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Chapter 2
RC and RL Circuits with Piecewise Constant
Sources
2.1 Background
Consider a circuit consisting of a single capacitor, resistors, dc independent sources, and controlled sources (assumed to be linear), as shown in Fig. 2.1 (a). Using Thevenin’s theorem,
C C
Circuit (resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)
A
B
A
B (a) (b)
VT h
RT h
i i VC VC
Figure 2.1: A linear circuit with a single capacitor.
the circuit can be reduced to that shown in Fig. 2.1 (b), where VTh is a dc source. The behaviour of this circuit is governed by the Ordinary Differential Equation (ODE),
VTh = i RTh + VC = RTh C
dVC dt
+ VC , (2.1)
i.e.,
dVC dt
τ
(VTh − VC ) , (2.2)
where τ = RTh C. The homogeneous part of Eq. 2.1 is,
dVC dt
VC
τ
for which the solution is V (^) C(h )= A e−t/τ^. The complete solution is VC (t) = V (^) C(h )(t) + V (^) C(p )(t),
V (^) C(p )(t) being a particular solution. In Eq. 2.2, if we substitute VC = K (a constant), we get a particular solution, viz., VC = K = VTh. In general, the complete solution for VC (t) is therefore given by, VC (t) = A e−t/τ^ + B , (2.4)
2.2 M. B. Patil, IIT Bombay
where A and B are constants, and τ = RTh C is the circuit time constant. In fact, any current or voltage in the circuit is given by the general form of Eq. 2.4, the constants A and B depending on the variable under consideration. For a circuit containing a single inductor, resistors, dc independent sources, and controlled sources (assumed to be linear), we get (see Fig. 2.2),
Circuit (resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)
A
B
A
B (a) (b)
VL VT h VL
i i
RT h
Figure 2.2: A linear circuit with a single inductor.
VTh = RTh i + L
di dt
i.e.,
di dt
i τ
VTh L
where τ = L/RTh. The general solution for i (or any other current or voltage in the circuit) is of the form, i(t) = A e−t/τ^ + B. (2.7)
There are situations in which the independent sources in the circuit of Fig. 2.1 or Fig. 2.2 are not constant at all times but are piece-wise constant (see Fig. 2.3.) The general form, x(t) = A e−t/τ^ + B is applicable in such cases as well except that the constants A and B need
t
Vs Vs
t
Figure 2.3: Examples of a piece-wise constant voltage source.
to be computed in each interval where the source is constant. If there is a sudden change in the circuit (such as a voltage source changing from 0 V to 5 V or a switch opening or closing), we need to obtain new initial conditions for the quantity of interest, as illustrated in the following example. Consider the circuit shown in Fig. 2.4. Suppose the change in the source voltage from 0 V to
5 V occurs at time t 0. If the capacitor voltage at t− 0 and t+ 0 are different, then
VC (t+ 0 ) − VC (t− 0 ) t+ 0 − t− 0
would be a large quantity. This quantity is in fact
dVC dt
, as t+ 0 → t− 0 , which means that, if
VC (t+ 0 ) 6 = VC (t− 0 ), a large capacitor current would result. In the circuit shown in Fig. 2.4, for
2.4 M. B. Patil, IIT Bombay
1 2 3
0
C
1 2 3
0
1 3
0
2
0 V (^) VC 5 V 5 V
i 3 R 3
R 1 R 2
(a) t = 0−
i 3 iC R 3
R 1 R 2
0 V
(b) t = 0+
R 3 VC
R 1 R 2
i 3
(c) t → ∞
Figure 2.6: Circuit for Example 1 at t = 0−, t = 0+, and t → ∞. 0 V , and VC (t = 0+) = VC (t = 0−) = 0 V , as shown in Fig. 2.6 (b). As t → ∞, the circuit reaches steady state, the capacitor is replaced by an open circuit, and VC (∞) = 5 V ×
R 3
R 1 + R 3
= 2. 5 V (see Fig. 2.6 (c)).
Next, we obtain the circuit time constant τ = RTh C for t > 0 (see Fig. 2.7). To obtain
1 2
0
C
1 2 3
0
3
5 V 5 V R 3
R 1 R 2
R 3
R 1 R 2
RT h
Figure 2.7: Computation of time constant for the circuit of Example 1 for t > 0.
the Thevenin resistance as seen by the capacitor, we deactivate the independent source (i.e., short the voltage source), and get RTh = R 2 + (R 1 ‖ R 3 ) = 7 kΩ. The time constant of the circuit for t > 0 is therefore τ = RTh C = 0.7 ms. We are now in a position to obtain an analytical expression for VC (t) for t > 0. Let
VC (t) = A e−t/τ^ + B. (2.8)
Using VC (0+) = 0 V and VC (∞) = 2. 5 V , we get A = − 2 .5, B = 2.5. The result of Eq. 2. is plotted in Fig. 2.8 (a). Note that, in about five time constants (i.e., 3.5 ms), VC becomes a constant. This is generally true – if there are no further changes in the source voltages (or switch conditions), all transients vanish in about five time constants because e−^5 = 0. 0067 ≈ 0, or (1 − e−^5 ) = 0. 9933 ≈ 1. Using the above VC (t), we can find i 3 for the circuit as (see Fig. 2.5),
i 3 =
R 3
(iC R 2 + VC ) =
R 3
C
dVC dt
R 2 + VC
Fig. 2.8 (b) shows a plot of i 3 versus time using Eq. 2.9. We can also compute i 3 directly, i.e., without using VC (t). Let
i 3 (t) = A 1 e−t/τ^ + B 1. (2.10)
RC and RL circuits · · · 2.
(a) (b)
−1 0 1 2 3 4 5 time (msec)
−1 0 1 2 3 4 5
0
5
0
time (msec)
Vs
VC
i^3
(m
A
)
Vs
,^ V
C^
(Volts)
Figure 2.8: (a) Vs(t), VC (t), (b) i 3 (t) for the circuit of Example 1. The constants A 1 and B 1 are obtained using the values of i 3 (0+) = 0.071 mA and i 3 (∞) = 0.25 mA, which are found from Figs. 2.6 (b) and 2.6 (c), respectively. SEQUEL file: ee101 rc6.sqproj
- For the circuit shown in Fig. 2.9, R 1 = 10 Ω, R 2 = 40 Ω, L = 0. 8 H, t 0 = 0 s, and t 1 = 0. 1 s.
(a) Find iL(t) and is(t). (b) Simulate the circuit and verify that the total power absorbed is equal to the total power delivered at all times.
1 2
0
Vs
is iL V 0 V t 0 t 1
10 V
R 1
L^ R 2
Figure 2.9: Circuit for Example 2.
The above problem is equivalent to three problems, each with a DC source, as shown in Fig. 2.10. The circuit time constant is the same in all cases, since the Thevenin
1 2
0
1 2
0
1 2
0
0 V 10 V 0 V
is iL V
R 1
L^ R 2
(a) t < t 0
is iL V
R 1
L^ R 2
(c) t > t 1
is iL V
R 1
L^ R 2
(b) t 0 < t < t 1 Figure 2.10: Circuit of Fig. 2.9 for three intervals: (a) t < t 0 , (b) t 0 < t < t 1 , and (c) t > t 1.
resistance seen by the inductor remains the same, viz., RTh = R 1 ‖ R 2 = 8 Ω, giving
RC and RL circuits · · · 2.
- For the circuit shown in Fig. 2.12, R = 1 kΩ and C = 0. 1 μF.
(a) Obtain V 2 (t), assuming that the diode has a negligible on resistance, infinite off resistance, and a voltage drop of 0. 7 V while conducting. (b) Repeat (a) with the diode reversed.
1
0
2
Vs 0 V (^0)
5 V C
R
D
1 ms
Figure 2.12: Circuit for Example 3.
Since the diode is a non-linear circuit element, it would appear that our analysis would not be valid for this circuit. However, when the diode is replaced with the approximate equivalent circuits shown in Figs. 2.13 (a) and (b), the capacitor sees a linear circuit in both the on and off cases. Our expression x(t) = A e−t/τ^ + B is therefore valid in both
1
0
2 1
0
2
(a) (b)
Vs Vs
- 7 V (^) R on Roff
C
R C
R
Figure 2.13: Approximate description of the circuit of Fig. 2.12: (a) diode con- ducting, (b) diode not conducting. (Ron → 0 Ω, Roff → ∞ Ω).
cases, the major difference between the two cases being the magnitude of the time constants. When D conducts (Fig. 2.13 (a)), the Thevenin resistance seen by the capacitor is RTh = R ‖ Ron ≈ Ron, a very small resistance. When D does not conduct (Fig. 2.13 (b)), RTh = R ‖ Roff ≈ R. Thus, the time constants τ 1 = Ron C and τ 2 = R C in the two cases are vastly different, with τ 1 ≪ τ 2 , and on the time scale of the given circuit, we can say that τ 1 ≈ 0 sec. When the source voltage rises to 5 V , the capacitor charges instantaneously (since τ 1 ≈ 0 s) up to about 4. 3 V , i.e., 0. 7 V less than 5 V (see Fig. 2.14). Beyond this point, the diode turns off, and the capacitor charges slowly, with a time constant of τ 2 = R C = 0.1 ms. Since the pulse duration (1 ms) is larger than 5 τ = 0.5 ms, the charging process is completed, and V 2 reaches 5 V.
2.8 M. B. Patil, IIT Bombay
−0.5 0 0.5 1 1.5 2
0
5
time (msec)
V^1
,^ V
2 (Volts)
V 2 V 1
Figure 2.14: V 1 and V 2 versus time for the circuit of Example 3. At t = 1 ms, the capacitor discharges with a time constant τ 2 = 0.1 ms, and after about five time constants, V 2 reaches its steady-state value of 0 V. When the diode is reversed, the capacitor gets charged with τ = R C and discharged with τ ≈ 0 s. The reader is encouraged to work out the waveform for V 2 in this case and compare it with simulation results. SEQUEL file: ee101 diode circuit 7.sqproj
- For the circuit shown in Fig. 2.15 (a), the switch opens at t = 0 s. The component values are R 1 = 5 kΩ, R 2 = 1 kΩ, R 3 = 5 kΩ, and C = 5 μF. Find VC (t) and i(t).
C C
(a) (c)
AND
(b)
VC
R 3 iC
6 V
i
R 1 R^2
R 3 i iC
R 1
R 3
6 V
i
R 1 R^2
Figure 2.15: (a) Circuit for Example 4, (b) Circuit for t < 0, (c) Circuit for t > 0.
Assuming that the switch has been closed for a long time before t = 0 s, we have iC = C
dVC dt
= 0 A (i.e., the capacitor is an open circuit), leading to
i(0−) =
6 V
R 1 + R 2
= 1 mA, and VC (0−) = R 1 i = 5 V (see Fig. 2.15 (b)). By continuity of VC , we get VC (0+) = VC (0−) = 5 V. At t = 0, the switch opens, and the Thevenin resistance seen by the capacitor for t > 0 is RTh = R 1 + R 3 = 10 kΩ, resulting in a time constant τ = RTh C = 50 ms (see Fig. 2.15 (c)). As t → ∞, VC → 0 V , and i → 0 A. Let VC (t) = = A 1 e−t/τ^ + B 1 for t > 0. Substituting VC (0+) = 5 V and VC (∞) = 0 V , we can get A 1 and B 1.
2.10 M. B. Patil, IIT Bombay
To find the time constant of the circuit for t > 0 s (see Fig. 2.18 (b)), we redraw the circuit as shown Fig. 2.19 (a). The Thevenin resistance as seen by the capacitor can be
(^2 )
(^0) (a)
C
(^2 )
(^0) (b)
VX
R 3
R 2
- 1 VX
VX
R 3
R 2
- 1 VX
V 0
I 0
Figure 2.19: (a) Circuit of Fig. 2.18 (b) redrawn, (b) Circuit in (a) with capacitor replaced with a test voltage source.
obtained by applying a test source as shown in Fig. 2.19 (b) and computing RTh = V 0 /I 0. Writing KCL at node 3,
−I 0 − 0. 1 V 0
R 2
R 2 + R 3
V 0
R 2 + R 3
RTh =
V 0
I 0
R 2 + R 3
1 − 0. 1 R 2
The circuit time constant for t > 0 is then τ = RTh C = 16 ms. Using the conditions, VC (0+) = 26 V and v(∞) = 0 V , we get
VC (t) = 26
1 − e−t/τ^
V t > 0. (2.17)
The current iC (t) is therefore
iC (t) = C
dVC dt
= − 1. 625 e−t/τ^ A t > 0. (2.18)
Fig. 2.20 shows VC and iC versus time.
(^0) time (sec) 0.
0
10
20
30
(^0) time (sec) 0. −
−
0
VC
(Volts) iC
( A )
Figure 2.20: (a) VC (t), (b) iC (t) for the circuit of Example 5.
SEQUEL file: ee101 rc4.sqproj
RC and RL circuits · · · 2.
2.3 Exercise Set:
- For the circuit shown in Fig. 2.21,
(a) Find i(0−), i(0+), and i(∞). (b) Find RTh as seen by the inductor for t > 0. (c) Obtain an expression for i(t), using the results of (a) and (b). Verify your result with simulation.
20 Ω 20 Ω
R 1 R 2
t = 0 10 V Vs 2
L
- 1 H
i 5 V Vs 1
Figure 2.21: Circuit for Exercise 1.
SEQUEL file: ee101 rl2.sqproj
- For the circuit shown in Fig. 2.22, show that the following results hold in the steady state:
(a) Vmax = V 0
1 − k 1 1 − k 0
, Vmin = k 2 Vmax, where k 1 = e−T^1 /τ^ , k 2 = e−T^2 /τ^ , k 0 = k 1 k 2 , τ = R C.
Hint: Obtain VC (t) in the T 1 and T 2 intervals, use the condition of periodicity of VC in the steady state. (b) The average value of VC is the same as the average value of Vs. i.e., 1 T
∫ T
0
Vs dt =
T
∫ T
0
VC dt.
Hint: write KVL for the circuit and integrate.
With simulation, verify the above results for various values of T 1 and T 2.
0
1 2
0
Vs
V 0
T 1 T 2
VC
,^ V
s
VC Vs
VC C
R
t
Vmin
Vmax
Figure 2.22: Circuit for Exercise 2.
SEQUEL file: ee101 rc1b.sqproj
