irwin07_296-368hr.qxd 28-07-2010 11:34 Page 296 7 C H A PT E R FIRST- AND SECOND-ORDER TRANSIENT CIRCUITS THE LEARNING GOALS FOR THIS CHAPTER ARE: ■ Be able to calculate initial values for inductor cur- rents and capacitor voltages in transient circuits ■ Know how to calculate voltages and currents in first- order transient circuits ■ Know how to calculate voltages and currents in second-order transient circuits E.S.A./SIPA/NewsCom S Sea Ice Measurements Data clearly demonstrate an ongoing using radar backscatter to map transitions in sea ice thickness. rise in global temperature. NASA’s Goddard Institute for Space Sea ice growth and melt measurements from radar during all Studies has reported that we have just experienced the warmest summer-to-winter seasons confirm that the ice is disappearing. decade (2000 to 2009) on record. Measuring the environmental This chapter describes voltage and current transitions in cir- impact of this change – and deciding how to respond to it – is a cuits resulting from switching of constant-level sources from one major challenge facing scientists and policymakers today and value to another. Capacitors, inductors, and resistors are con- will be for years to come. nected in series or parallel, modeling transitions for simple to Electronic devices are essential to efforts to measure impacts more complex circuits. Extending to a square-wave pulse voltage such as the reduction in polar ice. Polar ice regulates global cli- source demonstrates multiple switching dynamics. The transi- mate because it reflects about 80% of the incoming sunlight. tions in circuit voltages and currents that you will study in this Radar altimetry data from satellites is being used to measure the chapter are principles that you can see applied in a wide array of effect of warming on polar glaciers. Low-flying aircraft making devices. While some seem mundane, others, such as those that repeated runs across Arctic, Antarctic, and Greenland ice beds are measure sea ice, have a major impact on our future. 296 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 297 SECTION 7.1 In this chapter we perform what is normally referred to as a transient analysis. We begin our analysis with first-order circuits—that is, those that contain only a single storage element. When only a single storage element is present in the network, the network can be described by a first-order differential equation. Our analysis involves an examination and description of the behavior of a circuit as a function of time after a sudden change in the network occurs due to switches opening or closing. Because of the presence of one or more storage elements, the circuit response to a sudden change will go through a transition period prior to settling down to a steady-state value. It is this transition period that we will examine carefully in our transient analysis. One of the important parameters that we will examine in our transient analysis is the circuit’s time constant. This is a very important network parameter because it tells us how fast the circuit will respond to changes. We can contrast two very different systems to obtain a feel for the parameter. For example, consider the model for a room air-conditioning system and the model for a single-transistor stage of amplification in a computer chip. If we change the setting for the air conditioner from 70 degrees to 60 degrees, the unit will come on and the room will begin to cool. However, the temperature measured by a thermometer in the room will fall very slowly and, thus, the time required to reach the desired temperature is long. However, if we send a trigger signal to a transistor to change state, the action may take only a few nanoseconds. These two systems will have vastly different time constants. Our analysis of first-order circuits begins with the presentation of two techniques for performing a transient analysis: the differential equation approach, in which a differential equation is written and solved for each network, and a step-by-step approach, which takes advantage of the known form of the solution in every case. In the second-order case, both an inductor and a capacitor are present simultaneously, and the network is described by a second-order differential equation. Although the RLC circuits are more complicated than the first-order single storage circuits, we will follow a development similar to that used in the first-order case. Our presentation will deal only with very simple circuits, since the analysis can quickly become complicated for networks that contain more than one loop or one nonreference node. Furthermore, we will demonstrate a much simpler method for handling these circuits when we cover the Laplace transform later in this book. We will analyze several networks in which the parameters have been chosen to illustrate the different types of circuit response. Finally, a number of application-oriented examples are presented and discussed. We begin our discussion by recalling that in Chapter 6 we found that capacitors and inductors were capable of storing electric energy. In the case of a charged capacitor, the energy is stored in the electric field that exists between the positively and negatively charged plates. This stored energy can be released if a circuit is somehow connected across the capacitor that provides a path through which the negative charges move to the positive charges. As we know, this movement of charge constitutes a current. The rate at which the energy is discharged is a direct function of the parameters in the circuit that is connected across the capacitor’s plates. As an example, consider the flash circuit in a camera. Recall that the operation of the flash circuit, from a user standpoint, involves depressing the push button on the camera that triggers both the shutter and the flash and then waiting a few seconds before repeating the process to take the next picture. This operation can be modeled using the circuit in Fig. 7.1a. The voltage source and resistor RS model the batteries that power the camera and flash. The capacitor models the energy storage, the switch models the push button, and finally the resistor R models the xenon flash lamp. Thus, if the capacitor is charged, when the switch is closed, the capacitor voltage drops and energy is released through the xenon lamp, producing the flash. In practice this energy release takes about a millisecond, and the discharge time is a function of the elements in the circuit. When the push button is released and the switch is then opened, the battery begins to recharge the capacitor. Once again, the time required to charge the capacitor is a function of the circuit elements. The discharge and charge cycles are graphically illustrated in Fig. 7.1b. Although the discharge time is very fast, it is not instantaneous. To provide further insight into this phenomenon, consider what INTRODUCTION 297 7.1 Introduction irwin07_296-368hr.qxd 298 28-07-2010 CHAPTER 7 11:34 Page 298 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S Figure 7.1 Discharge time vC(t) Diagrams used to describe a camera’s flash circuit. Vo RS + VS Charge time C vC(t) R - t [(R) Xenon lamp] (a) (b) + vC(t) - C R (c) we might call a free-body diagram of the right half of the network in Fig. 7.1a, as shown in Fig. 7.1c (that is, a charged capacitor that is discharged through a resistor). When the switch is closed, KCL for the circuit is C dvC(t) vC(t) + dt R = 0 or dvC(t) + dt 1 v (t) = 0 RC C In the next section we will demonstrate that the solution of this equation is vC(t) = Vo e-t兾RC Note that this function is a decaying exponential and the rate at which it decays is a function of the values of R and C. The product RC is a very important parameter, and we will give it a special name in the following discussions. 7.2 First-Order Circuits In our study of first-order transient circuits we will show that the solution of these circuits (i.e., finding a voltage or current) requires us to solve a first-order differential equation of the form G E N E R A L F O R M O F T H E R E S P O N S E EQ U AT I O N S dx(t) dt + ax(t) = f(t) 7.1 Although a number of techniques may be used for solving an equation of this type, we will obtain a general solution that we will then employ in two different approaches to transient analysis. A fundamental theorem of differential equations states that if x(t) = xp(t) is any solution to Eq. (7.1), and x(t) = xc(t) is any solution to the homogeneous equation dx(t) dt + ax(t) = 0 7.2 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 299 SECTION 7.2 F I R S T- O R D E R C I R C U I T S then x(t) = xp(t) + xc(t) 7.3 is a solution to the original Eq. (7.1). The term xp(t) is called the particular integral solution, or forced response, and xc(t) is called the complementary solution, or natural response. At the present time we confine ourselves to the situation in which f(t)=A (i.e., some constant). The general solution of the differential equation then consists of two parts that are obtained by solving the two equations dx p(t) dt dx c(t) dt + ax p(t) = A 7.4 + ax c(t) = 0 7.5 Since the right-hand side of Eq. (7.4) is a constant, it is reasonable to assume that the solution xp(t) must also be a constant. Therefore, we assume that xp(t) = K1 7.6 Substituting this constant into Eq. (7.4) yields K1 = A a 7.7 = -a 7.8 Examining Eq. (7.5), we note that dx c(t)兾dt xc(t) This equation is equivalent to d C ln xc(t)D = -a dt Hence, ln xc(t) = -at + c and therefore, xc(t) = K2 e-at 7.9 Thus, a solution of Eq. (7.1) is x(t) = xp(t) + xc(t) = A + K2 e-at a 7.10 The constant K2 can be found if the value of the independent variable x(t) is known at one instant of time. Eq. (7.10) can be expressed in general in the form x(t) = K1 + K2 e-t兾 7.11 Once the solution in Eq. (7.11) is obtained, certain elements of the equation are given names that are commonly employed in electrical engineering. For example, the term K1 is referred to as the steady-state solution: the value of the variable x(t) as t S q when the second term becomes negligible. The constant is called the time constant of the circuit. Note that the second term in Eq. (7.11) is a decaying exponential that has a value, if 7 0, of K2 for t=0 and a value of 0 for t=q. The rate at which this exponential decays is determined by the time constant . A graphical picture of this effect is shown in Fig. 7.2a. As can be seen from the figure, the value of xc(t) has fallen from K2 to a value of 0.368K2 in one time constant, a drop of 63.2%. In two time constants the value of xc(t) has fallen to 0.135K2 , 299 irwin07_296-368hr.qxd 300 28-07-2010 CHAPTER 7 11:34 Page 300 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S Figure 7.2 xc(t)=K2e–t/ Time-constant illustrations. K2 0.368K2 0.632 GG 0 2 3 4 3 4 t (a) e–t/ 1.0 0.8 0.6 0.4 0.2 =4 s =0.5 s 0 1 2 t (b) a drop of 63.2% from the value at time t = . This means that the gap between a point on the curve and the final value of the curve is closed by 63.2% each time constant. Finally, after five time constants, xc(t) = 0.0067K2 , which is less than 1%. An interesting property of the exponential function shown in Fig. 7.2a is that the initial slope of the curve intersects the time axis at a value of t = . In fact, we can take any point on the curve, not just the initial value, and find the time constant by finding the time required to close the gap by 63.2%. Finally, the difference between a small time constant (i.e., fast response) and a large time constant (i.e., slow response) is shown in Fig. 7.2b. These curves indicate that if the circuit has a small time constant, it settles down quickly to a steady-state value. Conversely, if the time constant is large, more time is required for the circuit to settle down or reach steady state. In any case, note that the circuit response essentially reaches steady state within five time constants (i.e., 5). Note that the previous discussion has been very general in that no particular form of the circuit has been assumed—except that it results in a first-order differential equation. A N A LYS I S T EC H N I Q U E S The Differential Equation Approach Eq. (7.11) defines the general form of the solution of first-order transient circuits; that is, it represents the solution of the differential equation that describes an unknown current or voltage anywhere in the network. One of the ways that we can arrive at this solution is to solve the equations that describe the network behavior using what is often called the state-variable approach. In this technique we write the equation for the voltage across the capacitor and/or the equation for the current through the inductor. Recall from Chapter 6 that these quantities cannot change instantaneously. Let us first illustrate this technique in the general sense and then examine two specific examples. Consider the circuit shown in Fig. 7.3a. At time t = 0, the switch closes. The KCL equation that describes the capacitor voltage for time t 7 0 is C v(t) - VS dv(t) dt + R = 0 or v(t) dv(t) dt + RC = VS RC irwin07_296-368hr.qxd 28-07-2010 11:34 Page 301 SECTION 7.2 +vR(t)- t=0 R VS ± – 301 v(t), vR(t) v(t) t=0 F I R S T- O R D E R C I R C U I T S VS R VS ± – C L i(t) t (a) (b) From our previous development, we assume that the solution of this first-order differential equation is of the form v(t) = K1 + K2 e-t兾 Substituting this solution into the differential equation yields - K2 -t兾 K2 -t兾 K1 VS + e + = e RC RC RC Equating the constant and exponential terms, we obtain K1 = VS = RC Therefore, v(t) = VS + K2 e-t兾RC where VS is the steady-state value and RC is the network’s time constant. K2 is determined by the initial condition of the capacitor. For example, if the capacitor is initially uncharged (that is, the voltage across the capacitor is zero at t = 0), then 0 = VS + K2 or K2 = -VS Hence, the complete solution for the voltage v(t) is v(t) = VS - VS e-t兾RC The circuit in Fig. 7.3b can be examined in a similar manner. The KVL equation that describes the inductor current for t 7 0 is L di(t) + Ri(t) = VS dt A development identical to that just used yields i(t) = R VS -a b t + K2 e L R where VS兾R is the steady-state value and L/R is the circuit’s time constant. If there is no initial current in the inductor, then at t = 0 0 = VS + K2 R and K2 = -VS R (c) Figure 7.3 (a) RC circuit, (b) RL circuit, (c) plot of the capacitor voltage in (a) and resistor voltage in (b). irwin07_296-368hr.qxd 302 28-07-2010 CHAPTER 7 11:34 Page 302 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S Hence, VS VS - RL t e R R i(t) = is the complete solution. Note that if we wish to calculate the voltage across the resistor, then vR(t) = Ri (t) = VS A 1 - e R -Lt B Therefore, we find that the voltage across the capacitor in the RC circuit and the voltage across the resistor in the RL circuit have the same general form. A plot of these functions is shown in Fig. 7.3c. EXAMPLE Consider the circuit shown in Fig. 7.4a. Assuming that the switch has been in position 1 for a long time, at time t=0 the switch is moved to position 2. We wish to calculate the current i(t) for t 7 0. 7.1 t=0 1 v(t) R1 i(t) 6 k 6 k 12 V ± V – S 2 C 100 F R2 3 k 12 V ± – vC(0–) 12 V 3 k - (b) t=0– (a) R1 + v(t) i(t) (mA) i(t) 4 — 3 ± – C R2 t (c) (d) Figure 7.4 Analysis of RC circuits. SOLUTION At t = 0- the capacitor is fully charged and conducts no current since the capacitor acts like an open circuit to dc. The initial voltage across the capacitor can be found using voltage division. As shown in Fig. 7.4b, vC(0-) = 12 a 3k b = 4V 6k + 3k The network for t 7 0 is shown in Fig. 7.4c. The KCL equation for the voltage across the capacitor is v(t) R1 + C v(t) dv(t) dt + R2 Using the component values, the equation becomes dv(t) dt + 5v(t) = 0 = 0 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 303 SECTION 7.2 F I R S T- O R D E R C I R C U I T S 303 The form of the solution to this homogeneous equation is v(t) = K2 e-t兾 If we substitute this solution into the differential equation, we find that = 0.2 s. Thus, v(t) = K2 e-t兾0.2 V Using the initial condition vC(0-) = vC(0+) = 4 V, we find that the complete solution is v(t) = 4e-t兾0.2 V Then i(t) is simply i(t) = v(t) R2 or i(t) = 4 -t兾0.2 e mA 3 The switch in the network in Fig. 7.5a opens at t=0. Let us find the output voltage vo(t) for t 7 0. At t = 0- the circuit is in steady state and the inductor acts like a short circuit. The initial current through the inductor can be found in many ways; however, we will form a Thévenin equivalent for the part of the network to the left of the inductor, as shown in Fig. 7.5b. From this network we find that I1 = 4 A and Voc = 4 V. In addition, RTh=1 ⍀. Hence, iL(0-) obtained from Fig. 7.5c is iL(0-) = 4兾3 A. The network for t 7 0 is shown in Fig. 7.5d. Note that the 4-V independent source and the 2-ohm resistor in series with it no longer have any impact on the resulting circuit. The KVL equation for the circuit is -VS1 + R1 i(t) + L di(t) dt + R3 i(t) = 0 which with the component values reduces to di(t) dt + 2i(t) = 6 The solution to this equation is of the form i(t) = K1 + K2 e-t兾 which when substituted into the differential equation yields K1 = 3 = 1兾2 Therefore, i(t) = A3 + K2 e-2t B A Evaluating this function at the initial condition, which is iL(0-) = iL(0+) = i(0) = 4兾3 A we find that K2 = -5 3 EXAMPLE 7.2 SOLUTION irwin07_296-368hr.qxd 304 28-07-2010 CHAPTER 7 11:34 Page 304 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S R1 L 2 2H + I1 t=0 ± VS 1 – 12 V R2 R3 2 2 vo(t) 12 V 4V + 2 – ± V S2 2 ± – Voc – ± - 4V - (a) (b) iL(0–) + 1 4V ± – 2 vo(0–) (c) i(t) L R1 2H 2 12 V ± VS 1 – 2 + R3 2 vo(t) vo(t) (V) 6 4V – ± 8 — 3 (d) Figure 7.5 (e) Hence, Analysis of an RL circuit. t i(t) = a 3 - 5 -2t e bA 3 vo(t) = 6 - 10 -2t e V 3 and then A plot of the voltage vo(t) is shown in Fig. 7.5e. irwin07_296-368hr.qxd 28-07-2010 11:34 Page 305 SECTION 7.2 F I R S T- O R D E R C I R C U I T S Learning Assessments E7.1 Find vC(t) for t 7 0 in the circuit shown in Fig. E7.1. ANSWER: vC(t) = 8e-t兾0.6 V. t=0 3 k 4 k + 12 V + ± – vc(t) 100 F - 2 k vo(t) - Figure E7.1 E7.2 Use the differential equation approach to find vo(t) for t>0 in Fig. E7.2. Plot the response. t=0 6 k 9V + – + vo(t) 6 k - 2.5 F 6 k ANSWER: vo(t)= 12-5e-t/0.015 V. 2 mA Figure E7.2 E7.3 In the circuit shown in Fig. E7.3, the switch opens at t = 0. Find i1(t) for t 7 0. t=0 12 V ANSWER: i1(t) = 1e-9t A. i1(t) 6 12 i2(t) 2H ± – Figure E7.3 E7.4 Use the differential equation approach to find i(t) for t>0 in Fig. E7.4. 3 k t=0 2 k 6 mH i(t) Figure E7.4 2 k 12 V + – + 6V ANSWER: i(t)= -6 -2+6e-t/5⫻10 mA. 305 irwin07_296-368hr.qxd 306 28-07-2010 CHAPTER 7 11:34 Page 306 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S The Step-by-Step Approach In the previous analysis technique, we derived the differential equation for the capacitor voltage or inductor current, solved the differential equation, and used the solution to find the unknown variable in the network. In the very methodical technique that we will now describe, we will use the fact that Eq. (7.11) is the form of the solution and we will employ circuit analysis to determine the constants K1 , K2 , and . From Eq. (7.11) we note that as t S q, e-at S 0 and x(t) = K1 . Therefore, if the circuit is solved for the variable x(t) in steady state (i.e., t S q) with the capacitor replaced by an open circuit [v is constant and therefore i = C(dv兾dt) = 0] or the inductor replaced by a short circuit [i is constant and therefore v = L(di兾dt) = 0], then the variable x(t) = K1 . Note that since the capacitor or inductor has been removed, the circuit is a dc circuit with constant sources and resistors, and therefore only dc analysis is required in the steady-state solution. The constant K2 in Eq. (7.11) can also be obtained via the solution of a dc circuit in which a capacitor is replaced by a voltage source or an inductor is replaced by a current source. The value of the voltage source for the capacitor or the current source for the inductor is a known value at one instant of time. In general, we will use the initial condition value since it is generally the one known, but the value at any instant could be used. This value can be obtained in numerous ways and is often specified as input data in a statement of the problem. However, a more likely situation is one in which a switch is thrown in the circuit and the initial value of the capacitor voltage or inductor current is determined from the previous circuit (i.e., the circuit before the switch is thrown). It is normally assumed that the previous circuit has reached steady state, and therefore the voltage across the capacitor or the current through the inductor can be found in exactly the same manner as was used to find K1 . Finally, the value of the time constant can be found by determining the Thévenin equivalent resistance at the terminals of the storage element. Then = RTh C for an RC circuit, and = L兾RTh for an RL circuit. Let us now reiterate this procedure in a step-by-step fashion. Problem-Solving Strategy Using the Step-byStep Approach Step 1. We assume a solution for the variable x(t) of the form x(t) = K1 + K2 e-t兾. Step 2. Assuming that the original circuit has reached steady state before a switch was thrown (thereby producing a new circuit), draw this previous circuit with the capacitor replaced by an open circuit or the inductor replaced by a short circuit. Solve for the voltage across the capacitor, vC(0-), or the current through the inductor, iL(0-), prior to switch action. Step 3. Recall from Chapter 6 that voltage across a capacitor and the current flowing through an inductor cannot change in zero time. Draw the circuit valid for t = 0+ with the switches in their new positions. Replace a capacitor with a voltage source vC(0+) = vC(0-) or an inductor with a current source of value iL(0+) = iL(0-). Solve for the initial value of the variable x(0+). Step 4. Assuming that steady state has been reached after the switches are thrown, draw the equivalent circuit, valid for t 7 5, by replacing the capacitor by an open circuit or the inductor by a short circuit. Solve for the steady-state value of the variable x(t)|t 7 5 ⬟ x(q) Step 5. Since the time constant for all voltages and currents in the circuit will be the same, it can be obtained by reducing the entire circuit to a simple series circuit containing a voltage source, resistor, and a storage element (i.e., capacitor or inductor) by forming a simple Thévenin equivalent circuit at the terminals of irwin07_296-368hr.qxd 28-07-2010 11:34 Page 307 SECTION 7.2 F I R S T- O R D E R C I R C U I T S 307 the storage element. This Thévenin equivalent circuit is obtained by looking into the circuit from the terminals of the storage element. The time constant for a circuit containing a capacitor is = RTh C, and for a circuit containing an inductor it is = L兾RTh . Step 6. Using the results of steps 3, 4, and 5, we can evaluate the constants in step 1 as x(0+) = K1 + K2 x(q) = K1 Therefore, K1 = x(q), K2 = x(0+) - x(q), and hence the solution is x(t) = x(q) + [x(0+) - x(q)]e-t兾 Keep in mind that this solution form applies only to a first-order circuit having dc sources. If the sources are not dc, the forced response will be different. Generally, the forced response is of the same form as the forcing functions (sources) and their derivatives. Consider the circuit shown in Fig. 7.6a. The circuit is in steady state prior to time t=0, when the switch is closed. Let us calculate the current i(t) for t 7 0. Step 1. i(t) is of the form K1 + K2 e-t兾. Step 2. The initial voltage across the capacitor is calculated from Fig. 7.6b as vC (0-) = 36 - (2)(2) = 32 V Step 3. The new circuit, valid only for t = 0+, is shown in Fig. 7.6c. The value of the voltage source that replaces the capacitor is vC(0-) = vC(0+) = 32 V. Hence, 32 6k 16 mA = 3 i(0+) = Step 4. The equivalent circuit, valid for t 7 5, is shown in Fig. 7.6d. The current i(q) caused by the 36-V source is 36 2k + 6k 9 = mA 2 i(q) = Step 5. The Thévenin equivalent resistance, obtained by looking into the open-circuit terminals of the capacitor in Fig. 7.6e, is RTh = (2k)(6k) 2k + 6k = 3 k⍀ 2 Therefore, the circuit time constant is = RTh C 3 = a b A103 B(100)A10-6 B 2 = 0.15 s EXAMPLE 7.3 SOLUTION irwin07_296-368hr.qxd 308 28-07-2010 CHAPTER 7 11:34 Page 308 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 2 k 6 k i(t) 2 k 4 k 6 k i(0–) 4 k + 36 V ± – t=0 100 F 12 V ± – ± – vC(0–) 36 V - (b) (a) 2 k 36 V ± – 6 k ± – i(0±) 32 V 2 k 4 k 12 V ± – ± – 12 V ± – t=0– 6 k i(∞) 36 V 4 k 12 V ± – (d) t=∞ (c) t=0± i(t) (mA) 2 k 6 k 4 k 16 — 3 9 — 2 RTh 2 0 0.1 (e) 0.2 0.3 0.4 t(s) (f) Figure 7.6 Analysis of an RC transient circuit with a constant forcing function. Step 6. K1 = i(q) = 9 mA 2 K2 = i(0+) - i(q) = i(0+) - K1 = 16 9 3 2 = 5 mA 6 Therefore, i(t) = 36 5 + e-t兾0.15 mA 8 6 The plot is shown in Fig. 7.7 and can be compared to the sketch in Fig. 7.6f. Examination of Fig. 7.6f indicates once again that although the voltage across the capacitor is continuous at t=0, the current i(t) in the 6-k⍀ resistor jumps at t=0 from 2 mA to 5 1兾3 mA, and finally decays to 4 1兾2 mA. irwin07_296-368hr.qxd 28-07-2010 11:34 Page 309 SECTION 7.2 F I R S T- O R D E R C I R C U I T S 309 Figure 7.7 Plot for Example 7.3. The circuit shown in Fig. 7.8a is assumed to have been in a steady-state condition prior to switch closure at t=0. We wish to calculate the voltage v(t) for t 7 0. SOLUTION Step 1. v(t) is of the form K1 + K2 e-t兾. Step 2. In Fig. 7.8b we see that iL (0-) = 4 + = 6 24 a b (6)(3) 6 + 3 6 + 3 8 A 3 Step 3. The new circuit, valid only for t = 0+, is shown in Fig. 7.8c, which is equivalent to the circuit shown in Fig. 7.8d. The value of the current source that replaces the inductor is iL(0-) = iL(0+) = 8兾3 A. The node voltage v1(0+) can be determined from the circuit in Fig. 7.8d using a single-node equation, and v(0+) is equal to the difference between the source voltage and v1(0+). The equation for v1(0+) is v1(0+) - 24 4 v1(0+) + EXAMPLE 6 + v1(0+) 8 + = 0 3 12 or v1(0+) = 20 V 3 7.4 irwin07_296-368hr.qxd 310 28-07-2010 CHAPTER 7 11:34 Page 310 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 12 12 + v(t)- 4 1 4H 1 iL(0–) 4 24 V ± – 6 t=0 2 24 V ± – 6 2 (b) t=0– (a) 12 k + v(0±)4 24 V ± – + v(0±)- v1(0±) 1 4 8 —A 3 6 2 24 V ± – 8 —A 3 (c) t=0± 1 2 (d) t=0± 12 + 12 6 12 v(∞) - 1 4 1 4 24 V ± – 6 2 6 2 RTh (e) t=∞ (f) v(t) (V) 24 17.33 16 0 1 2 3 4 5 t (s) (g) Figure 7.8 Analysis of an RL transient circuit with a constant forcing function. irwin07_296-368hr.qxd 28-07-2010 11:34 Page 311 SECTION 7.2 F I R S T- O R D E R C I R C U I T S Then v(0+) = 24 - v1(0+) = 52 V 3 Step 4. The equivalent circuit for the steady-state condition after switch closure is given in Fig. 7.8e. Note that the 6-, 12-, 1-, and 2-⍀ resistors are shorted, and therefore v(q)=24 V. Step 5. The Thévenin equivalent resistance is found by looking into the circuit from the inductor terminals. This circuit is shown in Fig. 7.8f. Note carefully that RTh is equal to the 4-, 6-, and 12-⍀ resistors in parallel. Therefore, RTh=2 ⍀, and the circuit time constant is = 4 L = = 2s RTh 2 Step 6. From the previous analysis we find that K1 = v(q) = 24 K2 = v(0+) - v(q) = - 20 3 and hence that v(t) = 24 - 20 -t兾2 V e 3 From Fig. 7.8b we see that the value of v(t) before switch closure is 16 V. This value jumps to 17.33 V at t=0. A plot of this function for t>0 is shown in Fig. 7.9. Figure 7.9 Plot for Example 7.4. 311 irwin07_296-368hr.qxd 312 28-07-2010 CHAPTER 7 11:34 Page 312 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S Learning Assessments E7.5 Consider the network in Fig. E7.5. The switch opens at t = 0. Find vo(t) for t 7 0. 1 + 2 ANSWER: vo(t) = 24 1 + e-(5兾8)t V. 5 5 t=0 12 V ± – 2F 2 vo(t) 2 ± – 8V - Figure E7.5 E7.6 Consider the network in Fig. E7.6. If the switch opens at t = 0, find the output voltage vo(t) for t 7 0. ANSWER: vo(t) = 6 - 2 10 -2t e V. 3 + 2H t=0 12 V ± – 2 vo(t) 2 + 4V - Figure E7.6 E7.7 Find v0(t) for t>0 in Fig. E7.7 using the step-by-step method. 4 k 24 V + – ANSWER: vo(t)=-3.33e-t/0.06 V. t=0 + vo(t) - 6 3 6 5 F 4 k 12 V Figure E7.7 + – irwin07_296-368hr.qxd 28-07-2010 11:34 Page 313 SECTION 7.2 F I R S T- O R D E R C I R C U I T S E7.8 Find i0(t) for t>0 Fig. E7.8 using the step-by-step method. 4 313 ANSWER: i0(t)=2.1-0.6e-t/0.001 A. 3 t=0 6 100 mH + – 12 V 3 3A io(t) Figure E7.8 The circuit shown in Fig. 7.10a has reached steady state with the switch in position 1. At time t=0 the switch moves from position 1 to position 2. We want to calculate vo(t) for t>0. Step 1. vo(t) is of the form K1 + K2 e-t兾. Step 2. Using the circuit in Fig. 7.10b, we can calculate iL(0-) iA = 12 = 3A 4 Then iL(0-) = 12 + 2iA 18 = = 3A 6 6 Step 3. The new circuit, valid only for t = 0+, is shown in Fig. 7.10c. The value of the current source that replaces the inductor is iL(0-) = iL(0+) = 3 A.Because of the current source vo(0+) = (3)(6) = 18 V Step 4. The equivalent circuit, for the steady-state condition after switch closure, is given in Fig. 7.10d. Using the voltages and currents defined in the figure, we can compute vo(q) in a variety of ways. For example, using node equations we can find vo(q) from vB vB + 2iAœ vB - 36 + + = 0 2 4 6 vB iAœ = 4 vo(q) = vB + 2iAœ or, using loop equations, 36 = 2Ai1 + i2 B + 4i1 36 = 2Ai1 + i2 B + 6i2 - 2i1 vo(q) = 6i2 Using either approach, we find that vo(q)=27 V. EXAMPLE 7.5 SOLUTION irwin07_296-368hr.qxd 314 28-07-2010 CHAPTER 7 11:34 Page 314 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S Step 5. The Thévenin equivalent resistance can be obtained via voc and isc because of the presence of the dependent source. From Fig. 7.10e we note that iAfl = 36 = 6A 2 + 4 Therefore, voc = (4)(6) + 2(6) = 36 V 2 t=0 iL(0–) 3H 2 + 1 36 V ± – 12 V 4 6 vo(t) - ± – + iA + 4 12 V 2iA + iA 2iA (b) t=0– 2 3A vB + 4 36 V 6 + + vo(0±) - ± – iA 4 36 V ± – i1 36 V isc 6 ± – 4 36 V i"A 2i'A (d) t= 2 + ± – i A'" i A'" (e) Figure 7.10 Analysis of an RL transient circuit containing a dependent source. 6 isc + 2i"A (f) vo(∞) - i2 i'A + voc +– 4 6 2iA (c) t=0± 2 vo(0–) - ± – (a) 2 6 '" 2i A irwin07_296-368hr.qxd 28-07-2010 11:34 Page 315 SECTION 7.2 F I R S T- O R D E R C I R C U I T S 315 From Fig. 7.10f we can write the following loop equations: Ô 36 = 2Ai Ô A + isc B + 4i A Ô 36 = 2Ai Ô A + isc B + 6isc - 2i A Solving these equations for isc yields isc = 9 A 2 Therefore, RTh = voc 36 = 8⍀ = isc 9兾2 Hence, the circuit time constant is = 3 L = s RTh 8 Step 6. Using the information just computed, we can derive the final equation for vo(t): K1 = vo(q) = 27 K2 = vo(0+) - vo(q) = 18 - 27 = -9 Therefore, vo(t) = 27 - 9e-t兾(3兾8) V Learning Assessments E7.9 If the switch in the network in Fig. E7.9 closes at t = 0, find vo(t) for t 7 0. ANSWER: vo(t) = 24 + 36e-(t兾12) V. 24 V 4 + + 2 vA 2F + 3A t=0 4 vA + vo(t) - - Figure E7.9 E7.10 Find io(t) for t>0 in Fig. E7.10 using the step-by-step method. ANSWER: io(t) = 1.5 + 0.2143e-(t兾0.7) mA. t=0 iA 6 k 6 k 3 k 6 k io(t) 12 V + - Figure E7.10 4 k 2iA 100 F 2F irwin07_296-368hr.qxd 316 28-07-2010 CHAPTER 7 11:34 Page 316 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S At this point, it is appropriate to state that not all switch action will always occur at time t=0. It may occur at any time t0 . In this case the results of the step-by-step analysis yield the following equations: x At0 B = K1 + K2 x (q) = K1 and x(t) = x(q) + Cx At0 B - x(q)D e-At - t0B兾 t 7 t0 The function is essentially time-shifted by t0 seconds. Finally, note that if more than one independent source is present in the network, we can simply employ superposition to obtain the total response. P U L S E R E S P O N S E Thus far we have examined networks in which a voltage or current source is suddenly applied. As a result of this sudden application of a source, voltages or currents in the circuit are forced to change abruptly. A forcing function whose value changes in a discontinuous manner or has a discontinuous derivative is called a singular function. Two such singular functions that are very important in circuit analysis are the unit impulse function and the unit step function. We will defer a discussion of the unit impulse function until a later chapter and concentrate on the unit step function. The unit step function is defined by the following mathematical relationship: u(t) = b 0 1 t 6 0 t 7 0 In other words, this function, which is dimensionless, is equal to zero for negative values of the argument and equal to 1 for positive values of the argument. It is undefined for a zero argument where the function is discontinuous. A graph of the unit step is shown in Fig. 7.11a. 2 Figure 7.11 Graphs and models of the unit step function. 1 1 Vo ± – t=0 t 0 (a) (b) t=0 Vou(t) ± – Io (c) (d) u(t-t0) 1 Iou(t-t0) 0 (e) t0 (f) t irwin07_296-368hr.qxd 28-07-2010 11:34 Page 317 SECTION 7.2 F I R S T- O R D E R C I R C U I T S The unit step is dimensionless, and therefore a voltage step of Vo volts or a current step of Io amperes is written as Vo u(t) and Iou(t), respectively. Equivalent circuits for a voltage step are shown in Figs. 7.11b and c. Equivalent circuits for a current step are shown in Figs. 7.11d and e. If we use the definition of the unit step, it is easy to generalize this function by replacing the argument t by t-t0 . In this case 0 u At - t0 B = b 1 v(t) t 6 t0 t 7 t0 A A graph of this function is shown in Fig. 7.11f. Note that u At - t0 B is equivalent to delaying u(t) by t0 seconds, so that the abrupt change occurs at time t=t0 . Step functions can be used to construct one or more pulses. For example, the voltage pulse shown in Fig. 7.12a can be formulated by initiating a unit step at t=0 and subtracting one that starts at t=T, as shown in Fig. 7.12b. The equation for the pulse is (a) v(t) Au(t) A T If the pulse is to start at t=t0 and have width T, the equation would be t v(t) = AEu At - t0 B - uCt - At0 + TB D F Using this approach, we can write the equation for a pulse starting at any time and ending at any time. Similarly, using this approach, we could write the equation for a series of pulses, called a pulse train, by simply forming a summation of pulses constructed in the manner illustrated previously. The following example will serve to illustrate many of the concepts we have just presented. –A –Au(t-T) (b) Figure 7.12 Construction of a pulse via two step functions. Consider the circuit shown in Fig. 7.13a. The input function is the voltage pulse shown in Fig. 7.13b. Since the source is zero for all negative time, the initial conditions for the v(t) ± – EXAMPLE 7.6 4 k + + vC(t) - 100 F v(t) (V) 9 vo(t) 8 k 0 0.3 (a) t(s) (b) 4 k 6 k 4 k + + vc(0)=0 8 k vo(0±)=0 ± – 9V 8 k t=0 6 k vo(∞) - - (c) t T v(t) = ACu(t) - u(t - T)D 6 k 317 (d) 4 k vo(t) (V) 4 8 k RTh 2.11 0 (e) 0.3 t(s) (f) Figure 7.13 Pulse response of a network. irwin07_296-368hr.qxd 318 28-07-2010 CHAPTER 7 11:34 Page 318 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S network are zero C i.e., vC(0-) = 0D. The response vo(t) for 0 6 t 6 0.3 s is due to the application of the constant source at t=0 and is not influenced by any source changes that will occur later. At t = 0.3 s the forcing function becomes zero, and therefore vo(t) for t 7 0.3 s is the source-free or natural response of the network. Let us determine the expression for the voltage vo(t). SOLUTION Since the output voltage vo(t) is a voltage division of the capacitor voltage, and the initial voltage across the capacitor is zero, we know that vo(0+) = 0, as shown in Fig. 7.13c. If no changes were made in the source after t=0, the steady-state value of vo(t) Ci.e., vo(q)D due to the application of the unit step at t=0 would be 9 (8k) 6k + 4k + 8k = 4V vo(q) = as shown in Fig. 7.13d. The Thévenin equivalent resistance is RTh = (6k)(12k) 6k + 12k = 4 k⍀ as illustrated in Fig. 7.13e. Therefore, the circuit time constant is = RTh C = (4)A103 B(100)A10-6 B = 0.4 s Therefore, the response vo(t) for the period 0<t<0.3 s is vo(t) = 4 - 4e-t兾0.4 V 0<t<0.3 s The capacitor voltage can be calculated by realizing that using voltage division, vo(t) = 2兾3 vC(t). Therefore, vC(t) = 3 A4 - 4e-t兾0.4 B V 2 Since the capacitor voltage is continuous, vC(0.3-) = vC(0.3+) and therefore, vo(0.3+) = 2 v (0.3-) 3 C = 4A1 - e-0.3兾0.4 B = 2.11 V Since the source is zero for t 7 0.3 s, the final value for vo(t) as t S q is zero. Therefore, the expression for vo(t) for t 7 0.3 s is vo(t) = 2.11e-(t - 0.3)兾0.4 V t>0.3 s The term e-(t - 0.3)兾0.4 indicates that the exponential decay starts at t=0.3 s. The complete solution can be written by means of superposition as vo(t) = 4A1 - e-t兾0.4 Bu(t) - 4A1 - e-(t - 0.3)兾0.4 Bu(t - 0.3) V irwin07_296-368hr.qxd 28-07-2010 11:34 Page 319 SECTION 7.3 SECOND-ORDER CIRCUITS 319 or, equivalently, the complete solution is 0 vo(t) = • 4A1 - e-t兾0.4 B V 2.11e-(t - 0.3)兾0.4 V t 6 0 0 6 t 6 0.3 s ¶ 0.3 s 6 t which in mathematical form is vo(t) = 4A1 - e-t兾0.4 B Cu(t) - u(t - 0.3)D + 2.11e-(t - 0.3)兾0.4 u(t - 0.3) V Note that the term Cu(t)-u(t-0.3)D acts like a gating function that captures only the part of the step response that exists in the time interval 0<t<0.3 s. The output as a function of time is shown in Fig. 7.13f. Learning Assessments ANSWER: vo(t) = 0 for t 6 0, 4A1 - e-(3兾2)t B V for 0 ⬉ t ⬉ 1 s, and 3.11e-(3兾2)(t - 1) V for 1 s<t. E7.11 The voltage source in the network in Fig. E7.11a is shown in Fig. E7.11b. The initial current in the inductor must be zero. (Why?) Determine the output voltage vo(t) for t 7 0. 2H 2 v(t) ± – + 2 2 v(t) (V) vo(t) 12 0 (a) 1 t(s) (b) Figure E7.11 7.3 T H E B A S I C C I R C U I T EQ U AT I O N To begin our development, let us consider the two basic RLC circuits shown in Fig. 7.14. We assume that energy may be initially stored in both the inductor and capacitor. The node equation for the parallel RLC circuit is Second-Order Circuits t v 1 dv + v(x) dx + iL At0 B + C = iS(t) R L 3t0 dt Similarly, the loop equation for the series RLC circuit is Ri + t 1 di i(x) dx + vC At0 B + L = vS(t) C 3t0 dt v(t) i(t) vC(t0) R + iL(t0) iS(t) R L C (a) Figure 7.14 Parallel and series RLC circuits. - C vS(t) ± – L (b) irwin07_296-368hr.qxd 320 28-07-2010 CHAPTER 7 11:34 Page 320 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S Note that the equation for the node voltage in the parallel circuit is of the same form as that for the loop current in the series circuit. Therefore, the solution of these two circuits is dependent on solving one equation. If the two preceding equations are differentiated with respect to time, we obtain C d2v 2 dt di S 1 dv v + = R dt L dt + and L d 2i dt 2 + R dvS di i + = dt C dt Since both circuits lead to a second-order differential equation with constant coefficients, we will concentrate our analysis on this type of equation. In concert with our development of the solution of a first-order differential equation that results from the analysis of either an RL or an RC circuit as outlined earlier, we will now employ the same approach here to obtain the solution of a second-order differential equation that results from the analysis of RLC circuits. As a general rule, for this case we are confronted with an equation of the form T H E R E S P O N S E EQ U AT I O N S d2x(t) dt 2 + a1 dx(t) dt + a2 x(t) = f(t) 7.12 Once again we use the fact that if x(t)=xp(t) is a solution to Eq. (7.12), and if x(t)=xc(t) is a solution to the homogeneous equation d2x(t) dt 2 + a1 dx(t) dt + a2 x(t) = 0 then x(t) = xp(t) + xc(t) is a solution to the original Eq. (7.12). If we again confine ourselves to a constant forcing function Ci.e., f(t)=AD, the development at the beginning of this chapter shows that the solution of Eq. (7.12) will be of the form x(t) = A + xc(t) a2 7.13 Let us now turn our attention to the solution of the homogeneous equation d 2x(t) dt 2 + a1 dx(t) dt + a2 x(t) = 0 where a1 and a2 are constants. For simplicity we will rewrite the equation in the form d2x(t) dt 2 + 20 dx(t) dt + 20 x(t) = 0 7.14 where we have made the following simple substitutions for the constants a1 = 20 and a2 = 20 . Following the development of a solution for the first-order homogeneous differential equation earlier in this chapter, the solution of Eq. (7.14) must be a function whose first- and second-order derivatives have the same form, so that the left-hand side of Eq. (7.14) will become identically zero for all t. Again we assume that x(t) = Kest Substituting this expression into Eq. (7.14) yields s2Kest + 20 sKest + 20 Kest = 0 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 321 SECTION 7.3 SECOND-ORDER CIRCUITS Dividing both sides of the equation by Kest yields s2 + 20 s + 20 = 0 7.15 This equation is commonly called the characteristic equation; is called the exponential damping ratio, and 0 is referred to as the undamped natural frequency. The importance of this terminology will become clear as we proceed with the development. If this equation is satisfied, our assumed solution x(t) = Kest is correct. Employing the quadratic formula, we find that Eq. (7.15) is satisfied if -20 ; 24220 - 420 2 s = = -0 ; 0 22 - 1 7.16 Therefore, two values of s, s1 and s2, satisfy Eq. (7.15): s1 = -0 + 0 22 - 1 7.17 s2 = -0 - 0 22 - 1 In general, then, the complementary solution of Eq. (7.14) is of the form xc(t) = K1 es1 t + K2 es2 t 7.18 K1 and K2 are constants that can be evaluated via the initial conditions x(0) and dx(0)兾dt. For example, since x(t) = K1 es1 t + K2 es2 t then x(0) = K1 + K2 and dx(t) dt dx(0) 2 = t=0 dt = s1 K1 + s2 K2 Hence, x(0) and dx (0)兾dt produce two simultaneous equations, which when solved yield the constants K1 and K2 . Close examination of Eqs. (7.17) and (7.18) indicates that the form of the solution of the homogeneous equation is dependent on the value . For example, if >1, the roots of the characteristic equation, s1 and s2 , also called the natural frequencies because they determine the natural (unforced) response of the network, are real and unequal; if <1, the roots are complex numbers; and finally, if =1, the roots are real and equal. Let us now consider the three distinct forms of the unforced response—that is, the response due to an initial capacitor voltage or initial inductor current. Case 1, >1 This case is commonly called overdamped. The natural frequencies s1 and s2 are real and unequal; therefore, the natural response of the network described by the second-order differential equation is of the form xc(t) = K1 e- A0 - 0 2 2 - 1Bt + K2 e- A0 + 0 2 2 - 1Bt 7.19 where K1 and K2 are found from the initial conditions. This indicates that the natural response is the sum of two decaying exponentials. 321 irwin07_296-368hr.qxd 322 28-07-2010 CHAPTER 7 11:34 Page 322 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S Case 2, <1 This case is called underdamped. Since <1, the roots of the characteristic equation given in Eq. (7.17) can be written as s1 = -0 + j0 21 - 2 = - + jd s2 = -0 - j0 21 - 2 = - - jd where j = 1-1 , = 0, and d = 0 21 - 2 . Thus, the natural frequencies are complex numbers (briefly discussed in the Appendix). The natural response is then of the form xc(t) = e-0 t AA 1 cos 0 21 - 2 t + A 2 sin 0 21 - 2 tB 7.20 where A1 and A2 , like K1 and K2 , are constants, which are evaluated using the initial conditions x(0) and dx(0)兾dt. This illustrates that the natural response is an exponentially damped oscillatory response. Case 3, =1 This case, called critically damped, results in s1 = s2 = -0 In the case where the characteristic equation has repeated roots, the general solution is of the form xc(t) = B1 e-0 t + B2 te-0 t 7.21 where B1 and B2 are constants derived from the initial conditions. It is informative to sketch the natural response for the three cases we have discussed: overdamped, Eq. (7.19); underdamped, Eq. (7.20); and critically damped, Eq. (7.21). Figure 7.15 graphically illustrates the three cases for the situations in which xc(0)=0. Note that the critically damped response peaks and decays faster than the overdamped response. The underdamped response is an exponentially damped sinusoid whose rate of decay is dependent on the factor . Actually, the terms ; e-0 t define what is called the envelope of the response, and the damped oscillations (i.e., the oscillations of decreasing amplitude) exhibited by the waveform in Fig. 7.15b are called ringing. Figure 7.15 Comparison of overdamped, critically damped, and underdamped responses. xc(t) xc(t) Critically damped Underdamped e–t Overdamped t t (a) (b) Learning Assessments E7.12 A parallel RLC circuit has the following circuit parameters: R=1 ⍀, L=2 H, and C=2 F. Compute the damping ratio and the undamped natural frequency of this network. E7.13 A series RLC circuit consists of R=2 ⍀, L=1 H, and a capacitor. Determine the type of response exhibited by the network if (a) C = 1兾2 F, (b) C=1 F, and (c) C=2 F. ANSWER: = 0.5; 0 = 0.5 rad兾s. ANSWER: (a) underdamped; (b) critically damped; (c) overdamped. We will now analyze a number of simple RLC networks that contain both nonzero initial conditions and constant forcing functions. Circuits that exhibit overdamped, underdamped, and critically damped responses will be considered. THE NETWORK RESPONSE irwin07_296-368hr.qxd 28-07-2010 11:34 Page 323 SECTION 7.3 SECOND-ORDER CIRCUITS 323 Problem-Solving Strategy Step 1. Write the differential equation that describes the circuit. Step 2. Derive the characteristic equation, which can be written in the form s2 + 20 s + 20 = 0, where is the damping ratio and 0 is the undamped natural frequency. Step 3. The two roots of the characteristic equation will determine the type of response. If the roots are real and unequal (i.e., >1), the network response is overdamped. If the roots are real and equal (i.e., =1), the network response is critically damped. If the roots are complex (i.e., <1), the network response is underdamped. Step 4. The damping condition and corresponding response for the aforementioned three cases outlined are as follows: 2 2 Overdamped: x(t) = K1 e-A0 - 0 2 - 1Bt + K2 e-A0 + 0 2 - 1Bt Critically damped: x(t) = B1 e-0 t + B2 te-0 t Underdamped: x(t) = e-t AA 1 cos d t + A 2 sin d tB, where = 0 , and d = 0 21 - 2 Step 5. Two initial conditions, either given or derived, are required to obtain the two unknown coefficients in the response equation. Second-Order Transient Circuits The following examples will demonstrate the analysis techniques. Consider the parallel RLC circuit shown in Fig. 7.16. The second-order differential equation that describes the voltage v(t) is d2v EXAMPLE 7.7 v 1 dv + + = 0 RC dt LC dt 2 v(t) Figure 7.16 + R L C iL(0) Parallel RLC circuit. vC(0) - A comparison of this equation with Eqs. (7.14) and (7.15) indicates that for the parallel RLC circuit the damping term is 1/2 RC and the undamped natural frequency is 1兾1LC . If the circuit parameters are R=2 ⍀, C = 1兾5 F, and L=5 H, the equation becomes d2v dt 2 + 2.5 dv + v = 0 dt Let us assume that the initial conditions on the storage elements are iL(0)=-1 A and vC(0)=4 V. Let us find the node voltage v(t) and the inductor current. The characteristic equation for the network is s2 + 2.5s + 1 = 0 and the roots are s1 = -2 s2 = -0.5 SOLUTION irwin07_296-368hr.qxd 324 28-07-2010 CHAPTER 7 11:34 Page 324 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S Since the roots are real and unequal, the circuit is overdamped, and v(t) is of the form v(t) = K1 e-2t + K2 e-0.5t The initial conditions are now employed to determine the constants K1 and K2 . Since v(t)=vC(t), vC(0) = v(0) = 4 = K1 + K2 The second equation needed to determine K1 and K2 is normally obtained from the expression dv(t) dt = -2K1 e-2t - 0.5K2 e-0.5t However, the second initial condition is not dv(0)兾dt. If this were the case, we would simply evaluate the equation at t=0. This would produce a second equation in the unknowns K1 and K2 . We can, however, circumvent this problem by noting that the node equation for the circuit can be written as v(t) dv(t) C + dt R + iL(t) = 0 or iL(t) -1 = v(t) RC C dv(t) dt At t=0, dv(0) dt = -1 1 v(0) - iL(0) RC C = -2.5(4) - 5(-1) = -5 However, since dv(t) dt = -2K1 e-2t - 0.5K2 e-0.5t then when t=0 -5 = -2K1 - 0.5K2 This equation, together with the equation 4 = K1 + K2 produces the constants K1 = 2 and K2 = 2 . Therefore, the final equation for the voltage is v(t) = 2e-2t + 2e-0.5t V Note that the voltage equation satisfies the initial condition v(0)=4 V. The response curve for this voltage v(t) is shown in Fig. 7.17. Figure 7.17 Overdamped response. v(t) (V) 4.8 4.2 3.6 3.0 2.4 1.8 1.2 0.6 0.0 t(s) 0.0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3.0 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 325 SECTION 7.3 SECOND-ORDER CIRCUITS 325 The inductor current is related to v(t) by the equation iL(t) = 1 v(t) dt L 3 Substituting our expression for v(t) yields iL(t) = or 1 C 2e-2t + 2e-0.5t D dt 5 3 iL(t) = - 1 -2t 4 e - e-0.5t A 5 5 Note that in comparison with the RL and RC circuits, the response of this RLC circuit is controlled by two time constants. The first term has a time constant of 1兾2 s, and the second term has a time constant of 2 s. The series RLC circuit shown in Fig. 7.18 has the following parameters: C=0.04 F, L=1 H, R=6 ⍀, iL(0)=4 A, and vC(0)=-4 V. The equation for the current in the circuit is given by the expression 2 d i dt 2 EXAMPLE 7.8 R di i + = 0 L dt LC + A comparison of this equation with Eqs. (7.14) and (7.15) illustrates that for a series RLC circuit the damping term is R/2L and the undamped natural frequency is 1兾1LC . Substituting the circuit element values into the preceding equation yields d2i dt 2 + 6 di + 25i = 0 dt Let us determine the expression for both the current and the capacitor voltage. The characteristic equation is then SOLUTION s2 + 6s + 25 = 0 and the roots are s1 = -3 + j4 s2 = -3 - j4 Since the roots are complex, the circuit is underdamped, and the expression for i(t) is i(t) = K1 e-3t cos 4t + K2 e-3t sin 4t Using the initial conditions, we find that i(0)=4=K1 i(t) iL(0) Figure 7.18 Series RLC circuit. L R + C vC(0) - irwin07_296-368hr.qxd 326 28-07-2010 CHAPTER 7 11:34 Page 326 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S and di = -4K1 e-3t sin 4t - 3K1 e-3t cos 4t dt + 4K2 e-3t cos 4t - 3K2 e-3t sin 4t and thus di(0) dt = -3K1 + 4K2 Although we do not know di(0)兾dt, we can find it via KVL. From the circuit we note that Ri(0) + L di(0) dt + vC(0) = 0 or di(0) dt vC(0) R i(0) L L 6 4 = - (4) + 1 1 = -20 = - Therefore, -3K1 + 4K2 = -20 and since K1=4, K2=–2, the expression then for i(t) is i(t) = 4e-3t cos 4t - 2e-3t sin 4t A Note that this expression satisfies the initial condition i(0)=4. The voltage across the capacitor could be determined via KVL using this current: Ri(t) + L di(t) dt + vC(t) = 0 or vC(t) = -Ri(t) - L di(t) dt Substituting the preceding expression for i(t) into this equation yields vC(t) = -4e-3t cos 4t + 22e-3t sin 4t V Note that this expression satisfies the initial condition vC(0)=–4 V. A plot of the function is shown in Fig. 7.1.9: Figure 7.19 Underdamped response. v(t) (V) 10.0 8.0 6.0 4.0 2.0 0 –2.0 –4.0 t(s) 0.0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3.0 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 327 SECTION 7.3 SECOND-ORDER CIRCUITS Let us examine the circuit in Fig. 7.20, which is slightly more complicated than the two we have already considered. The two equations that describe the network are L di(t) dt EXAMPLE SOLUTION + R1 i(t) + v(t) = 0 v(t) dv(t) i(t) = C dt + R2 Substituting the second equation into the first yields d2v dt 2 + a R1 dv R1 + R2 1 + b + v = 0 R2 C L dt R2 LC If the circuit parameters and initial conditions are 1 F 8 R1 = 10 ⍀ C = R2 = 8 ⍀ L = 2H vC(0) = 1 V iL(0) = 1 A 2 the differential equation becomes d2 v dt 2 + 6 dv + 9v = 0 dt We wish to find expressions for the current i(t) and the voltage v(t). The characteristic equation is then s2 + 6s + 9 = 0 and hence the roots are s1 = -3 s2 = -3 Since the roots are real and equal, the circuit is critically damped. The term v(t) is then given by the expression v(t) = K1 e-3t + K2 te-3t Since v(t)=vC(t), v(0) = vC(0) = 1 = K1 i(t) Figure 7.20 Series-parallel RLC circuit. R1 iL(0) L + + vC(0) - C v(t) - R2 327 7.9 irwin07_296-368hr.qxd 328 28-07-2010 CHAPTER 7 11:34 Page 328 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S In addition, dv(t) dt = -3K1 e-3t + K2 e-3t - 3K2 te-3t However, i(t) dv(t) dt = C v(t) - R2 C Setting these two expressions equal to one another and evaluating the resultant equation at t=0 yields 1 1兾2 - = -3K1 + K2 1 1兾8 3 = -3K1 + K2 K1=1, K2=6, and the expression for v(t) is v(t) = e-3t + 6te-3t V Note that the expression satisfies the initial condition v(0)=1. The current i(t) can be determined from the nodal analysis equation at v(t): i(t) = C v(t) dv(t) dt + R2 Substituting v(t) from the preceding equation, we find i(t) = 1 1 C -3e-3t + 6e-3t - 18te-3t D + C e-3t + 6te-3t D 8 8 or i(t) = 3 1 -3t e - te-3t A 2 2 If this expression for the current is employed in the circuit equation, v(t) = -L di(t) dt - R1 i(t) we obtain v(t) = e-3t + 6te-3t V which is identical to the expression derived earlier. A plot of this critically damped function is shown in Fig. 7.21. Figure 7.21 Critically damped response. v(t) (V) 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 –0.2 t(s) 0.0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3.0 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 329 SECTION 7.3 SECOND-ORDER CIRCUITS 329 Learning Assessments E7.14 The switch in the network in Fig. E7.14 opens at t=0. Find i(t) for t>0. 6 3 t=0 1F ± – i(t) Figure E7.14 ANSWER: i(t) = -2e - t兾2 + 4e-t A. 2H 12 V E7.15 The switch in the network in Fig. E7.15 moves from position 1 to position 2 at t=0. Find vo(t) for t>0. 1 ANSWER: vo(t) = 2Ae-t - 3e-3t B V. 2 t=0 2 —F 3 1 —H 2 2A + vo(t) 2 - Figure E7.15 E7.16 Find vC(t) for t>0 in Fig. E7.16. ANSWER: vC(t)=-2e–2t - 1 vc(t) + cos t-1.5e–2t sin t+24 V. 0.8 F t=0 24 V 0.25 H 5 12 V + – +V – S2 Figure E7.16 Consider the circuit shown in Fig. 7.22. This circuit is the same as the one analyzed in Example 7.8, except that a constant forcing function is present. The circuit parameters are the same as those used in Example 7.8: C = 0.04 F iL(0) = 4 A L = 1H vC(0) = -4 V R = 6⍀ EXAMPLE 7.10 irwin07_296-368hr.qxd 330 28-07-2010 CHAPTER 7 11:34 Page 330 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S Figure 7.22 iL(0) R Series RLC circuit with a step function input. L 12u(t) V ± – + C vC (0) - We want to find an expression for vC(t) for t>0. SOLUTION From our earlier mathematical development we know that the general solution of this problem will consist of a particular solution plus a complementary solution. From Example 7.8 we know that the complementary solution is of the form K3 e-3t cos 4t + K4 e-3t sin 4t. The particular solution is a constant, since the input is a constant and therefore the general solution is vC(t) = K3 e-3t cos 4t + K4 e-3t sin 4t + K5 An examination of the circuit shows that in the steady state, the final value of vC(t) is 12 V, since in the steady-state condition, the inductor is a short circuit and the capacitor is an open circuit. Thus, K5 = 12. The steady-state value could also be immediately calculated from the differential equation. The form of the general solution is then vC(t) = K3 e-3t cos 4t + K4 e-3t sin 4t + 12 The initial conditions can now be used to evaluate the constants K3 and K4 : vC(0) = -4 = K3 + 12 -16 = K3 Since the derivative of a constant is zero, the results of Example 7.8 show that dvC(0) i(0) = dt C = 100 = -3K3 + 4K4 and since K3 = -16, K4 = 13. Therefore, the general solution for vC(t) is vC(t) = 12 - 16e-3t cos 4t + 13e-3t sin 4t V Note that this equation satisfies the initial condition vC(0)=–4 and the final condition vC(q)=12 V. EXAMPLE 7.11 Let us examine the circuit shown in Fig. 7.23. A close examination of this circuit will indicate that it is identical to that shown in Example 7.9 except that a constant forcing function is present. We assume the circuit is in steady state at t = 0-. The equations that describe the circuit for t 7 0 are L di(t) dt + R1 i(t) + v(t) = 24 i(t) = C v(t) dv(t) dt + R2 Combining these equations, we obtain d2v(t) dt 2 + a R1 dv(t) R1 + R2 1 24 + b + v(t) = R2 C L dt R2 LC LC If the circuit parameters are R1 = 10 ⍀, R2 = 2 ⍀, L = 2 H, and C = 1兾4 F, the differential equation for the output voltage reduces to d2v(t) dt 2 + 7 dv(t) dt + 12v(t) = 48 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 331 SECTION 7.3 SECOND-ORDER CIRCUITS 331 Figure 7.23 t=0 24 V + iL(0) ± – ± – Series-parallel RLC circuit with a constant forcing function. R1 i(t) L + vC(0) 12 V - R2 C v(t) - Let us determine the output voltage v(t). The characteristic equation is s2 + 7s + 12 = 0 and hence the roots are s1 = -3 s2 = -4 The circuit response is overdamped, and therefore the general solution is of the form v(t) = K1 e-3t + K2 e-4t + K3 The steady-state value of the voltage, K3 , can be computed from Fig. 7.24a. Note that v(q) = 4 V = K3 The initial conditions can be calculated from Figs. 7.24b and c, which are valid at t = 0- and t = 0+, respectively. Note that v(0+) = 2 V and, hence, from the response equation v(0+) = 2 V = K1 + K2 + 4 -2 = K1 + K2 Fig. 7.24c illustrates that i(0+) = 1. From the response equation we see that dv(0) dt i(∞)=2 A = -3K1 - 4K2 10 i(0–)=1 A 10 + 24 V ± – 2 + v(∞)=4 V 12 V ± – + vC(0–)=2 V - 2 - - (a) t=∞ i(0±)=1 A (b) t=0– 10 i(0±) + iC(0±) 24 V ± – 2V ± – 2 v(0±)=2 V - (c) t=0+ Figure 7.24 v(0–)=2 V Equivalent circuits at t = q , t = 0- , and t = 0+ for the circuit in Fig. 7.23. irwin07_296-368hr.qxd 332 28-07-2010 CHAPTER 7 11:34 Page 332 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S and since i(0) dv(0) dt = C v(0) - R2 C = 4 - 4 = 0 then 0 = -3K1 - 4K2 Solving the two equations for K1 and K2 yields K1 = -8 and K2 = 6. Therefore, the general solution for the voltage response is v(t) = 4 - 8e-3t + 6e-4t V Note that this equation satisfies both the initial and final values of v(t). Learning Assessments E7.17 The switch in the network in Fig. E7.17 moves from position 1 to position 2 at t=0. Compute io(t) for t>0 and use this current to determine vo(t) for t>0. io(t) 1 6 + 2H 2 t=0 18 1 —F 36 24 V vo(t) ± – ± – ± – 4V 12 V - Figure E7.17 E7.18 Find i(t) for t>0 in Fig. E7.18. ANSWER: i(t)=0.4144e–17.07t – 2.414e–2.93t+3 A. 10 t=0 i(t) 10 10 V + – 2H 20 V Figure E7.18 ANSWER: 11 14 -6t io(t) = - e-3t + e A; 6 6 vo(t)=12+18io(t) V. + – 0.01F irwin07_296-368hr.qxd 28-07-2010 11:34 Page 333 SECTION 7.4 A P P L I C AT I O N E X A M P L E S 333 7.4 There are a wide variety of applications for transient circuits. The following examples demonstrate some of them. Application Examples Let us return to the camera flash circuit, redrawn in Fig. 7.25, which was discussed in the introduction of this chapter. The Xenon flash has the following specifications: Voltage required for successful flash: b APPLICATION EXAMPLE 7.12 minimum 50 V maximum 70 V Equivalent resistance: 80 ⍀ For this particular application, a time constant of 1 ms is required during flash time. In addition, to minimize the physical size of the circuit, the resistor R1 must dissipate no more than 100-mW peak power. We want to determine values for VS , CF , and R1 . Furthermore, we wish to determine the recharge time, the flash bulb’s voltage, current, power, and total energy dissipated during the flash. R1 VS ± – iB(t) A model for a camera flash charging circuit. + + vCF(t) Figure 7.25 CF - vB(t) RB - We begin by selecting the source voltage, VS . Since the capacitor is applied directly to the Xenon bulb during flash, and since at least 50 V is required to flash, we should set VS higher than 50 V. We will somewhat arbitrarily split the difference in the bulb’s required voltage range and select 60 V for VS . Now we consider the time constant during the flash time. From Fig. 7.25, during flash, the time constant is simply F = RB CF 7.22 Given tF = 1 ms and RB = 80 ⍀, we find CF = 12.5 F. Next, we turn to the value of R1 . At the beginning of the charge time, the capacitor voltage is zero and both the current and power in R1 are at their maximum values. Setting the power to the maximum allowed value of 100 mW, we can write PRmax = V2S 3600 = = 0.1 R1 R1 7.23 and find that R1 = 36 k⍀. The recharge time is the time required for the capacitor to charge from zero up to at least 50 V. At that point the flash can be successfully discharged. We will define t = 0 as the point at which the switch moves from the bulb back to R1 . At t = 0, the capacitor voltage is zero, and at t = q, the capacitor voltage is 60 V; the time constant is simply R1 CF . The resulting equation for the capacitor voltage during recharge is vCF(t) = K1 + K2 e-t兾 = 60 - 60e-t兾R1 CF V 7.24 At t = tcharge , vCF(t) = 50 V. Substituting this and the values of R1 and CF into Eq. (7.24) yields a charge time of tcharge = 806 ms—just less than a second. As a point of interest, let us SOLUTION • irwin07_296-368hr.qxd 334 28-07-2010 CHAPTER 7 11:34 Page 334 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S reconsider our choice for VS . What happens if VS is decreased to only 51 V? First, from Eq. (7.23), R1 changes to 26.01 k⍀. Second, from Eq. (7.24), the charge time increases only slightly to 1.28 s. Therefore, it appears that selection of VS will not have much effect on the flash unit’s performance, and thus there exists some flexibility in the design. Finally, we consider the waveforms for the flash bulb itself. The bulb and capacitor voltage are the same during flash and are given by the decaying exponential function vB(t) = 60e-1000t V 7.25 where the time constant is defined in Eq. (7.22), and we have assumed that the capacitor is allowed to charge fully to VS (i.e., 60 V). Since the bulb’s equivalent resistance is 80 ⍀, the bulb current must be iB(t) = 60e-1000t = 750e-1000t mA 80 7.26 As always, the power is the v-i product. pB(t) = vB(t)iB(t) = 45e-2000t W 7.27 Finally, the total energy consumed by the bulb during flash is only q wB(t) = APPLICATION EXAMPLE 7.13 30 q pB(t) dt = 30 45e-2000t dt = 45 -2000t 0 45 2 = e = 22.5 mJ 2000 2000 q One very popular application for inductors is storing energy in the present for release in the future. This energy is in the form of a magnetic field, and current is required to maintain the field. In an analogous situation, the capacitor stores energy in an electric field, and a voltage across the capacitor is required to maintain it. As an application of the inductor’s energy storage capability, let us consider the high-voltage pulse generator circuit shown in Fig. 7.26. This circuit is capable of producing high-voltage pulses from a small dc voltage. Let’s see if this circuit can produce an output voltage peak of 500 V every 2 ms, that is, 500 times per second. pos. 1 Figure 7.26 A simple high-voltage pulse generator. Vin ± 5V – pos. 2 i(t) L 1 mH SOLUTION 7.28 R - vo(t) 100 + At the heart of this circuit is a single-pole, double-throw switch—that is, a single switch (single-pole) with two electrically connected positions (double-throw), 1 and 2. As shown in Fig. 7.27a, when in position 1, the inductor current grows linearly in accordance with the equation i(t) = T1 1 V dt L 30 in Then the switch moves from position 1 to position 2 at time T1 . The peak inductor current is ip(t) = Vin T1 L While at position 1, the resistor is isolated electrically, and therefore its voltage is zero. irwin07_296-368hr.qxd 28-07-2010 11:34 Page 335 SECTION 7.4 Vin ± 5V – i(t) L R - Vin ± 5V – vo(t) 100 + 1 mH L i(t) 1 mH (a) A P P L I C AT I O N E X A M P L E S R 335 - vo(t) 100 + (b) Figure 7.27 (a) Pulse generator with switch in position 1. Inductor is energized. (b) Switch in position 2. As energy is drained from the inductor, the voltage and current decay toward zero. At time t 7 T1 , when the switch is in position 2, as shown in Fig. 7.27b, the inductor current flows into the resistor producing the voltage vo At - T1 B = iAt - T1 BR t 7 T1 At this point, we know that the form of the voltage vo(t), in the time interval t 7 T1 , is vo(t) = Ke-(t - T1)兾 t 7 T1 And = L兾R. The initial value of vo(t) in the time interval t 7 T1 is K since at time T1 the exponential term is 1. According to the design specifications, this initial value is 500 and therefore K = 500. Since this voltage is created by the peak inductor current IP flowing in R, K = 500 = AVin T1 RB兾L = 5T1(100)兾10-3 and thus T1 is 1 ms and IP is 5 A. The equation for the voltage in the time interval t 7 T1 , or t 7 1 ms, is vo(t - 1 ms) = 500e-100,000(t - 1 ms) V At the end of the 2-ms period, that is, at t = 2 ms, the voltage is 500e-100 or essentially zero. The complete waveform for the voltage is shown in Fig. 7.28. It is instructive at this point to consider the ratings of the various components used in this pulse generator circuit. First, 500 V is a rather high voltage, and thus each component’s voltage rating should be at least 600 V in order to provide some safety margin. Second, the inductor’s peak current rating should be at least 6 A. Finally, at peak current, the power losses in the resistor are 2500 W! This resistor will have to be physically large to handle this power load without getting too hot. Fortunately, the resistor power is pulsed rather than continuous; thus, a lower power rated resistor will work fine, perhaps 500 W. In later chapters we will address the issue of power in much more detail. Figure 7.28 Output Voltage (V) 600 The output voltage of the pulse generator. 500 400 300 200 100 0 0 10 20 30 Time (ms) 40 50 irwin07_296-368hr.qxd 336 28-07-2010 CHAPTER 7 11:34 Page 336 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S APPLICATION EXAMPLE 7.14 A heart pacemaker circuit is shown in Fig. 7.29. An SCR (silicon-controlled rectifier) is a solid-state device that has two distinct modes of operation. When the voltage across the SCR is increasing but less than 5 V, the SCR behaves like an open circuit, as shown in Fig. 7.30a. Once the voltage across the SCR reaches 5 V, the device functions like a current source, as shown in Fig. 7.30b. This behavior will continue as long as the SCR voltage remains above 0.2 V. At this voltage, the SCR shuts off and again becomes an open circuit. Assume that at t=0, vC(t) is 0 V and the 1-F capacitor begins to charge toward the 6-V source voltage. Find the resistor value such that vC(t) will equal 5 V (the SCR firing voltage) at 1 s. At t=1 s, the SCR fires and begins discharging the capacitor. Find the time required for vC(t) to drop from 5 V to 0.2 V. Finally, plot vC(t) for the three cycles. R Figure 7.29 Heart pacemaker equivalent circuit. + vC(t) - C V=6 V 1 F SCR Figure 7.30 Equivalent circuits for an SCR. SCR I=50 A SCR (a) SOLUTION (b) For t<1 s, the equivalent circuit for the pacemaker is shown in Fig. 7.31. As indicated earlier, the capacitor voltage has the form vC(t) = 6 - 6e-t兾RC V A voltage of 0.2 V occurs at t1 = 0.034RC whereas a voltage of 5 V occurs at t2 = 1.792RC We desire that t2 - t1 = 1 s. Therefore, t2 - t1 = 1.758RC = 1 s and RC = 0.569 s R = 569 k⍀ R Figure 7.31 Pacemaker equivalent network during capacitor charge cycle. and + V=6 V C 1 F vC(t) - irwin07_296-368hr.qxd 28-07-2010 11:34 Page 337 SECTION 7.4 A P P L I C AT I O N E X A M P L E S R Figure 7.32 + C V=6 V 337 vC(t) 1 F I=50 A Pacemaker equivalent network during capacitor discharge cycle. - At t=1 s the SCR fires and the pacemaker is modeled by the circuit in Fig. 7.32. The form of the discharge waveform is v(t) = K1 + K2 e-(t - 1)兾RC The term (t-1) appears in the exponential to shift the function 1 s, since during that time the capacitor was charging. Just after the SCR fires at t = 1+ s, vC(t) is still 5 V, whereas at t=q, vC(t)=6-IR. Therefore, K1 + K2 = 5 and K1 = 6 - IR Our solution, then, is of the form vC(t) = 6 - IR + (IR - 1)e-(t - 1)兾RC Let T be the time beyond 1 s necessary for v(t) to drop to 0.2 V. We write vC(T + 1) = 6 - IR + (IR - 1)e-T兾RC = 0.2 Substituting for I, R, and C, we find T = 0.11 s The output waveform is shown in Fig. 7.33. v(t) (V) Figure 7.33 6 Heart pacemaker output voltage waveform. 4 2 t(s) 0 0 1 2 3 4 Consider the simple RL circuit, shown in Fig. 7.34, which forms the basis for essentially every dc power supply in the world. The switch opens at t = 0. VS and R have been chosen simply to create a 1-A current in the inductor prior to switching. Let us find the peak voltage across the inductor and across the switch. APPLICATION EXAMPLE 7.15 • irwin07_296-368hr.qxd 338 28-07-2010 CHAPTER 7 11:34 Page 338 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S SOLUTION We begin the analysis with an expression for the inductor current. At t = 0, the inductor current is 1 A. At t = q, the current is 0. The time constant is simply L兾R, but when the switch is open, R is infinite and the time constant is zero! As a result, the inductor current is R iL(t) = 1e-␣t A 1 L 100 H VS ± 1V – + 7.29 where ␣ is infinite. The resulting inductor voltage is vL(t) vL(t) = L - di L(t) = -␣e-␣t dt 7.30 iL(t) At t = 0, the peak inductor voltage is negative infinity! This voltage level is caused by the attempt to disrupt the inductor current instantaneously, driving di兾dt through the roof. Employing KVL, the peak switch voltage must be positive infinity (give or take the supply v t=0 switch voltage). This phenomenon is called inductive kick, and it is the nemesis of power supply designers. Given this situation, we naturally look for a way to reduce this excessive voltage and, more importantly, predict and control it. Let’s look at what we have and what we know. We Figure 7.34 have a transient voltage that grows very quickly without bound. We also have an initial curThe switched inductor rent in the inductor that must go somewhere. We know that capacitor voltages cannot network at the heart of change quickly and resistors consume energy. Therefore, let’s put an RC network around modern power supplies. the switch, as shown in Fig. 7.35, and examine the performance that results from this change. + Figure 7.35 R 1 Conversion of a switched inductor circuit to an RLC network in an attempt to control inductive kick. L 100 H VS ± 1V – + vL(t) iL(t) t=0 + C vswitch - R The addition of the RC network yields a series circuit. We need the characteristic equation of this series RLC network when the switch is open. From Eq. (7.15), we know that the characteristic equation for the series RLC circuit is s2 + 20 s + 20 = s2 + c R + 1 1 ds + = 0 L LC 7.31 To retain some switching speed, we will somewhat arbitrarily choose a critically damped system where = 1 and 0 = 106 rad兾s. This choice for 0 should allow the system to stabilize in a few microseconds. From Eq. (7.31) we can now write expressions for C and R. 20 = 1012 = 1 1 = -4 LC 10 C 20 = 2 * 106 = R + 1 R + 1 = L 10-4 7.32 Solving these equations yields the parameter values C = 10 nF and R = 199 ⍀. Now we can focus on the peak switch voltage. When the switch opens, the inductor current, set at 1 A by the dc source and the 1-⍀ resistor, flows through the RC circuit. Since the capacitor was previously discharged by the closed switch, its voltage cannot change immediately and irwin07_296-368hr.qxd 28-07-2010 11:34 Page 339 SECTION 7.4 A P P L I C AT I O N E X A M P L E S 339 its voltage remains zero for an instant. The resistor voltage is simply IL R where IL is the initial inductor current. Given our IL and R values, the resistor voltage just after opening the switch is 199 V. The switch voltage is then just the sum of the capacitor and resistor voltages (i.e., 199 V). This is a tremendous improvement over the first scenario! A plot of the switch voltage, shown in Fig. 7.36, clearly agrees with our analysis. This plot illustrates the effectiveness of the RC network in reducing the inductive kick generated by opening the switch. Note that the switch voltage is controlled at a 199-V peak value and the system is critically damped; that is, there is little or no overshoot, having stabilized in less than 5 s. Because of its importance, this R-C network is called a snubber and is the engineer’s solution of choice for controlling inductive kick. Figure 7.36 Output Voltage (V) 600 Plot of the switch voltage when the snubber circuit is employed to reduce inductive kick. 200 200 200 200 100 0 0 10 20 30 40 50 Time (ms) AC voltage waveform One of the most common and necessary subcircuits that appears in a wide variety of electronic systems—for example, stereos, TVs, radios, and computers—is a quality dc voltage source or power supply. The standard wall socket supplies an alternating current (ac) voltage waveform shown in Fig. 7.37a, and the conversion of this voltage to a desired dc level is done as illustrated in Fig. 7.37b. The ac waveform is converted to a quasi-dc voltage by an inexpensive ac–dc converter whose output contains remnants of the ac input and is unregulated. A higher quality dc output is created by a switching dc–dc converter. Of the several versions of dc–dc converters, we will focus on a topology called the boost converter, shown in Fig. 7.38. Let us develop an equation relating the output voltage to the switching characteristics. + AC–DC converter Vin - Time (a) (b) DC–DC switching power supply APPLICATION EXAMPLE 7.16 Figure 7.37 (a) The ac voltage waveform at a standard wall outlet and (b) a block diagram of a modern dc power supply. + Vo - • irwin07_296-368hr.qxd 340 28-07-2010 CHAPTER 7 11:34 Page 340 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S Figure 7.38 + vL(t) - The boost converter with switch settings for time intervals (a) ton and (b) toff. S2 + iL(t) Vin C R S1 Vo - (a) + vL(t) - S2 + iL(t) Vin C R Vo - S1 (b) SOLUTION Consider the boost converter in Fig. 7.38a, where switch 1 (S1) is closed and S2 is open for a time interval ton . This isolates the inductor from the capacitor, creating two subcircuits that can be analyzed independently. Note that during ton the inductor current and stored energy are increasing while at the output node, the capacitor voltage discharges exponentially into the load. If the capacitor’s time constant (=RC) is large, then the output voltage will decrease slowly. Thus, during ton energy is stored in the inductor and the capacitor provides energy to the load. Next, we change both switch positions so that S1 is open and S2 is closed for a time interval t off , as seen in Fig. 7.38b. Since the inductor current cannot change instantaneously, current flows into the capacitor and the load, recharging the capacitor. During toff the energy that was added to the inductor during ton is used to recharge the capacitor and drive the load. When toff has elapsed, the cycle is repeated. Note that the energy added to the inductor during ton must go to the capacitor and load during toff ; otherwise, the inductor energy would increase to the point that the inductor would fail. This requires that the energy stored in the inductor must be the same at the end of each switching cycle. Recalling that the inductor energy is related to the current by w(t) = 1 2 Li (t) 2 we can state that the inductor current must also be the same at the end of each switching cycle, as shown in Fig. 7.39. The inductor current during ton and toff can be written as iL(t) = ton Vin 1 ton 1 vL(t) dt = Vin dt = c d ton + I0 L 30 L 30 L iL(t) = ton + toff ton + toff 1 1 vL(t) dt = AVin - Vo B dt = L 3ton L 3ton c Vin - Vo d toff + I0 L ton 6 t 6 toff 0 6 t 6 ton 7.33 where I0 is the initial current at the beginning of each switching cycle. If the inductor current is the same at the beginning and end of each switching cycle, then the integrals in Eq. (7.33) must sum to zero. Or Vin ton = AVo - Vin Btoff = AVo - Vin B AT - ton B irwin07_296-368hr.qxd 28-07-2010 11:34 Page 341 SECTION 7.4 A P P L I C AT I O N E X A M P L E S 341 Figure 7.39 iL(t) I0 Vin vL(t) 0 Vin-Vo ton toff T 2T 3T Figure 7.40 30 Effect of duty cycle on boost converter gain. 25 Vo / Vin Waveform sketches for the inductor voltage and current. 20 15 10 5 0 0 20 40 60 Duty cycle (percent) 80 100 where T is the period AT = ton + toff B. Solving for Vo yields Vo = Vin c T 1 d = Vin c d = T - ton AT - ton B兾T Vin c 1 A1 - ton兾TB d = Vin c 1 d 1 - D where D is the duty cycle AD = ton兾TB. Thus, by controlling the duty cycle, we control the output voltage. Since D is always a positive fraction, Vo is always bigger than Vin —thus the name boost converter. A plot of Vo兾Vin versus duty cycle is shown in Fig. 7.40. An experimental schematic for a railgun is shown in Fig. 7.41. With switch Sw-2 open, switch Sw-1 is closed and the power supply charges the capacitor bank to 10 kV. Then switch Sw-1 is opened. The railgun is fired by closing switch Sw-2. When the capacitor discharges, the current causes the foil at the end of the gun to explode, creating a hot plasma that accelerates down the tube. The voltage drop in vaporizing the foil is negligible, and, therefore, more than 95% of the energy remains available for accelerating the plasma. The current flow establishes a magnetic field, and the force on the plasma caused by the magnetic field, which is proportional to the square of the current at any instant of time, accelerates the plasma. A higher initial voltage will result in more acceleration. The circuit diagram for the discharge circuit is shown in Fig. 7.42. The resistance of the bus (a heavy conductor) includes the resistance of the switch. The resistance of the foil and resultant plasma is negligible; therefore, the current flowing between the upper and lower conductors is dependent on the remaining circuit components in the closed path, as specified in Fig. 7.41. APPLICATION EXAMPLE 7.17 • irwin07_296-368hr.qxd 342 28-07-2010 CHAPTER 7 11:34 Page 342 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S Sw–1 Figure 7.41 Sw–2 Experimental schematic for a railgun. Charging resistor Upper conductor Power supply 10 kV 100 kW Capacitor bank 53.6 F Foil Insulator Lower conductor Sw-2 Figure 7.42 Rbus=12 m Railgun discharge circuit. Lbus=32 nH i(t) + 10 kV - C=53.6 F Rfoil The differential equation for the natural response of the current is d2i(t) dt + 2 i(t) Rbus di(t) + = 0 Lbus dt Lbus C Let us use the characteristic equation to describe the current waveform. SOLUTION Using the circuit values, the characteristic equation is s2 + 37.5 * 104s + 58.3 * 1010 = 0 and the roots of the equation are s1 , s2 = (-18.75 ; j74) * 104 and hence the network is underdamped. The roots of the characteristic equation illustrate that the damped resonant frequency is d = 740 krad兾s Therefore, fd = 118 kHz and the period of the waveform is T = 1 = 8.5 s fd An actual plot of the current is shown in Fig. 7.43, and this plot verifies that the period of the damped response is indeed 8.5 s. Figure 7.43 300 Current (kA) Load current with a capacitor bank charged to 10 kV. 200 100 0 –100 –200 0 10 20 Time (microsecond) 30 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 343 SECTION 7.5 343 DESIGN EXAMPLES 7.5 Design Examples We wish to design an efficient electric heater that operates from a 24-V dc source and creates heat by driving a 1-⍀ resistive heating element. For the temperature range of interest, the power absorbed by the heating element should be between 100 and 400 W. An experienced engineer has suggested that two quite different techniques be examined as possible solutions: a simple voltage divider and a switched inductor circuit. In the first case, the required network is shown in Fig. 7.44. The variable resistance element is called a rheostat. Potentiometers are variable resistors that are intended for low power (i.e., less than 1 W) operation. Rheostats, on the other hand, are devices used at much higher power levels. We know from previous work that the heating element voltage is Vo = VS Rhe Rhe + Radj 7.34 Po = V2o Rhe = V2S 2 Rhe ARhe + Radj B 7.35 By substituting the maximum and minimum values of the output power into Eq. (7.35), we can determine the range of resistance required for the rheostat: Radj, min = Radj, max (242)(1) V2S Rhe - Rhe = - 1 = 0.2 ⍀ C Po, max C 400 7.36 (242)(1) V2S Rhe = - Rhe = - 1 = 1.4 ⍀ C Po, min C 100 So, a 2-⍀ rheostat should work just fine. But what about the efficiency of our design? How much power is lost in the rheostat? The rheostat power can be expressed as Padj = AVS - Vo B Radj 2 = V2S SOLUTION Radj VS 24 V Changing Radj will change the voltage across, and the power dissipated by, the heating element. The power can be expressed as Radj ARhe + Radj B 2 7.37 We know from our studies of maximum power transfer that the value of Radj that causes maximum power loss, and thus the worst-case efficiency for the circuit, occurs when Radj = Rhe = 1 ⍀. Obviously, the resistances consume the same power, and the efficiency is only 50%. Now that we understand the capability of this voltage-divider technique, let’s explore the alternative solution. At this point, it would at least appear that the use of a switched inductor is a viable alternative since this element consumes no power. So, if we could set up a current in an inductor, switch it into the heating element, and repeat this operation fast enough, the heating element would respond to the average power delivered to it and maintain a constant temperature. Consider the circuit in Fig. 7.45, where the switch moves back and forth, energizing the inductor with current, and then directing that current to the heating element. Let’s examine this concept to determine its effectiveness. We begin by assuming that the inductor current • DESIGN EXAMPLE 7.18 ± – Rhe 1 + Vo - Figure 7.44 A simple circuit for varying the temperature of a heating element. irwin07_296-368hr.qxd 344 28-07-2010 CHAPTER 7 11:34 Page 344 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S Figure 7.45 iL(t) A switched inductor solution to varying the heating element’s temperature. L Rhe VS ± 24 V – 1 pos. 1 - vo(t) + pos. 2 is zero and the switch has just moved to position 1. The inductor current will begin to grow linearly in accordance with the fundamental equation iL(t) = VS 1 1 vL(t) dt = VS dt = t L 3 L 3 L 7.38 Note that VS兾L is the slope of the linear growth. Since VS is set at 24 V, we can control the slope with our selection for L. The inductor current increases until the switch moves at time t = t1, at which point the peak current is Ipeak = VS t L 1 7.39 This inductor current will discharge exponentially through the heating element according to the equation iL At¿B = Ipeak e-t¿兾 7.40 where t¿ is zero when the switch moves to position 2 and = L兾Rhe . If the switch is maintained in position 2 for about 5 time constants, the inductor current will essentially reach zero and the switch can return to position 1 under the initial condition—zero inductor current. A sketch of the inductor current over a single switching cycle is shown in Fig. 7.46. Repeated switching cycles will transfer power to the heating element. If the switching period is much shorter than the element’s thermal time constant—a measure of how quickly the element heats up—then the element’s temperature will be determined by the average power. This is a concept we don’t understand at this point. However, we will present Average Power in Chapter 9, and this example provides at least some motivation for its examination. Nevertheless, we should recognize two things: the load current is just the exponential decaying portion of the inductor current, and the initial value of that exponential is Ipeak, as defined in Eq. (7.39). Increasing Ipeak will increase the element’s power and temperature. As Eq. (7.39) indicates, this is easily done by controlling t1 ! It is impossible to proceed with the design until we can accurately predict the average power at the load. Figure 7.46 Ipeak Inductor current A single switching cycle for the inductor-based solution. The value of Ipeak is directly proportional to t1. VS /L t1 Time irwin07_296-368hr.qxd 28-07-2010 11:34 Page 345 SECTION 7.5 DESIGN EXAMPLES 345 Returning to our original concern, have we improved the efficiency at all? Note that there are no power-consuming components in our new circuit other than the heating element itself. Therefore, ignoring resistance in the inductor and the switch, we find that our solution is 100% efficient! In actuality, efficiencies approaching 95% are attainable. This is a drastic improvement over the other alternative, which employs a rheostat. • Consider the circuit in Fig. 7.47a, where a dc power supply, typically fed from a wall outlet, is modeled as a dc voltage source in series with a resistor RS . The load draws a constant current and is modeled as a current source. We wish to design the simplest possible circuit that will isolate the load device from disturbances in the power supply voltage. In effect, our task is to improve the performance of the power supply at very little additional cost. A standard solution to this problem involves the use of a capacitor CD, as shown in Fig. 7.47b. The two voltage sources and the single-pole double-throw switch model the input disturbance sketched in Fig. 7.47c. Engineers call CD a decoupling capacitor since it decouples disturbances in the input voltage from the output voltage. In typical electronic circuits, we find liberal use of these decoupling capacitors. Thus, our task is to develop a design equation for CD in terms of RS , VS , Vo , ¢VS , ¢Vo , and t¿. Our result will be applicable to any scenario that can be modeled by the circuit in Fig. 7.47b. RS t=t' t=0 RS IL t=t' IL t=0 + VS ± – ± – VS+ΔVS DESIGN EXAMPLE 7.19 CD vo(t) VS ± – ± V +ΔV S S – - VS+ΔVS + vo(t) - VS Vo+ΔVo Vo t' 0 (a) (b) (c) Figure 7.47 (a) A simple dc circuit that models disturbances in the source voltage. (b) The use of a decoupling capacitor to reduce disturbances in the load voltage. (c) Definitions of the input and output voltage disturbances. SOLUTION The voltage across CD can be expressed in the standard form as vo(t) = K1 + K2 e -t兾 7.41 Equivalent circuits for t = 0 and t = q are shown in Fig. 7.48a and b, respectively. At these two time extremes we find vo(0) = K1 + K2 = VS - IL RS vo(q) = K1 = VS + ¢VS - IL RS 7.42 To determine the time constant’s equivalent resistance, we return to the circuit in Fig. 7.48b, reduce all independent sources to zero, and view the resulting circuit from the capacitor’s terminals. It is easy to see that the time constant is simply RS CD . Thus, vo(t) = VS + ¢VS - IL RS - ¢VS e-t兾RS CD 7.43 t irwin07_296-368hr.qxd CHAPTER 7 11:34 Page 346 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S RS RS IL + VS ± – VS-ILRS vo(t) CD - - IL + + VS+ VS ± – VS+VS-ILRS + CD vo(t) - - (a) (b) Figure 7.48 The circuit in Fig. 7.47b at t = 0 just after the switch has moved. (b) The same circuit at t = q . At exactly t = t¿, the output voltage is its original value, Vo(0) plus ¢Vo . Substituting this condition into Eq. (7.43) yields vo At¿B = VS - IL RS + ¢Vo = VS + ¢VS - IL RS - ¢VS e-t¿兾RS CD which can be reduced to the expression ¢VS - ¢Vo = ¢VS e-t¿兾RS CD 7.44 Notice that the value of CD depends not on the input and output voltages, but rather on the changes in those voltages! This makes Eq. (7.44) very versatile indeed. A simple algebraic manipulation of Eq. (7.44) yields the design equation for CD: CD = t¿ RS ln c 7.45 ¢VS d ¢VS - ¢Vo Examining this expression, we see that CD is directly related to t¿ and inversely related to RS . If t¿ doubles or if RS is halved, then CD will double as well. This result is not very surprising. The dependence on the voltage changes is more complex. Let us isolate this term and express it as f = 1 1 -1 = = ¢VS ¢Vo 1 ln c d ln c d ln c 1 d 1 - ¢Vo兾¢VS ¢VS - ¢Vo ¢VS 7.46 Figure 7.49 shows a plot of this term versus the ratio ¢Vo兾¢VS . Note that for very small ¢Vo (i.e., a large degree of decoupling), this term is very large. Since this term is multiplied with t¿ in Eq. (7.45), we find that the price for excellent decoupling is a very large capacitance. Figure 7.49 A plot of the function f versus ¢Vo 兾¢VS. 100 Logarithmic term, f 346 28-07-2010 75 50 25 0 0 0.25 0.5 ΔVo /ΔVS 0.75 1 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 347 SECTION 7.5 DESIGN EXAMPLES Finally, as an example, consider the scenario in which VS is 5 V, RS is 20 ⍀, and the input disturbance is characterized by ¢VS = 1 V and t¿ = 0.5 ms. If the output changes are to be limited to only 0.2 V, the required capacitance would be CD = 112.0 F. Such a capacitor rated for operation at up to 16 V costs less than $0.20 and should be slightly smaller than a peanut M&M. This very simple, but very important, application demonstrates how an engineer can apply his or her basic circuit analysis skills to attack and describe a practical application in such a way that the result is broadly applicable. Remember, the key to this entire exercise is the creation of a circuit model for the decoupling scenario. The network in Fig. 7.50 models an automobile ignition system. The voltage source represents the standard 12-V battery. The inductor is the ignition coil, which is magnetically coupled to the starter (not shown). The inductor’s internal resistance is modeled by the resistor, and the switch is the keyed ignition switch. Initially, the switch connects the ignition circuitry to the battery, and thus the capacitor is charged to 12 V. To start the motor, we close the switch, thereby discharging the capacitor through the inductor. Assuming that optimum starter operation requires an overdamped response for iL(t) that reaches at least 1 A within 100 ms after switching and remains above 1 A for between 1 and 1.5 s, let us find a value for the capacitor that will produce such a current waveform. In addition, let us plot the response, including the time interval just prior to moving the switch, and verify our design. Switch t=0 VS=12 V Figure 7.50 vC (t) + C DESIGN EXAMPLE 7.20 iL(t) Circuit model for ignition system. L=200 mH R=4 Before the switch is moved at t=0, the capacitor looks like an open circuit, and the inductor acts like a short circuit. Thus, iL A0- B = iL A0+ B = 0 A and vC A0- B = vC A0+ B = 12 V After switching, the circuit is a series RLC unforced network described by the characteristic equation s2 + R 1 s + = 0 L LC with roots at s=–s1 and –s2 . The characteristic equation is of the form As + s1 B As + s2 B = s2 + As1 + s2 Bs + s1 s2 = 0 Comparing the two expressions, we see that and R = s1 + s2 = 20 L 1 = s1 s2 LC SOLUTION 347 irwin07_296-368hr.qxd 348 28-07-2010 CHAPTER 7 11:34 Page 348 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S Since the network must be overdamped, the inductor current is of the form iL(t) = K1 e-s1 t + K2 e-s2 t Just after switching, iL A0+ B = K1 + K2 = 0 or K2 = -K1 Also, at t = 0+, the inductor voltage equals the capacitor voltage because iL=0 and therefore iL R=0. Thus, we can write vL A0+ B = L di L A0+ B dt 1 - s1 K1 + s2 K1 = 12 L or K1 = 60 s2 - s1 At this point, let us arbitrarily choose s1=3 and s2=17, which satisfies the condition s1+s2=20, and furthermore, K1 = C = Hence, iL(t) is 60 60 = = 4.29 s2 - s1 14 1 1 = 98 mF = Ls1 s2 (0.2)(3)(17) iL(t) = 4.29Ce-3t - e-17t D A Figure 7.51a shows a plot of iL(t). At 100 ms, the current has increased to 2.39 A, which meets the initial magnitude specifications. However, 1 second later, at t=1.1 s, iL(t) has fallen to only 0.16 A—well below the magnitude-over-time requirement. Simply put, the current falls too quickly. To make an informed estimate for s1 and s2 , let us investigate the effect the roots exhibit on the current waveform when s2>s1 . Since s2>s1 , the exponential associated with s2 will decay to zero faster than that associated with s1 . This causes iL(t) to rise—the larger the value of s2 , the faster the rise. After 5A1兾s2 B seconds have elapsed, the exponential associated with s2 is approximately zero and iL(t) decreases exponentially with a time constant of =1/s1 . Thus, to slow the fall of iL(t) iL(t) 3.0 A 3.0 A 2.0 A 2.0 A 1.0 A 1.0 A 0.0 A –0.0 s 0.5 s 1.0 s 1.5 s Time 2.0 s 2.5 s 2.9 s (a) Figure 7.51 Ignition current as a function of time. 0.0 A –0.0 s 0.5 s 1.0 s 1.5 s 2.0 s 2.5 s 3.0 s 3.5 s 4.0 s Time (b) irwin07_296-368hr.qxd 28-07-2010 11:34 Page 349 SECTION 7.5 DESIGN EXAMPLES 349 iL(t), we should reduce s1 . Hence, let us choose s1=1. Since s1+s2 must equal 20, s2=19. Under these conditions C = 1 1 = 263 mF = Ls1 s2 (0.2)(1)(19) and K1 = Thus, the current is 60 60 = = 3.33 s2 - s1 18 iL(t) = 3.33 C e-t - e-19t D A which is shown in Fig. 7.51b. At 100 ms the current is 2.52 A. Also, at t=1.1 s, the current is 1.11 A—above the 1-A requirement. Therefore, the choice of C=263 mF meets all starter specifications. A defibrillator is a device that is used to stop heart fibrillations—erratic uncoordinated quivering of the heart muscle fibers—by delivering an electric shock to the heart. The Lown defibrillator was developed by Dr. Bernard Lown in 1962. Its key feature, shown in Fig. 7.52a, is its voltage waveform. A simplified circuit diagram that is capable of producing the Lown waveform is shown in Fig. 7.52b. Let us find the necessary values for the inductor and capacitor. DESIGN EXAMPLE 7.21 vo(t) (V) L 3000 i(t) t=0 VS 6000 V 0 5 + R=50 patient C vo(t) - 10 Time (milliseconds) (a) (b) Figure 7.52 Lown defibrillator waveform and simplified circuit. Reprinted with permission from John Wiley & Sons, Inc., Introduction to Biomedical Equipment Technology. Since the Lown waveform is oscillatory in nature, we know that the circuit is underdamped ( 6 1) and the voltage applied to the patient is of the form vo(t) = K1 e-o t sin[t] where o = R 2L = o 21 - 2 and o = 1 1LC SOLUTION • irwin07_296-368hr.qxd 350 28-07-2010 CHAPTER 7 11:34 Page 350 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S for the series RLC circuit. From Fig. 7.52a, we see that the period of the sine function is T = 10 ms Thus, we have one expression involving o and = 2 = o 21 - 2 = 200 rad兾s T A second expression can be obtained by solving for the ratio of vo(t) at t = T兾4 to that at t = 3T兾4. At these two instants of time, the sine function is equal to +1 and -1, respectively. Using values from Fig. 7.52a, we can write vo(t兾4) -vo(3T兾4) = K1 e-o(T兾4) K1 e-o(3T兾4) = eo(T兾2) L 3000 = 12 250 or o = 497.0 Given R = 50 ⍀, the necessary inductor value is L = 50.3 mH Using our expression for , 2 = (200)2 = 2o - (o)2 or = (200)2 = 1 - (497.0)2 LC Solving for the capacitor value, we find C = 31.0 F Let us verify our design using the circuit shown in Fig. 7.53a. The output voltage plot shown in Fig. 7.53b matches the Lown waveform in Fig. 7.52a; thus, we can consider the design to be a success. While this solution is viable, it is not the only one. Like many design problems, there are often various ways to satisfy the design specifications. Figure 7.53 vo(t) Circuit and output plot for the Lown defibrillator. 4.0 kV L=50.3 mH 3.0 kV i(t) + vC(t)= 5000 V - + R=50 C=31 F vo(t) - 2.0 kV 1.0 kV 0 kV –1.0 kV 0s 2 ms 4 ms 6 ms Time (a) (b) 8 ms 10 ms irwin07_296-368hr.qxd 28-07-2010 11:34 Page 351 PROBLEMS 351 SUMMARY • First-Order Circuits ■ ■ ■ An RC or RL transient circuit is said to be first order if it contains only a single capacitor or single inductor. The voltage or current anywhere in the network can be obtained by solving a first-order differential equation. Second-Order Circuits ■ The form of a first-order differential equation with a constant forcing function is x(t) dx(t) dt + 2 ■ • The time constant for an RC circuit is RTh C and for an RL circuit is L/RTh , where RTh is the Thévenin equivalent resistance looking into the circuit at the terminals of the storage element (i.e., capacitor or inductor). The two approaches proposed for solving first-order transient circuits are the differential equation approach and the step-by-step method. In the former case, the differential equation that describes the dynamic behavior of the circuit is solved to determine the desired solution. In the latter case, the initial conditions and the steady-state value of the voltage across the capacitor or current in the inductor are used in conjunction with the circuit’s time constant and the known form of the desired variable to obtain a solution. + 20 ■ The characteristic equation for a second-order circuit is s2 + 20 s + 20 = 0, where is the damping ratio and 0 is the undamped natural frequency. ■ If the two roots of the characteristic equation are where A is referred to as the steady-state solution and is called the time constant. The function e-t兾 decays to a value that is less than 1% of its initial value after a period of 5. Therefore, the time constant, , determines the time required for the circuit to reach steady state. dx(t) + 20 x(t) = f(t) dt dt where f(t) is the network forcing function. = A x(t) = A + K2 e-t兾 ■ The voltage or current in an RLC transient circuit can be described by a constant coefficient differential equation of the form d2x(t) and the solution is ■ The response of a first-order transient circuit to an input pulse can be obtained by treating the pulse as a combination of two step-function inputs. ■ ■ ■ ■ real and unequal, then >1 and the network response is overdamped real and equal, then =1 and the network response is critically damped complex conjugates, then <1 and the network response is underdamped The three types of damping together with the corresponding network response are as follows: 1. Overdamped: 2 2 x(t) = K1 e-A0 - 0 2 - 1Bt + K2 e-A0 + 0 2 - 1Bt 2. Critically damped: x(t) = B1 e-0 t + B2 te-0 t 3. Underdamped: x(t) = e-t AA 1 cos d t + A 2 sin d tB, where = 0 and d = 0 21 - 2 ■ Two initial conditions are required to derive the two unknown coefficients in the network response equations. PROBLEMS 7.1 Use the differential equation approach to find i(t) for t 7 0 in the network in Fig. P7.1. 7.2 Use the differential equation approach to find io(t) for t 7 0 in the network in Fig. P7.2. t=0 4 k 6 12 V ± – 1 k t=0 i(t) 2H 6 2 k 4 mA 300 F 2 k io(t) Figure P7.1 Figure P7.2 irwin07_296-368hr.qxd 352 28-07-2010 CHAPTER 7 11:34 Page 352 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7.7 Use the differential equation approach to find io(t) for 7.3 Use the differential equation approach to find vo(t) for t 7 0 in the circuit in Fig. P7.3 and plot the response including the time interval just prior to switch action. 12 k t=0 t 7 0 in the circuit in Fig. P7.7 and plot the response including the time interval just prior to opening the switch. 2 k 50 F t=0 12 V + + 6V ± – 100 F 4 k vo(t) 3 k 4 k - 4 k 12 k 12 k 8 k io(t) Figure P7.3 Figure P7.7 7.4 Use the differential equation approach to find vC(t) for 7.8 Use the differential equation approach to find io(t) for t 7 0 in the circuit in Fig. P7.4. t 7 0 in the circuit in Fig. P7.8 and plot the response including the time interval just prior to closing the switch. t=0 6 k 12 k 200 F 6 k + vC(t) 3 k ± – 100 F ± – 6V - t=0 12 k 12 V 6 k io(t) Figure P7.8 Figure P7.4 7.5 Use the differential equation approach to find vC(t) for 7.9 Use the differential equation approach to find vo(t) for t 7 0 in the circuit in Fig. P7.9 and plot the response including the time interval just prior to opening the switch. t 7 0 in the circuit in Fig. P7.5. 9 k 100 F 4 k 3 k 6V ± – + vC(t) t=0 100 F + 6 k - t=0 12 V ± – 6 k 6 k vo(t) - Figure P7.5 Figure P7.9 7.6 Use the differential equation approach to find vC(t) for t 7 0 in the circuit in Fig. P7.6 and plot the response including the time interval just prior to closing the switch. 2 k 4 k 7.10 In the network in Fig. P7.10, find io(t) for t 7 0 using the differential equation approach. 6 4 k io(t) + 6V ± – vC(t) t=0 Figure P7.6 t=0 100 F 2 k 2H - Figure P7.10 4 2A 12 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 353 PROBLEMS 7.11 Use the differential equation approach to find iL(t) for 7.14 Using the differential equation approach, find io(t) for t 7 0 in the circuit in Fig. P7.11 and plot the response, including the time interval just prior to opening the switch. 6 6 + 6 iL(t) t=0 6V 12 t 7 0 in the circuit in Fig. P7.14 and plot the response, including the time interval just prior to opening the switch. 2H + 353 6V 2H 3 12 V ± – 12 4 Figure P7.11 t=0 io(t) Figure P7.14 7.12 Use the differential equation approach to find vo(t) for t 7 0 in the circuit in Fig. P7.12 and plot the response, including the time interval just prior to opening the switch. 7.15 Use the step-by-step technique to find io(t) for t 7 0 in the network in Fig. P7.15. 100 F 4 k 2 k io(t) 6V 12 k + ± – ± – 12 V + 200 F 6 k t=0 12 V 8 k t=0 Figure P7.15 vo(t) 7.16 Use the step-by-step method to find io(t) for t 7 0 in the 4 k circuit in Fig. P7.16. 2 k Figure P7.12 Figure P7.16 7.13 Use the differential equation approach to find i(t) for 5 k t=0 2 k 7.17 Find vo(t) for t>0 in the network in Fig. P7.17. t=0 1 k i(t) 5 mA 1 k 4 k 2 k 12 V t=0 io(t) t 7 0 in the circuit in Fig. P7.13 and plot the response, including the time interval just prior to opening the switch. ± – 2H 6 k 1 k 4 k 1 mH 6 mA 4 k + 4 k 100 F vo(t) - Figure P7.13 Figure P7.17 irwin07_296-368hr.qxd 354 28-07-2010 CHAPTER 7 11:34 Page 354 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7.23 Find vo(t) for t>0 in the circuit in Fig. P7.23. 7.18 Find vo(t) for t>0 in the network in Fig. P7.18. t=0 3 k 6 k 6 k 6 k 2 k 50 F t=0 12 mA 6 k 4 k vo(t) 12 V + 3 k 8 k 2 k + - 100 F vo(t) 1 k - Figure P7.23 Figure P7.18 7.19 Use the step-by-step method to find io(t) for t 7 0 in the circuit in Fig. P7.19. 7.24 Find vo(t) for t>0 in the network in Fig. P7.24. t=0 6 ± – 2H io(t) 6V + - 6 12 k 6 k 24 V + 8 k 50 F vo(t) 4 k - t=0 Figure P7.24 Figure P7.19 7.25 Use the step-by-step technique to find io(t) for t 7 0 in the network in Fig. P7.25. 7.20 Use the step-by-step method to find io(t) for t 7 0 in the circuit in Fig. P7.20. ±– 4 k 12 V 4 k t=0 4 k 200 F 6 k 12 k 12 mA 200 F 4 k t=0 io(t) 12 k Figure P7.25 io(t) 7.26 Use the step-by-step method to find vo(t) for t 7 0 in Figure P7.20 the network in Fig. P7.26. t=0 7.21 Find vo(t) for t>0 in the circuit in Fig. P7.21. 80 F 6 k + t=0 4 mA 6 k 6 k + 12 mA 100 F vo(t) - Figure P7.21 7.27 Find io(t) for t>0 in the circuit in Fig. P7.27. 9 k 2 k 12 mA 60 F 4 k 6 k t=0 io(t) t=0 2 k + vo(t) 4 VA + - - Figure P7.22 vo(t) - Figure P7.26 7.22 Find vo(t) for t>0 in the circuit in Fig. P7.22. 3 k 6 k 6 k Figure P7.27 1 k 150 F 1 k + - 6V irwin07_296-368hr.qxd 28-07-2010 11:34 Page 355 PROBLEMS 7.28 Find io(t) for t>0 in the circuit in Fig. P7.28. 355 7.32 Use the step-by-step method to find vo(t) for t 7 0 in the network in Fig. P7.32. 6V –+ 1 —H 3 t=0 2 k + 4 t=0 2 k 100 F 2 k 2 io(t) ± – Figure P7.28 12 V 7.29 Find io(t) for t>0 in the circuit in Fig. P7.29. – ± 12 V 4 k vo(t) 2 - Figure P7.32 4 mA 8 k 120 F 4 k t=0 7.33 Find vc(t) for t 7 0 in the network in Fig. P7.33 using the step-by-step method. it (t) 2 k 2 k Figure P7.29 t=0 7.30 Find io(t) for t>0 in the network in Fig. P7.30. + 50 F vC(t) 2 k 100 F - ± – 6 k 6 k + - t=0 6V 12 V io(t) 6 k Figure P7.33 7.34 Find vo(t) for t 7 0 in the network in Fig. P7.34 using Figure P7.30 the step-by-step method. 7.31 Find io(t) for t>0 in the circuit in Fig. P7.31. t=0 2 k 150 F 50 F io(t) ± – 2 k 24 V + 2 k 10 k 4 k 6 mA t=0 4 k vo(t) - Figure P7.31 Figure P7.34 4 k ± – 12 V irwin07_296-368hr.qxd 356 28-07-2010 CHAPTER 7 11:34 Page 356 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7.35 Find vo(t) for t 7 0 in the circuit in Fig. P7.35 using the 7.38 Find vo(t) for t 7 0 in the circuit in Fig. P7.38 step-by-step method. using the step-by-step method. t=0 4 k 3 6 vo(t) 4 k + + 50 F 8 k 4 12 V 6 6A + vo(t) t=0 2H Figure P7.38 - 7.39 The switch in the circuit in Fig. P7.39 is opened at t=0. Find i(t) for t>0. Figure P7.35 0.5 H t=0 4 7.36 Use the step-by-step method to find io(t) for t 7 0 in the circuit in Fig. P7.36. 4 k 24 V 4 k 200 F 6V ± – t=0 4 k i(t) 3 2 k + – 2 – + io(t) 9V Figure P7.36 Figure P7.39 7.40 The switch in the circuit in Fig. P7.40 is moved at t=0. Find io(t) for t>0. 7.37 Find vo(t) for t 7 0 in the network in Fig. P7.37 using the step-by-step technique. 4 k 1 k 200 F 4 k + 24 V ± – 4 k t=0 2 k vo(t) - 18 V 100 F + t=0 4 k 3 k io(t) 4 k Figure P7.40 Figure P7.37 7.41 The switch in Fig. P7.41 closes at t=0. Find vo(t) for t>0. t=0 9 k + 6 k 6V + - 3 k 4 mA 120 F 4 k vo(t) - Figure P7.41 4 k irwin07_296-368hr3.qxd 2-09-2010 12:13 Page 357 357 PROBLEMS 7.42 The switch in the circuit in Fig. P7.42 has been closed for a long time and is opened at t=0. Find i(t) for t>0. 1 k⍀ 7.46 Find io(t) for t 7 0 in the network in Fig. P7.46 using the step-by-step method. 12 V 1 k⍀ + 2 k⍀ t=0 + - 30 V 200 F 0.3 H i(t) + - 10 V t=0 2 k⍀ 2 k⍀ 2 k⍀ 2 k⍀ 4 k⍀ 2 k⍀ io(t) Figure P7.42 7.43 The switch in the circuit in Fig. P7.43 has been closed for a long time and is moved at t=0. Find io(t) for t>0. Figure P7.46 7.47 Use the step-by-step method to find io(t) for t 7 0 in the t=0 5 k⍀ network in Fig. P7.47. io(t) - 10 V + 5 mA 5 k⍀ 6 k⍀ 2 k⍀ 6 k⍀ 2/3 H 6 k⍀ 5 k⍀ + - io(t) t=0 12 V 50 F Figure P7.43 7.44 The switch in the circuit in Fig. P7.44 has been in position A for a long time and is moved to position B at t=0. Calculate vo(t) for t>0. t=0 A Figure P7.47 7.48 Find vo(t) for t>0 in the network in Fig. P7.48. 5 k⍀ 3 k⍀ B + 24 V + - 1 k⍀ vo(t) 10 F + - + - t=0 8 k⍀ 60 F 6V 6 k⍀ 12 V + vo(t) 4 k⍀ 80 F - - Figure P7.44 Figure P7.48 7.45 The switch in the circuit in Fig. P7.45 has been closed for a long time and is opened at t=0. Determine vC(t) for t>0. 5 k⍀ 5 k⍀ 7.49 Find vo(t) for t>0 in the network in Fig. P7.49. 3 k⍀ 9 k⍀ t=0 30 V + + 10 k⍀ + - Figure P7.45 t=0 2 k⍀ + 18 V 20 F vC (t) 4V 6 k⍀ - + 8 k⍀ 200 F 4 k⍀ vo(t) - Figure P7.49 irwin07_296-368hr.qxd 358 28-07-2010 CHAPTER 7 11:34 Page 358 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7.50 Find vo(t) for t>0 in the network in fig. P7.50. 7.54 Find vo(t) for t>0 in the network in Fig. P7.54. 12 V t=0 6 k 3 k 2 k 6 k 3 k + - 160 F 3 k vo(t) 200 F 3 k + 12 V + 4 k 12 k + 6V - + - 2 k vo(t) 1 k - Figure P7.50 t=0 7.51 Determine vo(t) for t>0 in the network in Fig. P7.51 Figure P7.54 7.55 Use the step-by-step technique to find vo(t) for t 7 0 in t=0 6 k 5 k the circuit in Fig. P7.55. + - + 6V 50 F 6 k 3 k + 12 V + - 200 F 6 k vo(t) 1 k - 6 k + - 4 k 7.56 Use the step-by-step technique to find io(t) for t 7 0 in + - 3 k the network in Fig. P7.56. 6V 4 mA 100 F + 12 V + - 12 V Figure P7.55 1 k 2 k - t=0 Figure P7.51 7.52 Find vo(t) for t>0 in the circuit in Fig. P7.52. vo(t) 6 k t=0 100 F 4 k 2 k vo(t) 4 k 2 k 2 k 12 k 2 k - t=0 ± – 4V io(t) Figure P7.52 Figure P7.56 7.53 Find vo(t) for t>0 in the network in Fig. P7.53. + 150 F 6 k 3 k vo(t) - t=0 6 mA 7.57 Find vo(t) for t 7 0 in the network in Fig. P7.57 using the step-by-step technique. + 2 2H 3 12 V 12 k 2 ± – 6 4 k t=0 Figure P7.53 vo(t) Figure P7.57 - irwin07_296-368hr2.qxd 2-08-2010 16:43 Page 359 359 PROBLEMS 7.62 Find vo(t) for t>0 in the circuit in Fig. P7.62. 7.58 Find io(t) for t>0 in the network in Fig. P7.58. t=0 1 k⍀ 4 k⍀ 150 F 6V t=0 3 k⍀ + - 1 k⍀ 4 mA 1 k⍀ + - 6V 2 k⍀ + 1 k⍀ io(t) 200 F 1 k⍀ 1 k⍀ Figure P7.58 vo(t) - 7.59 Find vc(t) for t 7 0 in the circuit in Fig. P7.59 using Figure P7.62 the step-by-step method. 5⍀ 7.63 Find io(t) for t>0 in the network in Fig. P7.63. t=0 io(t) 5⍀ 20 ⍀ 6⍀ t=0 0.5 F 6A - vC(t) 12 V + 10 ⍀ 3⍀ + – 12 ⍀ + 0.2 H 10 ⍀ Figure P7.63 5⍀ 7.64 Find vo(t) for t>0 in the circuit in Fig. P7.64. Figure P7.59 6⍀ 7.60 Find i(t) for t 7 0 in the circuit in Fig. P7.60 using the step-by-step method. 6⍀ t=0 5⍀ t=0 12 V 2⍀ + 0.5 H ± – i(t) 2⍀ –30 V + t=0 2⍀ 30 V ± – 6⍀ 0.1 H vo(t) 6⍀ - Figure P7.64 Figure P7.60 7.61 Find i(t) for t 7 0 in the circuit in Fig. P7.61 using the step-by-step method. 7.65 Find io(t) for t>0 in the circuit in Fig. P7.65. 2H 12 ⍀ 10 ⍀ 105 V 15 ⍀ t=0 1H ± – 6V + 6⍀ i(t) 6⍀ 6⍀ 4⍀ t=0 2⍀ io(t) Figure P7.61 Figure P7.65 6V irwin07_296-368hr.qxd 360 28-07-2010 CHAPTER 7 11:34 Page 360 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7.66 Find io(t) for t>0 in the network in Fig. P7.66. 3 + – 12 V 2 1H t=0 6 4 4A io(t) Figure P7.66 7.67 Find io(t) for t>0 in the circuit in Fig. P7.67. 2H 6 6V + – t=0 4 12 2 2A io(t) Figure P7.67 7.68 Find vo(t) for t>0 in the network in Fig. P7.68. 1H + 6 3A vo(t) 2 t=0 3 + – 2 24 A - Figure P7.68 7.71 Use the step-by-step method to find vo(t) for t 7 0 in 7.69 Find io(t) for t>0 in the circuit in Fig.P7.69. the network in Fig. P7.71. io(t) 2 6 k 6 12 V 6 vo(t) 4 k t=0 3 1H 6 F 6 12 V + – + 2 k t=0 ± – 6 F 6 F - Figure P7.71 Figure P7.69 7.70 Use the step-by-step method to find io(t) for t 7 0 in the network in Fig. P7.70. 7.72 Find iL(t) for t 7 0 in the circuit in Fig. P7.72 using the step-by-step method. 4 t=0 t=0 6 12 V ± – 4 12 io(t) 10 V ± – 3H 3A 2H 12 + 4A v1(t) - 2 2H ±– 5v1(t) Figure P7.70 Figure P7.72 iL(t) irwin07_296-368hr.qxd 28-07-2010 11:34 Page 361 PROBLEMS 7.73 Find io(t) for t 7 0 in the network in Fig. P7.73 using the step-by-step method. iA 2 k 7.76 The current source in the network in Fig. P7.76a is defined in Fig. P7.76b. The initial voltage across the capacitor must be zero. (Why?) Determine the current io(t) for t 7 0. t=0 2 k io(t) 10 mA 2 k 2 2 mH 361 2 iA 2 k i(t) 2F 2 2 io(t) Figure P7.73 (a) i(t) (A) 6 7.74 Use the step-by-step technique to find vo(t) for t 7 0 in the network in Fig. P7.74. 24 V 3 k ±– 6V 200 F 3 k 2 k 0 4.5 (b) + – ± vo(t) t=0 - t (s) 6 k 2000 iA ± – iA Figure P7.76 7.77 The voltage v(t) shown in Fig. P7.77a is given by the graph shown in Fig. P7.77b. If iL(0) = 0, answer the following questions: (a) How much energy is stored in the inductor at t = 3 s? Figure P7.74 (b) How much power is supplied by the source at t = 4 s? (c) What is i(t = 6 s)? (d) How much power is absorbed 7.75 Determine the equation for the voltage vo(t) for t 7 0 in by the inductor at t = 3 s? Fig. P7.75a when subjected to the input pulse shown in Fig. P7.75b. 3 k 2 k + v(t) ± – i(t) vo(t) 200 F 6 k v(t) ± – 2 2H iL(t) (a) (a) v(t) v(t) (V) 10 V 12 5 t(s) 2 0 1 t (s) –10 V (b) Figure P7.75 (b) Figure P7.77 irwin07_296-368hr.qxd 362 28-07-2010 CHAPTER 7 11:34 Page 362 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7.78 Find the output voltage vo(t) in the network 7.82 The switch in the circuit in Fig. P7.82 is closed at t = 0. If i1(0-) = 2 A, determine i2(0+), vR(0+), and i1(t = q) . in Fig. P7.78 if the input voltage is vi(t) = 5(u(t) - u(t - 0.05))V. t=0 i1(t) vi(t) ± – + + 1 F vR(t) 2H vo(t) 100 k i2(t) 3H 4 - - Figure P7.82 Figure P7.78 7.83 The switch in the circuit in Fig. P7.83 has been closed for a long time and is opened at t = 0. If vC(t) = 20 - 8e-0.05t V, find R1, R2, and C. 7.79 Given that vC1(0-) = -10 V and vC2(0-) = 20 V in the circuit in Fig. P7.79, find i(0+). + R2 t=0 4 + i(t) vC1(t) 2F 4F - + vC(t) C R1 2A vC2(t) t=0 - - Figure P7.79 Figure P7.83 7.84 Given that i(t) = 2.5 + 1.5e-4t A for t 7 0 in the 7.80 In the network in Fig. P7.80, find i(t) for t 7 0. If circuit in Fig. P7.84, find R1, R2, and L. t=0 vC1(0-) = -10 V, calculate vC2(0-). i(t) i(t) vC1(t) + - 0.3 F ± – 10e–5tu(t) V R1 + ± – vC2(t) 0.6 F - R2 L 20 V Figure P7.84 7.85 Given that i(t) = 13.33e-t - 8.33e-0.5tA for t 7 0 in Figure P7.80 the network in Fig. P7.85, find the following: (a) vc(0), 7.81 In the circuit in Fig. P7.81, vR(t) = 100e-400t V for t 6 0. Find vR(t) for t 7 0. (b) vc(t = 1 s), (c) the capacitance C. t=0 + 25 100 vR(t) 4 L ± – 40e–tu(t) V - Figure P7.81 v (t) + C - i(t) Figure P7.85 C 2 irwin07_296-368hr.qxd 28-07-2010 11:34 Page 363 363 PROBLEMS 7.86 The differential equation that describes the current io(t) 7.92 For the underdamped circuit shown in Fig. P7.92, determine the voltage v(t) if the initial conditions on the storage elements are iL(0) = 1 A and vC(0) = 10 V. in a network is 2 d io(t) dt 2 + 6 dio(t) dt + 4io(t) = 0 + iL(0) Find v(t) 2H (a) the characteristic equation of the network, (b) the network’s natural frequencies dt + 4 dv1(t) dt + 5v1(t) = 0 1 —F 40 vC (0) - Figure P7.92 7.87 The voltage v1(t) in a network is defined by the equation 2 5 - (c) the expression for io(t). d 2v1(t) + 7.93 Find vC(t) for t 7 0 in the circuit in Fig. P7.93. t=0 Find (a) the characteristic equation of the network (b) the circuit’s natural frequencies 0.04 F 8 1A (c) the expression for v1(t) . + iL(t) vC(t) 1H - 7.88 The voltage v1(t) in a network is defined by the equation d2v1(t) dt 2 + 2c dv1(t) dt Figure P7.93 d + 5v1(t) = 0 7.94 Find vC(t) for t 7 0 in the circuit in Fig. P7.94 if Find vC(0) = 0. (a) the characteristic equation of the network. t=0 (b) the circuit’s natural frequencies. 1 k 100 mH (c) the expression for v1(t). + 7.89 A parallel RLC circuit contains a resistor R = 1 ⍀ and ± – 1 F 12 V vC(t) - an inductor L = 2H. Select the value of the capacitor so that the circuit is critically damped. Figure P7.94 7.90 A series RLC circuit contains a resistor of R = 2 ⍀ and a capacitor C = 1兾2 F. Select the value of the inductor so that the circuit is critically damped. 7.95 Find vo(t) for t 7 0 in the circuit in Fig. P7.95 and plot the response, including the time interval just prior to closing the switch. 2 7.91 In the critically damped circuit shown in Fig. P7.91, the t=0 initial conditions on the storage elements are iL(0) = 2 A and vC(0) = 5 V. Determine the voltage v(t). + vC (0) + 0.01 F - Figure P7.91 v(t) iL(0) 10 + 12 V ± – 4H - Figure P7.95 1 —F 400 1H 10 vo(t) - irwin07_296-368hr.qxd 364 28-07-2010 CHAPTER 7 11:34 Page 364 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7.96 The switch in the circuit in Fig. P7.96 has been closed for 7.97 The switch in the circuit in Fig P7.97 has been closed for a a long time and is opened at t = 0. Find i(t) for t 7 0. long time and is opened at t = 0. Find i(t) for t 7 0. 6 t=0 5 1.5 5 t=0 5 20 V 1H – ± 12 V 0.04 F ± – 0.1 F ± – i(t) ± – i(t) 6V 10 V 1.25 H Figure P7.96 Figure P7.97 7.98 Find vo(t) for t 7 0 in the circuit in Fig. P7.98 and plot the response, including the time interval just prior to closing the switch. 1 —H 8 8 1 —H 8 t=0 + 1 —F 25 4 1.5 A 6 vo(t) 1 - Figure P7.98 7.99 In the circuit shown in Fig. P7.99, find v(t) 7 0. 7.101 Find vo(t) for t 7 0 in the circuit in Fig. P7.101 and plot the response, including the time interval just prior to moving the switch. 12 H — 5 t=0 t=0 2.5 mH 8 k + ± – 1 F — 12 2 8V v(t) 2 — nF 15 1 k 6 k ± – 7.100 Find vo(t) for t 7 0 in the circuit in Fig. P7.100 and plot the response, including the time interval just prior to moving the switch. t=0 6 k 12 V 0.05 nF vo(t) Figure P7.101 7.102 Find vo(t) for t 7 0 in the network in Fig. P7.102 and plot the response, including the time interval just prior to moving the switch. 1 mH 1 k 4 k 3 mA - 10 k 2.5 mA 4 k + Figure P7.99 100 V ± – Figure P7.100 5 k t=0 + 10 k 10 F vo(t) - 100 mH Figure P7.102 2 k irwin07_296-368hr.qxd 28-07-2010 11:34 Page 365 PROBLEMS 7.103 Find vo(t) for t 7 0 in the circuit in Fig. P7.103 and plot the response, including the time interval just prior to moving the switch. the characteristic equation s 2 + 4 * 107s + 3 * 1014 = 0 ± t=0 ± – 7.104 Design a parallel RLC circuit with R ⭌ 1 k⍀ that has 1 mH 2 k 12 V 365 1 k 6.25 nF vo(t) 7.105 Design a parallel RLC circuit with R ⭌ 1 k⍀ that has – the characteristic equation s2 + 4 * 107s + 4 * 1014 = 0 Figure P7.103 Solve the remaining problems using computational methods. 7.106 Given the network in Fig. P7.106 to plot vo(t) over a 10-s interval starting at t = 0, using a 100-ms step size. 2 t=0 4 2H ± – 12 V 6 2A 1F + 2 vo(t) - Figure P7.106 7.107 Given the network in Fig. P7.107, plot vo(t) over a 10-s interval starting at t = 0, using a 100-ms step size. –± t=0 20 k 6V age shown in Fig. P7.108b, plot the voltage vo(t) over the interval 0 ⱕ t ⱕ 4 s, using a 20-ms step size. 12 V 100 k 10 k ± – + 10 mH 7.108 Given the network in Fig. P7.108a and the input volt- 5 k 50 F vin(t) ± – + 1M vo(t) 0.1 F 2 mH - vo(t) - (a) vin(t) V Figure P7.107 10 0 1 t(s) (b) Figure P7.108 irwin07_296-368hr.qxd 366 28-07-2010 CHAPTER 7 11:34 Page 366 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7.109 Given the network in Fig. P7.109a, and the input voltage shown in Fig. P7.109b, plot vo(t) over the interval 0 ⱕ t ⱕ 10 s, using a 100-ms step size. –± vin(t) 2 2 2 2A 1F 1 1H + vo(t) 4 - (a) vin(t) V 10 0 1 t (s) (b) Figure P7.109 7.110 Given the network in Fig. P7.110a and the input in Fig. P 7.110b, plot vo(t) in the interval 0 ⱕ t ⱕ 4 s, using a 20-ms step size. 10 k 5 k vs(t) ± – + 100 F 10 mH 10 k vo(t) - (a) vs(t) V 10 0 1 2 3 (b) Figure P7.110 4 t (s) irwin07_296-368hr.qxd 28-07-2010 11:34 Page 367 TYPICAL PROBLEMS FOUND ON THE FE EXAM • TYPICAL PROBLEMS FOUND ON THE FE EXAM 7PFE-1 In the circuit in Fig. 7PFE-1, the switch, which has been closed for a long time, opens at t = 0. Find the value of the capacitor voltage vc(t) at t = 2s. a. 0.936 V b. 0.756 V c. 0.264 V d. 0.462 V t=0 8 k 12 V 6 k + ± – vC(t) 6 k 100 F 6 k - Figure 7PFE-1 7PFE-2 In the network in Fig. 7PFE-2, the switch closes at t = 0. Find v0(t) at t = 1s. a. 5.62 V b. 1.57 V c. 4.25 V d. 3.79 V 12 k 4 k + t=0 ± – 12 V 100 F 12 k vo(t) - Figure 7PFE-2 7PFE-3 Assume that the switch in the network Fig. 7PFE-3 has been closed for some time. At t = 0 the switch opens. Determine the time required for the capacitor voltage to decay to one-half of its initially charged value. a. 0.416 s b. 0.625 s c. 0.235 s d. 0.143 s 12 k t=0 12 V ± – + vC(t) - Figure 7PFE-3 100 F 6 k 367 irwin07_296-368hr.qxd 368 28-07-2010 CHAPTER 7 11:34 Page 368 F I R S T- A N D S E C O N D - O R D E R T R A N S I E N T C I R C U I T S 7PFE-4 Find the inductor current iL(t) for t 7 0 in the circuit in Fig. 7PFE-4. a. iL(t) = 3 - 2e-t兾6 A, t 7 0 b. iL(t) = 1 + 2e-2t兾3 A, t 7 0 c. iL(t) = 6 - e-t兾6 A, t 7 0 d. iL(t) = 3 - e-2t兾3 A, t 7 0 t=0 2 iL(t) ± – 10 V 2 2 4H 1A Figure 7PFE-4 7PFE-5 Find the inductor current iL(t) for t 7 0 in the circuit in Fig. 7PFE-5. a. iL(t) = 1.4 + 0.4e-4t兾3 A, t 7 0 b. iL(t) = 1.2 + 0.4e-5t兾3 A, t 7 0 c. iL(t) = 0.4 + 0.2e-4t兾3 A, t 7 0 d. iL(t) = 2.4 + 0.6e-5t兾3 A, t 7 0 t=0 iL(t) 3H 3 12 V ± – Figure 7PFE-5 2 4 4