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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
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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
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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
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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
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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
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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).
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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
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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
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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.
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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
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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
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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
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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.
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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
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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.
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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
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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
+
–
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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
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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
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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
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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
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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.
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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
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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)
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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
+ 2␨␻0
dx(t)
dt
+ ␻20 x(t) = 0
7.14
where we have made the following simple substitutions for the constants a1 = 2␨␻0 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 + 2␨␻0 sKest + ␻20 Kest = 0
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Dividing both sides of the equation by Kest yields
s2 + 2␨␻0 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
-2␨␻0 ; 24␨2␻20 - 4␻20
2
s =
= -␨␻0 ; ␻0 2␨2 - 1
7.16
Therefore, two values of s, s1 and s2, satisfy Eq. (7.15):
s1 = -␨␻0 + ␻0 2␨2 - 1
7.17
s2 = -␨␻0 - ␻0 2␨2 - 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- A␨␻0 - ␻0 2␨
2
- 1Bt
+ K2 e- A␨␻0 + ␻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
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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 + j␻0 21 - ␨2 = -␴ + j␻d
s2 = -␨␻0 - j␻0 21 - ␨2 = -␴ - j␻d
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
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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 + 2␨␻0 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-A␨␻0 - ␻0 2␨ - 1Bt + K2 e-A␨␻0 + ␻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
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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
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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)
-
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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
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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
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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
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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
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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
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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.
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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
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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
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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.
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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
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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)
-
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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
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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 + 2␨␻0 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
2␨␻0 = 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
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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
-
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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
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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
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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
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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.
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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
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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
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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
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75
50
25
0
0
0.25
0.5
ΔVo /ΔVS
0.75
1
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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
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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)
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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
•
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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)
= e␨␻o(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
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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.
+ 2␨␻0
■
The characteristic equation for a second-order circuit is
s2 + 2␨␻0 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-A␨␻0 - ␻0 2␨ - 1Bt + K2 e-A␨␻0 + ␻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
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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
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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
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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
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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
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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
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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
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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
-
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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
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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)
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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
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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
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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)
-
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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
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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
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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)
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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
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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