An op amp consisting of a complex arrangement of resistors, transistors, capacitors, and diodes. Here, we ignore the details.

Transcription

1 CHAPTER 5 Operational Amplifiers In this chapter, we learn how to use a new circuit element called op amp to build circuits that can perform various kinds of mathematical operations. Op amp is a building block of modern electronic instrumentation. Therefore, mastery of operational amplifier fundamentals is paramount to any practical application of electronic circuits. They are popular in practical circuit degigns because they are versatile, inexpensive, easy to use, and fun to work with Introduction to Op Amp Operational amplifiers (or Op Amp) is an active circuit element that can perform mathematical operations (e.g., amplification, summation, subtraction, multiplication, division, integration, differentiation) between signals. The ability of the op amp to perform these mathematical operations is the reason it is called an operational amplifier. It is also the reason for the widespread use of op amps in analog design An op amp consisting of a complex arrangement of resistors, transistors, capacitors, and diodes. Here, we ignore the details. 59

2 V CC OPERATIONAL AMPLIFIERS The circuit symbol for the op amp is shown below. Inverting input 2 Noninverting input 3 6 Output V CC Offset Null It has two inputs and one output. The inputs are marked with minus (-) and plus () to specify inverting and noninverting 1 inputs, respectively As an active element, the op amp must be powered by a voltage supply: 2 3 i - i 7 4 i 7 i 4 i o 6 V CC V CC Although, in this class, the power supplies are often ignored in op amp circuit diagrams for the sake of simplicity, the power supply currents must not be overlooked: i o = i 7 i 4 i i. Caution: Even when pins 7 and 4 are not shown explicitly, they are always there and the corresponding currents are also there. 1 An input applied to the noninverting terminal will appear with the same polarity at the output, while an input applied to the inverting terminal will appear inverted at the output.

3 5.1. INTRODUCTION TO OP AMP The equivalent circuit of non-ideal op amp is shown below. Note that the output section consists of a voltage-controlled source in series with the output resistance R o. v - v d R i R o v o Av d v Input-output relations: where v o = Av d = A(v v ) v o = voltage between the output terminal and ground v = voltage between the inverting terminal and ground v = voltage between the noninverting terminal and ground v d = v v = differential input voltage A = open-loop voltage gain In words, the op amp senses the difference between the two inputs, multiplies it by the gain A, and causes the resulting voltage to appear at the output. A is called the open-loop voltage gain because it is the gain of the op amp without any external feedback from output to input The concept of feedback is crucial to our understanding of op amp circuits. A negative feedback is achieved when the output is fed back to the inverting terminal of the op amp.

4 62 5. OPERATIONAL AMPLIFIERS Example Consider the circuit below. There is a feedback path from output to input. The ratio of the output voltage to the input voltage is called the closed-loop gain. R f R f v i R 1 v o v i R 1 v d R o R i Av d v o Typical ranges for op amp parameters are shown in the following table Working with a nonideal op amp is tedious because it involves dealing Parameter Typical range Ideal values Open-loop gain, A 10 5 to 10 8 Input resistance, R i 10 5 to Ω Ω Output resistance, R o 10 to 100 Ω 0 Ω Supply voltage, V CC 5 to 24 V with very large numbers. Example Consider, again, the circuit in Example Suppose R 1 = 10 kω and R f = 20 kω. Assume that the op amp has an open-loop voltage gain of , input resistance R i of 2 MΩ, and output resistance R o of 50 Ω. Find the closed-loop gain v o /v i. It can be shown that the closed-loop gain is almost insensitive to the open-loop gain A of the op amp. For this reason, op amps are used in circuits with feedback paths.

5 5.2. IDEAL OP-AMP Ideal Op-Amp To facilitate understanding, we assume ideal op amps with the ideal values above. Definition An ideal op amp is an amplifier with infinite openloop gain, infinite input resistance, and zero output resistance. Unless stated otherwise, we will assume from now on that every op amp is ideal Two important characteristics of the ideal op-amp: (a) The current into both input terminals are zero. i = 0 and i = 0. (b) The voltage across the input terminals is negligibly small. or v d = v v 0 v v. i - = 0 v - i = 0 v = v - i o v d v o Caution:

6 64 5. OPERATIONAL AMPLIFIERS Example An ideal op amp is used in the circuit below. Find the closed-loop gain 2 v o /v s. Determine current i o when v s = 1 V. i 0 v s 5 kω 40 kω 20 kω v o 2 closed-loop gain = ratio of the output voltage to the input voltage.

7 5.3. INVERTING AMPLIFIER Inverting Amplifier Op amp can be used in circuits as modules for creating more complex circuits. The first of such op-amp circuits is the inverting amplifier which reverses the polarity of the input signal while amplifying it. A key feature of the inverting amplifier is that both the input signal and the feedback are applied at the inverting terminal of the op amp. i f R f i 1 R 1 v i v o

8 66 5. OPERATIONAL AMPLIFIERS The equivalent circuit for the inverting amplifier is: v i R 1 R f R 1 v i v o The voltage gain is A v = v o /v i = R f /R Noninverting Amplifier A noninverting amplifier amplifies a signal by a constant positive gain (no inversion of polarity). The circuit for a noninverting amplifier is i f R f R 1 i 1 v i v o

9 5.4. NONINVERTING AMPLIFIER 67 The voltage gain is A v = v o v i = 1 R f R 1, which does not have a negative sign. Thus the output has the same polarity as the input. Special case: If R f = 0 or R 1 =, or both, the gain becomes 1. Under this conditions, the circuit becomes a voltage follower (The output follows the input). v i v o = v i A voltage follower is used to isolate two cascaded stages of a circuit. First stage v i v o Second stage

10 68 5. OPERATIONAL AMPLIFIERS Example Calculate the output voltage v o for the op amp circuit below. 10 kω 4 kω 6 V 4 V v o

11 5.5. SUMMING AMPLIFIER Summing Amplifier A summing amplifier is an op-amp circuit that combines several inputs and produces an output that is the weighted sum of the inputs. For this reason, the circuit is called a summer. v 1 R 1 i 1 R f i f v 2 v 3 R 2 R 3 i 2 i 3 i a v o ( Rf v o = v 1 R f v 2 R ) f v 3. R 1 R 2 R 3 Needless to say, the summer can have more that three inputs.

12 70 5. OPERATIONAL AMPLIFIERS 5.6. Difference Amplifier A difference amplifier is a device that amplifies the difference between two inputs but rejects any signals common to the two inputs. R 2 R 1 0 v a R 3 v b 0 v 1 v 2 R 4 v o v o = (1 R 2/R 1 ) (1 R 3 /R 4 ) v 2 R 2 v 1 = R 2(1 R 1 /R 2 ) R 1 R 1 (1 R 3 /R 4 ) v 2 R 2 v 1 R 1 Since a difference amplifier must reject a signal common to the two inputs, the amplifier must have the property that v o = 0 when v 1 = v 2. This property exists when R 1 R 2 = R 3 R 4.

13 5.6. DIFFERENCE AMPLIFIER 71 Thus, v o = R 2 R 1 (v 2 v 1 ). If R 2 = R 1 and R 3 = R 4, the difference amplifier becomes a subtractor, with the output v o = v 2 v 1. Example Design an op amp circuit with inputs v 1 and v 2 such that v o = 5v 1 3v 2.

14

15 72 5. OPERATIONAL AMPLIFIERS 5.7. Cascaded of Op Amp Circuits In practice, we can connect op amp circuits in cascade (i.e., head to tail) to achieve a large overall gain. Each circuit in the cascade is called stage. The output of one stage is the input to the next stage. Op amp circuits have the advantage that they can be cascaded without changing their input-output relationships. This is due to the fact that each (ideal) op amp circuit has infinite input resistance and zero output resistance. Stage 1 Stage 2 Stage 3 v 1 v A 2 = A 1v 1 v 1 A 3 = A 2v 2 2 A 3 v o = A 3v Application: Digital-to-Analog Converter (DAC) The digital-to-analog converter (DAC) transforms digital signals into analog form. A typical example of a four-bit DAC is shown in (a) below. Digital Input ( ) Four-bit DAC (a) Analog output V 1 V 2 V 3 V 4 R 1 R 2 R 3 R 4 R f MSB LSB V o (b)

16 5.9. APPLICATION: INSTRUMENTATION(AL) AMPLIFIERS (IA) Application: Instrumentation(al) Amplifiers (IA) One of the most useful and versatile op amp circuits for precision measurement and process control. IA amplifies the difference between the input signals. Inverting input Gain set v 1 1 R R R G R R 3 v o Output Gain set Noninverting input v 2 2 R R (a) (b) v o = ( 1 2R ) (v 2 v 1 ). R G

17 74 5. OPERATIONAL AMPLIFIERS The instrumentation amplifier amplifies small differential signal voltages superimposed on larger common-mode voltages. Since the common-mode voltages are equal, they cancel each other. The IA has three major characteristics: (a) The voltage gain is adjusted by one external resistor R G. (b) The input impedance of both inputs is very high and does not vary as the gain is adjusted. (c) The output v o depends on the difference between the inputs, not on the voltage common to them. Typical example of IA has gain from 1 to 1000.

18 CHAPTER 6 Energy Storage Elements: Capacitors and Inductors To this point in our study of electronic circuits, time has not been important. The analysis and designs we have performed so far have been static, and all circuit responses at a given time have depended only on the circuit inputs at that time. In this chapter, we shall introduce two important passive circuit elements: the capacitor and the inductor Introduction and a Mathematical Fact Capacitors and inductors, which are the electric and magnetic duals of each other, differ from resistors in several significant ways. Unlike resistors, which dissipate energy, capacitors and inductors do not dissipate but store energy, which can be retrieved at a later time. They are called storage elements. Furthermore, their branch variables do not depend algebraically upon each other. Rather, their relations involve temporal derivatives and integrals. Thus, the analysis of circuits containing capacitors and inductors involve differential equations in time An important mathematical fact: Given d f(t) = g(t), dt 75

19 76 6. ENERGY STORAGE ELEMENTS: CAPACITORS AND INDUCTORS 6.2. Capacitors A capacitor is a passive element designed to store energy in its electric field. The word capacitor is derived from this element s capacity to store energy When a voltage source v(t) is connected across the capacitor, the amount of charge stored, represented by q, is directly proportional to v(t), i.e., q(t) = Cv(t) where C, the constant of proportionality, is known as the capacitance of the capacitor. The unit of capacitance is the farad (F) in honor of Michael Faraday. 1 farad = 1 coulomb/volt Circuit symbol for capacitor of C farads: i C i C v (a) v (b) Since i = dq dt, then the current-voltage relationship of the capacitor is (6.2) i = C dv dt. Note that in (6.2), the capacitance value C is constant (time-invariant) and that the current i and voltage v are both functions of time (time-varying). So, in fact, the full form of (6.2) is i(t) = C d dt v(t). Hence, the voltage-current relation is v(t) = 1 C t t o i(τ)dτ v(t o )

20 6.2. CAPACITORS 77 i Slope = C 0 dv/dt where v(t o ) is the voltage across the capacitor at time t o. Note that capacitor voltage depends on the past history of the capacitor current. Hence, the capacitor has memory The instantaneous power delivered to the capacitor is p(t) = i(t) v(t) = (C ddt ) v(t) v(t). The energy stored in the capacitor is w(t) = t p(τ)dτ = 1 2 Cv2 (t). In the above calculation, we assume v( ) = 0, because the capacitor was uncharged at t = Typical values (a) Capacitors are commercially available in different values and types. (b) Typically, capacitors have values in the picofarad (pf) to microfarad (µf) range. (c) For comparison, two pieces of insulated wire about an inch long, when twisted together, will have a capacitance of about 1 pf Two important implications of (6.2): (a) A capacitor is an open circuit to dc. When the voltage across a capacitor is not changing with time (i.e., dc voltage), its derivative wrt. time is dv dt = 0 and hence the current through the capacitor is i(t) = C dv dt = C 0 = 0.

21 78 6. ENERGY STORAGE ELEMENTS: CAPACITORS AND INDUCTORS (b) The voltage across a capacitor cannot jump (change abruptly) Because i = C dv dt, a discontinuous change in voltage requires an infinite current, which is physically impossible. v v t t Remark: An ideal capacitor does not dissipate energy. It takes power from the circuit when storing energy in its field and returns previously stored energy when delivering power to the circuit. Example If a 10 µf is connected to a voltage source with v(t) = 50 sin 2000t determine the current through the capacitor. V Example Determine the voltage across a 2-µF capacitor if the current through it is i(t) = 6e 3000t ma Assume that the initial capacitor voltage (at time t = 0) is zero.

22 6.2. CAPACITORS 79 Example Obtain the energy stored in each capacitor in the figure below under dc conditions. 2 mf 2 kω 5 kω 6 ma 3 kω 4 kω 4 mf

23 80 6. ENERGY STORAGE ELEMENTS: CAPACITORS AND INDUCTORS 6.3. Series and Parallel Capacitors We know from resistive circuits that series-parallel combination is a powerful tool for simplifying circuits. This technique can be extended to series-parallel connections of capacitors, which are sometimes encountered. We desire to replace these capacitors by a single equivalent capacitor C eq The equivalent capacitance of N parallel-connected capacitors is the sum of the individual capacitance. C eq = C 1 C 2 C N i i 1 i 2 i 3 i N v C 1 C2 C3 CN The equivalent capacitance of N series-connected capacitors is the the reciprocal of the sum of the reciprocals of the individual capacitances. 1 = C eq C 1 C 2 C N v i C 1 C 2 C 3 C N v 1 v 2 v 3 v N Example Find the C eq. 5 µf 60 µf a 20 µf 6 µf 20 µf C eq b

24 6.4. INDUCTORS Inductors An inductor is a passive element designed to store energy in its magnetic field Inductors find numerous applications in electronic and power systems. They are used in power supplies, transformers, radios, TVs, radars, and electric motors Circuit symbol of inductor: i i i v L v L v L If a current is allowed to pass through an inductor, the voltage across the inductor is directly proportional to the time rate of change of the current, i.e., (6.3) v(t) = L d dt i(t), where L is the constant of proportionality called the inductance of the inductor. The unit of inductance is henry (H), named in honor of Joseph Henry. 1 henry equals 1 volt-second per ampere By integration, the current-voltage relation is i(t) = 1 L t where i(t o ) is the current at time t o. t o v(τ) dτ i(t o ), The instantaneous power delivered to the inductor is p(t) = v(t) i(t) = (L ddt ) i(t) i(t)

25 82 6. ENERGY STORAGE ELEMENTS: CAPACITORS AND INDUCTORS v Slope = L The energy stored in the inductor is 0 di/dt w(t) = t p(τ) dτ = 1 2 Li2 (t) Like capacitors, commercially available inductors come in different values and types. Typical practical inductors have inductance values ranging from a few microhenrys (µh), as in communication systems, to tens of henrys (H) as in power systems Two important implications of (6.3): (a) An inductor acts like a short circuit to dc. When the current through an inductor is not changing with time (i.e., dc current), its derivative wrt. time is di dt = 0 and hence the voltage across the inductor is v(t) = L di dt = L 0 = 0. (b) The current through an inductor cannot change instantaneously. This opposition to the change in current is an important property of the inductor. A discontinuous change in the current through an inductor requires an infinite voltage, which is not physically possible. i i (a) t Remark: The ideal inductor does not dissipate energy. The energy stored in it can be retrieved at a later time. The inductor takes (b) t

26 6.4. INDUCTORS 83 power from the circuit when storing energy and delivers power to the circuit when returning previously stored energy. Example If the current through a 1-mH inductor is i(t) = 20 cos 100t ma, find the terminal voltage and the energy stored. Example Find the current through a 5-H inductor if the voltage across it is { 30t 2, t > 0 v(t) = 0, t < 0. In addition, find the energy stored within 0 < t < 5 s.

27 84 6. ENERGY STORAGE ELEMENTS: CAPACITORS AND INDUCTORS Example The terminal voltage of a 2-H inductor is v(t) = 10(1 t) V. Find the current flowing through it at t = 4 s and the energy stored in it within 0 < t < 4 s. Assume i(0) = 2 A. Example Determine v C, i L and the energy stored in the capacitor and inductor in the following circuit under dc conditions. i 1 Ω 5 Ω i L 4 Ω 12 V 2 H v C 1 F

28 6.5. SERIES AND PARALLEL INDUCTORS 85 Example Determine v C, i L and the energy stored in the capacitor and inductor in the following circuit under dc conditions. i L 6 H 4 A 6 Ω 2 Ω v C 4 F 6.5. Series and Parallel Inductors The equivalent inductance of N series-connected inductors is the sum of the individual inductances, i.e., L eq = L 1 L 2 L N i L 2 L 3 L N L 1 v v 1 v 2 v 3 v N

29 86 6. ENERGY STORAGE ELEMENTS: CAPACITORS AND INDUCTORS The equivalent inductance of N parallel inductors is the reciprocal of the sum of the reciprocals of the individual inductances, i.e., 1 = L eq L 1 L 2 L N i :59 AM Page 232 v i 1 i 2 i 3 i N L 1 L 2 L 3 L N Remark: Note that (a) inductors in series are combined in exactly the same way as resistors in series and (b) inductors Chapter 6 Capacitors in parallel and Inductors are combined in the same way as resistors in parallel. TABLE 6.1 Important characteristics of the basic elements. Relation Resistor (R) Capacitor (C) Inductor (L) v-i: i-v: p or w: v i R i v R p i 2 R v2 R v 1 C t i dt v(t 0 ) t 0 dv i C dt w 1 2 Cv2 Series: C eq C 1C 2 R eq R 1 R 2 L eq L 1 L 2 C 1 C 2 Parallel: L eq L 1L 2 R eq R 1R 2 C eq C 1 C 2 R 1 R 2 L 1 L 2 At dc: Same Open circuit Short circuit Circuit variable that cannot change abruptly: Not applicable v i di v L dt i 1 L t w 1 2 Li2 t 0 v dt i(t 0 ) Passive sign convention is assumed. It is appropriate at this point to summarize the most important characteristics of the three basic circuit elements we have studied. The summary is given in Table 6.1. The wye-delta transformation discussed in Section 2.7 for resistors can be extended to capacitors and inductors.

30 6.6. APPLICATIONS: INTEGRATORS AND DIFFERENTIATORS 87 Example Find the equivalent inductance L eq of the circuit shown below. 4 H 20 H a L eq 7 H 12 H 8 H 10 H b 6.6. Applications: Integrators and Differentiators Capacitors and inductors possess the following three special properties that make them very useful in electric circuits: (a) The capacity to store energy makes them useful as temporary voltage or current sources. Thus, they can be used for generating a large amount of current or voltage for a short period of time. (b) Capacitors oppose any abrupt change in voltage, while inductors oppose any abrupt change in current. This property makes inductors useful for spark or arc suppression and for converting pulsating dc voltage into relatively smooth dc voltage. (c) Capacitors and inductors are frequency sensitive. This property makes them useful for frequency discrimination. The first two properties are put to use in dc circuits, while the third one is taken advantage of in ac circuits. In this final part of the chapter, we will consider two applications involving capacitors and op amps: integrator and differentiator.

31 88 6. ENERGY STORAGE ELEMENTS: CAPACITORS AND INDUCTORS An integrator is an op amp circuit whose output is proportional to the integral of the input signal. We obtain an integrator by replacing the feedback resistor R f in the inverting amplifier by a capacitor. i C C i R R a v i v 0 This gives which implies d dt v o(t) = 1 RC v i(t), t v o (t) = 1 v i (τ)dτ v o (0). RC 0 To ensure that v o (0) = 0, it is always necessary to discharge the integrators capacitor prior to the application of a signal. In practice, the op amp integrator requires a feedback resistor to reduce dc gain and prevent saturation. Care must be taken that the op amp operates within the linear range so that it does not saturate.

32 6.6. APPLICATIONS: INTEGRATORS AND DIFFERENTIATORS A differentiator is an op amp circuit whose output is proportional to the differentiation of the input signal. We obtain a differentiator by replacing the input resistor in the inverting amplifier by a capacitor. This gives i R R i C C a v i v 0 v o (t) = RC d dt v i(t). Differentiator circuits are electronically unstable because any electrical noise within the circuit is exaggerated by the differentiator. For this reason, the differentiator circuit above is not as useful and popular as the integrator. It is seldom used in practice.

33 CHAPTER 7 Sinusoids and Phasors Recall that, for capacitors and inductors, the branch variables (current values and voltage values) are related by differential equations. Normally, to analyze a circuit containing capacitor and/or inductor, we need to solve some differential equations. The analysis can be greatly simplifies when the circuit is driven (or excited) by a source (or sources) that is sinusoidal. Such assumption will be the main focus of this chapter Prelude to Second-Order Circuits The next example demonstrates the complication normally involved when analyzing a circuit containing capacitor and inductor. This example and the analysis presented is not the main focus of this chapter. Example The switch in the figure below has been open for a long time. It is closed at t = 0. 4 Ω i 1 H 12 V t = 0 2 Ω 1 - F 2 v (a) Find v(0) and dv dt (0). (b) Find v(t) for t > 0. (c) Find v( ) and dv dt ( ). (d) Find v(t) for t > 0 when the source is v s (t) = { 12, t < 0, 12 cos (t), t 0. 91

34 92 7. SINUSOIDS AND PHASORS From MATLAB, v = dsolve('d2v 5*Dv 6*v = 24','v(0) = 12','Dv(0) = 12') gives v(t) = 4 12e 2t 4e 3t. Similarly, v = dsolve('d2v 5*Dv 6*v = 2*12*cos(t)','v(0) = 12','Dv(0) = 12','t') gives v(t) = 72 5 e 2t 24 5 e 3t 12 5 cos(t) 12 5 sin(t) t t

35 7.2. SINUSOIDS Sinusoids Definition Some terminology: (a) A sinusoid is a signal (, e.g. voltage or current) that has the form of the sine or cosine function. Turn out that you can express them all under the same notation using only cosine (or only sine) function. We will use cosine. (b) A sinusoidal current is referred to as alternating current (AC). (c) We use the term AC source for any device that supplies a sinusoidally varying voltage (potential difference) or current. (d) Circuits driven by sinusoidal current or voltage sources are called AC circuits Consider the sinusoidal signal (in cosine form) x(t) = X m cos(ωt φ) = X m cos(2πft φ), where X m : the amplitude of the sinusoid, ω: the angular frequency in radians/s (or rad/s), φ: the phase. First, we consider the case when φ = 0: t 3 When φ 0, we shift the graph of X m cos(ωt) to the left by φ. t 2

36 94 7. SINUSOIDS AND PHASORS The period (the time of one complete cycle) of the sinusoid is T = 2π ω. The unit of the period is in second if the angular frequency unit is in radian per second. The frequency f (the number of cycles per second or hertz (Hz)) is the reciprocal of this quantity, i.e., f = 1 T Standard form for sinusoid: In this class, when you are asked to find the sinusoid representation of a signal, make sure that your answer is in the form x(t) = X m cos(ωt φ) = X m cos(2πft φ), where X m is nonnegative and φ is between 180 and Conversions to standard form When the signal is given in the sine form, it can be converted into its cosine form via the identity sin(x) = cos(x 90 ). t 1 In particular, X m sin(ωt φ) = X m cos(ωt φ 90 ). X m is always non-negative. We can avoid having the negative sign by the following conversion: cos(x) = cos(x ± 180 ).

37 7.3. PHASORS 95 In particular, A cos(ωt φ) = A cos(2πft φ ± 180 ). Note that usually you do not have the choice between 180 or 180. The one that you need to use is the one that makes φ ± 180 falls somewhere between 180 and For any 1 linear AC circuit, the steady-state voltage and current are sinusoidal with the same frequency as the driving source(s). Although all the voltage and current are sinusoidal, their amplitudes and phases can be different. These can be found by the technique discussed in this chapter Phasors Sinusoids are easily expressed in terms of phasors, which are more convenient to work with than sine and cosine functions. The tradeoff is that phasors are complex-valued The idea of phasor representation is based on Euler s identity: e jφ = cos φ j sin φ, From the identity, we may regard cos φ and sin φ as the real and imaginary parts of e jφ : cos φ = Re { e jφ}, sin φ = Im { e jφ}, where Re and Im stand for the real part of and the imaginary part of e jφ. Definition A phasor is a complex number that represents the amplitude and phase of a sinusoid. Given a sinusoid x(t) = X m cos(ωtφ), then } x(t) = X m cos(ωtφ) = Re {X m e j(ωtφ) = Re { X m e jφ e jωt} = Re { Xe jωt}, where X = X m e jφ = X m φ. The complex number X is called the phasor representation of the sinusoid v(t). Notice that a phasor captures information about amplitude and phase of the corresponding sinusoid. 1 When there are multiple sources, we assume that all sources are at the same frequency.

38 96 7. SINUSOIDS AND PHASORS Whenever a sinusoid is expressed as a phasor, the term e jωt is implicit. It is therefore important, when dealing with phasors, to keep in mind the frequency f (or the angular frequency ω) of the phasor Given a phasor X, to obtain the time-domain sinusoid corresponding to a given phasor, there are two important routes. (a) Simply write down the cosine function with the same magnitude as the phasor and the argument as ωt plus the phase of the phasor. (b) Multiply the phasor by the time factor e jωt and take the real part Any complex number z (including any phasor) can be equivalently represented in three forms. (a) Rectangular form: z = x jy. (b) Polar form: z = r φ. (c) Exponential form: z = re jφ where the relations between them are r = x 2 y 2, φ = tan 1 y x ± 180. x = r cos φ, y = r sin φ. Note that for φ, the choice of using 180 or 180 in the formula is determined by the actual quadrant in which the complex number lies. As a complex quantity, a phasor may be expressed in rectangular form, polar form, or exponential form. In this class, we focus on polar form Summary: By suppressing the time factor, we transform the sinusoid from the time domain to the phasor domain. This transformation is summarized as follows: x(t) = X m cos(ωt φ) X = X m φ. Time domain representation Phasor domain representation

39 7.3. PHASORS 97 Definition Standard form for phasor: In this class, when you are asked to find the phasor representation of a signal, make sure that your answer is a complex number in polar form, i.e. r φ where r is nonnegative and φ is between 180 and 180. Example Transform these sinusoids to phasors: (a) i = 6 cos(50t 40 ) A (b) v = 4 sin(30t 50 ) V Example Find the sinusoids represented by these phasors: (a) I = 3 j4 A (b) V = j8e j20 V The differences between x(t) and X should be emphasized: (a) x(t) is the instantaneous or time-domain representation, while X is the frequency or phasor-domain representation. (b) x(t) is time dependent, while X is not. (c) x(t) is always real with no complex term, while X is generally complex Adding sinusoids of the same frequency is equivalent to adding their corresponding phasors. To see this, A 1 cos (ωt φ 1 ) A 2 cos (ωt φ 2 ) = Re { A 1 e jωt} Re { A 2 e jωt} = Re { (A 1 A 2 ) e jωt}. Because A 1 A 2 is just another complex number, we can conclude a (surprising) fact: adding two sinusoids of the same frequency gives another sinusoids.

40 98 7. SINUSOIDS AND PHASORS Example x(t) = 4 cos(2t) 3 sin(2t) Properties involving differentiation and integration: (a) Differentiating a sinusoid is equivalent to multiplying its corresponding phasor by jω. In other words, dx(t) jωx. dt To see this, suppose x(t) = X m cos(ωt φ). Then, dx dt (t) = ωx m sin(ωt φ) = ωx m cos(ωt φ ) = Re { ωx m e jφ e j90 e jωt} = Re { jωxe jωt} Alternatively, express v(t) as } x(t) = Re {X m e j(ωtφ). Then, d { } dt x(t) = Re X m jωe j(ωtφ). (b) Integrating a sinusoid is equivalent to dividing its corresponding phasor by jω. In other words, x(t)dt X jω. Example Find the voltage v(t) in a circuit described by the intergrodifferential equation 2 dv 5v 10 vdt = 50 cos(5t 30 ) dt using the phasor approach.

41 7.4. PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS Phasor relationships for circuit elements Resistor R: If the current through a resistor R is i(t) = I m cos(ωt φ) I = I m φ, the voltage across it is given by v(t) = i(t)r = RI m cos(ωt φ). i I v R V R v = ir The phasor of the voltage is V = RI m φ. V = IR Hence, V = IR. We note that voltage and current are in phase and that the voltage-current relation for the resistor in the phasor domain continues to be Ohms law, as in the time domain. Im V I f 0 Re

42 SINUSOIDS AND PHASORS Capacitor C: If the voltage across a capacitor C is v(t) = V m cos(ωt φ) V = V m φ, the current through it is given by i(t) = C dv(t) I = jωcv = ωcv m (φ 90 ). dt i I v C V C dv i = C I = jwcv dt The voltage and current are 90 out of phase. Specifically, the current leads the voltage by 90. I Im w V f 0 Re Mnemonic: CIVIL In a Capacitive (C) circuit, I leads V. In an inductive (L) circuit, V leads I.

43 7.4. PHASOR RELATIONSHIPS FOR CIRCUIT ELEMENTS Inductor L: If the current through an inductor L is i(t) = I m cos(ωt φ) I = I m φ, the voltage across it is given by v(t) = L di(t) V = jωli = ωli m (φ 90 ). dt i I v L V L di v = L dt V = jw LI The voltage and current are 90 out of phase. Specifically, the current lags the voltage by 90. relations for the capacitor; Fig. Im9.14 gives the phasor diagram. Table 9.2 w summarizes the time-domain V and phasor-domain representations of the circuit elements. TABLE 9.2 I f CHAPTER 9 Summary of voltage-current 0 Re relationships. Element Time domain Frequency domain R v = Ri V = RI L C v = L di dt i = C dv dt V = jωli V = I jωc Sinusoids and Phasors I Figure 9 E X A M P L E 9. 8 The voltage v = 12 cos(60t 45 ) is applied to a 0.1-H inductor. Find the steady-state current through the inductor.

44 SINUSOIDS AND PHASORS 7.5. Impedance and Admittance In the previous part, we obtained the voltage current relations for the three passive elements as V = IR, V = jωli, I = jωcv. These equations may be written in terms of the ratio of the phasor voltage to the phasor of current as V I = R, V I = jωl, V = 1 I jωc. From these equations, we obtain Ohm s law in phasor form for any type of element as Z = V or V = IZ. I Definition The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms (Ω). As a complex quantity, the impedance may be expressed in rectangular form as Z = R jx = Z θ, with Z = R 2 X 2, θ = tan 1 X, R = Z cos θ, X = Z sin θ. R R = Re {Z} is called the resistance and X = Im {Z} is called the reactance. The reactance X may be positive or negative. We say that the impedance is inductive when X is positive or capacitive when X is negative. Definition The admittance (Y) is the reciprocal of impedance, measured in Siemens (S). The admittance of an element(or a circuit) is the ratio of the phasor current through it to phasor voltage across it, or Y = 1 Z = I V.

45 7.5. IMPEDANCE AND ADMITTANCE Kirchhoff s laws (KCL and KVL) hold in the phasor form. To see this, suppose v 1, v 2,..., v n are the voltages around a closed loop, then v 1 v 2 v n = 0. If each voltage v i is a sinusoid, i.e. v i = V mi cos(ωt φ i ) = Re { V i e jωt} with phasor V i = V mi φ i = V mi e jφ i, then Re { (V 1 V 2 V n ) e jωt} = 0, which must be true for all time t. To satisfy this, we need V 1 V 2 V n = 0. Hence, KVL holds for phasors. Similarly, we can show that KCL holds in the frequency domain, i.e., if the currents i 1, i 2,..., i n are the currents entering or leaving a closed surface at time t, then i 1 i 2 i n = 0. If the currents are sinusoids and I 1, I 2,..., I n are their phasor forms, then I 1 I 2 I n = Major Implication: Since Ohm s Law and Kirchoff s Laws hold in phasor domain, all resistance combination formulas, volatge and current divider formulas, analysis methods (nodal and mesh analysis) and circuit theorems (linearity, superposition, source transformation, and Thevenin s and Norton s equivalent circuits) that we have previously studied for dc circuits apply to ac circuits!!! Just think of impedance as a complex-valued resistance!! The three-step analysis in the next chapter is based on this insight. In addition, our ac circuits can now effortlessly include capacitors and inductors which can be considered as impedances whose values depend on the frequency ω of the ac sources!!

46 SINUSOIDS AND PHASORS 7.6. Impedance Combinations Consider N series-connected impedances as shown below. I Z 2 Z N Z 1 V V 1 V 2 V N Z eq The same current I flows through the impedances. Applying KVL around the loop gives V = V 1 V 2 V N = I(Z 1 Z 2 Z N ) The equivalent impedance at the input terminals is Z eq = V I = Z 1 Z 2 Z N. In particular, if N = 2, the current through the impedance is I Z 1 V V 1 V 2 Z 2 I = Because V 1 = Z 1 I and V 2 = Z 2 I, Z 1 V Z 1 Z 2. V 1 = V, V 2 = Z 1 Z 2 which is the voltage-division relationship. Z 2 Z 1 Z 2 V

47 7.6. IMPEDANCE COMBINATIONS 105 Now, consider N parallel-connected impedances as shown below. I I 1 I 2 I N I V Z 1 Z 2 Z N Z eq The voltage across each impedance is the same. Applying KCL at the top node gives ( 1 I = I 1 I 2 I N = V 1 1 ). Z 1 Z 2 Z N The equivalent impedance Z eq can be found from When N = 2, 1 = I Z eq V = Z 1 Z 2 Z N Z eq = Z 1Z 2 Z 1 Z 2. Because V = IZ eq = I 1 Z 1 = I 2 Z 2, we have Z 2 I 1 = I, I 2 = Z 1 Z 2 which is the current-division principle. Z 1 Z 1 Z 2 I I 1 I 2 I V Z 1 Z 2

48 SINUSOIDS AND PHASORS Example Find the input impedance of the circuit below. Assume that the circuit operates at ω = 50 rad/s. 2 mf 0.2 H Z in 3 Ω 10 mf 8 Ω Example Determine v o (t) in the circuit below. 60 Ω 20 cos(4t 15 ) 10 mf 5 H v 0

49 CHAPTER 8 Sinusoidal Steady State Analysis 8.1. General Approach In the previous chapter, we have learned that the steady-state response of a circuit to sinusoidal inputs can be obtained by using phasors. In this chapter, we present many examples in which nodal analysis, mesh analysis, Thevenin s theorem, superposition, and source transformations are applied in analyzing ac circuits Steps to analyze ac circuits, using phasor domain: Step 1. Transform the circuit to the phasor or frequency domain. Not necessary if the problem is specified in the frequency domain. Step 2. Solve the problem using circuit techniques (e.g., nodal analysis, mesh analysis, Thevenin s theorem, superposition, or source transformations ) The analysis is performed in the same manner as dc circuit analysis except that complex numbers are involved. Step 3. Transform the resulting phasor back to the time domain ac circuits are linear (they are just composed of sources and impedances) The superposition theorem applies to ac circuits the same way it applies to dc circuits. This is the case when all the sources in the circuit operate at the same frequency. If they are operating at different frequency, see Section

50 SINUSOIDAL STEADY STATE ANALYSIS Source transformation: V s = Z s I s, I s = V s Z s. Z s a a V s I s Z s V s = Z s I s b I s = V s Z s b Thevenin and Norton Equivalent circuits: a Z Th a Linear circuit V Th b a a b Linear circuit I N Z N b b V Th = Z N I N, Z Th = Z N

51 8.1. GENERAL APPROACH 109 Example Compute V 1 and V 2 in the circuit below using nodal analysis V V 1 4 Ω 1 2 V A j3 Ω j6 Ω 12 Ω Example Determine current I o in the circuit below using mesh analysis. 4 Ω I A j2 Ω j10 Ω I 2 I V 8 Ω I 1 j2 Ω

52 The voltage-to-current converter, as shown in Figure 8-2, produces an output current that depends on the input voltage and the resistor R. In particular, the output current I out = V i /R independent of the loading resistance R L SINUSOIDAL STEADY STATE ANALYSIS The current-to-voltage converter, as shown in Figure 8-3, produces an output voltage that depends on the input current and the resistor R. In particular, the output voltage Example Find the Thevenin equivalent at terminals a-b of the V o = -I in R circuit below. independent of the size of the loading resistance R L. 6 Ω j2 Ω V i V V I out I in a j4 Ω b R V 10 Ω R R L V- V- R L V o - Figure 8-2: Voltage-to-current converter. Figure 8-3: Current-to-voltage converter. 4. An integrating amplifier is shown in Figure 8-4a. Example Op Amp AC Circuits: Find the (closed-loop) gain of the circuit below. v C - i C C v i i in R X V V- v o - Figure 8-4a: Integrating amplifier 2

53 8.2. CIRCUIT WITH MULTIPLE SOURCES OPERATING AT DIFFERENT FREQUENCIES Circuit With Multiple Sources Operating At Different Frequencies A special care is needed if the circuit has multiple sources operating at different frequencies. In which case, one must add the responses due to the individual frequencies in the time domain. In other words, the superposition still works but (a) We must have a different frequency-domain circuit for each frequency. (b) The total response must be obtained by adding the individual response in the time domain Since the impedance depend on frequency, it is incorrect to try to add the responses in the phasor or frequency domain. To see this note that the exponential factor e jωt is implicit in sinusoidal analysis, and that factor would change for every angular frequency ω. In particular, although V mi cos(ωt φ i ) = Re { {( ) } V i e jωt} = Re V i e jωt, i i i when we allow ω to be different for each sinusoid, generally V mi cos(ω i t φ i ) = Re { V i e } {( ) } jω it Re V i e jω it. i i i Therefore, it does not make sense to add responses at different frequencies in the phasor domain The Thevenin or Norton equivalent circuit (if needed) must be determined at each frequency and we have one equivalent circuit for each frequency.

54 SINUSOIDAL STEADY STATE ANALYSIS Example Find v o in the circuit below using the superposition theorem. 2 H 1 Ω 4 Ω v 0 10 cos 2t V 2 sin 5t A 0.1 F 5 V 1 Ω 4 Ω j4 Ω 1 Ω 4 Ω I 1 1 Ω v1 5 V 10 0 V V2 j5 Ω j10 Ω V A j2 Ω 4 Ω (a) (b) (c)

55 CHAPTER 9 AC Power Analysis Our effort in ac circuit analysis so far has been focused mainly on calculating voltage and current. The major concern in this chapter is power analysis Power is the most important quantity in electric utilities, electronic and communication systems because such systems involve transmission of power (or energy) from one point to another. Every industrial and household electrical device (every fan, motor, lamp, pressing iron, TV, personal computer) has a power rating that indicates how much power the equipment requires; exceeding the power rating can do permanent damage to an appliance The most common form of electric power is 50-Hz (Thailand) or 60-Hz (United States) ac power. The choice of ac over dc allowed high-voltage power transmission from the power generating plant to the consumer. (DC attenuation is high.) 9.1. Instantaneous Power Definition The instantaneous power p(t) absorbed by an element is the product of the instantaneous voltage v(t) across the element and the instantaneous current i(t) through it. Assuming the passive sign convention as shown in Figure 1, p(t) = v(t)i(t). The instantaneous power is the power at any instant of time. It is the rate at which an element absorbs energy. 113

56 AC POWER ANALYSIS Consider the general case of instantaneous power absorbed by an arbitrary combination of circuit elements under sinusoidal excitation. i(t) Sinusoidal source v(t) Passive Linear network Figure 1. Sinusoidal source and passive linear circuit Let the voltage and current at the terminals of the circuit be v(t) = V m cos(ωt θ v ) and i(t) = I m cos(ωt θ i ) where V m and I m are the amplitudes, and θ v and θ i are the phase of the voltage and current, respectively. The instantaneous power absorbed by the circuit is (9.4) p(t) = v(t)i(t) = V m I m cos(ωt θ v ) cos(ωt θ i ) e j(ωtθv) e j(ωtθ v) e j(ωtθi) e j(ωtθ i) (9.5) = V m I m ( ) = V m I m e j(2ωtθ vθ i ) e j(θ i θ v ) e j(θ v θ i ) e j(2ωtθ vθ i ) (9.6) 4 ( 1 e j(θ v θ i ) e j(θ i θ v ) = V m I m ej(2ωtθ vθ i ) e j(2ωtθ vθ i ) ) (9.7) (9.8) = V m I m 2 (cos (θ v θ i ) cos (2ωt θ v θ i )). Alternatively, from (9.4), we can apply the trigonometric identity cos A cos B = 1 {cos(a B) cos(a B)} 2 to directly arrive at (9.8) which is

57 9.1. INSTANTANEOUS POWER 115 (9.9) p(t) = 1 2 V mi m cos(θ v θ i ) 1 }{{} 2 V mi m cos(2ωt θ v θ i ). }{{} constant term time-dependent term The instantaneous power, when expressed in the form of (9.9), has two parts: (a) First Part: a constant or time-independent term. Its value depends on the phase difference between the voltage and the current. (b) Second Part: a sinusoidal function whose angular frequency is 2ω, which is twice the angular frequency of the voltage or current. p(t) 1 V mi m 2 1 V mi m cos(q v - q i) 2 0 T t T Consider the sketch of p(t), we observe that (a) p(t) is periodic and has a period of T o = T 2, where T = 2π ω is the period of the voltage and the current (b) p(t) may become positive for some part(s) of each cycle and negative for the rest of the cycle. When p(t) is positive, power is absorbed by the circuit. When p(t) is negative, power is absorbed by the source. In this case, power is transferred from the circuit to the source. This is possible because of the storage elements (capacitors and inductors) in the circuit.

58 AC POWER ANALYSIS 9.2. Average Power The instantaneous power changes with time and is therefore difficult to measure. The average power is more convenient to measure. Definition The average power (of any periodic signal) is the average of the instantaneous power over one period. Thus, the average power is given by P = 1 T T 0 p(t)dt = 1 2 V mi m cos(θ v θ i ). Since cos(θ v θ i ) = cos(θ i θ v ), what is important is the difference in the phases of the voltage and the current. Note also that p(t) is time varying while P does not depend on time Using the phasor forms of v(t) and i(t), which are V = V m θ v and I = I m θ i, we obtain P = 1 2 V mi m cos(θ v θ i ) = 1 2 Re{VI } Two special cases: Case 1: When θ v = θ i, the voltage and the current are in phase. This implies a purely resistive circuit or resistive load R, and P = 1 2 V mi m This shows that a purely resistive circuit (e.g. resistive load (R)) absorbs power all times. Case 2: When θ v θ i = ±90, we have a purely reactive circuit, and P = 1 2 V mi m cos(90 ) = 0 showing that a purely reactive circuit (e.g. a reactive load L or C) absorbs no average power.

59 9.2. AVERAGE POWER 117 Example Calculate the average power absorbed by an impedance Z = 30 j70 Ω when a voltage V = is applied across it. Example A current I = flows through an impedance Z = Ω. Find the average power delivered to the impedance From Ohm s law, we have two more formula. (a) The average power absorbed by an impedance Z when a voltage V is applied across it is P = 1 2 Re{VI } = 1 2 Re{VV Z } = 1 2 V 2 Re{ 1 Z } = 1 Re{Z} 2 V 2 Z 2. (b) The average power absorbed by an impedance Z when a current I flows through it is P = 1 2 Re{VI } = 1 2 Re{IZI } = 1 2 I 2 Re{Z}. Example For the circuit shown below, find the average power supplied by the source and the average power absorbed by the resistor. I 4 Ω 5 30 V j2 Ω

60 AC POWER ANALYSIS 9.3. Maximum Average Power Transfer Recall that, in an earlier chapter, we solved the problem of maximizing the power delivered by a power-supplying resistive network to a load R L. Representing the circuit by its Thevenin equivalent, we proved that the maximum power would be delivered to the load if the load resistance is equal to the Thevenin resistance R L = R T h Optimal Load Impedance: We now extend that result to ac circuits. Linear circuit Z L Consider an ac circuit which is connected to a load Z L and is represented by its Thevenin equivalent. I Z Th V Th Z L In a rectangular form, the Thevenin impedance Z Th and load impedance Z L are Z Th = R T h jx T h Z L = R L jx L The current through the load is I = V Th, Z L Z Th and the average power delivered to the load is P = 1 2 I 2 Re {Z L } = 1 2 I 2 R L = 1 2 VTh R L 2 (R L R T h ) 2 (X L X T h ) 2 Our objective is to adjust the load parameter R L and X L so that P is maximum. To do this we set P R L and P X L equal to zero.

61 9.3. MAXIMUM AVERAGE POWER TRANSFER 119 Setting P X L Setting P R L = 0 gives = 0 gives R L = X L = X T h. R 2 T h (X T h X L ) 2 Hence, to get the maximum average power transfer, the load impedance Z L must selected so that i.e., X L = X T h and R L = R T h, Z L = R L jx L = R T h jx T h = Z Th That is, for the maximum average power transfer, the load impedance Z L must be equal to the complex conjugate of the Thevenin impedance Z Th. When Z L is set to be Z Th, the corresponding maximum average power that can be transferred to the load is P max = V Th 2 8R T h. Example Determine the load impedance Z L that maximizes the average power drawn from the circuit below. What is the maximum average power? 4 Ω j5 Ω 10 0 V 8 Ω j6 Ω Z L

62 AC POWER ANALYSIS Optimal Purely Resistive Load: In a situation in which the load must be purely real; that is X L must be 0. Then, P = 1 2 I 2 R L = 1 V 2 2 T h R L (R L R T h ) 2 (X T h ) 2 Setting P R L = 0 gives R L = R 2 T h X2 T h = Z Th. Hence, for maximum average power transfer to a purely resistive load, the load impedance is equal to the magnitude of the Thevenin impedance Z Th. In which case, the maximum average power is P = 1 V 2 Th 1 4 Z Th R T h Note that Z Th R T h R T h R T h = 2R T h Hence, 1 V 2 Th VTh. 4 Z Th R T h 8 R T h

63 9.4. EFFECTIVE OR RMS VALUE Effective or RMS Value The idea of effective value arises from the need to measure the effectiveness of an ac voltage or current source in delivering power to a resistive load. We start by considering the more general case of periodic signals and then consider the special case of sinusoidal signals in Definition The effective value I eff of a periodic current i(t) is the dc current that delivers the same average power to a resistor as the periodic current Consider the following ac and dc circuits, i(t) v(t) R our objective is to find the current I eff that will transfer the same power to the resistor R as the sinusoid current i I eff V eff R The average power absorbed by the resistor in the ac circuit is P = 1 T T 0 i 2 (t)rdt = R T T 0 i 2 (t)dt. Q: Where don t we have the factor of 1 2? The power absorbed by the resistor in the dc circuit is P = I 2 effr.

64 AC POWER ANALYSIS Equating the two expressions and solving for I eff, we obtain 1 T I eff = i T 2 (t)dt Similarly, the effective value of the periodic voltage is found in the same way as current; that is, 1 T V eff = v T 2 (t)dt This indicates that the effective value is the square root of the mean (or average) of the square of the periodic signal. Thus, the effective value is often know as the root mean square, or rms value for short. We write I eff = I rms, V eff = V rms Note that the rms value of a constant is the constant itself AC Circuit: For a sinusoid x(t) = X m cos(ωtθ x ), the effective value or rms value is 1 T X rms = Xm T 2 cos 2 (ωt θ x )dt = X m. 2 0 In particular, for i(t) = I m cos(ωt θ i ) and v(t) = V m cos(ωt θ v ), we have I rms = Im 2 and V rms = V m 2. and the average power can be written in terms of the rms values as P = 1 2 V mi m cos(θ v θ i ) = V rms I rms cos(θ v θ i ) Similarly, the average power absorbed by a resistor R can be written as P = I rms V rms = IrmsR 2 = V rms 2 R. 0 0

65 9.5. APPARENT POWER AND POWER FACTOR Apparent Power and Power Factor Definition The apparent power S (in VA) is the product of the rms values of voltage and current. S = V rms I rms Hence, the average power P = S cos(θ v θ i ). Definition The power factor (pf) is the ratio of the average power to the apparent power. Hence, pf = P S = cos(θ v θ i ). average power P = apparent power S power factor pf. The angle θ v θ i is called the power factor angle which is equal to the angle of the load impedance if V = V m θ v is the voltage across the load and I = I m θ i is the current through it. This is evident from the fact that Z = V I = V m θ v = V m (θ v θ i ). I m θ i I m Alternatively, define V rms = V 2 = V rms Θ v and I rms = I 2 = I rms Θ i. The impedance can then be written as Z = V I = V rms = V rms (θ v θ i ). I rms I rms The power factor is the cosine of the phase difference between the voltage and current. It is also the cosine of the angle of the load impedance The value of the power factor pf ranges between 0 and 1. For a purely resistive load, the voltage and current are in phase, so that θ v θ i = 0 and pf =1. This implies that the average power is equal to the apparent power. For a purely reactive load, θ v θ i = ±90. Hence, pf = 0. In this case the average power is zero. In between these two extreme cases, pf is said to be leading or lagging. Leading power factor means that current leads voltage which implies a capacitive load. Lagging power factor means that current lags voltage, implying an inductive load.