ej(180°), Z2 = 2ej(90°) Z1 × Z2 = 5 (2) ej(180° + 90°)             = 10 ej(270°) Example: Z1 = 8ej(180°), Z2 = 2ej(60°) Polar coordinate form (shorthand) z1 = r1 ∠ θ1 z2 = r2 ∠ θ2 Does not make much sense to add or subtract in this form because the result will not be in simplified form, except if r1 and r2 are equal; it is better to first convert to rectangular form and then add or subtract. Z1 × Z2 = r1r2 ∠ (θ1 + θ2)* Example:      Z1 = 5 ∠ 180°, Z2 = 2 ∠ 90° Z1 × Z2           = 5(2) ∠ (180° + 90°)           = 10 ∠ 270° Example: Z1 = 8 ∠ 180°, Z2 = 2 ∠ 60° = 4 ∠ 120° *Usually the most efficient form to use when doing your calculations—other forms are too difficult or provide unintuitive results.

Here are some useful relationships to know when dealing with complex numbers:

X (in degrees) =  X (in radians), X (in radians) = X (in degrees)

ej(0°) = 1, ej(90°) = j, ej(180°) = −1, ej(270°) = −j, ej(360°) = 1

1 ∠ 0° = 1, 1 ∠ 90° = j, 1 ∠ 180° = −1, 1 ∠ 270° = −j, 1 ∠ 360° = 1

Z2 = (rejθ)2 = r2ej or Z2 = (r ∠ θ)2 = r2 ∠ 2θ

The following example shows a calculation that makes use of both rectangular and polar forms of a complex number to simplify a calculation involving addition, multiplication, and division of complex numbers:

The result can easily be converted into trigonometric or rectangular forms if the need exists:

0.17 ∠ −84.1° = 0.17 cos (−84.1°) + j0.17 sin (−84.1°) = 0.017 − j0.17

Notice that when multiplication or division is involved, it is easier to first convert the complex terms into exponential form (shorthand version). In essence, for addition and subtraction you’ll probably use rectangular form (though trigonometric form isn’t difficult in this case), and use exponential (shorthand) form for multiplication and division. If you understand the preceding calculation, you should find the ac theory to come easy.

Note that sometimes the following notation is used to express a complex number:

 (2.66)

where |Z| is the magnitude or modulus of a complex number (or r), Re Z is the real part of the complex number, and Im Z is the imaginary part of the number, while arg(Z) (or phase θ) represents the argument of Z or the phase angle (θ). For example, if Z = 3 + j4, then:

Re Z = 3       Im Z = 4       |Z| =        arg(Z) = = 53.1°

2.27 Circuit with Sinusoidal Sources

Suppose that you are given the two circuits shown in Fig. 2.157 containing linear elements (resistors, capacitors, inductors) driven by sinusoidal voltage sources.

FIGURE 2.157

To analyze the simpler of the two circuits, you could apply Kirchhoff’s voltage law to get the following:

which reduces to:

This expression, as we discovered earlier, is a linear second-order nonhomogeneous differential equation. To find the solution to this equation, you could apply, say, the technique of variation of parameters or the method of undetermined coefficients. After the solution for the current is found, finding the voltages across the resistor, capacitor, and inductor is a simple matter of plugging the current into the characteristic voltage/current equation for that particular component. However, coming up with the solution for the current in this case is not easy because it requires advanced math (and the work is tedious).

Now, as if things were not bad enough, let’s consider the more complex circuit in Fig. 2.157. To analyze this mess, you could again apply Kirchhoff’s voltage and current laws to a number of loops and junctions within the circuit and then come up with a system of differential equations. The math becomes even more advanced, and finding the solution becomes ridiculously difficult.

Before we scare you too much with these differential equations, let us tell you about an alternative approach, one that does away with differential equations completely. This alternative approach makes use of what are called complex impedances—something that will use the complex numbers we talked about in the last section.

2.27.1 Analyzing Sinusoidal Circuits with Complex Impedances

To make solving sinusoidal circuits easier, it’s possible to use a technique that enables you to treat capacitors and inductors like special kinds of resistors. After that, you can analyze any circuit containing resistors, capacitors, and inductors as a “resistor” circuit. By doing so, you can apply all the dc circuit laws and theorems that were presented earlier. The theory behind how the technique works is a bit technical, even though the act of applying it is not hard at all. For this reason, if you do not have the time to learn the theory, we suggest simply breezing through this section and pulling out the important results. Here’s a look at the theory behind complex impedances.

In a complex, linear, sinusoidally driven circuit, all currents and voltages within the circuit will be sinusoidal in nature after all transients have died out. These currents and voltages will be changing with the same frequency as the source voltage (the physics makes this so), and their magnitudes will be proportional to the magnitude of the source voltage at any particular moment in time. The phase of the current and voltage patterns throughout the circuit, however, most likely will be shifted relative to the source voltage pattern. This behavior is a result of the capacitive and inductive effects brought on by the capacitors and inductors.

As you can see, there is a pattern within the circuit. By using the fact that the voltages and currents will be sinusoidal everywhere, and considering that the frequencies of these voltages and currents will all be the same, you can come up with a mathematical trick to analyze the circuit—one that avoids differential equations. The trick involves using what is called the superposition theorem. The superposition theorem says that the current that exists in a branch of a linear circuit that contains several sinusoidal sources is equal to the sum of the currents produced by each source independently. The proof of the superposition theorem follows directly from the fact that Kirchhoff’s laws applied to linear circuits always result in a set of linear equations that can be reduced to a single linear equation with a single unknown. The unknown branch current thus can be written as a linear superposition of each of the source terms with an appropriate coefficient. (Figure 2.158 shows the essence of superimposing of sine waves.)

FIGURE 2.158 (a) Shows two sine waves and the resulting sum—another sine wave of the same frequency, but shifted in phase and amplitude. This is the key feature that makes it easy to deal with sinusoidally driven linear circuits containing resistors, capacitors, and inductors. Note that if you were to try this with sine waves of different frequencies, as shown in (b), the resultant waveform isn’t sinusoidal. Superimposing nonsinusoidal waveforms of the same frequency, such as squarewaves, isn’t guaranteed to result in a similar waveform, as shown in (c).

What this all means is that you do not have to go to the trouble of calculating the time dependence of the unknown current or voltage within the circuit because you know that it will always be of the form cos (ωt + ϕ). Instead, all you need to do is calculate the peak value (or RMS value) and the phase, and apply the superposition theorem. To represent currents and voltages and apply the superposition theorem, it would seem obvious to use sine or cosine functions to account for magnitude, phase, and frequency. However, in the mathematical process of superimposing (adding, multiplying, etc.), you would get messy sinusoidal expressions in terms of sines and cosines that would require difficult trigonometric rules and identities to convert the answers into something you could understand. Instead, what you can do to represent amplitudes and phase of voltages and currents in a circuit is to use complex numbers.

Recall from the section on complex numbers that a complex number exhibits sinusoidal behavior—at least in the complex plane. For example, the trigonometric form of a complex number z1 = r1 cos θ1 + jr1 sin θ1 will trace out a circular path in the complex plane when θ runs from 0 to 360°, or from 0 to 2π radians. If you graph the real part of z versus θ, you get a sinusoidal wave pattern. To change the amplitude of the wave pattern, you simply change the value of r. To set the frequency, you simply multiply θ by some number. To induce a phase shift relative to another wave pattern of the same frequency, you simply add some number (in degrees or radians) to θ. If you replace θ with ωt, where (ω = 2πf), replace the r with V0, and leave a place for a term to be added to ωt (a place for phase shifts), you come up with an expression for the voltage source in terms of complex numbers. You could do the same sort of thing for currents, too.

Now, the nice thing about complex numbers, as compared with sinusoidal functions, is that you can represent a complex number in various ways, within rectangular, polar-trigonometric, or polar-exponential forms (standard or shorthand versions). Having these different options makes the mathematics involved in the superimposing process easier. For example, by converting a number, say, into rectangular form, you can easily add or subtract terms. By converting the number into polar-exponential form (or shorthand form), you can easily multiply and divide terms (terms in the exponent will simply add or subtract).

It should be noted that, in reality, currents and voltages are always real; there is no such thing as an imaginary current or voltage. But then why are there imaginary parts? The answer is that when you start expressing currents and voltages with real and imaginary parts, you are simply introducing a mechanism for keeping track of the phase. (The complex part is like a hidden part within a machine; its function does not show up externally but does indeed affect the external output—the “real,” or important, part, as it were.) What this means is that the final answer (the result of the superimposing) always must be converted back into a real quantity. This means that after all the calculations are done, you must convert the complex result into either polar-trigonometric or polar-exponential (shorthand form) and remove the imaginary part. For example, if you come across a resultant voltage expressed in the following complex form:

V(t) = 5 V + j 10 V

where the voltages are RMS, we get a meaningful real result by finding the magnitude, which we can do simply by converting the complex expression into polar exponential or shorthand form:

= (11.2 V) ej(63.5°) = 11.2 V ∠ 63.5°

Whatever is going on, be it reactive or resistive effects, there is really 11.2 V RMS present. If the result is a final calculation, the phase often isn’t important for practical purposes, so it is often ignored.

You may be scratching your head now and saying, “How do I really do the superimposing and such? This all seems too abstract or wishy-washy. How do I actually account for the resistors, capacitors, and inductors in the grand scheme of things?” Perhaps the best way to avoid this wishy-washiness is to begin by taking a sinusoidal voltage and converting it into a complex number representation. After that, you can apply it individually across a resistor, a capacitor, and then an inductor to see what you get. Important new ideas and concrete analysis techniques will surface in the process.

2.27.2 Sinusoidal Voltage Source in Complex Notation

Let’s start by taking the following expression for a sinusoidal voltage:

V0 cos (ωt)      (ω = 2πf)

and converting it into a polar-trigonometric expression:

V0 cos (ωt) + jV0 sin (ωt)

What about the jV0 sin (ωt) term? It is imaginary and does not have any physical meaning, so it does not affect the real voltage expression (you need it, however, for the superimposing process). To help with the calculations that follow, the polar-trigonometric form is converted into the polar-exponential form using Euler’s relation ejθ = r cos (θ) + jr sin (θ):

 V0ej(ωt) (2.67)

In polar-exponential shorthand form, this would be:

 V0 ∠ (ωt) (2.68)

Graphically, you can represent this voltage as a vector rotating counterclockwise with angular frequency ω in the complex plane (recall that ω = dθ/dt, where ω = 2πf). The length of the vector represents the maximum value of V—namely, V0—while the projection of the vector onto the real axis represents the real part, or the instantaneous value of V, and the projection of the vector onto the imaginary axis represents the imaginary part of V.

FIGURE 2.159

Now that you have an expression for the voltage in complex form, you can place, say, a resistor, a capacitor, or an inductor across the source and come up with a complex expression for the current through each component. To find the current through a resistor in complex form, you simply plug V0ejt) into V in I = V/R. To find the capacitor current, you plug V0ejt) into I = C dV/dt. Finally, to find the inductor current, you plug V0ejt) into I = 1/L Vdt. The results are shown in Fig. 2.160.

FIGURE 2.160

Comparing the phase difference between the current and voltage through and across each component, notice the following:

·    Resistor: The current and voltage are in phase, ϕ = 0°, as shown in the graph in Fig. 2.160. This behavior can also be modeled within the complex plane, where the voltage and current vectors are at the same angle with respect to each other, both of which rotate around counterclockwise at an angular frequency ω = 2πf.

·    Capacitor: The current is out of phase with the applied voltage by +90°. In other words, the current leads the voltage by 90°. By convention, unless otherwise stated, the phase angle ϕ is referenced from the current vector to the voltage vector. If ϕ is positive, then current is leading (further counterclockwise in rotation); if ϕ is negative, current is lagging (further clockwise in rotation).

·    Inductor: The current is out of phase with the applied voltage by −90°. In other words, the current lags the voltage by 90°.

We call the complex plane model, showing the magnitude and phase of the voltages and currents, a phasor diagram—where the term phasor implies phase comparison. Unlike a time-dependent mathematical function, a phasor provides only a snapshot of what’s going on. In other words, it only tells you the phase and amplitude at a particular moment in time.

Now comes the important trick to making ac analysis easy to deal with. If we take the voltage across each component and divide it by the current, we get the following (see Fig. 2.161):

FIGURE 2.161

As you can see, the nasty V0ejt) terms cancel, giving us resistance, capacitive reactance, and inductive reactance in complex form. Notice that the resulting expressions are functions only of frequency, not of time. This is part of the trick to avoiding the nasty differential equations.

Now that we have a way of describing capacitive reactance and inductive reactance in terms of complex numbers, we can make an important assumption. We can now treat capacitors and inductors like frequency-sensitive resistors within sinusoidally driven circuits. These frequency-sensitive resistors take the place of normal resistors in dc circuit analysis. We must also replace dc sources with sinusoidal ones. If all voltages, currents, resistances, and reactances are expressed in complex form when we are analyzing a circuit, when we plug them into the old circuit theorems (Ohm’s law, Kirchhoff’s law, Thevinin’s theorem, etc.) we will come up with equations whose solutions are taken care of through the mathematical operations of the complex numbers themselves (the superposition theorem is built in).

For example, ac Ohm’s law looks like this:

 V(ω) = I(ω) × Z(ω) (2.69)

What does the Z stand for? It’s referred to as complex impedance, which is a generic way of describing resistance to current flow, in complex form. The complex impedance may simply be resistive, it may be only capacitive, it may be inductive, or it could be a combination of resistive and reactive elements (e.g., an RLC circuit element). For example:

Resistor: VR = IR × R

Capacitor:

Inductor: VL = IL × XL = IL (jωL) = jILωL = ILωL ∠ + 90°

Any complex impedance Z: VZ = IZ × Z

Figure 2.162 shows what we just discussed, along with the phasor expression for a sinusoidal source, and a complex impedance Z composed of a resistor, a capacitor, and an inductor.

FIGURE 2.162

Continuing along the line of treating complex impedances as frequency-sensitive resistors, we can make use of the resistors in series equation, except now we take impedances in series:

 Ztotal = Z1 + Z2 + Z3 + … + ZN (2.70) N in series

Likewise, the old voltage divider now becomes the ac voltage divider:

FIGURE 2.163

To find the equivalent impedance for a larger number of impedances in parallel:

 (2.71) N in parallel
 (2.72) Two in parallel

The ac current divider is consequently:

FIGURE 2.164

And perhaps, most important, we can throw complex impedances in Kirchhoff’s voltage and loop equations and solve complex circuits with many nodes:

FIGURE 2.165

FIGURE 2.166

Example 1: Find the complex impedances of the following networks.

(a)  (b)  (c)  (d)

You can simplify those results that have a complex number in the denominator by using:

Example 2: Express networks (a) and (c) in the previous example in polar coordinate form.

(a)

(c)

What a mathematical nightmare, you may say, after doing the last example. As you can see, the expressions can get ugly if we don’t start plugging values into the variables from the start. However, the act of finding a complex impedance and using it in ac analysis is vastly easier than using the characteristic equations of resistors, capacitors, and inductors; inserting them into Kirchhoff’s laws; and solving the differential equations.

Example 3: The series RL circuit in Fig. 2.167 is driven by a 12-VAC (RMS), 60-Hz source. L = 265 mH, R = 50 Ω. Find IS, IR, IL, VR, and VL and the apparent power, real power, reactive power, and power factor.

FIGURE 2.167

VS(t) = 17.0 V sin (ωt)

VR(t) = 7.6 V sin (ωt − 63.4°)

VL(t) = 15.1 V sin (ωt + 26.6°)

IS(t) = 0.151 A sin (ωt − 63.4°)

* Peak voltages and currents used in these functions are the RMS equivalents multiplied by 1.414.

First, calculate the reactance of the inductor:

XL = jωL = j(2π × 60 Hz × 265 × 10−3 H) = j 100 Ω

Since the resistor and inductor are in series, the math is easy—simply add complex numbers in rectangular form:

Z = R + XL = 50 Ω + j100 Ω

In polar form, the result is:

Don’t let the imaginary part or phase angle fool you. The impedance is real; it provides 112 ohms’ worth, though only a portion of this is real resistance—the rest is inductive reactance.

The current can now be found using ac Ohm’s law:

The −63.4° result means the current lags the applied voltage or total voltage across the network by 63.4°. Since this is a series circuit I = IR = IL. The voltage across the resistance and inductor can be found using ac Ohm’s law or the ac voltage divider:

Note that the preceding calculations were for only an instant in time t = 0, as is the case by stating the starting condition: VS = 12 VAC ∠ 0°. But that’s all we need, since we know the phases and the voltages at those phases, which link amplitude. So to create an accurate picture of how the whole system behaves over time, we simply plug the more general ωt into the source voltage, and convert the RMS values into true values by multiplying by 1.414. We get VS = 17.0 V ∠ ωt, which isn’t a snapshot but a continuous mathematic description for all times. To make a graph, we must convert into trigonometric form, ignoring the imaginary part in the process: VS = 17.0 V cos (ωt). Since we are only concerned with phase, we can express VS in terms of a sine function, VS = 17.0 V sin (ωt) and reference all other voltage and current waveforms from it by adding their phase terms and including their peak values, as shown in the following equations.

The apparent power due to the total impedance:

VA = IRMSVRMS = (0.107A)(12 VAC) = 1.284 VA

Only resistance consumes power, however:

PR = IRMS2R = (0.107A)2 (50 Ω) = 0.572 W

The reactive power due to the inductor is:

VAR = IRMS2XL = (0.107A)2 (100 Ω) = 1.145 VA

Power factor (real power/apparent power) is:

where ϕ is the phase angle between VS and IS. Note that we haven’t actually discussed apparent and reactive power yet—that’s coming up.

2.27.3 Odd Phenomena in Reactive Circuits

In reactive circuits, circulation of energy accounts for seemingly odd phenomena. In our example in Fig. 2.167, it appears that Kirchhoff’s law doesn’t add up, since the arithmetical sum of the resistor and inductor voltages is:

5.35 VAC + 10.70 VAC = 16.05 VAC

This is greater than the 12-VAC source voltage. The problem here is that we haven’t taken into account the phase.

Their actual result, when phase is taken into account, is:

Vtotal = VR + VL = 5.35 VAC ∠ −63.4° + 10.70 VAC ∠ 26.6°

= 2.4 VAC − j4.8 VAC + 9.6 VAC + j4.8 VAC = 12 VAC

Figure 2.168 graphically illustrates this point.

FIGURE 2.168

Note that the use of “VAC” to represent the fact that the voltages are expressed in RMS values is not always followed. Some people like to simply write “V” and assume that all voltages are sinusoidal and thus given in RMS form. The actual peak voltages are related to the RMS voltage value by VP = 1.414 × VRMS.

In other cases, such as a series circuit with capacitance and inductance, the voltages across the components may exceed the supply voltage. This condition can exist because, while energy is being stored by the inductor, the capacitor is returning energy to the circuit from its previously charged state, and vice versa. In a parallel circuit with inductive and capacitive branches, the currents circulating through the components may exceed the current drawn from the source. Again, the phenomenon occurs because the inductor’s collapsing field supplies current to the capacitor and the discharging capacitor provides current to the inductor. We will take a look at these cases in a moment.

2.28 Power in AC Circuits (Apparent Power, Real Power, Reactive Power)

In a complex circuit containing resistors, inductors, and capacitors, how do you determine what kind of power is being used? The best place to start is with the generalized power law P = IRMSVRMS. However, for now, let’s replace P with VA, and call VA the apparent power:

 VA = IRMSVRMS (2.74) Apparent power

In light of our RL series circuit in Fig. 2.167, we find the apparent power to be:

VA = IRMSVRMS = (0.107A)(12 V) = 1.284 VA

The apparent power VA is no different from the power we calculate using the generalized ac power expression. The reason for using VA instead of P is simply a convention used to help distinguish the fact that the calculated power isn’t purely real and is not expressed in watts, as real power is. Instead, apparent power is expressed in volt-amperes, or VA, which happens to be the same letters used for the variable for apparent power. (This is analogous to the variable for voltage being similar to the unit volt, though the variable is distinguished from the unit by being italicized.) In essence, the apparent power takes into account both resistive power losses as well as reactive power. The reactive power, however, isn’t associated with power loss, but is instead associated with energy storage in the form of magnetic fields within inductors and electric fields within capacitors. This energy is later returned to the circuit as the magnetic field in an inductor collapses, or the electric field vanishes as a capacitor discharges later in the ac cycle. Only if a circuit is purely resistive can we say that the apparent power is in watts.

So how do we distinguish what portions of the apparent power are real power and reactive power? Real power is associated with power loss due to heating through an ohmic material, so we can define real power using ac Ohm’s law within the generalized power expression:

 PR = IRMS2R (2.75) Real power

In our series RL circuit in Fig. 2.167, we find the real power to be:

PR = IRMS2R = (0.107 A)2 (50 Ω) = 0.572 W

Notice that real power is always measured in watts.

Now, to determine the reactive power due to capacitance and inductance within a circuit, we create the notion of reactive power. The reactive power is given in volt-ampere reactive, or VAR. We can define the reactive power by using the Ohm’s power law, replacing resistance (or impedance) with a generic reactance X:

 VAR = IRMS2X (2.76) Reactive power

At no time should watts be associated with reactive power. In fact, reactive power is called wattless power, and therefore is given in volt-amperes, or VA.

Considering the RL series circuit in Fig. 2.167, the reactive power is:

VAR = IRMS2XL = (0.107 A)2 (100 Ω) = 1.145 VA

You may be saying, great, we can now add up the real power and reactive power, and this will give us the apparent power. Let’s try it out for our RL circuit in Fig. 2.167:

0.572 + 1.145 = 1.717

But wait, the calculated apparent power was 1.284 VA, not 1.717 VA. What’s wrong? The problem is that a simple arithmetic operation on variables that are reactive can’t be done without considering phase (as we saw with adding voltages). Considering phase for our RL series circuit:

VAR = I2RMSXL = (0.107 A ∠ −63.4°)2 (100 Ω ∠ 90°) = 1.145 VA ∠ −36.8°

VAR = 0.917 VA − j0.686 VA

PR = I2RMS R = (0.107 A ∠ −63.4°)2 (50 Ω ∠ 0°) = 0.573 ∠ −126.8°

PR = −0.343 W − j0.459 W

Adding the reactive and real power together now gives us the correct apparent power:

VA = VAR + PR = 0.574 VA − j1.145 VA = 1.281 VA ∠ −63.4°

To avoid doing such calculations, we note the following relationship:

 (2.77)

FIGURE 2.169

Here, we don’t have to worry about phase angles—the Pythagorean theorem used in the complex plane, as shown in Fig. 2.169, takes care of that. Using the RL series example again and inserting our values into Eq. 2.77, we get an accurate expression relating real, apparent and reactive powers:

2.28.1 Power Factor

Another way to represent the amount of apparent power to reactive power within a circuit is to use what’s called the power factor. The power factor of a circuit is the ratio of consumed power to apparent power:

 (2.78)

In the example in Fig. 2.167:

Power factor is frequently expressed as a percentage, in this case 45 percent.

An equivalent definition of power factor is:

 PF = cos ϕ (2.79)

where ϕ is the phase angle (between voltage and current). In the example in Fig. 2.167, the phase angle is −63.4°, so:

PF = cos (−63.4°) = 0.45

as the earlier calculation confirms.

The power factor of a purely resistive circuit is 100 percent, or 1, while the power factor of a purely reactive circuit is 0.

Since the power factor is always rendered as a positive number, the value must be followed by the words “leading” or “lagging” to identify the phase of the voltage with respect to the current. Specifying the numerical power factor is not always sufficient. For example, many dc-to-ac power inverters can safely operate loads having large net reactance of one sign but only a small reactance of the opposite sign. The final calculation of the power factor in the example in Fig. 2.167 yields the value of 0.45 lagging.

In ac equipment, the ac components must handle reactive power as well as real power. For example, a transformer connected to a purely reactive load must still be capable of supplying the voltage and be able to handle the current required by the reactive load. The current in the transformer windings will heat the windings as a result of I2R losses in the winding resistances.

As a final note, there is another term to describe the percentage of power used in reactance, which is called the reactive factor. The reactive factor is defined by:

 (2.80)

Using the example in Fig. 2.167:

Example 4: The LC series circuit shown in Fig. 2.170 is driven by a 10-VAC (RMS), 127,323-Hz source. L = 100 µH, C = 62.5 nF. Find IS, IR, IL, VL, and VC and the apparent power, real power, reactive power, and power factor.

FIGURE 2.170

VS(t) = 14.1 V sin (ωt)

VL(t) = 18.90 V sin (ωt)

VC(t) = 4.72 V sin (ωt − 180°)

IS(t) = 0.236 A sin (ωt − 90°)

* Peak voltages and currents used in these functions are the RMS equivalents multiplied by 1.414.

First we calculate the inductive and capacitive reactances:

XL = jωL = j(2π × 127,323 Hz × 100 × 10−6 H)

= j 80 Ω

Since the inductor and capacitor are in series, the math is easy—simply add complex numbers in rectangular form:

Z = XL + XC = j80 Ω + (−j20 Ω) = j60 Ω

In polar form, this is 60 Ω ∠ 90°. The fact that the phase is 90°, or the result is positive imaginary, means the impedance is net inductive. Don’t let the imaginary part fool you—the impedance is actually felt—it provides 60 Ω of impedance, even though it’s not resistive.

The current can now be found using ac Ohm’s law:

Note that we used 60 Ω ∠ 90° (polar) to make the division easy. The −90° result means the source current (total current) lags the source voltage by 90°. Since this is a series circuit: IS = IL = IC.

The voltage across the inductor and capacitor can be found using ac Ohm’s law (or the ac voltage divider):

VL = IS × XL = (0.167 A ∠ −90°)(80 Ω ∠ 90°)

= 13.36 VAC ∠ 0°

VC = IS × XC = (0.167 A ∠ −90°)(20 Ω ∠ −90°)

= 3.34 VAC ∠ −180°

Notice that the voltage across the inductor is greater than the supply voltage; the capacitor supplies current to the inductor as it discharges.

To convert these snapshots into a continuous set of functions, we convert all RMS values to true value (× 1.414), add the ωt term to the phase angles, and convert to trigonometric form, then delete the imaginary part. Our results would all be in terms of cosines, but to make things pretty, we select sines. Doing this doesn’t make any practical difference. See graphs and equations to left.

The apparent power due to the total impedance is:

VA = IRMSVRMS = (0.167 A)(10 VAC) = 1.67 VA

The real (true) power consumed by the circuit is:

PR = IRMS2R = (0.167 A)2(0 Ω) = 0 W

Only reactive power exists, and for the inductor and capacitor:

VARL = IRMS2XL = (0.167 A)2(80 Ω) = 2.23 VA

VARC = IRMS2XC = (0.167 A)2(20 Ω) = 0.56 VA

Power factor is:

A power factor of 0 means the circuit is purely reactive.

Example 5: The LC parallel circuit in Fig. 2.171 is driven by a 10-VAC (RMS), 2893.7-Hz source. L = 2.2 mH, C = 5.5 µF. Find IS, IL, IC, VL, and VC and the apparent power, real power, reactive power, and power factor.

FIGURE 2.171

VS(t) = VL(t) = VC(t) = 14.1V sin (ωt)

IS(t) = 1.061 A sin (ωt + 90°)

IL(t) = 0.354 A sin (ωt − 90°)

IC(t) = 1.414 A sin (ωt + 90°)

* Peak voltages and currents used in these functions are the RMS equivalents multiplied by 1.414.

First we find the inductive and capacitive reactances:

XL = jωL = j(2π × 2893.7 Hz × 2.2 × 10−3 H)

= j40 Ω

Since the inductor and capacitor are in parallel, the math is relatively easy—use two components in parallel formula, and multiply and add complex numbers:

(Tricks used to solve: j × j = −1, 1/j = −j)

In polar form, the result is 13.33 Ω ∠ −90°. The fact that the phase is −90°, or negative imaginary, means the impedance is net capacitive. Don’t let the imaginary part fool you—the impedance is actually felt—it provides 13.3 Ω of impedance, even though it’s not resistive.

The total current can now be found using ac Ohm’s law:

Note that we used 13.3 Ω ∠ −90° (polar form) to make the division easy. The +90° result means the total current leads the source voltage by 90°. Since this is a parallel circuit: VS = VL = VC.

The current through each component can be found using ac Ohm’s law (or the current divider relation):

Notice that the capacitor current is larger than the supply current; the collapsing magnetic field of the inductor supplies current to the capacitor.

To convert these snapshots into a continuous set of functions, we convert all RMS values to true value (× 1.414), add the ωt term to the phase angles, and convert to trigonometric form, then delete the imaginary part. Our results would all be in terms of cosines, but to make things pretty, we select sines. Doing this doesn’t make any practical difference. See graphs and equations in Fig. 2.171.

The apparent power is:

VA = IRMSVRMS = (0.750 A)(10 VAC) = 7.50 VA

Only reactive power exists, and for the inductor and capacitor:

VARL = IL2XL = (0.25A)2(40 Ω) = 2.50 VA

VARC = IC2XC = (1.00 A)2(10 Ω) = 10.00 VA

Power factor is:

Again, a power factor of 0 means the circuit is purely reactive.

Example 6: The LCR series circuit in Fig. 2.172 is driven by a 1.00-VAC (RMS), 1000-Hz source. L = 25 mH, C = 1 µF, and R = 1.0 Ω. Find the total impedance Z, VL, VC, VR, and IS, and the apparent power, real power, reactive power, and power factor.

FIGURE 2.172

VS(t) = 1.41 V sin (ωt)

VL(t) = 95.32 V sin (ωt + 154.5°)

VC(t) = 96.60 V sin (ωt − 25.5°)

VR(t) = 0.61 V sin (ωt + 64.5°)

IS(t) = 0.607 A sin (ωt + 64.5°)

* Peak voltages and currents used in these functions are the RMS equivalents multiplied by 1.414.

First let’s find the reactances of the inductor and capacitor:

XL = jωL = j(2π × 1000 Hz × 25 × 10−3 H)

= j157.1 Ω

= −j159.2 Ω

To find the total impedance, take the L, C, and R in series:

Z = R + XL + XC = 1 Ω + j157.1 Ω − j159.2 Ω

= 1 Ω − j(2.1 Ω)

In polar form, the result is 2.33 Ω ∠ −64.5°. The fact that the phase is −64.5°, or the imaginary part is negative, means the impedance is net capacitive. Don’t let the imaginary part fool you—the impedance is actually felt—it provides 2.33 Ω of impedance, even though it’s not entirely resistive.

The total current can now be found using ac Ohm’s law:

Note that we used 2.33 Ω ∠ −64.5° (polar form) to make the division easy. The +64.5° result means the total current leads the source voltage by 64.5°. Since this is a series circuit: IS = IL = IC = IR.

The voltage across each component can be found using ac Ohm’s law:

VL = ISXL = (0.429 A ∠ 64.5°)(157.1 Ω ∠ 90°)

= 67.40 VAC ∠ 154.5°

VC = ISXC

= (0.429 A ∠ 64.5°)(159.2 Ω ∠ −90°)

= 68.30 VAC ∠ −25.5°

VR = ISR = (0.429 A ∠ 64.5°)(1 Ω ∠ 0°)

= 0.429 VAC ∠ 64.5°

Notice that the voltage across the inductor and capacitor are huge at this particular phase when compared to the supply voltage; the capacitor supplies current to the inductor as it discharges, while the inductor supplies current to the capacitor as its magnetic field collapses. To convert these snapshots into a continuous set of functions, we convert all RMS values to true value (× 1.414), add the ωt term to the phase angles, and convert to trigonometric form, then delete the imaginary part. Our results would all be in terms of cosines, but to make things pretty, we select sines. Doing this doesn’t make any practical difference. See graphs and equations in Fig. 2.172.

The apparent power is:

VA = IRMSVRMS = (0.429 A)(1.00 VAC) = 0.429 VA

The real (true) power, or power dissipated by resistor is:

PR = IS2R = (0.429 A)2(1 Ω) = 0.18 W

The reactive powers for the inductor and capacitor:

VARL = IL2XL = (0.429 A)2(157.1 Ω) = 28.91 VA

VARC = IC2XC = (0.429 A)2(159.2 Ω) = 29.30 VA

Power factor is:

It should have become apparent from the last example that the VARs for the inductor and the capacitor became surprisingly large. When dealing with real-life components, the VAR values become important. Even though volt-amps do not contribute to the overall true power loss, this does not mean that the volt-amps aren’t felt by the reactive components. Components like inductors and transformers (ideally reactive components) are usually given a volt-amp rating that provides the safety limit to prevent overheating the component. Again, it is the internal resistances within the inductor or transformer that must be considered.

Example 7: The LCR parallel circuit in Fig. 2.173 is driven by a 12.0-VAC (RMS), 600-Hz source. L = 1.061 mH, C = 66.3 µF, and R = 10 Ω. Find ZtotVL, VC, VR, and IS, and the apparent power, real power, reactive power, and power factor. First let’s find the reactances of the inductor and capacitor:

XL = jωL = j(2π × 600 Hz × 1.061 × 10−3 H)

= j4.0 Ω

= −j4.0 Ω

Since the inductor and capacitor are in parallel, the math is relatively easy—use two components in general formula, and multiply and add complex numbers:

The fact that there is no reactive part to the total impedance is quite interesting and makes our life easy in terms of calculations. Before getting into the interesting matter, let’s finish solving.

The total current can now be found using ac Ohm’s law:

Since there is no phase angle, the source current and voltage are in phase. Since this is a parallel circuit, VS = VL = VC = VR.

The current through each component can be found using ac Ohm’s law:

You can convert the voltages and current back to true sinusoidal form to get the graph shown in Fig. 2.173.

FIGURE 2.173

VS(t) = VL(t) = VC(t) = VR(t) = 16.9 V sin (ωt)

IS(t) = 1.70 A sin (ωt)

IL(t) = 4.24 A sin (ωt −90°)

IC(t) = 4.24 A sin (ωt + 90°)

IR(t) = 1.70 A sin (ωt)

* Peak voltages and currents used in these functions are the RMS equivalents multiplied by 1.414.

The apparent power is:

VA = IRMSVRMS = (1.20 A)(12 VAC) = 14.4 VA

The real (true) power, or power dissipated by resistor, is:

PR = IS2R = (1.20 A)2(10 Ω) = 14.4 W

The reactive powers for the inductor and capacitor:

VARL = IL2XL = (1.20 A)2(4 Ω) = 4.8 VA

VARC = IC2XC = (1.20 A)2(4 Ω) = 4.8 VA

Power factor is:

A power factor of 1 indicates the circuit is purely resistive. How can this be? In this case we have a special condition where a circulating current is “trapped” within the LC section. This only occurs at a special frequency called the resonant frequency. We’ll cover resonant circuits in a moment.

2.29 Thevenin’s Theorem in AC Form

Thevenin’s theorem, like the other dc theorems, can be modified so that it can be used in ac linear circuits analysis. The revised ac form of Thevenin’s theorem reads: Any complex network of resistors, capacitors, and inductors can be represented by a single sinusoidal power source connected to a single equivalent impedance. For example, if you want to find the voltage across two points in a complex, linear, sinusoidal circuit or find the current and voltage through and across a particular element within, you remove the element, find the Thevenin voltage VTHEV(t), replace the sinusoidal sources with a short, find the Thevenin impedance ZTHEV(t), and then make the Thevenin equivalent circuit. Figure 2.174 shows the Thevenin equivalent circuit for a complex circuit containing resistors, capacitors, and inductors. The following example will provide any missing details.

FIGURE 2.174

Example: Suppose that you are interested in finding the current through the resistor in the circuit in Fig. 2.175.

FIGURE 2.175

VS(t) = VC(t) = 14.1 V sin (ωt)

IR(t) = IS(t) = 4.64 mA sin (ωt − 24.3°)

* Peak voltages and currents used in these functions are the RMS equivalents multiplied by 1.414.

Answer: First, you remove the resistor in order to free up two terminals to make a black box. Next, you calculate the open circuit, or Thevenin voltage VTHEV, by using the ac voltage divider equation. First, however, let’s determine the reactances of the capacitor and inductor:

XL = jωL = j(2π × 1000 Hz × 200 × 10−3 H)

= j 1257 Ω

Using the ac voltage divider:

To find ZTHEV, you short the source with a wire and take the impedance of the inductor and capacitor in parallel:

Next, you reattach the load resistor to the Thevenin equivalent circuit and find the current by combining VTHEV and R in series:

ZTOTAL = R + ZTHEV = 3300 Ω + j1493 Ω

= 3622 Ω ∠ 24.3°

Using ac Ohm’s law, we can find the current:

Don’t let the complex expression fool you; the resistor current is indeed 3.28 mA, but lags the source voltage by 24.3°.

To turn our snapshots into real functions with respect to time, we add ωt to every phase angle expression, and convert from RMS to true values—see graphs and equations with Fig. 2.175.

Apparent power, real (true) power, reactive power, and power factor are:

VA = IS2 × ZTOTAL = (0.00328 A)2(3622 Ω) = 0.039 VA

PR = IR2 × R = (0.00328 A)2(3300 Ω) = 0.035 W

VAR = IR2 × ZTHEV = (0.00328 A)2(1493 Ω) = 0.016 VA

2.30 Resonant Circuits

When an LC circuit is driven by a sinusoidal voltage source at a special frequency called the resonant frequency, an interesting phenomenon occurs. For example, if you drive a series LC circuit (shown in Fig. 2.176) at its resonant angular frequency ω0 = , or equivalently its resonant frequency f0 = 1/(), the effective impedance across the LC network goes to zero. In effect, the LC network acts like a short. This means that the sourced current flow will be at a maximum. Ideally, it will go to infinity, but in reality it is limited by internal resistances of all the components in the circuit. To get an idea of how the series LC resonant circuit works, let’s take a look at the following example.

FIGURE 2.176

Example: To get an idea of how the LC series circuit works, we find the equivalent impedance of the circuit—taking L and C in series. Unlike the previous examples, the frequency is unknown, so it must be left as a variable:

In polar form:

Note that we got the phase angle by assuming that the arc tangent of anything divided by 0 is 90°.

The current through the series reactance is then:

If you plug in the L = 100 µH and C = 62.5 nF, and ω = 2πf, the total impedance and current, ignoring phase angle, become:

Both the impedance and the current as a function of frequency are graphed in Fig. 2.176. Notice that as we approach the resonant frequency:

the impedance goes to zero while the current goes to infinity. In other words, if you plug the resonant frequency into the impedance and current equations, the result gives you zero and infinity, respectively. In reality, internal resistance prevents infinite current.

Inductive and capacitive reactances at resonance are equal but opposite in phase, as depicted by the equations with Fig. 2.176.

Intuitively, you can imagine that the voltage across the capacitor and the voltage across the inductor are exactly equal but opposite in phase at resonance, within the LC series circuit. This means the effective voltage drop across the series pair is zero; therefore, the impedance across the pair must also be zero.

Resonance occurs in a parallel LC circuit as well. The angular resonant frequency is  or equivalently its resonant frequency . This is the same resonant frequency expression for the series LC circuit; however, the circuit behavior is exactly opposite. Instead of the impedance going to zero and the current going to infinity at resonance, the impedance goes to infinity while the current goes to zero. In essence, the parallel LC network acts like an open circuit. Of course, in reality, there is always some internal resistance and parasitic capacitance and inductance within the circuit that prevent this from occurring. To get an idea of how the parallel LC resonant circuit works, let’s take a look at the following example.

Example: For an LC parallel-resonant circuit, we take the inductor and capacitor in parallel (applying Eq. 2.72):

In polar form:

Note that we got the phase angle by assuming that the negative arc tangent of anything over zero is −90°.

The current through the parallel reactance is then:

If you plug in the L = 100 µH and C = 62.5 nF, and ω = 2πf, the total impedance and current, ignoring phase angle, are:

Both the impedance and the current as a function of frequency are graphed in Fig. 2.177. Notice that as we approach the resonant frequency:

FIGURE 2.177

Inductive and capacitive reactances at resonance are equal but opposite in phase:

the impedance goes to infinity while the current goes to zero. In other words, if you plug the resonant frequency into the impedance and current equations, you get infinity and zero, respectively. In reality, internal resistances and parasitic inductances and capacitors within the circuit prevent infinite current. Notice that as the frequency approaches zero, the current increases toward infinity, since the inductor acts like a short at dc. On the other hand, if the frequency increases toward infinity, the capacitor acts like a short, and the current goes toward infinity.

Intuitively, we can imagine that at resonance, the impedance and voltage across L are equal in magnitude but opposite in phase (direction) with respect to C. From this you can infer that an equal but opposite current will flow through L and C. In other words, at one moment a current is flowing upward through L and downward through C. The current through L runs into the top of C, while the current from C runs into the bottom of the inductor. At another moment the directions of the currents reverse (energy is “bounced” back in the other direction; L and C act as an oscillator pair that passes the same amount of energy back and forth, and the amount of energy is determined by the sizes of L and C). This internal current flow around the LC loop is referred to as a circulating current. Now, as this is going on, no more current will be supplied through the network by the source. Why? The power source doesn’t “feel” a potential difference across it. Another way to put this would be to say that if an external current were to be supplied through the LC network, it would mean that one of the elements (L or C) would have to be passing more current than the other. However, at resonance, this does not happen because the L and C currents are equal and pointing in the opposite direction.

Example 1: What is the resonant frequency of a circuit containing an inductor of 5.0 µH and a capacitor of 35 pF?

Example 2: What is the value of capacitance needed to create a resonant circuit at 21.1 MHz, if the inductor is 2.00 µH?

For most electronics work, these previous formulas will permit calculation of frequency and component values well within the limits of component tolerances. Resonant circuits have other properties of importance in addition to the resonant frequency, however. These include impedance, voltage drop across components in series-resonant circuits, circulating currents in parallel-resonant circuits, and bandwidth. These properties determine such factors as the selectivity of a tuned circuit and the component ratings for circuits handling considerable power. Although the basic determination of the tuned-circuit resonant frequency ignored any resistance in the circuit, that will play a vital role in the circuit’s other characteristics.

2.30.1 Resonance in RLC Circuits

The previous LC series- and parallel-resonant circuits are ideal in nature. In reality, there is internal resistance or impedance within the components that leads to a deviation from the true resonant effects we observed. In most real LC resonant circuits, the most notable resistance is associated with losses in the inductor at high frequencies (HF range); resistive losses in a capacitor are low enough at those frequencies to be ignored. The following example shows how a series RLC circuit works.

FIGURE 2.178

Inductive and capacitive reactances at resonance are equal but opposite in phase:

Example: We start by finding the total impedance of the RLC circuit by taking R, L, and C in series:

In polar form:

The current through the total impedance is, ignoring phase:

If you plug in the L = 100 µH and C = 62.5 nF, and ω = 2πf, the current expressed as a function of frequency is:

Now, unlike the ideal LC series-resonant circuit, when we plug in the resonant frequency:

the total current doesn’t go to infinity; instead it goes to 2 A, which is just VS/R = 10 VAC/5 Ω = 2 A. The resistance thus prevents a zero impedance condition that would otherwise be present when the inductor and capacitor reactances cancel at resonance.

The unloaded Q is the reactance at resonance divided by the resistance:

As pointed out in Fig. 2.178, at resonance, the reactance of the capacitor cancels the reactance of the inductor, and thus the impedance is determined solely by the resistance. We can therefore deduce that at resonance the current and voltage must be in phase—recall that in a sinusoidal circuit with a single resistor, the current and voltage are in phase. However, as we move away from the resonant frequency (keeping component values the same), the impedance goes up due to increases in reactance of the capacitor or the inductor. As you go lower in frequency from resonance, the capacitor’s reactance becomes dominant—capacitors increasingly resist current as the frequency decreases. As you go higher in frequency from resonance, the inductor’s reactance becomes dominant—inductors increasingly resist current as the frequency increases. Far from resonance in either direction, you can see that resistance has an insignificant effect on current amplitude.

Now if you take a look at the graph in Fig. 2.178, notice how the current curve looks like a pointy hilltop. In electronics, describing the pointiness of the current curve is an important characteristic of concern. When the reactance of the inductor or capacitor is of the same order of magnitude as the resistance, the current decreases rather slowly as you move away from the resonant frequency in either direction. Such a curve, or “hilltop,” is said to be broad. Conversely, when the reactance of the inductor or capacitor is considerably larger than the resistance, the current decreases rapidly as you move away from the resonant frequency in either direction. Such a curve, or “hilltop,” is said to be sharp. A sharp resonant circuit will respond a great deal more readily to the resonant frequency than to frequencies quite close to resonance. A broad resonant circuit will respond almost equally well to a group or band of frequencies centered about the resonant frequency.

As it turns out, both sharp and narrow circuits are useful. A sharp circuit gives good selectivity. This means that it has the ability to respond strongly (in terms of current amplitude) at one desired frequency and is able to discriminate against others. A broad circuit, on the other hand, is used in situations where a similar response over a band of frequencies is desired, as opposed to a strong response at a single frequency.

Next, we’ll take a look at quality factor and bandwidth—two quantities that provide a measure of the sharpness of our RLC resonant circuit.

2.30.2 Q (Quality Factor) and Bandwidth

As mentioned earlier, the ratio of reactance or stored energy to resistance or consumed energy is by definition the quality factor Q. (Q is also referred to as the figure of merit, or multiplying factor.) As it turns out, within a series RLC circuit (where R is internal resistance of the components), the internal resistive losses within the inductor dominate energy consumption at high frequencies. This means the inductor Q largely determines the resonant circuit Q. Since the value of Q is independent of any external load to which the circuit might transfer power, we modify the resonant circuit Q, by calling it the unloaded Q or QU of the circuit. See Fig. 2.179.

FIGURE 2.179

In the example RLC series-resonant circuit from Fig. 2.178, we can determine the unloaded Q of the circuit by dividing the reactance of either the inductor or the capacitor (they have the same relative impedance at resonance) by the resistance:

As you can see, if we increase the resistance, the unloaded Q decreases, giving rise to broad current response curves about resonance, as shown in the graph in Fig. 2.179. With resistances of 10, 20, and 50 Ω, the unloaded Q decreases to 4, 2, and 0.8, respectively. Conversely, if the resistance is made smaller, the unloaded Q increases, giving rise to a sharp current response curve about resonance. For example, when the resistance is lowered to 2 Ω, the unloaded Q becomes 20.

2.30.3 Bandwidth

An alternative way of expressing the sharpness of a series-resonant circuit is using what is called bandwidth. Basically, what we do is take the quality factor graph in Fig. 2.179 and convert it to the bandwidth graph in Fig. 2.179. This involves changing the current axis to a relative current axis and moving the family of curves for varying Q values up so that all have the same peak current. By assuming that the peak current of each curve is the same, the rate of change of current for various values of Q and the associated ratios of reactance to resistance are more easily compared. From the curves, you can see that lower Q circuits pass frequencies over a greater bandwidth of frequencies than circuits with a higher Q. For the purpose of comparing tuned circuits, bandwidth is often defined as the frequency spread between the two frequencies at which the current amplitude decreases to 0.707 or  times the maximum value. Since the power consumed by the resistance R is proportional to the square of the current, the power at these points is half the maximum power at resonance, assuming that R is constant for the calculations. The half-power, or −3-dB, points are marked on the drawing.

For Q values of 10 or greater, the curves shown in Fig. 2.179 are approximately symmetrical. On this assumption, bandwidth (BW) can be easily calculated:

 (2.81)

where BW and f are in units of hertz.

Example: What is the bandwidth of the series-resonant circuit in Fig. 2.178 at 100 kHz and at 1 MHz?

2.30.4 Voltage Drop Across Components in RLC Resonant Circuit

The voltage drop across a given inductor or a capacitor within an RLC resonant circuit can be determined by applying ac Ohm’s law:

As we discovered earlier, these voltages may become many times larger than the source voltage, due to the magnetic and electric stored energy returned by the inductor and capacitor. This is especially true for circuits with high Q values. For example, at resonance, the RLC circuit of Fig. 2.178 has the following voltage drops across the capacitor and inductor:

VC = XCI = 40 Ω ∠ −90° × 2 A ∠ 0° = 80 VAC ∠ −90°

VL = XLI = 40 Ω ∠ +90° × 2 A ∠ 0° = 80 VAC ∠ +90°

The actual amplitude of the voltage when we convert from the RMS values is a factor of 1.414 higher, or 113 V. High-Q circuits such as those found in antenna couplers, which handle significant power, may experience arcing from high reactive voltages, even though the source voltage to the circuit is well within component ratings. When Qs of greater than 10 are considered, the following equation gives a good approximation of the reactive voltage within a series-resonant RLC circuit at resonance:

 VX = QUVS (2.82)

2.30.5 Capacitor Losses

Note that although capacitor energy losses tend to be insignificant compared to inductor losses up to about 30 MHz within a series-resonant circuit, these losses may affect circuit Q in the VHF range (30 to 300 MHz). Leakage resistance, principally in the solid dielectric between the capacitor plates, is not exactly like the internal wire resistive losses in an inductor coil. Instead of forming a series resistance, resistance associated with capacitor leakage usually forms a parallel resistance with the capacitive reactance. If the leakage resistance of a capacitor is significant enough to affect the Q of a series-resonant circuit, the parallel resistance must be converted to an equivalent series resistance before adding it to the inductor’s resistance. This equivalent series resistance is given by:

 (2.83)

where RP is the leakage resistance and XC is the capacitive reactance. This value is then added to the inductor’s internal resistance and the sum represents the R in an RLC resonant circuit.

Example: A 10.0-pF capacitor has a leakage resistance of 9,000 Ω at 40.0 MHz. What is the equivalent series resistance?

When calculating the impedance, current, and bandwidth of a series-resonant circuit, the series leakage resistance is added to the inductor’s coil resistance. Since an inductor’s resistance tends to increase with frequency due to what’s called the skin effect (electrons forced to the surface of a wire), the combined losses in the capacitor and the inductor can seriously reduce circuit Q.

Example 1: What is the unloaded Q of a series-resonant circuit with a loss resistance of 4 Ω and inductive and capacitive components having a reactance of 200 Ω each? What is the unloaded Q if the reactances are 20 Ω each?

Example 2: What is the unloaded Q of a series-resonant circuit operating at 7.75 MHz, if the bandwidth is 775 kHz?

2.30.6 Parallel-Resonant Circuits

Although series-resonant circuits are common, the vast majority of resonant circuits are parallel-resonant circuits. Figure 2.180 shows a typical parallel-resonant circuit. As with series-resonant circuits, the inductor internal coil resistance is the main source of resistive losses, and therefore we put the series resistor in the same leg as the inductor. Unlike the series-resonant circuit whose impedance goes toward a minimum at resonance, the parallel-resonant circuit’s impedance goes to a maximum. For this reason, RLC parallel-resonant circuits are often called antiresonant circuits or rejector circuits. (RLC series-resonant circuits go by the name acceptor). The following example will paint a picture of parallel resonance behavior.

FIGURE 2.180

Example: The impedance of the RLC parallel circuit is found by combining the inductor and resistor in series, and then placing it in parallel with the capacitor (using the impedances in parallel formula):

In polar form:

If you plug in L = 5.0 µH and C = 50 pF, R = 10.5 Ω and ω = 2πf, the total impedance, ignoring phase, is:

The total current (line current), ignoring phase, is:

Plugging these equations into a graphing program, you get the curves shown in Fig. 2.180. Note that at a particular frequency, the impedance goes to a maximum, while the total current goes to a minimum. This, however, is not at the point where XL = XCthe point referred to as the resonant frequency in the case of a simple LC parallel circuit or a series RLC circuit. As it turns out, the resonant frequency of a parallel RLC circuit is a bit more complex and can be expressed three possible ways. However, for now, we make an approximation that is expressed as before:

The unloaded Q for this circuit, using the reactance of L, is:

The lower graph in Fig. 2.180 shows the quality factor and how it is influenced by the size of the parallel resistance in the inductor leg.

Unlike the ideal LC parallel resonant circuit we saw in Fig. 2.177, the addition of R changes the conditions of resonance. For example, when the inductive and capacitive reactances are the same, the impedances of the inductive and capacitive legs do not cancel—the resistance in the inductive leg screws things up. When XL = XC, the impedance of the inductive leg is greater than XC and will not be 180° out of phase with XC. The resultant current is not at a true minimum value and is not in phase with the voltage. See line (A) in Fig. 2.181.

FIGURE 2.181

Now we can alter the value of the inductor a bit (holding Q constant) and get a new frequency where the current reaches a true minimum—something we can accomplish with the help of a current meter. We associate the dip in current at this new frequency with what is termed the standard definition of RLC parallel resonance. The point where minimum current (or maximum impedance) is reached is called the antiresonant point and is not to be confused with the condition where XL = XC. Altering the inductance to achieve minimum current comes at a price; the current becomes somewhat out of phase with the voltage—see line (B) in Fig. 2.181.

It’s possible to alter the circuit design of our RLC parallel-resonant circuit to draw some of the resonant points shown in Fig. 2.181 together—for example, compensating for the resistance of the inductor by altering the capacitance (retuning the capacitor). The difference among the resonant points tends to get smaller and converge to within a percent of the frequency when the circuit Q rises above 10, in which case they can be ignored for practical calculations. Tuning for minimum current will not introduce a sufficiently large phase angle between voltage and current to create circuit difficulties.

As long as we assume Qs higher than 10, we can use a single set of formulas to predict circuit performance. As it turns out, what we end up doing is removing the series inductor resistance in the leg with the inductor and substituting a parallel-equivalent resistor of the actual inductor loss series resistor, as shown in Fig. 2.182.

FIGURE 2.182

Series and parallel equivalents when both circuits are resonant. Series resistance RS in (a) is replaced by the parallel resistance RP in (b), and vice versa:

This parallel-equivalent resistance is often called the dynamic resistance of the parallel-resonant circuit. This resistance is the inverse of the series resistance; as the series resistance value decreases, the parallel-equivalent resistance increases. Alternatively, this means that the parallel-equivalent resistance will increase with circuit Q. We use the following formula to calculate the approximate parallel-equivalent resistance:

 (2.84)

Example: What is the parallel-equivalent resistance for the inductor in Fig. 2.182b, taking the inductive reactance to be 316 Ω and a series resistance to be 10.5 Ω at resonance? Also determine the unloaded Q of the circuit.

Since the coil QU remains the inductor’s reactance divided by its series resistance, we get:

Multiplying QU by the reactance also provides the approximate parallel-equivalent resistance of the inductor’s series resistance.

At resonance, assuming our parallel equivalent representation, XL = XC, and RP now defines the impedance of the parallel-resonant circuit. The reactances just equal each other, leaving the voltage and current in phase with each other. In other words, at resonance, the circuit demonstrates only parallel resistance. Therefore, Eq. 2.84 can be rewritten as:

 (2.85)

In the preceding example, the circuit impedance at resonance is 9510 Ω.

At frequencies below resonance, the reactance of the inductor is smaller than the reactance of the capacitor, and therefore the current through the inductor will be larger than that through the capacitor. This means that there is only partial cancellation of the two reactive currents, and therefore the line current is larger than the current with resistance alone. Above resonance, things are reversed; more current flows through the capacitor than through the inductor, and again the line current increases above a value larger than the current with resistance alone. At resonance, the current is determined entirely by RP; it will be small if RP is large and large if RP is small.

Since the current rises off resonance, the parallel-resonant-circuit impedance must fall. It also becomes complex, resulting in an increasing phase difference between the voltage and the current. The rate at which the impedance falls is a function of QUFigure 2.180 presents a family of curves showing the impedance drop from resonance for circuit Qs ranging from 10 to 100. The curve family for parallel-circuit impedance is essentially the same as the curve family for series-circuit current. As with series-tuned circuits, the higher the Q of a parallel-tuned circuit, the sharper the response peak. Likewise, the lower the Q, the wider the band of frequencies to which the circuit responds. Using the half-power (−3-dB) points as a comparative measure of circuit performance, we can apply the same equations for bandwidth for a series-resonant circuit to a parallel-resonant circuit, BW = f/QUTable 2.11 summarizes performance of parallel-resonant circuits at high and low Qs, above and below resonance.

TABLE 2.11 Performance of Parallel-Resonant Circuits

 A. High and Low Q Parallel-Resonant Circuits HIGH Q CIRCUIT LOW Q CIRCUIT Selectivity High Low Bandwidth Narrow Wide Impedance High Low Line current Low High Circulating current High Low
 B. Off-Resonance Performance for Constant Values of Inductance and Capacitance ABOVE RESONANCE BELOW RESONANCE Inductive reactance Increases Decreases Capacitive reactance Decreases Increases Circuit resistance Same* Same* Circuit impedance Decreases Decreases Line current Increases Increases Circulating current Decreases Decreases Circuit behavior Capacitive Inductive *True near resonance, but far from resonance skin effects within inductor alter the resistive losses.

Note on Circulating Current

When we covered the ideal LC parallel-resonant circuit, we saw that quite a large circulating current could exist between the capacitor and the inductor at resonance, with no line current being drawn from the source. If we consider the more realistic RLC parallel-resonant circuit, we also notice a circulating current at resonance (which, too, can be quite large compared to the source voltage), but now there exists a small line current sourced by the load. This current is attributed to the fact that even though the impedance of the resonant network is high, it isn’t infinite because there are resistive losses as the current circulates through the inductor and capacitor—most of which are attributed to the inductor’s internal resistance.

Taking our example from Fig. 2.183 and using the parallel-equivalent circuit, as shown to the right in Fig. 2.183, we associate the total line current as flowing through the parallel inductor resistance RP. Since the inductor, capacitor, and parallel resistor are all in parallel according to the parallel-equivalent circuit, we can determine the circulating current present between the inductor and capacitor, as well as the total line current now attributed to the parallel resistance:

FIGURE 2.183

The circuiting current is simply ICIR = IC = IL when the circuit is at the resonant frequency. For parallel-resonant circuits with an unloaded Q of 10 or greater, the circulating current is approximately equal to:

 ICIR = QU × ITOT (2.86)

Using our example, if we measure the total line current to be 1 mA, and taking the Q of the circuit to be 30, the approximated circulating current is (30)(1 mA) = 30 mA.

Example: A parallel-resonant circuit permits an ac line current of 50 mA and has a Q of 100. What is the circulating current through the elements?

IC = QU × IT = 100 × 0.05 A = 5 A

Circulating current in high-Q parallel-tuned circuits can reach a level that causes component heating and power loss. Therefore, components should be rated for the anticipated circulating current, and not just for the line currents.

It is possible to use either series- or parallel-resonant circuits to do the same work in many circuits, thus providing flexibility. Figure 2.184 illustrates this by showing both a series-resonant circuit in the signal path and a parallel-resonant circuit shunted from the signal path to ground. Assuming both circuits are resonant at the same frequency f and have the same Q: the series-tuned circuit has its lowest impedance at the resonant frequency, permitting the maximum possible current to flow along the signal path; at all other frequencies, the impedance increases causing a decrease in current. The circuit passes the desired signal and tends to impede signals at undesired frequencies. The parallel circuit, on the other hand, provides the highest impedance at resonance, making the signal path lowest in impedance for the signal; at all frequencies off resonance, the parallel circuit presents a lower impedance, providing signals a route to ground away from the signal path. In theory, the effects displayed for both circuits are the same. However, in actual circuit design, there are many other things to consider, which ultimately determine which circuit is best for a particular application. We will discuss such circuits later, when we cover filter circuits.

FIGURE 2.184

2.30.7 The Q of Loaded Circuits

In many resonant circuit applications, the only practical power lost is dissipated in the resonant circuit internal resistance. At frequencies below around 30 MHz, most of the internal resistance is within the inductor coil itself. Increasing the number of turns in an inductor coil increases the reactance at a rate that is faster than the accompanying internal resistance of the coil. Inductors used in circuits where high Q is necessary have large inductances.

When a resonant circuit is used to deliver energy to a load, the energy consumed within the resonant circuit is usually insignificant compared to that consumed by the load. For example, in Fig. 2.185, a parallel load resistor RLOAD is attached to a resonant circuit, from which it receives power.

FIGURE 2.185

If the power dissipated by the load is at least 10 times as great as the power lost in the inductor and capacitor, the parallel impedance of the resonant circuit itself will be so high compared with the resistance of the load that for all practical purposes the impedance of the combined circuit is equal to the load impedance. In these circumstances, the load resistance replaces the circuit impedance in calculating Q. The Q of a parallel-resonant circuit loaded by a resistance is:

 (2.87)

where QLOAD is the circuit-loaded Q, RLOAD is the parallel load resistance in ohms, and X is the reactance in ohms of either the inductor or the capacitor.

Example 1: A resistive load of 4000 Ω is connected across the resonant circuit shown in Fig. 2.185, where the inductive and capacitive reactances at resonance are 316 Ω. What is the loaded Q for this circuit?

The loaded Q of a circuit increases when the reactances are decreased. When a circuit is loaded with a low resistance (a few kiloohms) it must have low-reactance elements (large capacitance and small inductance) to have reasonably high Q.

Sometimes parallel load resistors are added to parallel-resonant circuits to lower the Q and increase the circuit bandwidth, as the following example illustrates.

Example 2: A parallel-resonant circuit needs to be designed with a bandwidth of 400.0 kHz at 14.0 MHz. The current circuit has a QU of 70.0 and its components have reactances of 350 Ω each. What is the parallel load resistor that will increase the bandwidth to the specified value?

Answer: First, we determine the bandwidth of the existing circuit:

The desired bandwidth, 400 kHz, requires a loaded circuit Q of:

Since the desired Q is half the original value, halving the resonant impedance or parallel-resistance value of the circuit will do the trick. The present impedance of the circuit is:

Z = QUXL = 70 × 350 Ω = 24,500 Ω

The desired impedance is:

Z = QUXL = 35.0 × 350 Ω = 12,250 Ω

or half the present impedance.

A parallel resistor of 24,500 Ω will produce the required reduction in Q as bandwidth increases. In real design situations, things are a bit more complex—one must consider factors such as shape of the bandpass curve.

2.31 Lecture on Decibels

In electronics, often you will encounter situations where you will need to compare the relative amplitudes or the relative powers between two signals. For example, if an amplifier has an output voltage that is 10 times the input voltage, you can set up a ratio:

Vout/Vin = 10 VAC/1 VAC = 10

and give the ratio a special name—called gain. If you have a device whose output voltage is 10 times smaller than the input voltage, the gain ratio will be less than 1:

Vout/Vin = 1 VAC/10 VAC = 0.10

In this case, you call the ratio the attenuation.

Using ratios to make comparisons between two signals or powers is done all the time—not only in electronics. However, there are times when the range over which the ratio of amplitudes between two signals or the ratio of powers between two signals becomes inconveniently large. For example, if you consider the range over which the human ear can perceive different levels of sound intensity (average power per area of air wave), you would find that this range is very large, from about 10−12 to 1 W/m2. Attempting to plot a graph of sound intensity versus, say, the distance between your ear and the speaker, would be difficult, especially if you are plotting a number of points at different ends of the scale—the resolution gets nasty. You can use special log paper to automatically correct this problem, or you can stick to normal linear graph paper by first “shrinking” your values logarithmically. For this we use decibels.

To start off on the right foot, a bel is defined as the logarithm of a power ratio. It gives us a way to compare power levels with each other and with some reference power. The bel is defined as:

 (2.88)

where P0 is the reference power and P1 is the power you are comparing with the reference power.

In electronics, the bel is often used to compare electrical power levels; however, what’s more common in electronics and elsewhere is to use decibels, abbreviated dB. A decibel is 110 of a bel (similar to a millimeter being 110 of a centimeter). It takes 10 decibels to make 1 bel. So in light of this, we can compare power levels in terms of decibels:

 (2.89) Decibelsin terms of power

Example: Express the gain of an amplifier (output power divided by input power) in terms of decibels, if the amplifier takes a 1-W signal and boosts it up to a 50-W signal.

Answer: Let P0 represent the 1-W reference power, and let P1 be the compared power:

The amplifier in this example has a gain of nearly 17.00 dB (17 decibels).

Sometimes when comparing signal levels in an electronic circuit, we know the voltage or current of the signal, but not the power. Though it’s possible to calculate the power, given the circuit impedance, we take a shortcut by simply plugging ac Ohm’s law into the powers in the decibel expression. Recall that P = V2/Z = I2Z. Now, this holds true only as long as the impedance of the circuit doesn’t change when the voltage or current changes. As long as the impedance remains the same, we get a comparison of voltage signals and current signals in terms of decibels:

 (2.90) Decibels interms of voltage andcurrent levels

In the expressions above we used the laws of logarithms to remove square terms. For example:

10 log (V12/V02) = 10(log V12 − log V02)

= 10(2 log V1 − 2 log V0) = 20(log V1 − log V2)

= 20 log (V1/V0).

Notice that the impedance terms cancel and the final result is a factor of 2 greater—a result of the square terms in the log being pushed out (see law of logarithms). The power and voltage and current expressions are all fundamentally the same—they are all based on the ratio of powers.

There are several power ratios you should learn to recognize and be able to associate with the corresponding decibel representations.

For example, when doubling power, the final power is always 2 times the initial (or reference) power—it doesn’t make a difference if you are going from 1 to 2 W, 40 to 80 W, or 500 to 1000 W, the ratio is always 2. In decibels, a power ratio of 2 is represented as:

dB = 10 log (2) = 3.01 dB

There is a 3.01-dB gain if the output power is twice the input power. Usually, people don’t care about the .01 fraction and simply refer to the power doubling as a 3-dB increase in power.

When the power is cut in half, the ratio is always 0.5 or 1/2. Again, it doesn’t matter if you’re going from 1000 to 500 W, 80 to 40 W, or 2 to 1 W, the ratio is still 0.5. In decibels, a power ratio of 0.5 is represented as:

dB = 10 log (0.5) = −3.01 dB

A negative sign indicates a decrease in power. Again, people usually ignore the .01 fraction and simply refer to the power being cut in half as a −3-dB change in power or, more logically, a 3-dB decrease in power (the term “decrease” eliminates the need for the negative sign).

Now if you increase the power by 4, you can avoid using the decibels formula and simply associate such an increase with doubling two times: 3.01 dB + 3.01 dB = 6.02 dB or around 6 dB. Likewise, if you increase the power by 8, you, in effect, double four times, so the power ratio in decibels is 3.01 × 4 = 12.04, or around 12 dB.

The same relationship is true of power decreases. Each time you cut the power in half, there is a 3.01-dB, or around 3-dB decrease. Cutting the power by four times is akin to cutting in half twice, or 3.01 + 3.01 dB = 6.02 dB, or around a 6-dB decrease. Again, you can avoid stating “decrease” and simply say that there is a change of −6 dB.

Table 2.12 shows the relationship between several common decibel values and the power change associated with those values. The current and voltage changes are also included, but these are valid only if the impedance is the same for both values.

TABLE 2.12 Decibels and Power Ratios

 COMMON DECIBEL VALUES AND POWER-RATIO EQUIVALENTS dB P2 /P1 V2 /V1 or I2 /I1 120 1012 106 60 106 103 20 102 10.0 10 10.00 3.162 6.0206 4.0000 2.0000 3.0103 2.0000 1.4142 1 1.259 1.122 0 1.000 1.000 −120 10−12 10−6 −60 10−6 10−3 −20 10−2 0.1000 −10 0.1000 0.3162 −6.0206 0.2500 0.5000 −3.0103 0.5000 0.7071 −1 0.7943 0.8913 0 1.000 1.000 * Voltage and current ratios hold only if the impedance remains the same.

2.31.1 Alternative Decibel Representations

It is often convenient to compare a certain power level with some standard reference. For example, suppose you measured the signal coming into a receiver from an antenna and found the power to be 2 × 10−13 mW. As this signal goes through the receiver, it increases in strength until it finally produces some sound in the receiver speaker or headphones. It is convenient to describe these signal levels in terms of decibels. A common reference power is 1 mW. The decibel value of a signal compared to 1 mW is specified as “dBm” to mean decibels compared to 1 mW. In our example, the signal strength at the receiver input is:

There are many other reference powers used, depending upon the circuits and power levels. If you use 1 W as the reference power, then you would specify dBW. Antenna power gains are often specified in relation to a dipole (dBd) or an isotropic radiator (dBi). Anytime you see another letter following dB, you will know that some reference power is being specified. For example, to describe the magnitudes of a voltage relative to a 1-V reference, you indicate the level in decibels by placing a “V” at the end of dB, giving units of dBV (again, impedances must stay the same). In acoustics, dB, SPL is used to describe the pressure of one signal in terms of a reference pressure of 20 µPa. The term decibel is also used in the context of sound (see Sec. 15.1).

2.32 Input and Output Impedance

2.32.1 Input Impedance

Input impedance ZIN is the impedance “seen” looking into the input of a circuit or device (see Fig. 2.186). Input impedance gives you an idea of how much current can be drawn into the input of a device. Because a complex circuit usually contains reactive components such as inductors and capacitors, the input impedance is frequency sensitive. Therefore, the input impedance may allow only a little current to enter at one frequency, while highly impeding current enters at another frequency. At low frequencies (less than 1 kHz) reactive components may have less influence, and the term input resistance may be used—only the real resistance is dominant. The effects of capacitance and inductance are generally more significant at high frequencies.

FIGURE 2.186

When the input impedance is small, a relatively large current can be drawn into the input of the device when a voltage of specific frequency is applied to the input. This typically has an effect of dropping the source voltage of the driver circuit feeding the device’s input. (This is especially true if the driver circuit has a large output impedance.) A device with low input impedance is an audio speaker (typically 4 or 8 Ω), which draws lots of current to drive the voice coil.

When the input impedance is large, on the other hand, a small current is drawn into the device when a voltage of specific frequency is applied to the input, and, hence, doesn’t cause a substantial drop in source voltage of the driver circuit feeding the input. An op amp is a device with a very large input impedance (1 to 10 MΩ); one of its inputs (there are two) will practically draw no current (in the nA range). In terms of audio, a hi-fi preamplifier that has a 1-MΩ radio input, a 500-kΩ CD input, and a 100-kΩ tap input all have high impedance input due to the preamp being a voltage amplifier not a current amplifier—something we will discuss later.

As a general rule of thumb, in terms of transmitting a signal, the input impedance of a device should be greater than the output impedance of the circuit supplying the signal to the input. Generally, the value should be 10 times as great to ensure that the input will not overload the source of the signal and reduce the strength by a substantial amount. In terms of calculations and such, the input impedance is by definition equal to:

For example, Fig. 2.187 shows how to determine what the input impedance is for two resistor circuits. Notice that in example 2, when a load is attached to the output, the input impedance must be recalculated, taking R2 and Rloadin parallel.

FIGURE 2.187

In Sec. 2.33.1, which covers filter circuits, you’ll see how the input impedance becomes dependent on frequency.

2.32.2 Output Impedance

Output impedance ZOUT refers to the impedance looking back into the output of a device. The output of any circuit or device is equivalent to an output impedance ZOUT in series with an ideal voltage source VSOURCEFigure 2.186 shows the equivalent circuit; it represents the combined effect of all the voltage sources and the effective total impedance (resistances, capacitance, and inductance) connected to the output side of the circuit. You can think of the equivalent circuit as a Thevenin equivalent circuit, in which case it should be clear that the VSOURCE present in Fig. 2.186 isn’t necessarily the actual supply voltage of the circuit but the Thevenin equivalent voltage. As with input impedance, output impedance can be frequency dependent. The term output resistance is used in cases where there is little reactance within the circuit, or when the frequency of operation is low (say, less than 1 kHz) and reactive effects are of little consequence; the effects of capacitance and inductance are generally most significant at high frequencies.

When the output impedance is small, a relatively large output current can be drawn from the device’s output without significant drop in output voltage. A source with an output impedance much lower than the input impedance of a load to which it is attached will suffer little voltage loss driving current through its output impedance. For example, a lab dc power supply can be viewed as an ideal voltage source in series with a small internal resistance. A decent supply will have an output impedance in the milliohm (mΩ) range, meaning it can supply considerable current to a load without much drop in supply voltage. A battery typically has a higher internal resistance and tends to suffer a more substantial drop in supply voltage as the current demands increase. In general, a small output impedance (or resistance) is considered a good thing, since it means that little power is lost to resistive heating in the impedance and a larger current can be sourced. Op amps, which, we saw, have large input impedance and tend to have low output impedance.

When the output impedance is large, on the other hand, a relatively small output current can be drawn from the output of a device before the voltage at the output drops substantially. If a source with a large output impedance attempts to drive a load that has a much smaller input impedance, only a small portion appears across the load; most is lost driving the output current through the output impedance.

Again, the rule of thumb for efficient signal transfer is to have an output impedance that is at least 110 that of the load’s input impedance to which it is attached.

In terms of calculations, the output impedance of a circuit is simply its Thevenin equivalent resistance RTHEV. The output impedance is sometimes called the source impedance. In terms of determining the output impedance in circuit analysis, it amounts to “killing” the source (shorting it) and finding the equivalent impedance between the output terminals—the Thevenin impedance.

For example, in Fig. 2.188, to determine the output impedance of this circuit, we effectively find the Thevinin resistance by “shorting the source” and removing the load, and determining the impedance between the output terminals. In this case, the output impedance is simply R1 and R2 in parallel.

FIGURE 2.188

In Sec. 2.3.1, when we cover filter circuits, you’ll see how the output impedance becomes dependent on frequency.

2.33 Two-Port Networks and Filters

2.33.1 Filters

By combining resistors, capacitors, and inductors in special ways, you can design networks that are capable of passing certain frequencies of signals while rejecting others. This section examines four basic kinds of filters: low-pass, high-pass, bandpass, and notch filters.

Low-Pass Filters

The simple RC filter shown in Fig. 2.189 acts as a low-pass filter—it passes low frequencies but rejects high frequencies.

FIGURE 2.189

Example: To figure out how this network works, we find the transfer function. We begin by using the voltage divider to find Vout in terms of Vin, and consider there is no load (open output or RL = ∞):

The transfer function is then found by rearranging the equation:

The magnitude and phase of H are:

Here, τ is called the time constant, and ωC is called the angular cutoff frequency of the circuit—related to the standard cutoff frequency by . The cutoff frequency represents the frequency at which the output voltage is attenuated by a factor of , the equivalent of half power. The cutoff frequency in this example is:

Intuitively, we imagine that when the input voltage is very low in frequency, the capacitor’s reactance is high, so it draws little current, thus keeping the output amplitude near the input amplitude. However, as the frequency of the input signal increases, the capacitor’s reactance decreases and the capacitor draws more current, which in turn causes the output voltage to drop. Figure 2.189 shows attenuation versus frequency graphs—one graph expresses the attenuation in decibels.

The capacitor produces a delay, as shown in the phase plot in Fig. 2.189. At very low frequency, the output voltage follows the input—they have similar phases. As the frequency rises, the output starts lagging the input. At the cutoff frequency, the output voltage lags by 45°. As the frequency goes to infinity, the phase lag approaches 90°.

Figure 2.190 shows an RL low-pass filter that uses inductive reactance as the frequency-sensitive element instead of capacitive reactance, as was the case in the RC filter.

FIGURE 2.190

The input impedance can be found by definition  while the output impedance can be found by “killing the source” (see Fig. 2.191):

and

and

FIGURE 2.191

Now what happens when we put a finite load resistance RL on the output? Doing the voltage divider stuff and preparing the voltage transfer function, we get:

FIGURE 2.192

where

This is similar to the transfer function for the unterminated RC filter, but with resistance R being replaced by R′. Therefore:

and

As you can see, the load has the effect of reducing the filter gain  and shifting the cutoff frequency to a higher frequency as .

The input and output impedance with load resistance become:

and

and

FIGURE 2.193

As long as  or  (condition for good voltage coupling),  and the terminated RC filter will look exactly like an unterminated filter. The filter gain is one, the shift in cutoff frequency disappears, and the input and output resistance become the same as before.

Example: To find the transfer function or attenuation of the RL circuit with no load, we find Vout in terms of Vin using the voltage divider:

The magnitude and phase become:

Here, ωC is called the angular cutoff frequency of the circuit—related to the standard cutoff frequency by . The cutoff frequency represents the frequency at which the output voltage is attenuated by a factor of , the equivalent of half power. The cutoff frequency in this example is:

Intuitively we imagine that when the input voltage is very low in frequency, the inductor doesn’t have a hard time passing current to the output. However, as the frequency gets big, the inductor’s reactance increases, and the signal becomes more attenuated at the output. Figure 2.190 shows attenuation versus frequency graphs—one graph expresses the attenuation in decibels.

The inductor produces a delay, as shown in the phase plot in Fig. 2.190. At very low frequency, the output voltage follows the input—they have similar phases. As the frequency rises, the output starts lagging the input. At the cutoff frequency, the output voltage lags by 45°. As the frequency goes to infinity, the phase lag approaches 90°.

The input impedance can be found using the definition of the input impedance:

The value of the input impedance depends on the frequency ω. For good voltage coupling, the input impedance of this filter should be much larger than the output impedance of the previous stage. The minimum value of Zin is an important number, and its value is minimum when the impedance of the inductor is zero ω → 0:

The output impedance can be found by shorting the source and finding the equivalent impedance between output terminals:

where the source resistance is ignored. The output impedance also depends on the frequency ω. For good voltage coupling, the output impedance of this filter should be much smaller than the input impedance of the next stage. The maximum value of Zout is also an important number, and it is maximum when the impedance of the inductor is infinity ω → ∞:

When the RL low-pass filter is terminated with a load resistance RL, the voltage transfer function changes to:

where

The input impedance becomes:

The output impedance becomes:

The effect of the load is to shift the cutoff frequency to a lower value. Filter gain is not affected. Again, for  or  (condition for good voltage coupling), the shift in cutoff frequency disappears, and the filter will look exactly like an unterminated filter.

High-Pass Filters

Example: To figure out how this network works, we find the transfer function by using the voltage divider equation and solving in terms of Vout and Vin:

or

The magnitude and phase of H are:

Here τ is called the time constant and ωC is called the angular cutoff frequency of the circuit—related to the standard cutoff frequency by ωC = 2πfC. The cutoff frequency represents the frequency at which the output voltage is attenuated by a factor of , the equivalent of half power. The cutoff frequency in this example is:

Intuitively, we imagine that when the input voltage is very low in frequency, the capacitor’s reactance is very high, and hardly any signal is passed to the output. However, as the frequency rises, the capacitor’s reactance decreases, and there is little attenuation at the output. Figure 2.194 shows attenuation versus frequency graphs—one graph expresses the attenuation in decibels.

FIGURE 2.194

In terms of phase, at very low frequency the output leads the input in phase by 90°. As the frequency rises to the cutoff frequency, the output leads by 45°. When the frequency goes toward infinity, the phase approaches 0, the point where the capacitor acts like a short.

Input and output impedances of this filter can be found in a way similar to finding these impedances for low-pass filters:

and

With a terminated load resistance, the voltage transfer function becomes:

where

This is similar to the transfer function for the unterminated RC filter, but with resistance R being replaced by R’:

and

The load has the effect of shifting the cutoff frequency to a higher frequency ().

The input and output impedances are:

As long as  or  R (condition for good voltage coupling),  and the terminated RC filter will look like an unterminated filter. The shift in cutoff frequency disappears, and input and output resistance become the same as before.

RL High-Pass Filter

Figure 2.195 shows an RL high-pass filter that uses inductive reactance as the frequency-sensitive element instead of capacitive reactance, as was the case in the RC filter.

FIGURE 2.195

Example: To find the transfer function or attenuation of the RL circuit, we again use the voltage divider equation and solve for the transfer function or attenuation of the RL circuit in terms of Vout and Vin:

The magnitude and phase of H are:

Here ωC is called the angular cutoff frequency of the circuit—related to the standard cutoff frequency by ωC = 2πfC. The cutoff frequency represents the frequency at which the output voltage is attenuated by a factor of , the equivalent of half power. The cutoff frequency in this example is:

Intuitively, we imagine that when the input voltage is very low in frequency, the inductor’s reactance is very low, so most of the current is diverted to ground—the signal is greatly attenuated at the output. However, as the frequency rises, the inductor’s reactance increases and less current is passed to ground—the attenuation decreases. Figure 2.195 shows attenuation versus frequency graphs—one graph expresses the attenuation in decibels.

In terms of phase, at very low frequency the output leads the input in phase by 90°. As the frequency rises to the cutoff frequency, the output leads by 45°. When the frequency goes toward infinity, the phase approaches 0, the point where the inductor acts like an open circuit.

The input and output impedances are:

For a terminated RL high-pass filter with load resistance, we do a similar calculation as we did with the RC high-pass filter, replacing the resistance with R′:

The input and output impedances are:

The load has the effect of lowering the gain, K = R′/R < 1, and it shifts the cutoff frequency to a lower value. As long as  or  (condition for good voltage coupling),  and the terminated RC filter will look like an unterminated filter.

Bandpass Filter

The RLC bandpass filter in Fig. 2.196 acts to pass a narrow range of frequencies (band) while attenuating or rejecting all other frequencies.

FIGURE 2.196 The parallel bandpass filter shown here yields characteristics similar to those of the previous bandpass filter. However, unlike the previous filter, as you approach the resonant frequency of the tuned circuit, the LC (RL coil) section’s impedance gets large, not allowing current to be diverted away from the load. On either side of resonance, the impedance goes down, diverting current away from load.

Example: To find the transfer function or attenuation of the unloaded RLC circuit, we set up equations for Vin and Vout:

Vout = R × I

The transfer function becomes:

This is the transfer function for an unloaded output. However, now we get more realistic and have a load resistance attached to the output. In this case we must replace R with RT, which is the parallel resistance of R and RLOAD:

Placing this in the unloaded transfer function and solving for the magnitude, we get:

Plugging in all component values and setting ω = 2πf, we get:

The attenuation versus frequency graph based on this equation is shown in Fig. 2.196. (Three other families of curves are provided for loads of 4 Ω, 100 Ω, and an infinite resistance load.)

The resonant frequency, Q, bandwidth, and upper and lower cutoff frequencies are given by:

f1 = f0 − BW/2 = 2055 − 172/2 = 1969 Hz

f2 = f0 + BW/2 = 2055 + 172/2 = 2141 Hz

Notch Filter

The notch filter in Fig. 2.197 acts to pass a wide range of frequencies, while attenuating (rejecting) a narrow band of frequencies.

FIGURE 2.197 The notch filter shown here yields stopband characteristics similar to those of the previous notch filter. However, unlike the previous filter, as you approach the resonant frequency of the tuned circuit, the LC (RL coil) section’s impedance gets large, not allowing current to be sent to the load. On either side of resonance, the impedance goes down, allowing current to reach load.

Example: To find the transfer function or attenuation of the unloaded RLC circuit, we set up equations for Vin and Vout:

The transfer function becomes:

This is the transfer function for an unloaded output. Now we get more realistic and have a load resistance attached to the output. However, in this case, the load resistance is so large that we can assume it draws inconsequential current, so we don’t need to place it into the equation:

The magnitude of the transfer function is:

Plugging in all component values and setting ω = 2πf, we get:

The attenuation versus frequency graph based on this equation is shown in Fig. 2.197.

The resonant frequency, Q, bandwidth, and upper and lower cutoff frequencies are given by:

f1 = f0 − BW/2 = 18,960 − 1053/2 = 18,430 Hz

f2 = f0 + BW/2 = 18,960 + 1053/2 = 19,490 Hz

2.33.2 Attenuators

Often it is desirable to attenuate a sinusoidal voltage by an amount that is independent of frequency. We can do this by using a voltage divider, since its output is independent of frequency. Figure 2.198 shows a simple voltage divider attenuator network inserted between a source and a load circuit to decrease the source signal’s magnitude before it reaches the load. As we go through this example, you will pick up additional insight into input and output impedances.

FIGURE 2.198

Example: In (a), the source has an output impedance equal to the internal resistance of the source:

Zout = RS

In (b), the attenuator network has input and output impedances of:

Zin = R1 + R2

Zout = R2

We can come up with a transfer function for this, taking the same current to flow through R2:

If you rearrange terms, you can see this is simply a voltage divider.

In (c), the load has an input impedance of:

Zin = RL

However, when we assemble the circuit, the useful input and output impedances change. Now, looking at things from the point of view of the source, as shown in (d), the source sees an input impedance of the attenuator combined with the load impedance, which is R1 in series with the parallel combination of R2 and RL:

But now, from the point of view of the load, as shown in (e), the output impedance of the attenuator and source combined is R2 in parallel with the series combination of R1 and RS:

(The input impedance to the load is still RL.)

This output impedance is equivalent to the Thevinen equivalent impedance ZTHEV, as shown in (f). If we substitute R1 = 100 Ω, R2 = 3300 Ω, RS = 1 Ω, and VS = 10 VAC, we get the graph shown in (f). If we set the load RL = 330 Ω, using the Thevenin circuit, we find:

Compensated Attenuator

We just saw how a voltage divider could be used to attenuate a signal in a manner that is independent of frequency. However, in practice, there is always stray capacitance in a real circuit, and eventually a frequency is reached at which the voltage divider behaves like either a low- or a high-pass filter. This problem can be overcome by using a compensated attenuator circuit, as shown in Fig. 2.199.

FIGURE 2.199

At low frequencies the circuit behaves like an ordinary resistive divider, but at high frequencies the capacitive reactance dominates, and the circuit behaves like a capacitive voltage divider. The attenuation is independent of frequency, provided that:

 R1C1 = R2C2 (2.91)

In practice, one of the capacitors is usually variable, so that the attenuator can be adjusted to compensate for any stray capacitance.

Such compensated attenuators are often used at the input of an oscilloscope to raise the input resistance and lower the input capacitance so as to make the oscilloscope into a more nearly ideal voltmeter. However, this results in a decrease in sensitivity of the scope to input voltage.

2.34 Transient Circuits

Transients within circuits remove any steady-state condition. They represent a sudden change in voltage introduced by an external agent, such as a switch being thrown or a transistor switching states. During a transient, the voltages and currents throughout the circuit readjust to a new dc value in a brief but nonnegligible time interval immediately following the transient event. The initial condition is a dc circuit; the final condition is a different dc circuit; but the interval in between—while the circuit is readjusting to the new conditions—may exhibit complex behavior. The introduction of a transient into a circuit containing reactive components usually requires solving differential equations, since the response is time dependent. The following simple example is an exception to the rule (no reactances—no differential equations), but is a good illustration of transient behavior, nevertheless. Refer to Fig. 2.200.

FIGURE 2.200

Example:

1.    Initially the switch S is open. The instant it is closed (t = 0), VS is applied across R and current flows immediately, according to Ohm’s law. If the switch remains closed thereafter (t > 0), the current remains the same as it was at the instant S was closed:

2.    Initially S is closed. The instant it is open (t = 0), the voltage across and the current through the resistor go to zero. The voltage and current remain zero thereafter (t > 0), until the switch is opened again.

3.    Initially S is in position A (t < 0), and the voltage across R is V1, thus making the current I = V1/R. Neglecting the time delay for the switch element to move to position B, the instant S is thrown to position B (t = 0), the voltage immediately changes and the voltage across R is V2, thus making I = V2/R.

The preceding example may have seemed like child’s play, but there is an important thing to notice from it. Just before the transient event of switching, the voltage across and the current through the resistor were both zero (or a constant value). Immediately after the switching event, the voltage and current jumped to new levels instantaneously. The resistor’s natural response after the event was time independent as indicated by Ohm’s law, V = IR. In other words, under a forced response (source is now in the picture), the resistor voltage can change instantly to a new steady state; under a forced response, the resistor current can change instantly to a new steady state.

When we consider capacitors and inductors, a forced response (applying or removing source influence) does not result in voltages and currents instantly jumping to a new steady state. Instead, there is a natural response after the forced response where the voltage and currents vary with time. With the help of Kirchhoff’s laws, transient circuits can be modeled; the resulting differential equations take into account transient events by applying initial conditions. The following two examples, which were actually covered earlier in the capacitor and inductor sections, show what happens when voltage is suddenly applied to an RL and an RC circuit.

Example: The circuit in Fig. 2.201 can be modeled using Kirchhoff’s voltage equation:

FIGURE 2.201

This is a first-order nonhomogeneous differential equation. To solve this equation, we separate the variables and integrate:

The solution of integration:

Using the initial conditions t(0) = 0, I(0) = 0, we can determine the constant of integration:

Substituting this back into the solution, we get:

Using the component values shown in Fig. 2.201, this equation becomes:

Once the current is known, the voltage across the resistor and inductor can be easily determined:

VR = IR = VS (1 − eRt/L) = 10 V (1 − e10,000t)

The graph in Fig. 2.201 shows how these voltages change with time. The section on inductors explains some important details of the RL circuit not mentioned here and also explains how the RL deenergizing circuit works.

Example: The circuit in Fig. 2.202 can be modeled using Kirchhoff’s voltage equation:

FIGURE 2.202

The first step in solving such an equation is to eliminate the integral by differentiating each term:

This is an example of a linear first-order homogeneous differential equation. It is linear because the unknown appears only once to the first power in each term. It is first-order because the highest derivative is the first, and it is homogeneous because the right-hand side, which would contain any terms not dependent on the unknown I, is zero. The solution to all linear first-order homogeneous differential equations is of the form:

I = I0eαt

where the constant α is determined by substituting the solution back into the differential equation and solving the resulting algebraic equation:

In this case the solution is:

The constant I0 is determined from the initial condition at t = 0, at which time the voltage across the capacitor cannot change abruptly, and thus if the capacitor has zero voltage across it before the switch is closed, it will also have zero voltage immediately after the switch is closed. The capacitor initially behaves like a short circuit, and the initial current is:

Therefore, the complete solution for the transient series RC circuit for an initially discharged capacitor is:

Using the component values shown in Fig. 2.202, this equation becomes:

I = 0.001 A e−0.1t

Once the current is known, the voltage across the resistor and capacitor can be easily determined:

VR = IR = VSet/RC = 10 V e−0.1t

The graph in Fig. 2.202 shows how these voltages change with time. The section on capacitors explains some important details of the RC circuit not mentioned here and also explains how the discharging RC circuit works.

Once you’ve memorized the response equations for a charging and discharging RC as well as the equations for an energizing and a deenergizing RL circuit (see following equations), it’s often possible to incorporate these equations into transient circuits without starting from scratch and doing the differential equations.

 RC charging RC discharging RL energizing RL deenergizing VR = VSe−t/RC VR = VSe−t/RC VR = VS (1 − e−Rt/L) VR = VSe−Rt/L VC = VS (1 − e−t/RC) VC = VSe−t/RC VL = VSe−Rt/L VL = −VSe−Rt/L

The following example problems illustrate how this is done.

Example 1: The circuit in Fig. 2.203 was under steady state before the switch was opened. Determine the current flow I2 the instant the switch is opened, and then determine both I2 and the voltage across R2 1 ms after that.

FIGURE 2.203

Answer: Unlike a resistor’s voltage, a capacitor’s voltage cannot change instantaneously (instead, it takes time). Because of this, the instant the switch is opened, the capacitor voltage remains as it was prior to the event:

VC (0+) = VC (0) = 24 V

where 0+ means the instant after the switch is flipped, and 0 means the instant before the switch is flipped.

The instant after the switch is opened (t = 0+), the source is no longer part of the circuit, and we are left with the capacitor and two resistors all in series. Using Kirchhoff’s law for this new circuit:

VC + VR1 + VR2 = 0

24 V + I(10 Ω + 20 Ω) = 0

I = −24 V/30 Ω = 0.800 A

This is the same current through R2 the instant the switch is opened. To determine the current 1 ms (t = 0.001 s) after the switch is opened, simply treat the circuit as an RC discharge circuit (see Eq. 2.44), taking R = R1 + R2:

The voltage across R2 at this time is:

VR2 = IR2 = (0.573 A)(20 Ω) = 11.46 V

Example 2: In the circuit in Fig. 2.204, determine the current I when the switch is opened (t = 0). Also determine the voltage across the resistor and inductor at t = 0.1 s.

FIGURE 2.204

Answer: When the switch is opened, the inductor voltage cannot change instantaneously, so it resembles a short, which gives rise to the forced response:

where the inductor’s resistance is in series with R.

For any time after t = 0, the natural, or source-free, response is simply a deenergizing RL circuit, where R is the combined resistance of the fixed resistor and the resistance of the inductor:

In = CeRt/L = Ce−30t/5 = Ce−6t

The total current is the sum of the forced and natural responses:

I = If + In = 0.80 A + Ce−6t

The trick then is to find C by finding the current I(0+):

I(0+) = VS/RL = 24 V/10 Ω = 2.4 A

Thus:

I(t) = (0.8 + 1.60 e−6t) A

The voltage across the resistor and the inductor at t = 0.1 s can be found by first calculating the current at this time:

I(0.1) = (0.8 + 1.6e(−6 × 0.1))A = 1.68 A

Then:

VR = IR = 1.68 A × 20 Ω = 33.6 V

VL = 24 V − VR = 24 V − 33.6 V = −9.6 V

Example 3: Calculate the current IL in the circuit in Fig. 2.205 at t = 0.3 s.

FIGURE 2.205

Answer: Notice that the 12-Ω resistor has no effect on the current IL. Therefore, you get a simple RL energizing circuit with RL in series with L:

So at t = 0.3 s:

IL = (3 A)(1 − e−1.3(0.3 s)) = 0.99 A

Here are some important things to notice during a forced response in regard to resistors, capacitors, and inductors:

·    Resistor: Under a forced response, a voltage is instantly placed across a resistor and a current immediately flows. There is no delay in voltage or current response (ideally).

·    Capacitor: Under a forced response, the voltage across a capacitor cannot change instantly, so at the instant a transition occurs it acts like an open circuit or constant voltage source. The voltage at instant t = 0 or t = 0+ is a constant—the voltage that was present before the event. Also, at the instant t = 0 or t = 0+ the current is zero, since no time transpires for charge to accumulate. However, after t = 0+, the capacitor voltage and current have a natural response that is a function of time.

·    Inductor: Under a forced response, an inductor voltage cannot change instantaneously, so it acts like a short, meaning there is no voltage across it at t = 0 or t = 0+. The current, however, at t = 0 or t = 0+ will be a constant—the value of the current prior to the transient event. However, after t = 0+, the inductor voltage and current have a natural response that is a function of time.

(Recall that t = 0 means the instant prior to the transient event, and t = 0+ is the instant immediately after the transient event.)

Sometimes determining the voltage and currents within a transient circuit is a bit tricky, and requires a different approach than we encountered in the last three examples. The following few examples provide a good illustration of this.

Example 4: In the circuit in Fig. 2.206, determine the current through the inductor and the voltage across the capacitor the instant before the switch is closed (t = 0) and the instant after the switch is closed (t = 0+). Find IC and ILwhen the switch has been closed for t = 0.5 s.

FIGURE 2.206

Answer: Before the switch is closed, the capacitor acts as an open circuit, preventing current from flowing. The instant after the switch is closed, the capacitor cannot change in voltage instantaneously, so the current is still zero. The instant before and the instant after the switch is closed, the voltage across the capacitor is equal to the supply voltage. All this can be expressed by:

IL (0) = IL (0+) = 0

VC (0) = VC (0+) = 18 V

When the switch is closed, we have a source-free RC circuit for which:

IC = Bet/R1C = Bet/(1.5)

At t = 0+,

IC (0+)R1 + VC (0+) = 18 V

So,

Plugging this back in to find B we get:

So the complete expression for IC is:

IC = (−6 A)et/(1.5)

At t = 0.5 s,

IC = (−6 A)e−0.5/(1.5) = −4.3 A

To find IL at t = 0.5 s, we consider an RL circuit excited by an 18-V source (forced response) plus the natural response of an RL circuit:

Example 5: State all the initial conditions for the circuit in Fig. 2.207, which is under steady state for t < 0, and the switch is opened at t = 0. Determine the current in the circuit 0.6 s after the switch is opened. What is the induced voltage in the inductor at t = 0.4 s?

FIGURE 2.207

Answer: Since the switch is initially closed, the capacitor is shorted out, there is no voltage across it, and the current is simply equal to the supply voltage divided by the resistance of the inductor:

VC (0) = VC (0+) = 0 V

By Kirchhoff’s voltage law, we get the following equation for the circuit:

from which the characteristic equation is:

60 p2 + 20 p + 1 = 0

The characteristic roots are:

p = −0.46, −0.04

The complete current response is the sum of the forced and natural responses:

I = If + In = 0 + A1e−0.46t + A2e−0.04t

I(0+) = 4 implies that A1 + A2 = 4.

At t = 0,

Solving for A1 and A2 gives A1 = −0.38, A2 = 4.38. Thus:

I = −0.38e−0.46t + 4.38e−0.04t

The induced voltage in the inductor at t = 0.4 s is simply found by plugging I into the definition of voltage for an inductor:

At t = 0.4 s,

VL = 60(0.145 − 0.172) = −1.62 V

Example 6: In the circuit in Fig. 2.208, the switch is moved from 1 to 2 at t = 0. Determine I as a function of time thereafter.

FIGURE 2.208

Answer: The complete current response when the switch is thrown is the sum of the forced response and the natural response:

2.34.1 Series RLC Circuit

There’s another transient example, which is a bit heavy in the math but is a classic analog of many other phenomena found in science and engineering. See Fig. 2.209.

FIGURE 2.209

Assume the capacitor is charged to a voltage V0, and then at t = 0, the switch is closed. Kirchhoff’s voltage law for t ≥ 0:

Rewriting in the standard form gives:

This is an example of a linear second-order homogeneous differential equation. It is reasonable to guess that the solution is of the same form as for the first-order homogeneous differential equation encountered earlier:

I = I0eαt

Substituting this into the differential equation gives:

Note that a solution of the form eαt always reduces a linear homogeneous differential equation to an algebraic equation in which first derivatives are replaced by α and second derivates by α2, and so forth. A linear second-order homogeneous differential equation then becomes a quadratic algebraic equation, and so on. This particular algebraic equation has the following solutions:

 (2.92)

Since either value of α represents a solution to the original differential equation, the most general solution is one in which the two possible solutions are multiplied by arbitrary constants and added together:

I = I1eα1t + I2eα2t

The constants I1 and I2 must be determined from the initial conditions. An nth-order differential equation will generally have n constants that must be determined from initial conditions. In this case the constants can be evaluated from a knowledge of I(0) and dI/dt(0). Since the current in the inductor was zero for t < 0, and since it cannot change abruptly, we know that:

I(0) = 0

The initial voltage across the inductor is the same as across the capacitor, so that:

The solution for the current in the series RLC circuit is thus:

 (2.93)

where α1 and α2 are given in Eq. 2.92.

The solution to the preceding equation has a unique character, depending on whether the quantity under the square root in Eq. 2.92 is positive, zero, or negative. We consider these three cases.

Case 1: Overdamped

For R2 > 4L/C, the quantity under the square root is positive, and both values of α are negative with |α2| > |α1|. The solution is the sum of a slowly decaying positive term and a more rapidly decaying negative term of equal initial magnitude. The solution is sketched in Fig. 2.210. An important limiting case is the one in which R2 >> 4L/C. In that limit, the square root can be approximated as:

FIGURE 2.210

and the corresponding values of α are:

Then the current in Eq. 2.93 is:

 (2.94)

In this limit, the current rises very rapidly (in a time ∼L/R) to a value near V0/R and then decays very slowly back to zero. The overdamped response curve is shown in Fig. 2.210.

Case 2: Critically Damped

For R2 = 4L/C, the quantity under the square root is zero, and α1 = α2Equation 2.93 is then zero divided by zero, which is undefined. Therefore, the method of solution outlined here fails. A more productive approach is to let

and take the limit of Eq. 2.93 as ε → 0. Then:

and Eq. 2.93 becomes:

Using the expansion ex ≈ 1 + x, for |x| << 1, the preceding equation becomes:

 (2.95)

The shape of the curve is not very different from the overdamped case, except that it approaches zero as fast as possible without overshooting the t axis and going negative. The critically damped response curve is shown in Fig. 2.211.

FIGURE 2.211

Critical damping is difficult to achieve: only a small change in R moves from this point. Small temperature change will cause that to occur. Energy transfer from C to L is now smaller than loss in R.

Case 3: Underdamped

For R2 < 4L/C, the quantity under the square root is negative, and α can be written as:

where

It is useful to define another symbol, ω, which we call the angular frequency:

 (2.96)

When R2 << 4L/C, the angular frequency is:

and this approximation will usually suffice for most cases of interest. With these substitutions, Eq. 2.93 becomes:

We now make use of Euler’s equation ejθ = cos θ + j sin θ to express the current as follows:

 (2.97)

Notice that the solution is very different from the others, since it is oscillatory, with the oscillation amplitude decaying exponentially in time, as shown in Fig. 2.212. Although ω is referred to as the angular frequency, note that it has units of radians per second, and it is related to the usual frequency f, which has units of cycles per second or hertz ω = 2πf. The period of oscillation is T = 1/f = 2π/ω.

FIGURE 2.212

The underdamped case is very interesting. At t = 0, all the energy is stored in the capacitor. As the current increases, energy is dissipated in the resistor and stored in the inductor until one-quarter of a cycle has passed, at which time there is no energy left in the capacitor. But as time goes on, the energy in the inductor decreases and the energy in the capacitor increases until one-half cycle has elapsed, at which time all the energy except that dissipated in the resistor is back in the capacitor. The energy continues to slosh back and forth, until it is eventually all dissipated by the resistor. A series LC circuit without any resistance would oscillate forever without damping.

This type of differential equation that we have seen for a series RLC circuit appears in many areas of science and engineering. It is referred to as a damped harmonic oscillator. Shock absorbers on an automobile, for example, are part of a mechanical harmonic oscillator designed to be nearly critically damped.

2.35 Circuits with Periodic Nonsinusoidal Sources

Suppose that you are given a periodic nonsinusoidal voltage (e.g., a squarewave, triangle wave, or ramp) that is used to drive a circuit containing resistors, capacitors, and inductors. How do you analyze the circuit? The circuit is not dc, so you cannot use dc theorems on it. The circuit is not sinusoidal, so you cannot directly apply complex impedances on it. What do you do?

If all else fails, you might assume that the only way out would be to apply Kirchhoff’s laws on it. Well, before going any further, how do you mathematically represent the source voltage in the first place? That is, even if you could set up Kirchhoff’s equations and such, you still have to plug in the source voltage term. For example, how do you mathematically represent a squarewave? In reality, coming up with an expression for a periodic nonsinusoidal source is not easy. However, for the sake of argument, let’s pretend that you can come up with a mathematical representation of the waveform. If you plug this term into Kirchhoff’s laws, you would again get differential equations (you could not use complex impedance then, because things are not sinusoidal).

To solve this dilemma most efficiently, it would be good to avoid differential equations entirely and at the same time be able to use the simplistic approach of complex impedances. The only way to satisfy both these conditions is to express a nonsinusoidal wave as a superposition of sine waves. In fact, a man by the name of Fourier discovered just such a trick. He figured out that a number of sinusoidal waves of different frequencies and amplitudes could be added together in a special manner to produce a superimposed pattern of any nonsinusoidal periodic wave pattern. More technically stated, a periodic nonsinusoidal waveform can be represented as a Fourier series of sines and cosines, where the waveform is a summation over a set of discrete, harmonically related frequencies.

2.35.1 Fourier Series

A time-dependent voltage or current is either periodic or nonperiodic. Figure 2.213 shows an example of a periodic waveform with period T.

FIGURE 2.213

The wave is assumed to continue indefinitely in both the +t and the −t directions. A periodic function can be displaced by one period, and the resulting function is identical to the original function:

V(t ± T) = V(t)

A periodic waveform can be represented as a Fourier series of sines and cosines:

 (2.98)

where ω0 is called the fundamental angular frequency,

 (2.99)

0 is called the second harmonic, and so on. The constants an and bn are determined from:

(2.100)

(2.101)

The constant term a0/2 is the average value of V(t). The superposition theorem then allows you to analyze any linear circuit having periodic sources by considering the behavior of the circuit for each of the sinusoidal components of the Fourier series. Although most of the examples that we will use have voltage or current as the dependent variable and time as the independent variable, the Fourier methods are very general and apply to any sufficiently smooth function f(t).

Using Euler’s expression, ejθ = cos θ + j sin θ, we can convert Eq. 2.98 into a general expression for a periodic waveform as the sum of complex numbers:

 (2.102)

By allowing both positive and negative frequencies (n > 0 and n < 0), it is possible to choose the Cn in such a way that the summation is always a real number. The value of Cn can be determined by multiplying both sides of Eq. 2.102 by ejmω0t, where m is an integer, and then integrating over a period. Only the term with m = n survives, and the result is:

 (2.103)

Note that Cn is the complex conjugate of Cn, and so the imaginary parts of Eq. 2.102 will always cancel, and the resulting V(t) is real. The n = 0 term has a particularly simple interpretation; it is simply the average value of V(t):

 (2.104)

and corresponds to the dc component of the voltage. Whether the integrals in the preceding expression are over the interval −T/2 to T/2 or some other interval such as 0 to T is purely a matter of convenience, so long as the interval is continuous and has duration T.

The following example illustrates a Fourier series of a squarewave, as depicted in Fig. 2.214.

FIGURE 2.214

To create a mathematical expression in terms of a series of complex numbers, as Eq. 2.102 requests, we first determine the constants from Eq. 2.103 by breaking the integral into two parts for which V(t) is constant:

Since ω0T = 2π, the preceding equation can be written as

With the use of equation ejθ = cos Θ + j sin Θ, the preceding equation becomes

Note that cos nπ is +1 when n is even (0,2,4, …) and −1 when n is odd (1,3,5, …), so that all the even values of Cn are zero. Any periodic function, when displaced in time by half a period, is identical to the negative of the original function:

In this case V(t) is said to have half-wave symmetry, and its Fourier series will contain only odd harmonics. The squarewave is an example of such a function. If the wave remained at +V0 and −V0 for unequal times, the half-wave symmetry would be lost, and its Fourier series would then contain even as well as odd harmonics.

In addition to its half-wave symmetry, the squarewave shown in Fig. 2.214 is an odd function, because it satisfies the relation:

V(t) = −V(-t)

This property is not a fundamental property of the wave but arises purely out of the choice of where, with respect to the wave, the time origin (t = 0) is assumed. For example, if the squarewave in Fig. 2.214 were displaced by a time of T/4, the resulting squarewave would be an even function, because it would then satisfy the relation:

V(t) = V(-t)

It’s important to note that an odd function can have no dc component, since the negative parts exactly cancel the positive parts on opposite sides of the time axis. The cosine is an even function, and the sine is an odd function. Any even function can be written as a sum of cosines (bn = 0 in Eq. 2.98), and any odd function can be written as a sum of sines (an = 0 in Eq. 2.98). Most periodic functions (such as the one in Fig. 2.213) are neither odd nor even.

The Fourier series calculation can often be simplified by adding or subtracting a constant to the value of the function, or by displacing the time origin so that the function is even or odd or so that it has half-wave symmetry.

The odd-numbered coefficients of the Fourier series representation of the squarewave are given by

and the Fourier series is

With the use of Euler’s equation and the fact that sin θ = −sin(-θ) and cos θ = cos(-θ), the preceding equation becomes

The first three terms of the preceding series (n = 1,3,5) along with their sum are plotted in Fig. 2.215. Note that the series, even with as few as three terms, is beginning to resemble the squarewave.

FIGURE 2.215

For waveforms more complicated than a squarewave, the integrals are more difficult to perform, but it is still usually easier to calculate a Fourier series for a periodic voltage than to solve a differential equation in which the same time-dependent voltage appears. Furthermore, tables of Fourier series for the most frequently encountered waveforms are available and provide a convenient shortcut for analyzing many circuits. Some common waveforms and their Fourier series are listed in Fig. 2.216.

FIGURE 2.216

Example: Squarewave RC Circuit. The following example demonstrates the use of the Fourier series to analyze a circuit with a periodic source. Here a squarewave source is connected to a simple RC circuit. Refer to Fig. 2.217.

FIGURE 2.217

Since the source is periodic, the current I(t) is also periodic with the same period, and it can be written as a Fourier series:

Each Cn is a phasor current representing one frequency component of the total current in the same way that each Cn represents a component of the phasor voltage in the previous section. The relationship between the two phasors is determined by dividing by the circuit impedance:

Substituting the value of Cn derived earlier for the squarewave gives:

for n odd. For n even, Cn is zero, since Cn is zero for even n. The corresponding current is then

With the use of Euler’s relation, the preceding current can be written as:

The voltage across the resistor and capacitor can be determined from the definitions of an ideal resistor and an ideal capacitor:

The sum of the first three terms (n = 1, n = 3, n = 5) of the Fourier series for VC(t) and VR(t) are shown in Fig. 2.217b. As n approaches infinity, the waveforms approach the real deal.

Note that circuits with squarewave sources can also be analyzed as transient circuits. In the circuit in Fig. 2.217a, during a half period (such as 0 < t < T/2) when the source voltage is constant, the voltage across the capacitor is expressed as:

VC(t) = A + Bet/RC

The constants A and B can be determined from

VC (∞) = A = V0

VC (T/2) = A + BeT/2RC = −VC(0) = −A − B

The first equation results from knowing that if the source remains at +V0 forever, the capacitor would charge to voltage V0. The second equation is required to ensure that the function has half-wave symmetry. Hence,

The capacitor voltage is then:

for 0 < t < T/2. The waveform repeats itself for t > T/2 with each half cycle alternating in sign.

2.36 Nonperiodic Sources

Nonperiodic voltages and currents can also be represented as a superposition of sine waves as with the Fourier series. However, instead of a summation over a set of discrete, harmonically related frequencies, the waveforms have a continuous spectrum of frequencies. It is possible to think of a nonperiodic function as a periodic function with an infinite period. The fundamental angular frequency, which was ω0 = 2π/T for the Fourier series, approaches zero as the period approaches infinity. In this case, to remind us that we’re now dealing with an infinitesimal quantity, we represent the angular frequency as Δω. The various harmonics within the waveform are separated by the infinitesimal Δω, so that all frequencies are present. The waveform as a summation, as was done with Fourier series, is:

where we have used the fact that ω = nω0 and TΔω = 2π. Since Δω is infinitesimal, the preceding summation can be replaced with an integral (dω = Δω):

As before, Cn is given by

However, since T is infinite, we can write

Although T is infinite, the term CnT is usually finite in value. The CnT term is referred to as the Fourier transform of V(t) and is rewritten as (ω). After integration, it is only a function of angular frequency ω. The following two equations are called a Fourier transform pair:

(2.105)

(2.106)

These two equations are symmetric. ((ω) is sometimes defined as  to make the symmetry even more perfect.)

Like the coefficients of the Fourier series, the Fourier transform (ω) is generally complex, unless V(t) is an even function of time. When V(t) is an odd function of time, the Fourier transform (ω) is entirely imaginary. For this reason, when plotting a Fourier transform, it is customary to plot either the magnitude |(ω)| or the square of the magnitude |(ω)|2—referred to as the power spectrum—as a function of ω.

As an example, we will calculate the Fourier transform of the square pulse shown in Fig. 2.218a and given by:

FIGURE 2.218

From Eq. 2.106, the Fourier transform is:

The magnitude |(ω)| is plotted as a function of ω in Fig. 2.218b. As before, most of the Fourier spectrum is a band of frequencies within about 1/τ of zero.

In practice, if we were to attach a nonperiodic pulse, such as our square pulse, to some complex circuit with total impedance Z(ω), we could determine the current as a function of time (and thus individual voltages and currents within the circuit) by first using the Fourier transform voltage for the square pulse,

and using this to find the Fourier transform of the current by dividing by the impedance:

 (2.107)

Once this is found, the current as a function of time could be determined using the inverse Fourier transform:

Using the Fourier transforms for a problem like this seems incredibly difficult—try placing a simple RCL network into the impedance and solving the integrals. However, the Fourier transform, as nasty as it can get, provides the easiest method of solution for the nonperiodic problems.

In summary, analyzing a circuit by this method involves first converting to the frequency domain by calculating the Fourier transform of the sources from Eq. 2.106, then using the impedance to determine the Fourier transform of the unknown Eq. 2.107, and finally converting back to the time domain by calculating the inverse Fourier transform of the unknown from Eq. 2.105. Solving things this way, using the difficult integrals, is actually usually much easier than solving the corresponding differential equation with a time-dependent source.

Note that special devices called spectrum analyzers can display the Fourier transform |(ω)| of a voltage as a function of frequency.

Enough of the difficult stuff; let’s let a simulator do the thinking for us.

2.37 SPICE

SPICE is a computer program that simulates analog circuits. It was originally designed for the development of integrated circuits, from which it derives its name: Simulation Program with Integrated Circuit Emphasis, or SPICE for short.

The origin of SPICE is traced back to another circuit simulation program called CANCER (Computer Analysis of Non-Linear Circuits Excluding Radiation), developed by Ronald Rohrer, of University of California–Berkeley, along with some of his students. CANCER was able to perform dc, ac, and transient analysis, and included special linear-companion models for basic active devices like diodes (Shockley equation) and bipolar transistors (Ebers-Moll equations).

When Rohrer left Berkeley, CANCER was rewritten and renamed SPICE, released as version 1 to the public in 1972. SPICE 1 was based on nodal analysis and included revised models for bipolar transistors (using Gummel-Poon equations), as well as new models for JFET and MOSFET devices.

In 1975, SPICE 2 was introduced, with modified nodal analysis (MNA) that replaced the old nodal analysis, and now supported voltage sources and inductors. Many things were added and many alterations were made to SPICE 2. The last version of SPICE 2 to be written in FORTRAN, version SPICE 2G.6, came out in 1983.

In 1985, SPICE 3 appeared on the scene, written in the C programming language rather than FORTRAN. It included a graphical interface for viewing results and also polynomial capacitors and inductors and voltage-controlled sources, as well as models for MESFETs, lossy transmission lines, and nonideal switches. SPICE 3 also had improved semiconductor models and was designed to eliminate many convergence problems found in previous versions. From this time on, commercial versions of SPICE have appeared: HSPICE, IS_SPICE, and MICROCAP and PSPICE (MicroSim’s PC version of SPICE).

Today, there are many user-friendly simulator programs out there that use SPICE as the brains behind the analysis. These high-level simulator programs allow you to click, drag, and drop components onto a page and draw wire connections. Test instruments, such as voltmeters, power meters, oscilloscopes, and spectrum analysis, can be dragged and connected to the circuit. Almost any type of source is available as well as any type of device (passive, active, digital, etc.)—MultiSim from Electronics Workbench contains a component library of over 13,000 models. Figure 2.219 shows an example screen shot depicting the various elements that come with a simulator. Three popular commercial simulators include MicroSim, TINAPro, and CircuitMaker. There are also online simulation programs such as CircuitLab (www.circuitlab.com) that are quick and easy to use.

FIGURE 2.219 Screen shot of a simulator user interface, showing the various devices, sources, test equipment, and analysis capabilities. The simulator shown is MultiSim from Electronics Workbench.

Basic devices: Passive devices, diodes, LEDs, thyristors, transistors, analog amplifiers and comparators, TTL logic devices, CMOS logic devices, miscellaneous digital (e.g., TIL, VHDL, VERILOG HDL), mixed signal devices, indicators, RF devices, electromechanical devices, and so on.

Sources: dc and ac voltage and current sources; clock source; AM source and FM voltage and current sources; FSK source; voltage-controlled sine, square, and triangle sources as well as current-controlled ones; pulse voltage and current sources; exponential voltage and current sources; piecewise linear voltage and current sources; controlled one-shot, polynomial, and nonlinear dependent sources.

Analysis techniques: dc operating point, ac analysis, transient analysis, Fourier analysis, noise analysis, distortion analysis, dc sweep, sensitivity, parameter sweep, temperature sweep, pole zero, transfer function, worst case, Monte Carlo, trace width analysis, batched analysis, user-defined analysis, noise figure analysis, and so on.

Test equipment: Multimeter, function generator, wattmeter, oscilloscope, bode plotter, word generator, logic analyzer, logic converter, distortion analyzer, spectrum analyzer, network analyzer, and so on.

2.37.1 How SPICE Works

Here we take a look at the heart of simulators—SPICE. What are the mathematical tricks the code in SPICE uses to simulate electrical circuits described by nonlinear differential equations? At the core of the SPICE engine is a basic technique called nodal analysis. It calculates the voltage at any node, given all of the circuit resistances (or, inversely, their conductances) and the current sources. Whether the program is performing dc, ac, or transient analysis, SPICE ultimately casts its components (linear, nonlinear, and energy-storage elements) into a form where the innermost calculation is nodal analysis.

Kirchhoff discovered that the total current entering a node is equal to the total current leaving a node. Stated another way, the sum of currents in and out of a node is zero. These currents can be described by equations in terms of voltages and conductances. If you have more than one node, then you get more than one equation describing the same system (simultaneous equations). The trick then is finding the voltage at each node that satisfies all the equations simultaneously.

For example, let’s consider the simple circuit in Fig. 2.220. In this circuit there are three nodes: node 0 (which is always ground), node 1, and node 2.

FIGURE 2.220

Using Kirchhoff’s current law in the form of “the sum of current in and out of a node is zero,” we can create two equations for the two nodes 1 and 2 (0 is by default ground):

Since the mission here is to calculate the node voltages, we rewrite these equations in order to separate V1 and V2:

The trick now becomes finding values of V1 and V2 that satisfy both equations. Although we could solve one variable in terms of the other from one equation and plug it into the second equation to find the other variable, things get messy (so many R variables to deal with). Instead, we’ll take a cleaner approach—one that involves conductance G, where G = 1/R. This will make the bookkeeping easier, and it becomes especially important when the nodal number increases with complex circuits.

Writing resistors in terms of total conductance:

G11 = 1/R1 + 1/R2G21 = −1/R2G12 = −1/R2G22 = 1/R2 + 1/R3.

Thus, the system of equations is transformed into:

G11V1 + G12V2 = IS

G21V1 + G22V2 = 0

Solving the second equation for V1:

we then stick this into the first equation and solve for V2:

See how much cleaner the manipulations were using conductances? V2 is described by circuit conductances and IS alone. We still must find the numerical value of V2 and stick it back into the V1 equation while, in the process, calculating the numerical values of the conductances. However, we have found circuit voltages V1 and V2 that satisfy both system equations.

Although the last approach wasn’t that hard, or messy, there are times when the circuits become big and the node and device counts so large that bookkeeping terms become a nightmare. We must go a step further in coming up with a more efficient and elegant approach. What we do is use matrices.

In matrix form, the set of nodal equations is written:

Or in terms of total conductances and source currents:

 (2.108)

Treating each matrix as a variable, we can rewrite the preceding equation in compact form:

 G · v = i (2.109)

In matrix mathematics, you can solve for a variable (almost) as you would in any other algebraic equation. Solving for v you get:

 v = G−1 · i (2.110)

where G−1 is the matrix inverse of G. (1/G does not exist in the matrix world.) This equation is the central mechanism of the SPICE algorithm. Regardless of the analysis—ac, dc, or transient—all components or their effects are cast into the conductance matrix G and the node voltages are calculated by v = G−1 · i, or some equivalent method.

Substituting the component values present in the circuit in Fig. 2.218 into the conductance matrix and current matrix (we could use an Excel spreadsheet and apply formulas to keep things clean), we get:

Hence, the voltage is:

V1 = 9.9502 and V2 = 4.9751

2.37.2 Limitations of SPICE and Other Simulators

Simulation of a circuit is only as accurate as the behavior models in the SPICE devices created for it. Many simulations are based on simplified models. For more complex circuits or subtle behaviors, the simulation can be misleading or incorrect. This can mean disaster if you’re relying entirely on SPICE (or an advanced simulator based on SPICE) when developing circuits. SPICE can also be deceiving, since simulations are free of noise, crosstalk, interference, and so on—unless you incorporate these behaviors into the circuit. Also, SPICE is not the best predictor of component failures. You must know what dangers to look for and which behaviors are not modeled in your SPICE circuit. In short, SPICE is not a prototype substitute—the performance of the actual breadboard provides the final answer.

2.37.3 A Simple Simulation Example

As an example, we will look at simulating a simple RLC crossover network (which we will meet again in Chap. 15) with the free and easy-to-use online simulator CircuitLab (www.circuitlab.com).

The first step is to draw the schematic using the editing tool (see Fig. 2.221). The LRC network is responsible for separating an audio signal into two parts: a low-pass filter to drive the low-frequency woofer speaker and a high-pass filter to drive the tweeter.

FIGURE 2.221 CircuitLab schematic diagram.

As well as adding the appropriate components to the design and connecting them, we have added an ac voltage source to represent the input signal and attached two labels (Tweeter and Woofer), so that we can see the results of the simulation (there will be labels next to our graph plots).

When we are ready, we can run a simulation. In this case, we are going to carry out a frequency-domain simulation so that we can determine the crossover frequency of the network. To do this, we specify V1 as being the input as well as start and end frequencies. We also specify the outputs that we wish to see plotted (see Fig. 2.222).

FIGURE 2.222 CircuitLab simulation parameters.

When we click the Simulate button, CircuitLab will run the simulation and produce an output plot, as shown in Fig. 2.223.

FIGURE 2.223 CircuitLab simulation results.

From Fig. 2.223, we can see that the crossover frequency where both speakers are receiving the same magnitude of signal is at around 1.6 kHz.

CircuitLab is a good starting point for understanding simulators. We strongly encourage you to try some of the sample schematics on the website and experiment with a few simulations.

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