Chap. 7 Response of First-Order
RL and RC Circuits
Contents
7.1
7.2
7.3
7.4
The Natural Response of an RL Circuit
The Natural Response of an RC Circuit
The Step Response of RL and RC Circuits
A General Solution for Step and Natural
Responses
7.5 Sequential Switching
7.6 Unbounded Response
7.7 The Integrating Amplifier
Objectives
1. 能定出RL 和RC 電路的自然響應。
2. 能定出RL 和RC 電路的步階響應。
3. 知道如何分析具有順序切換的電路。
4. 能分析含有電阻和單一電容的運算放大器電路。
1
First-Order Circuits
步階響應
(step response) ：
當電感器或電容器
突然獲得來自直流
電壓源或電流源的
能量時，該電路電
流和電壓的變化。
Two forms of the
circuits for
natural response.
Four possible first-order circuits.
自然響應
(natural response)：
當電感器或電容器所
儲存的能量突然釋放
給電阻性網路時，該
電路電流和電壓的變
化。
2
7.1 The Natural Response of an RL Circuit
t0
By KVL:
RL 電路的自然響應
3
3
Power and Energy Delivered to the Resistor
4
The Significance of the Time Constant
時間常數(time constant) ：
在e -(R/L)t 的項次中，t 的係數R/L將會決定電壓或電流趨近
於零的速率，此係數的倒數稱為電路的時間常數。
Steady-state response
穩態響應
Transient response
暫態響應
對單一時間常數電路而言，經過了一個長時
間(a long time) 或指經過了五倍以上的時間
常數，這時電流或電壓幾乎已經到達終值。
5
EX 7.1 Determining the Natural Response of an RL Circuit
Hint:
1. 找出流經電感器的初始電流 I0。
The switch has been closed for a long time
before it is opened at t = 0.

iL (0 )  iL (0 )  20 A
_
2. 找出電路的時間常數，τ= L/R 。
3. 從I0 和τ得到i (t) = I0 e - t /τ。

L
 0.2 s
Req
By current division,
The initial energy stored in the inductor:
6
EX 7.2 Determining the Natural Response of an RL Circuit
with Parallel Inductors
L1 //L 2 
(15// 10  4)//40
15 10
(
 4)//40
15  10
10  40

8
10  40
5  20
4H
5  20
I0  8  4  12 A
  4  0.5 s
8
i(t )  12e- 2t A,
As t  ,
i1  1.6 A and i2  -1.6 A.
t  0 v(t )  8i  96e- 2t V, t  0 & v(0- )  0
The initial energy stored in the inductors:
1
1
w(0  )  (5)(8) 2  (20)(4) 2  320 J
2
2
w() 
1
1
(5)(1.6) 2  (20)(-1.6)2  32 J
2
2
7
7.2 The Natural Response of an RC Circuit
By KCL:
RC 電路的自然響應
Initial voltage
Time constant
8
8
EX 7.3 Determining the Natural Response of an RC Circuit
Hint:
1. 找出跨在電容器的初始電壓 V0
 
 
vC 0-  vC 0   V0
2. 找出電路的時間常數，τ= RC
3. 從V0 和τ得到v (t) = V0 e - t /τ。
The switch has been in position x for a long time.
At t = 0, the switch moves to position y.
60// 240  32 
 
  80k  0.5 μ  40 ms & V0  vC 0-  100 V
By voltage division, vo (t ) 
vC (t )  100e - 25t V, t  0
48
vC (t )  60e - 25t V, t  0  & vo (0- )  0
80
io (t )  vo (t )/60 k  e - 25t mA,
The power dissipated in
the 60k-resistor:
60  240
 32  80 
60  240
t  0
p60kΩ (t )  io2 (t )  60k  60e -50t mW, t  0 
The total energy dissipated
in the 60k-resistor:



w60kΩ   io2 (t )  60k dt  1.2 mJ
0
9
EX 7.4 Determining the Natural Response of an RC Circuit
with Series Capacitors
C1 //C 2 
5  20
 4 μF
5  20
V0  24 - 4  20 V
  250k  4 μ  1 s
v(t )  20e V, t  0
-t
i (t ) 
v(t )
 80e -t μA, t  0 
250k
The initial energy stored in C1: w1 
1
(5 μ)(4) 2  40 μ J
2
The initial energy stored in C2: w2 
1
(20 μ )(24) 2  5760 μ J
2
As t  ,
v1  -20 V and v2  20 V.
w 
1
1
(5 μ)(20) 2  (20 μ)(-20) 2  5000 μ J
2
2
10
7.3 The Step Response of RL & RC Circuits
The Step Response of an RL Circuit
By KVL:
RL 電路的步階響應
VS  -  R/Lt

  I0 e
R 
R
VS


it 
vL
di
dt
11
The Step Response of an RL Circuit
i t  
VS  VS  -  R/Lt
  I 0-  e
R 
R
If initial I0 = 0,
At t = ,
斜率
v  VS - I 0 R  e -  R/Lt
If initial I0 = 0,
At t = , vτ   VS e -1  0.368 VS
斜率
dv
R
dv
R
 - VS e - ( R/L)t  (0)  - VS
dt
L
dt
L
12
EX 7.5 Determining the Step Response of an RL Circuit
VS  VS  -  R/Lt
i t  
  I 0-  e
R 
R
I 0  -8 A
  200m/2  0.1 s
v(t )  L
it   12  -8-12 e-t/ 0.1  12-20e-10t , t  0
di
 40e -10t V, t  0 
dt
v(0  )  40 V, v(0- )  0 V
t = ? when v = 24 V.
13
The Step Response of an RC Circuit
By KCL:
RC 電路的步階響應
iC
dvC
OR
dt
整式乘以C再微分
14
EX 7.6 Determining the Step Response of an RL Circuit
The switch has been in position 1 for a long time.
At t = 0, the switch moves to position 2.
RTH  40k//160k  8k  40k 
vOC  -75
160k
 -60 V
40k  160k
V0  40
-60 V
40kΩ
 -1.5 A
60k
 30 V For t  0,
60k  20k
Norton
Equivalent
iSC 
vo (t )  -60  30--60e -100t V  -60  90e -100t V, t  0
30  -100t

io (t )   -1.5m -2.25e -100t mA,
e
40k 

OR

dvo
 0.25  - 9000e -100t
dt
 -2.25e -100t mA
io (t )  C
t  0

15
7.4 A General Solution for Step and Natural
Responses
When x reach its final value
xf (a constant), dx/dt =0.
變數iL t 或vC t   變數之終值
 變數之初值-變數之終值  e
- t-切換時間 
時間常數
自然和步階響應一般解
16
RL 和RC 電路的自然和步階響應之計算步驟
1. 確定電路中想要的變數，對RC 電路而言，選
擇電容電壓最為方便；對RL 電路而言，則最好
選擇電感電流。
2. 決定變數的初始值，也就是t0 時的值。
注意，如果選擇了電容電壓或電感電流作變數，
則不須區分t = t0- 和t = t0+ 有何不同，因為它們
都是連續變數；如果選擇了其他變數，則必須
記住它的初始值定義於t = t0+ 。
3. 計算出變數的最終值，亦即 t →∞ 的值。
4. 計算出電路的時間常數。
變數iL t 或vC t   變數之終值  變數之初值-變數之終值 e
-t-切換時間
時間常數
17
EX 7.7 Using the General Solution Method to Find an RC
Circuit’s Step Response
The switch has been in position a for a long time.
At t = 0, the switch moves to position b.
vC 0  -40
60
 -30 V
60  20
vC   90 V
  RC  400k  0.5 μ  0.2 s
vC (t )  90  - 30 - 90 e -5t V  90 - 120 e -5t V, t  0


dvC
i (t )  C
 0.5  600e -5t  300e -5t μA, t  0 
dt
18
EX 7.8 Using the General Solution Method with Zero Initial Conditions
vC 0  0
The switch has been open for a long time.
 
i 0 
20
7.5m   3 mA
20  30
i   i f  0 A
  RC  20k  30k  0.1μ  5 ms
i(t )  0  3m - 0 e-t/0.005  3 e- 200t mA, t  0
Also, vC 0  0
vC   7.5m  20k  150 V Capacitor : open
vC (t )  150  0 - 150 e - 200t V  150 - 150 e - 200t V, t  0
v(t )  vC t   30k  it   150 - 150 e - 200t  90 V e - 200t  150 - 60 e - 200t , t  0 
19
EX 7.9 Using the General Solution Method to Find an RL
Circuit’s Step Response
The switch has been open for a long time.
i 0  
20
5A
1 3
20
i    i f 
 20 A
1
  L/R  80m/1  80 ms
i(t )  20  5 - 20 e-t/ 0.08  20-15 e-12.5t A, t  0

Also, vL 0   20-5 1  15 V vL   0 Inductor : short
vL (t )  0  15 - 0 e-12.5t V  15 e-12.5t V, t  0
20
EX 7.10 Determining Step Response of a Circuit
with Magnetically Coupled Coils
The switch has been open for a long time.
io 0  0 A
io    i f  120/7.5  16 A
L1 L2-M 2
45-36
Leq 

 1.5 H (Prob. 6.46)
L1  L2-2M 18-12
  L/R  1.5/7.5  0.2 s
io (t )  16  0 - 16 e -t/ 0.2  16 - 16 e -5t A, t  0

Also, vo 0   120-0  7.5  120 V vo   0 Inductor : short
vo (t )  0  120 - 0 e -5t V  120 e -5t V, t  0 
Since vo :
Also,
By KCL,
21
7.5 Sequential Switching
順序切換(sequential switching):
指電路中的切換動作超過一次以上，重點在於求得初始值x(t0)。
EX 7.11 Analyzing an RL Circuit that has Sequential Switching
The two switches in the circuit
have been closed for a long
time. At t = 0, switch 1 is
opened. Then, 35 ms later,
switch 2 is opened.
t<0
6// 12//3  4  4//3  4  40/7 
 
iL 0 - 
60
4

6A
40/7 4  3
0  t  35 ms
18// (3  6)  6  iL   0 A
  150m/6  25 ms
iL (t )  6 e- 40t A, 0  t  35 ms
22
EX 7.11 RL-Sequential Switching (Contd.)
t  35 ms
iL (35 ms)  6 e-1.4  1.48 A
t  35 ms
  150m/9  50/3 ms
iL   0 A
iL (t )  1.48 e- 60t -0.035 A, t  35 ms
What percentage of the initial energy stored in the inductor is dissipated in R3?
For t  35 ms
For 0 < t < 35 ms
vL (t )  0.15 


d
6 e - 40t  -36 e - 40t V
dt
563.51m  54.73m
 22.9 %
2
1
2 0.15  6
 
23
EX 7.12 Analyzing an RC Circuit that has Sequential Switching
At t = 0, switch is moved to
position b. Then, 15 ms later,
switch is moved to position c.
For 0  t  15 ms
v0  0 v  400 V
Capacitor : open
  100k  0.1μ  10 ms
v(t )  400  0 - 400 e-100t V  400 - 400 e-100t V, 0  t  15ms
For 15ms  t
v(15ms)  400 - 400 e-1.5  310.75 V v  0 V
  50k  0.1μ  5 ms
v(t )  0  310.75 - 0 e- 200t -0.015 V  310.75 e- 200t -0.015 V, t  15ms
?
?
24
7.6 Unbounded Response
電路的響應可以是指數增長型而非衰減型，這種響應稱為無限制響應
(unbounded response)，可能發生在含有相依電源的電路中。
EX 7.13 Finding the Unbounded Response in an RC Circuit
When the switch is closed, the
voltage on the capacitor is 10 V.
Find the expression for vo for t ≥ 0.
RTH = vT / iT
For 0  t
vo (t )  10 e 40t A, t  0
25
7.7 The Integrating Amplifier
Ideal OP-Amp
輸入電壓的積分乘上(1 / RsCf) 和負值後，
再加上電容器的初始電壓值。
26
7.7 The Integrating Amplifier (Contd.)
27