236 Response of First-Order RL and RC Circuits no energy is stored in the circuit at the instant the switch is closed. Next we observe v o (0 + ) = 120 V, which is consistent with the fact that i o (0) = 0. Now we observe the solutions for ^ and i 2 are consistent with the solution for v 0 by observing dii di 2 v a = 3— + 6— dt dt = 360*T 5 ' - 240<T 5 ' = 120«?-* V, t > 0 + , or dt dt = 720e~ 5 ' - 600e _5/ = 120e" 5 ' V, t > 0 + . The final values of i\ and / 2 can be checked using flux linkages. The flux linking the 3 H coil (Aj) must be equal to the flux linking the 15 H coil (A 2 ), because v„ = Now and rfAj ~dt dki dt' A, = 3ij + 6/ 2 Wb-turns A 2 = 6i] + 15/ 2 Wb-turns. Regardless of which expression we use, we obtain A, = A 2 = 24 - 24<T 5 ' Wb-turns. Note the solution for A t or A 2 is consistent with the solution for v D . The final value of the flux linking either coil 1 or coil 2 is 24 Wb-turns, that is, A t (oo) = A 2 (oo) = 24 Wb-turns. The final value of iy is ^(oo) = 24 A and the final value of i 2 is / 2 (oo) = -8 A. The consistency between these final values for jj and i 2 and the final value of the flux link- age can be seen from the expressions: A^oo) = 3/2(00) + 6/ 2 (oo) = 3(24) + 6(-8) = 24 Wb-turns, A 2 (oo) = 6^(00) + 15/ 2 (oo) = 6(24) + 15(-8) = 24 Wb-turns. It is worth noting that the final values of ij and / 2 can only be checked via flux linkage because at t — 00 the two coils are ideal short circuits. The division of current between ideal short circuits cannot be found from Ohm's law. NOTE: Assess your understanding of this material by using the general solution method to solve Chapter Problems 7.68 and 7.69. 7.5 Sequential Switching Whenever switching occurs more than once in a circuit, we have sequential switching. For example, a single, two-position switch may be switched back and forth, or multiple switches may be opened or closed in sequence. The time reference for all switchings cannot be f = 0. We determine the volt- ages and currents generated by a switching sequence by using the tech- niques described previously in this chapter. We derive the expressions for v(t) and i(t) for a given position of the switch or switches and then use these solutions to determine the initial conditions for the next position of the switch or switches. With sequential switching problems, a premium is placed on obtaining the initial value x(t 0 ). Recall that anything but inductive currents and capacitive voltages can change instantaneously at the time of switching. Thus solving first for inductive currents and capacitive voltages is even more pertinent in sequential switching problems. Drawing the circuit that pertains to each time interval in such a problem is often helpful in the solution process. 7.5 Sequential Switching 237 Examples 7.11 and 7.12 illustrate the analysis techniques for circuits with sequential switching. The first is a natural response problem with two switching times, and the second is a step response problem. Example 7.11 Analyzing an RL Circuit that has Sequential Switching The two switches in the circuit shown in Fig. 7.31 have been closed for a long time. At t - 0, switch 1 is opened. Then, 35 ms later, switch 2 is opened. a) Find i L {t) for 0 < t < 35 ms. b) Find i L for t a 35 ms. c) What percentage of the initial energy stored in the 150 mH inductor is dissipated in the 18 ft resistor? d) Repeat (c) for the 3 17 resistor. e) Repeat (c) for the 6 D, resistor. t = 0 t = 35 ms 4H KVr^fr^— 60 V £12 (1 £6(1 <M 150 mH $18(1 Figure 7.31 • The circuit for Example 7.11. Solution a) For t < 0 both switches are closed, causing the 150 mH inductor to short-circuit the 18 D, resis- tor. The equivalent circuit is shown in Fig. 7.32. We determine the initial current in the inductor by solving for ii£0~) in the circuit shown in Fig. 7.32. After making several source transformations, we find i L (0~) to be 6 A. For 0 < t < 35 ms, switch 1 is open (switch 2 is closed), which disconnects the 60 V voltage source and the 4 H and 12 £l resis- tors from the circuit. The inductor is no longer behaving as a short circuit (because the dc source is no longer in the circuit), so the 18 O resistor is no longer short-circuited. The equivalent circuit is shown in Fig. 7.33. Note that the equivalent resist- ance across the terminals of the inductor is the parallel combination of 9 O and 18 0, or 6 i\. The time constant of the circuit is (150/6) X 10~ 3 , or 25 ms. Therefore the expression for i L is i L = 6e- A{]l A, 0 < t < 35 ms. 4(1 3(1 60 V two") Figure 7.32 • The circuit shown in Fig. 7.31, for t < 0. Figure 7.33 • The circuit shown in Fig. 7.31, for 0 < t ^35 ms. b) When t = 35 ms, the value of the inductor current is i L = 6e~ u = 1.48 A. Thus, when switch 2 is opened, the circuit reduces to the one shown in Fig. 7.34, and the time constant changes to (150/9) x 10 _ \ or 16.67 ms.The expression for i L becomes i,= 1.486>- 6(,( '- a()35 >A, t >35 ms. Note that the exponential function is shifted in time by 35 ms. 3(. *'/. 6(1 v L ] 150 mH _? |i L {0.035)s 1.48 A Figure 7.34 • The circuit shown in Fig. 7.31, for t > 35 ms. c) The 18 n resistor is in the circuit only during the first 35 ms of the switching sequence. During this interval, the voltage across the resistor is v, = 0.154(6<?" 40 ') dt = -36e~ 40 ' V, 0 < t < 35 ms. 238 Response of First-Order RL and RC Circuits The power dissipated in the 18 ft resistor is p = -^ = 72<r 80 ' W, 0 < t < 35 ms. 18 Hence the energy dissipated is /.().035 W = 72<T hl V/ 80/ .7(1 72 -SO/ 0.035 0 -80 = 0.9(1 - ef 2 - 8 ) = 845.27 rnJ. Tlie initial energy stored in the 150 mH inductor is Wi = j(0.15)(36) = 2.7 J = 2700 mj. Therefore (845.27/2700) x 100, or 31.31% of the initial energy stored in the 150 mH inductor is dissipated in the 18 ft resistor. d) For 0 < i < 35 ms, the voltage across the 3 ft resistor is *>m »L (3) = r L 40/ Therefore the energy dissipated in the 3 ft resis- tor in the first 35 ms is .().035 WMl "I44e -SO/ dt Jo 3 = 0.6(1 - e' 2 *) = 563.51 mJ. For t > 35 ms, the current in the 3 O resistor is ha = <L = (6e-^)e~W-^ A. Hence the energy dissipated in the 3 ft resistor for t > 35 ms is t%i = I iiti X3dt /().035 f 3(36)e- 2 *e- ia *- tt,B5 >A .7().035 108e - - 8 x ,-120(/-0.035) 120 0.035 108 _•) o _ . __, —,»- 54.73mJ. The total energy dissipated in the 3 ("1 resistor is w 3(l (total) = 563.51 + 54.73 = 618.24 mJ. The percentage of the initial energy stored is 618.24 2700 X 100 = 22.90%. e) Because the 6 ft resistor is in series with the 3 12 resistor, the energy dissipated and the percent- age of the initial energy stored will be twice that of the 3 ft resistor: w 6n (total) = 1236.48 mJ, and the percentage of the initial energy stored is 45.80%. We check these calculations by observ- ing that 1236.48 4- 618.24 + 845.27 = 2699.99 mJ and 31.31 + 22.90 + 45.80 = 100.01%. The small discrepancies in the summations are the result of roundoff errors. 7.5 Sequential Switching 239 Example 7.12 Analyzing an RC Circuit that has Sequential Switching The uncharged capacitor in the circuit shown in Fig. 7.35 is initially switched to terminal a of the three-position switch. At t — 0, the switch is moved to position b, where it remains for 15 ms. After the 15 ms delay, the switch is moved to position c, where it remains indefinitely. a) Derive the numerical expression for the voltage across the capacitor. b) Plot the capacitor voltage versus time. c) When will the voltage on the capacitor equal 200 V? Solution a) At the instant the switch is moved to position b, the initial voltage on the capacitor is zero. If the switch were to remain in position b, the capacitor would eventually charge to 400 V. The time con- stant of the circuit when the switch is in position b is 10 ms. Therefore we can use Eq. 7.59 with t () = 0 to write the expression for the capacitor voltage: v = 400 + (0 - 400)e -100/ = (400 - 400e" m ") V, 0 =s t < 15 ms. Note that, because the switch remains in posi- tion b for only 15 ms, this expression is valid only for the time interval from 0 to 15 ms. After the switch has been in this position for 15 ms, the voltage on the capacitor will be y(15ms) = 400 - 400e~ 15 = 310.75 V. Therefore, when the switch is moved to position c, the initial voltage on the capacitor is 310.75 V. With the switch in position c, the final value of the capacitor voltage is zero, and the time con- stant is 5 ms. Again, we use Eq. 7.59 to write the expression for the capacitor voltage: v = 0 + (310.75 - ()) e -200(/-o.oi5) = 310.75e- 20t) ('- 0()15 >V, 15ms si/. 400 V( ) L v(t)^:0AfjLF Figure 7.35 A The circuit for Example 7.12. In writing the expression for ?;, we recognized that r<) = 15 ms and that this expression is valid only for t ^ 15 ms. b) Figure 7.36 shows the plot of v versus t. c) The plot in Fig. 7.36 reveals that the capacitor voltage will equal 200 V at two different times: once in the interval between 0 and 15 ms and once after 15 ms. We find the first time by solving the expression 200 = 400 - 400<T 10,, \ which yields t\ = 6.93 ms. We find the second time by solving the expression 200 = 310.756>- 2(,0( ';-°- ,),5) . In this case, u = 17.20 ms. v = 4()0-4()0t'" ,m/ v = 3U).75e 2m ' {m5) t (ms) Figure 7.36 • The capacitor voltage for Example 7.12. 240 Response of First-Order RL and RC Circuits ^ASSESSMENT PROBLEMS Objective 3—Know how to analyze circuits with sequential switching 7.7 In the circuit shown, switch 1 has been closed and switch 2 has been open for a long time. At t = 0, switch 1 is opened. Then 10 ms later, switch 2 is closed. Find a) v c (t) for 0 < f < 0.01 s, b) v c (t) for t > 0.01 s, c) the total energy dissipated in the 25 kft resistor, and d) the total energy dissipated in the 100 kO resistor. (U 60kn r=10m " )10mAf40kft 25kft£lAtF Answer: (a) 80e~ 40/ V; (b) 53.63e- 5 °('- a01 W; (c) 2.91 mJ; (d) 0.29 mJ. NOTE: Also try Chapter Problems 7.71 and 7.78. 7.8 Switch a in the circuit shown has been open for a long time, and switch b has been closed for a long time. Switch a is closed at t = 0 and, after remaining closed for 1 s, is opened again. Switch b is opened simultaneously, and both switches remain open indefinitely. Determine the expression for the inductor current i that is valid when (a)0sf<h and (b) t > 1 s. Answer: (a) (3 - 3e _a5 ') A, 0 < f < 1 s; (b) (-4.8 + 5.98tf~ l - 25( '~ 1> ) A, t > 1 s. 7.6 Unbounded Response A circuit response may grow, rather than decay, exponentially with time. This type of response, called an unbounded response, is possible if the cir- cuit contains dependent sources. In that case, the Thevenin equivalent resistance with respect to the terminals of either an inductor or a capacitor may be negative. This negative resistance generates a negative time con- stant, and the resulting currents and voltages increase without limit. In an actual circuit, the response eventually reaches a limiting value when a component breaks down or goes into a saturation state, prohibiting fur- ther increases in voltage or current. When we consider unbounded responses, the concept of a final value is confusing. Hence, rather than using the step response solution given in Eq. 7.59, we derive the differential equation that describes the circuit con- taining the negative resistance and then solve it using the separation of variables technique. Example 7.13 presents an exponentially growing response in terms of the voltage across a capacitor. 7.7 The Integrating Amplifier 241 Example 7.13 Finding the Unbounded Response in an RC Circuit a) When the switch is closed in the circuit shown in Fig. 7.37, the voltage on the capacitor is 10 V. Find the expression for v a for t > 0. b) Assume that the capacitor short-circuits when its terminal voltage reaches 150 V. How many milliseconds elapse before the capacitor short- circuits? 20 kn Figure 7.37 • The circuit for Example 7.13. Solution a) To find the Thevenin equivalent resistance with respect to the capacitor terminals, we use the test- source method described in Chapter 4. Figure 7.38 shows the resulting circuit, where v r is the test voltage and i T is the test current. For Vj expressed in volts, we obtain ir = TT: ~ 7( ) + —- mA. ' 10 W 20 Solving for the ratio Vj/ir yields the Thevenin resistance: Km = — = -5 kn. i T With this Thevenin resistance, we can simplify the circuit shown in Fig. 7.37 to the one shown in Fig. 7.39. 'T Figure 7.38 • The test-source method used to find i? Th . -5kO Figure 7.39 A A simplification of the circuit shown in Fig. 7.37. For t S: 0, the differential equation describing the circuit shown in Fig. 7.39 is (5 X 10 -6 )—^ - -r X 1()- J = 0. dt 5 Dividing by the coefficient of the first derivative yields dv 0 dt 4(h>„ = 0. We now use the separation of variables technique to find v ( ,(t): v () (t) = lOe 40 ' V, t>0. b) v a = 150 V when e m = 15. Therefore, 40r = In 15, and t = 67.70 ms. NOTE: Assess your understanding of this material by trying Chapter Problems 7.85 and 7.87. The fact that interconnected circuit elements may lead to ever- increasing currents and voltages is important to engineers. If such inter- connections are unintended, the resulting circuit may experience unexpected, and potentially dangerous, component failures. 7.7 The Integrating Amplifier Recall from the introduction to Chapter 5 that one reason for our interest in the operational amplifier is its use as an integrating amplifier. We are now ready to analyze an integrating-amplifier circuit, which is shown in Fig. 7.40. The purpose of such a circuit is to generate an output voltage proportional to the integral of the input voltage. In Fig. 7.40, we added the branch cur- rents if and / v , along with the node voltages v n and v p , to aid our analysis. Figure 7.40 • An integrating amplifier. 242 Response of First-Order RL and RC Circuits »i K„ - 2f, Figure 7.41 • An input voltage signal. Figure 7.42 • The output voltage of an integrating amplifier. We assume that the operational amplifier is ideal. Thus we take advantage of the constraints Because v„ = 0, if + i s = 0, v n = v p . i =^- l ^^~dt Hence, from Eqs. 7.61,7.63, and 7.64, dv a _ 1 dt ~ R s C f v s . (7.61) (7.62) (7.63) (7.64) (7.65) Multiplying both sides of Eq. 7.65 by a differential time dt and then inte- grating from f () to t generates the equation Vo(0 \ R<C v s dy + v 0 (t () ). (7.66) /•/A, In Eq. 7.66, t () represents the instant in time when we begin the integration. Thus u„(?o) is the value of the output voltage at that time. Also, because v n = v p = 0, v o (t 0 ) is identical to the initial voltage on the feedback capacitor C/. Equation 7.66 states that the output voltage of an integrating ampli- fier equals the initial value of the voltage on the capacitor plus an inverted (minus sign), scaled (l/R s Cf) replica of the integral of the input voltage. If no energy is stored in the capacitor when integration commences, Eq. 7.66 reduces to vM = - i R S C V s dy. (7.67) f A, If v s is a step change in a dc voltage level, the output voltage will vary lin- early with time. For example, assume that the input voltage is the rectan- gular voltage pulse shown in Fig. 7.41. Assume also that the initial value of v a (t) is zero at the instant v s steps from 0 to V m . A direct application of Eq. 7.66 yields v„ = 1 V m t + 0, 0 < t < t h (7.68) When t lies between t\ and 2t u 1 v, t = - #sQ Jt {-V, n )dy - R S C, V m h RcCi 2V m R x c t t h t L < t < 2t v (7.69) Figure 7.42 shows a sketch of v ( ,(t) versus t. Clearly, the output voltage is an inverted, scaled replica of the integral of the input voltage. The output voltage is proportional to the integral of the input voltage only if the op amp operates within its linear range, that is, if it doesn't sat- urate. Examples 7.14 and 7.15 further illustrate the analysis of the inte- grating amplifier. 7.7 The Integrating Amplifier 243 Example 7.14 Analyzing an Integrating Amplifier Assume that the numerical values for the signal voltage shown in Fig. 7.41 are V m = 50 mV and t\ = 1 s. This signal voltage is applied to the integrating-amplifier circuit shown in Fig. 7.40. The circuit parameters of the amplifier are R s = 100 kfl, Cf = 0.1 ^iF, and Vcc = 6 V. The initial voltage on the capacitor is zero. a) Calculate v a (t). b) Plot v () (t) versus t. Solution a) For 0 < t < 1 s, -1 " (100 X 10 3 )(0.1 x 10" 6 ) = St V, 0 < f < 1 s. 50 X 10~ 3 f + 0 For 1 < t < 2 s, v <t = (5r - 10) V. b) Figure 7.43 shows a plot of v„(t) versus r. »„(0(V)* 2 t{t) Figure 7.43 • The output voltage for Example 7.14. Example 7.15 Analyzing an Integrating Amplifier that has Sequential Switching At the instant the switch makes contact with termi- nal a in the circuit shown in Fig. 7.44, the voltage on the 0.1 /AF capacitor is 5 V. The switch remains at terminal a for 9 ms and then moves instantaneously to terminal b. How many milliseconds after making contact with terminal b does the operational ampli- fier saturate? Figure 7.44 • The circuit for Example 7.15. Solution The expression for the output voltage during the time the switch is at terminal a is 1 io- 2 7o = (-5 + 10000 v (-10)^ Thus, 9 ms after the switch makes contact with ter- minal a, the output voltage is -5 + 9, or 4 V. The expression for the output voltage after the switch moves to terminal b is &dy 10" 2 9X 1()- = 4 - 800(t - 9 X 10" 3 ) = (11.2 - 8000 V. During this time interval, the voltage is decreas- ing, and the operational amplifier eventually satu- rates at —6 V. Therefore we set the expression for v a equal to —6 V to obtain the saturation time t s : 11.2 - 800/, = -6, or t s = 21.5 ms. Thus the integrating amplifier saturates 21.5 ms after making contact with terminal b. 244 Response of First-Order RL and RC Circuits From the examples, we see that the integrating amplifier can perform the integration function very well, but only within specified limits that avoid saturating the op amp. The op amp saturates due to the accumula- tion of charge on the feedback capacitor. We can prevent it from saturat- ing by placing a resistor in parallel with the feedback capacitor. We examine such a circuit in Chapter 8. Note that we can convert the integrating amplifier to a differentiating amplifier by interchanging the input resistance R s and the feedback capac- itor Cf. Then v a = -R S C, ~dt (7.70) We leave the derivation of Eq. 7.70 as an exercise for you. The differentiat- ing amplifier is seldom used because in practice it is a source of unwanted or noisy signals. Finally, we can design both integrating- and differentiating-amplifier circuits by using an inductor instead of a capacitor. However, fabricating capacitors for integrated-circuit devices is much easier, so inductors are rarely used in integrating amplifiers. ^ASSESSMENT PROBLEMS Objective 4—Be able to analyze op amp circuits containing resistors and a single capacitor 7.9 There is no energy stored in the capacitor at the time the switch in the circuit makes contact with terminal a. The switch remains at position a for 32 ms and then moves instantaneously to position b. How many milliseconds after mak- ing contact with terminal a does the op amp saturate? 7.10 a) When the switch closes in the circuit shown, there is no energy stored in the capacitor. How long does it take to saturate the op amp? b) Repeat (a) with an initial voltage on the capacitor of 1 V, positive at the upper terminal. 40 m 10 kO -AW 40 kn /VW v„ £ 6.8 kO, Answer: 262 ms. NOTE: Also try Chapter Problems 7.95 and 7.96. Answer: (a) 1.11 ms; (b) 1.76 ms. Practical Perspective 245 Practical Perspective A Flashing Light Circuit We are now ready to analyze the flashing light circuit introduced at the start of this chapter and shown in Fig. 7.45. The lamp in this circuit starts to conduct whenever the lamp voltage reaches a value V max . During the time the lamp conducts, it can be modeled as a resistor whose resist- ance is R L . The lamp will continue to conduct until the lamp voltage drops to the value V^ lin . When the lamp is not conducting, it behaves as an open circuit. Before we develop the analytical expressions that describe the behav- ior of the circuit, let us develop a feel for how the circuit works by noting the following. First, when the lamp behaves as an open circuit, the dc voltage source will charge the capacitor via the resistor R toward a value of V s volts. However, once the lamp voltage reaches V max , it starts con- ducting and the capacitor will start to discharge toward the Thevenin voltage seen from the terminals of the capacitor. But once the capacitor voltage reaches the cutoff voltage of the lamp (V min ), the lamp will act as an open circuit and the capacitor will start to recharge. This cycle of charging and discharging the capacitor is summarized in the sketch shown in Fig. 7.46. In drawing Fig. 7.46 we have chosen t = 0 at the instant the capacitor starts to charge. The time t 0 represents the instant the lamp starts to con- duct, and t c is the end of a complete cycle. We should also mention that in constructing Fig. 7.46 we have assumed the circuit has reached the repeti- tive stage of its operation. Our design of the flashing light circuit requires we develop the equation for Vjjj) as a function of V^ax, Vj^^, V s , R, C, and R L for the intervals 0 to t 0 and t Q to f c . To begin the analysis, we assume that the circuit has been in operation for a long time. Let t = 0 at the instant when the lamp stops conducting. Thus, at t = 0, the lamp is modeled as an open circuit, and the voltage drop across the lamp is V^ in , as shown in Fig. 7.47. From the circuit, we find vd°°) = V s , t>i,(0) = v min , T = RC. R ->VvV V. c: v L Lamp Figure 7.45 A A flashing light circuit. V L{t) y v max V • etc. Figure 7.46 • Lamp voltage versus time for the circuit in Fig. 7.45. + R 4- v L Figure 7.47 A The flashing light circuit at t = 0, when the lamp is not conducting. Thus, when the lamp is not conducting, V L (t) = Vs + (Knin ~ V s )e-'S RC . How long does it take before the lamp is ready to conduct? We can find this time by setting the expression for v L (t) equal to V max and solving for t. If we call this value t 0 , then t n = RC In V • - V y min v s V — V K max y s When the lamp begins conducting, it can be modeled as a resistance R L , as seen in Fig. 7.48. In order to find the expression for the voltage drop + T R C^ + ^v L 1 R, Figure 7.48 • The flashing light circuit at t = t 0 when the lamp is conducting.