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 flowthrough the circuit would jump immediately from zero to V/R, according to Ohm’slaw. However, according to Faraday’s law of electromagnetic induction, this isn’tentirely accurate. Instead, when the switch is initially closed, the current increasesrapidly. As the current increases with time, the magnetic flux through the loop risesrapidly. This increasing magnetic flux then induces an EMF in the circuit that opposesthe current flow, giving rise to an exponentially delayed rise in current. We call theinduced EMF a self-induced EMF.

EMF , , EMFddt B dA N ddtMMM  ∫= − Φ Φ = ⋅ = − Φ (for a coiled wire of N loops) (2.49)where ΦM is the magnetic flux threading through a closed- loop circuit (which is equalto the magnetic field B dotted with the direction surface area—both of which are vec-tors, and summed over the entire surface area—as the integral indicates). Accordingto the law, the EMF can be induced in the circuit in several ways: (1) the magnitudeof B can vary with time; (2) the area of the circuit can change with time; (3) the anglebetween B and the normal of A can change with time; (4) any combination of thesecan occur. See Fig. 2.116.The simple ac generator in Fig. 2.117 shows Faraday’s law in action. A simplerotating loop of wire in a constant magnetic field generates an EMF that can be usedto power a circuit. As the loop rotates, the magnetic flux through it changes withtime, inducing an EMF and a current in an external circuit. The ends of the loop areconnected to slip rings that rotate with the loop, while the external circuit is linked tothe generator by stationary brushes in contact with the slip rings.A simple dc generator is essentially the same as the ac generator, except that thecontacts to the rotating loop are made using a split ring or commutator. The result isa pulsating direct current, resembling the absolute value of a sine wave—there are nopolarity reversals.A motor is essentially a generator operating in reverse. Instead of generating a cur-rent by rotating a loop, a current is supplied to the loop by a battery, and the torqueFIGURE 2.116 Illustration of Faraday’s law of induction.

2.24 Inductors

2.23.15 Quality Factor

2.23.14 Capacitive Divider

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 thefrequency goes to 0, XC goes to infinity, and the capacitor acts like an open circuit;capacitors do not like to pass low frequencies

Within an ac circuit, the amount of charge movesback 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 capacitanceand frequency are considered together, they form a quantity similar to resistance.However, since no actual heat is being generated, the effect is termed capacitive reac-tance. The unit for reactance is the ohm, just as for resistors, and the formula for cal-culating the reactance of a capacitor at a particular frequency is given by:121X fC CC = π = ω (2.47) Capacitive reactance

2.23.12 Alternating Current in a Capacitor

When two or more capacitors are connected in series, the total capacitance is less thanthat of the smallest capacitor in the group. The equivalent capacitance is similar toresistors in parallel:1 1 1 1tot 1 2C C C CN

When capacitors are placed in parallel, their capacitances add, just like resistors inseries:Ctot = C1 + C2 + ... Cn

Intuitively, you can think of capacitors in parallel representing one single capaci-tor with increased plate surface area. It’s important to note that the largest voltagethat can be applied safely to a group of capacitors in parallel is limited to the volt-age rating of the capacitor with the lowest voltage rating.

Current and voltage equations for discharging an RC circuit∫= = −  = = = −  = = = −  τ =−−−I VR e tRCIRVV IR V e tRCVVV C Idt V e tRCVVRCs t RCsR s t RC RsC s t RC Cs, ln, ln1 , lntime constant///0(2.44)where I is the current in amps, VS is the source voltage involts, R is the resistance in ohms, C is the capacitance in far-ads, t is the time in seconds after the source voltage isremoved, e = 2.718, VR is the resistor voltage in volts, and VCis the capacitor voltage in volts. Graph shown to the left inFig. 2.103 is for circuit with R = 3 kΩ and C = 0.1 μF.

Current and voltage equations for RC charging circuit∫= = −  = = = −  = = − = − − τ =−−−I VR e tRCIRVV IR V e tRCVVV C Idt V e tRCV VVRCs t RCsR s t RC RsC s t RC s Cslnln1 (1 ) lntime constant///(2.43)where I is the current in amps, VS is the source voltage involts, R is the resistance in ohms, C is the capacitance in far-ads, t is the time in seconds after the source voltage is ap-plied, e = 2.718, VR is the resistor voltage in volts and VC is thecapacitor voltage in volts. Graph shown to the left is for cir-cuit with R = 10 kΩ, and C = 100 μF. Decreasing the resistancemeans the capacitor charges up more quickly and the voltageacross the capacitor rises more quickly.

E VIdt VC dVdt dt CVdV CV∫ ∫∫= = = = (2.42)

V C I dtC C∫= (2.41) Voltage across capacitor

t’s important to note that these equations are representative of what’s definedas an ideal capacitor. Ideal capacitors, as the equation suggests, have several curiousproperties that are misleading if you are dealing with real capacitors. First, if weapply a constant voltage across an ideal capacitor, the capacitor current would bezero, since the voltage isn’t changing (dV/dt = 0). In a dc circuit, a capacitor thus actslike an open circuit. On the other hand, if we try to change the voltage abruptly, from0 to 9 V, the quantity dV/dt = 9 V/0 V = infinity and the capacitor current would haveto be infinite (see Fig. 2.98). Real circuits cannot have infinite currents, due to resistiv-ity, available free electrons, inductance, capacitance, and so on, so the voltage acrossthe capacitor cannot change abruptly. A more accurate representation of a real capaci-tor, considering construction and materials, looks like the model shown in Fig. 2.93.

2.23.4 Maxwell’s Displacement Current

2.23.3 Voltage Rating and Dielectric Breakdown

The ratio of charge on one of the plates of a capacitor to the voltage that existsbetween the plates is called capacitance (symbolized C):C QV= (2.32) Capacitance relatedto charge and voltageC is always taken to be positive, and has units of farads (abbreviated F). One farad isequal to one coulomb per volt:1 F = 1 C/1 V

2.23 Capacitors

2.22 Mains Power

Conversion Factors for AC Voltage and CurrentFROM TO MULTIPLY BYPeak Peak-to-peak 2Peak-to-peak Peak 0.5Peak RMS 1/ 2 or 0.7071RMS Peak 2 or 1.4142Peak-to-peak RMS 1/( 2 2 ) or 0.35355RMS Peak-to-peak (2 × 2 ) or 2.828Peak Average* 2/o or 0.6366Average* Peak o/2 or 1.5708RMS Average* (2 × 2 )/o or 0.9003Average* RMS o/( 2 × 2 ) or 1.1107* Represents average over half a cycle.

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, withoutthe scary calculus:

The measurement that is used instead of the average value is the RMS or rootmean square value, which is found by squaring the instantaneous values of the acvoltage or current, then calculating their mean (i.e., their average), and finally takingthe 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 acequivalents of dc voltages and currents. The RMS values of ac voltage and currentare based upon equating the values of ac and dc power required to heat a resistiveelement to exactly the same degree

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:I t V tRVR f tP( ) ( ) sin(2

2.20.8 Phase

2.20.7 Frequency and Period

Amplitude

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

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 isvital to digital electronics, where states are either true (on) or false (off). Triangularand ramp waveforms—sometimes called sawtooth waves—are especially useful intiming circuits. As we’ll see later in the book, using Fourier analysis, you can createany desired shape of periodic waveform by adding a collection of sine waves together.

Besides combining ac and dc voltages and currents, we can also combine separateac voltages and currents. Such combinations will result in complex waveforms.Figure 2.81 shows two ac waveforms fairly close in frequency, and their resultantcombination. The figure also shows two ac waveforms dissimilar in both frequencyand wavelength, along with the resultant combined waveform.FIGURE 2.80FIGURE 2.81 (Left) Two ac waveforms of similar magnitude and close in frequency form a composite wave. Note thepoints where the positive peaks of the two waves combine to create high composite peaks: this is the phenomenon ofbeats. The beat note frequency is f2 − f1 = 500 Hz. (Right) Two ac waveforms of widely different frequencies and amplitudesform a composite wave in which one wave appears to ride upon the other.

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

The most common way to generate sinusoidal waveforms is by electromagneticinduction, by means of an ac generator (or alternator). For example, the simple acgenerator in Fig. 2.78 consists of a loop of wire that is mechanically rotated aboutan axis while positioned between the north and south poles of a magnet. As theloop rotates in the magnetic field, the magnetic flux through it changes, and chargesare 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 ofthe loop relative to the direction of the magnetic field. The resultant induced voltageis sinusoidal, with angular frequency ω (radians per second).Real ac generators are, of course, more complex than this, but they operate underthe same laws of induction, nevertheless. Other ways of generating ac include usinga transducer (e.g., a microphone) or even using a dc-powered oscillator circuit thatuses special inductive and capacitive effects to cause current to resonate back andforth between an inductor and a capacitor.Why Is AC Important?There are several reasons why sinusoidal waveforms are important in electronics. Thefirst obvious reason has to do with the ease of converting circular mechanical motioninto induced current via an ac generator. However, another very important reason forusing sinusoidal waveforms is that if you differentiate or integrate a sinusoid, you geta sinusoid. Applying sinusoidal voltage to capacitors and inductors leads to sinusoi-dal current. It also avoids problems on systems, a subject that we’ll cover later. But oneof the most important benefits of ac involves the ability to increase voltage or decreasevoltage (at the expense of current) by using a transformer. In dc, a transformer isuseless, and increasing or decreasing a voltage is a bit tricky, usually involving someFIGURE 2.78 Simple ac generator.

Another type of source that is fre-quently used in electronics is an alternating source that causes current to periodicallychange direction, resulting in an alternating current (ac). In an ac circuit, not onlydoes the current change directions periodically, the voltage also periodically reverses.

Norton’s theorem is another tool for analyzing complex networks. Like Thevenin’stheorem, it takes a complex two-terminal network and replaces it with a simple equiv-alent circuit. However, instead of a Thevenin voltage source in series with a Theveninresistance, the Norton equivalent circuit consists of a current source in parallel with aresistance—which happens to be the same as the Thevenin resistance.

Luckily, a man by the name of Thevenin came up with a theorem, or trick, tosimplify 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 thecircuit can be considered a black box. Now the only things left to work with are thesetwo 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 berepresented by a voltage source in series with a resistor. (This statement is referredto as Thevenin’s theorem.) The voltage source is called the Thevenin voltage VTHEV, andthe resistance is called the Thevenin resistance RTHEV; together, the two form what iscalled the Thevenin equivalent circuit. From this simple equivalent circuit you caneasily calculate the current flow through a load placed across its terminals by usingOhm’s law: I = VTHEV/(RTHEV + RLOAD)

The superposition is also an underlying mechanism that makespossible two important circuit theorems: Thevenin’s theorem and Norton’s theorem.These two theorems, which use some fairly ingenious tricks, are much more practicalto 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 analysistools rest.FIGURE 2.70 The circuit in (a) can be analyzed using the superposition theorem by considering the simpler circuits in(b) and (c).

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

Kirchhoff’s Voltage Law (or Loop Rule): The algebraic sum of the voltages aroundany loop of a circuit is zero:V V V Vnclosed path0

Kirchhoff’s Current Law (or Junction Rule): The sum of the currents that enter a junc-tion equals the sum of the currents that leave the junction:I Iin out∑∑ = (2.21)Kirchhoff’s current law is a statement about the conservation of charge flow througha circuit: at no time are charges created or destroyed.

Often, you will run across a circuit that cannot be analyzed with simple resistor cir-cuit reduction techniques alone. Even if you could find the equivalent resistance byusing circuit reduction, it might not be possible for you to find the individual cur-rents 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, aswell as the current and voltage divider equations, may not cut it. For this reason, weturn to Kirchhoff’s laws

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

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 areplaced end to end or in series; the terminal voltages of each battery add togetherto give a final supply voltage equal to the sum from the batteries. To create a sup-ply with added current capacity (increased operating time), batteries can be placed inparallel—positive terminals are joined together, as are negative terminals, as shown inFig. 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 net-work. Note that the ground symbol shown here acts simply as a 0-V reference, not asan actual physical connection to ground. Rarely would you ever connect a physicalground to a battery-operated device.

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

Intheory, an ideal voltmeter should draw no current as it measures a voltage betweentwo points in a circuit; it has infinite input resistance Rin. Likewise, an ideal ammetershould introduce no voltage drop when it is placed in series within the circuit; it haszero input resistance. An ideal ohmmeter should provide no additional resistancewhen making a resistance measurement.

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 sup-ply an infinite amount of current.

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 percentrule and Fig. 2.49 to construct the voltage divider.

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

The 10 Percent Rule: This rule is a standardmethod for selecting R1 and R2 that takes into accountthe load and minimizes unnecessary power losses inthe divider.The first thing you do is select R2 so that I2 is 10percent of the desired load current. This resistance andcurrent are called the bleeder resistance and bleeder cur-rent. The bleeder current in our example is:Ibleed = I2 = (0.10)(9.1 mA) = 0.91 mAUsing Ohm’s law, next we calculate the bleederresistance:Rbleed = R2 = 3 V/0.00091 A = 3297 ΩConsidering resistor tolerances and standard resistancevalues, we select a resistor in close vicinity—3300 Ω.Next, we need to select R1, so that the output ismaintained at 3 V.To do this, simply calculate the total currentthrough the resistor and use Ohm’s law:I1 = I2 + Iload = 0.91 mA + 9.1 mA = 10.0 mA = 0.01 A10 V 3 V0.01 A 7001R − = Ω

V IR VR R RV IR VR R R1 1in1 212 2in1 22= = + ×= = + ×(Voltage dividerequations)These equations are called voltage divider equations and are so useful in electronics thatit is worth memorizing them. (See Fig. 2.43.) Often V2 is called the output voltage Vout.

Example 1: If a 1000-Ω resistor is connected in parallel with a3000-Ω resistor, what is the total or equivalent resistance? Alsocalculate total current and individual currents, as well as thetotal and individual dissipated powers.R R RR R1000 30001000 30003, 000, 0004000 750total1 21 2= ×+ = Ω × ΩΩ + Ω = ΩΩ = ΩTo find how much current flows through each resistor, applyOhm’s law:I VRI VR12 V1000 0.012 A 12 mA12 V3000 0.004 A 4 mA111222= = Ω = == = Ω = =These individual currents add up to the total input current:Iin = I1 + I2 = 12 mA + 4 mA = 16 mAThis statement is referred to as Kirchhoff’s current law. Withthis law, and Ohm’s law, you come up with the current divider

Rtotal = R1 + R2 + R3 + R4 + ...

R R R R R1total 11 12 13 14 = + + + +

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

The single-point ground concept ensures that no ground loops are created. As thename implies, all circuit grounds are returned to a common point. While this approachlooks good on paper, it is usually not practical to implement. Even the simplest cir-cuits can have 10 or more grounds. Connecting all of them to a single point becomesa nightmare. An alternative is to use a ground bus.A ground bus, or bus bar found in breadboards and prototype boards, or whichcan be etched in a custom printed circuit board (PCB), serves as an adequate substi-tute for a single-point ground. A bus bar is simply a heavy copper wire or bar of lowresistance that can carry the sum total of all load currents back to the power supply.This bus can be extended along the length of the circuitry so that convenient con-nections can be made to various components spaced about the board. Figure 2.35eshows a bus bar return. Most prototype boards come with two or three lines of con-nected terminals extending along the length of the board. One of these continuousstrips should be dedicated as a circuit ground bus. All circuit grounds should betied directly to this bus. Care must be taken to make sure that all lead and wire con-nections to the bus are secure. For prototype boards, this means a good solder joint;for wire-wrap board, a tight wire wrap; for breadboard, proper gauge wire leads tosecurely fit within sockets. Bad connections lead to intermittent contact that leadsto noise.

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

For example, in Fig. 2.35a when a load circuit uses a metal enclosure asa chassis ground, resistive leakage paths (unwanted resistive paths) can exist, whichresult in high voltages between the enclosure and earth ground. If, inadvertently, anearth-grounded object, such as a grounded metal pipe, and the circuit chassis aresimultaneously touched, a serious shock will result. To avoid this situation, the chas-sis is simply wired to an earth ground connection, as shown in Fig. 2.35b. This placesthe metal pipe and the enclosure at the same potential, eliminating the shock hazard.

These two terminals are connected together, forming a common return pathfor load current, as shown in Fig. 2.32. The connection between the negative and thepositive terminals of the supplies results in a common or floating return. The floatingcommon may be connected to the earth ground terminal of a supply, if a particularcircuit requests this. Generally, it will neither help nor hinder circuit performance.

As previously mentioned, the ground symbol, in many cases, has been used as ageneric symbol in circuit diagrams to represent the current return path, even thoughno physical earth ground is used. This can be confusing for beginners when theyapproach a three-terminal dc power supply that has a positive (+), negative (−), andground terminal. As we have learned, the ground terminal of the supply is tied to thecase of the instrument, which in turn is wired to the mains earth ground system. Acommon 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 nocurrent will flow from the source; hence, the load current will be zero. The correctprocedure, of course, is to either connect the load between the positive and negativeterminals directly, thus creating a floating load, or, using a jumper wire between theground and negative supply, create a grounded load. Obviously, many dc circuits don’tneed to be grounded—it will generally neither help nor hinder performance (e.g.,battery-powered devices need no such connection).Circuits that require both positive and negative voltage require a power supplyto provide each polarity. The supply for the positive voltage will have the negativeFIGURE 2.31

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

Wiresize is expressed in gauge number—the common standard being the American WireGauge (AWG)—whereby a smaller gauge number corresponds to a larger-diameterwire (high current capacity).

For example, Fig. 2.28 shows various ways in which to define voltages by select-ing a ground—which in this case is simply a 0-V reference marker.

Example: Why shouldn’t you connect a wire across a voltage source? For example, ifyou connect a 12-gauge wire directly across a 120-V source (120-V mains outlet), whatdo 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 meltingthe wire and perhaps in the process receiving a nasty shock (if the wire isn’t insu-lated). But more likely, your circuit breaker in the home will trip, since the wire willdraw a huge current due to its low resistance—breakers trip when they sense a largelevel 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 orblow a fuse, or in a bad supply, ruin the inner circuitry. In the case of a battery, there isinternal resistance in the battery, which will result in heating of the battery. There willbe less severe levels of current due to the internal resistance of the battery, but thebattery will soon drain, possibly even destroying the battery, or in an extreme casecausing the battery to rupture.