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Chap. 7 Response of First-Order RL and RC Circuits

Contents7.1 The Natural Response of an RL Circuit 7.2 The Natural Response of an RC Circuit 7.3 The Step Response of RL and RC Circuits 7.4 A General Solution for Step and Natural Responses7.5 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。

Two forms of the circuits for

natural response.

2

Fo

ur p

ossi

ble

firs

t-o

rde

r ci

rcui

ts.

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。

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。

First-Order Circuits

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7.1 The Natural Response of an RL Circuit

RL 電路的自然響應

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t 0

By KVL:

Power and Energy Delivered to the Resistor

4

The Significance of the Time Constant

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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。

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。

Transient response暫態響應

Steady-state response穩態響應

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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 。

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 。The switch has been closed for a long time

before it is opened at t = 0.

A 20)0()0(_

LL ii s20.

R

L

eq

By current division,

The initial energy stored in the inductor:

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EX 7.2 Determining the Natural Response of an RL Circuit with Parallel Inductors

The initial energy stored in the inductors:

H 4205

20521

//LL

80410

0410

4)//400151

0151(

4)//4010(15//

A 1248I0

s 0.584 0 A, 12)( 2 teti t- 0)0( & 0 V, 968)( 2 -t- vteitv

A.61 andA 1.6

, As

21 .-ii

t

J 320(20)(4)2

1(5)(8)

2

1)(0 22 w

J 32(20)(-1.6)2

1(5)(1.6)

2

1)( 22 w

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7.2 The Natural Response of an RC Circuit

RC 電路的自然響應

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By KCL:

Initial voltage Time constant

Hint:

1.找出跨在電容器的初始電壓 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 。

000 Vvv C-

C

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EX 7.3 Determining the Natural Response of an RC Circuit

The switch has been in position x for a long time.At t = 0, the switch moves to position y.

By voltage division,

The power dissipated in the 60k-resistor:

803202460

024603224060//

ms405080 μ.k 0 V, 100)( 52 tetv t-C& V 10000 -

CvV

0)0( & 0 V, 60)(80

48)( 52 -

ot-

Co vtetvtv

0 mA, )/60()( 52 tektvti t-oo

0 mW, 6060)()( 50260 tektitp t-

okΩ

The total energy dissipated in the 60k-resistor:

mJ 1.260)(0

260

dtktiw okΩ

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EX 7.4 Determining the Natural Response of an RC Circuit with Series Capacitors

s14025 μk 0 V, 20)( tetv -t

F 4025

02521 μ//CC

V 204-240 V

0 A, 80250

)()( tμe

k

tvti -t

The initial energy stored in C1:

V. 20 and V 20

, As

21

v-v

t

J 40)(4)(52

1 21 μμw

J 500020))((202

1)(20)(5

2

1 22 μ-μμw

The initial energy stored in C2: J 5760)(24)(202

1 22 μμw

7.3 The Step Response of RL & RC Circuits

RL 電路的步階響應

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By KVL:

The Step Response of an RL Circuit

dt

diLv

tR/L-SS eR

V-I

R

Vti

0

The Step Response of an RL Circuit

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If initial I0 = 0,

At t = ,

斜率

If initial I0 = 0,

斜率S

tR/L-S V

L

R-

dt

dveV

L

R-

dt

dv )0( )(

S-

S VeVτv 0.3681 At t = ,

tR/L-S eRI-Vv 0

tR/L-SS eR

V-I

R

Vti

0

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EX 7.5 Determining the Step Response of an RL Circuit

s0.1200m/2

0 V, 40)( 10 tedt

diLtv t-

t = ? when v = 24 V.

A -80 I

tR/L-SS eR

V-I

R

Vti

0

0201212812 1010 , te- e--ti t-.-t/

V 0)0( V, 40)(0 -vv

The Step Response of an RC Circuit

RC 電路的步階響應

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By KCL:

dt

dvCi C OR 整式乘以 C 再微分

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EX 7.6 Determining the Step Response of an RL Circuit

0 V, 9060V 603060)( 100100 te-e---tv t-t-o

0 mA, 252

40

3051)( 100100 te.- e

km-.-ti t-t-

o

V 302060

60400

kk

kVThe switch has been in position 1 for a long time.

At t = 0, the switch moves to position 2., t 0For

Norton Equivalent 40k 816040 kkk//RTH

V 6016040

16075 -

kk

k-vOC

A 51

40

60

.-kΩ

V-iSC

OR

mA 252

9000-0.25)(

100

100

t-

t-oo

e.-

edt

dvCti

7.4 A General Solution for Step and Natural Responses

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When x reach its final valuexf (a constant), dx/dt =0.

自然和步階響應一般解

時間常數切換時間

變數之終值變數之初值

變數之終值或變數t--

CL

e-

tvti

RL 和 RC 電路的自然和步階響應之計算步驟

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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。

時間常數切換時間

變數之終值變數之初值變數之終值或變數t--

CL e-tvti

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EX 7.7 Using the General Solution Method to Find an RC Circuit’s Step Response

0 V, 120 90V 90 30 09)( 55 te-e--tv t-t-C

V302060

60400 --vC

The switch has been in position a for a long time.At t = 0, the switch moves to position b.

0 A, 3006000.5)( 55 tμeedt

dvCti t-t-C

V09 vC

s0.250040 μ.kRC

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EX 7.8 Using the General Solution Method with Zero Initial Conditions

0 mA, 3 0 3m0) 200/0.005 tee-i(t t--t

open :Capacitor V0152057 km.vC

ms51030k20k μ.RC

The switch has been open for a long time.

00 Cv

mA 37.5m3020

200

i A 0 fii

0 V, 051 015V 015 0015)( 200200 te-e-tv t-t-C

00 Cv

0 , 06 015 V 90 051 01530)( 200200200 te-ee-tiktvtv t-t-t-C

Also,

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EX 7.9 Using the General Solution Method to Find an RL Circuit’s Step Response

0 A,152020502)( 512080 t e- e - ti t.-.-t/

short :Inductor 0 vL

ms8080m/1 L/R

The switch has been open for a long time.

A 531

200

i A 20

1

20 fii

0 V, 51 V 0 150)( 12.512.5 tee-tv t-t-L

V 5115020 -vLAlso,

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EX 7.10 Determining Step Response of a Circuit with Magnetically Coupled Coils

0 A,61 1616016)( 520 t e- e - ti t-.-t/o

short :Inductor 0 vo

s0.21.5/7.5 L/R

The switch has been open for a long time.

A 00 oi A 16120/7.5 fo ii

0 V, 201 V 0 1200)( 55 tee-tv t-t-o

V 2017.500120 -voAlso,

H 1.51218

3645

221

221

-

-

M-LL

-MLLLeq (Prob. 6.46)

Since vo : Also,

By KCL,

7.5 Sequential Switching

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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.

A 634

4

40/7

600

-

Li

40/744//3412//36//

ms 350 t

t < 0

66)(318//

ms25150m/6

A 0Li

ms 350 A,6)( 40 t eti t-L

EX 7.11 RL-Sequential Switching (Contd.)

ms 35t A1.486ms) (35 1.4 ei -L

ms 35t

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ms50/3150m/9

A 0Li

ms 35 A,1.48)( 0.035-60 t eti t-L

What percentage of the initial energy stored in the inductor is dissipated in R3?

For t 35 ms

V -366150)( 4040 e edt

d.tv t-t-

L

For 0 < t < 35 ms

% 22.960.15

54.73m563.51m2

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EX 7.12 Analyzing an RC Circuit that has Sequential Switching

V0 v V 310.75 400-400(15ms) 1.5 -ev ms51005 μ.k

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ms1010010 μ.k For 0 t 15 ms

At t = 0, switch is moved to position b. Then, 15 ms later, switch is moved to position c.

open :Capacitor V400 v 00 v

15ms0 V, 400-400V 040 0040)( 100100 tee-tv t-t-

For 15ms t

15ms V, 310.75V 0 310.750)( 0.015-2000.015-200 tee-tv t-t-

? ?

7.6 Unbounded Response

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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.

0 A,10)( 40 t etv to

RTH = vT / iT

For 0 t

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。

7.7 The Integrating Amplifier

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Ideal OP-Amp

7.7 The Integrating Amplifier (Contd.)

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