EMLAB

1

Chapter 7. First and second order transient circuits

EMLAB

2Contents

1. Introduction

2. First order circuits

3. Second order circuits

4. Application examples

EMLAB

31. Introduction

Camera flash discharge :

EMLAB

4Analysis

0)()(

R

t

dt

tdC CC

0)(1)(

tRCdt

tdC

C

RC

t

C eVt

0)(

EMLAB

52. First order transient circuits

Solution to 1st order differential equation :

f (t) = 0 → homogeneous equationf (t) ≠ 0 → inhomogeneous equation

xh(t) or xc(t) → homogeneous or complementary solutionxp(t) → inhomogeneous or particular solution

)()()(

tftaxdt

tdx

0)()(

taxdt

tdxc

c

)()()( txtxtx cp

EMLAB

6

a

AKtxp 1)(

Particular solution :

For polynomial functions f (t) , x(t) should also be a polynomial func-tion of the order of equal or lower degree.

cbtattxtttfex p 22 )(232)()

1)()() KtxAtfex p

AaKtaxdt

tdxAtftax

dt

tdxp

p 1)()(

)()()(

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7

Homogeneous solution :

CatCatc eKeKetx

22)(

00)()(

cc

cc ax

dt

dxtax

dt

tdx

dtaxddtax

dxax

dt

dxc

c

cc

c ][ln

Cattxc )(ln

atatcp eK

a

AeKKtxtxtx 221)()()(

The constant K2 can be found, if the value of x(t) is known at one instant of time.

Integrating both sides with respect to the arguments,

EMLAB

8General solution : 1st order differential equation

K1 : steady-state solution: x(t) → K1 as t→∞ when the second term becomes negligible.

τ : time constant

aeKKtx t 1

)( /21

EMLAB

9Simple RC circuit

)0( t

Sp VKKt )(

01

)( ttth e

RCeet

RC

1

RC

t

Shp eKVttt

2)()()(

0)()(

R

Vt

dt

tdC S

RC

Vt

RCdt

td S )(1)(

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10

S

S

VK

KVt

2

20)0(

RC

t

SS eVVt

)(

EMLAB

11Simple RL circuit

L

Vti

L

R

dt

tdiVtRi

dt

tdiL S

S )()(

)()(

tL

RS eK

R

Vti

2)(

0)0()0( 2 KR

Viti S

R

VK S 2

tL

RS

tL

RSS e

R

Ve

R

V

R

Vti 1)(

tL

R

SR eVtiRt 1)()(

EMLAB

12Example 7.1

Consider the circuit shown in Fig. 7.4a. Assuming that the switch has been in posi-tion 1 for a long time, at time t=0 the switch is moved to position 2. We wish to cal-culate the current i(t) for t > 0.

Initial condition :

][436

312)0( V

kk

kC

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13

0)()()(

21

R

t

dt

tdC

R

t

FCkRkR 100,3,6 21

0)(5)(

tdt

td

tAet )(Suppose

0)5(0)(5)(

tt eeAtdt

td

5

tt eAet 55 4)( ][4)0()0( VCC

][3

4)()( 5

2

mAeR

tti t

EMLAB

14Example 7.2

The switch in the network in Fig. 7.5a opens at t=0. Let us find the output volt-age vo(t) for t > 0.

][3

4)0()0( Aii LL

Initial condition :

Thevenin’s equivalent

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15

0)(

)()( 311

dt

tdiLtiRRVS

HLRR 2,2,2 31

6)(2)(

tidt

tdi

teKti 223)(

][3

4)0()0( Aii LL 3

53

3

4

3

43)0( 22 KKi

teti 2

3

53)(

6)(2)(

tidt

tdi

to etit 2

3

106)(2)(

EMLAB

16Example 7.3Consider 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 > 0.

Initial condition :

][2462

1236)0(

][32122012)0()46()0(

mAkkk

i

VikkC

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17

06

)()(100

2

36)(

k

t

dt

td

k

t

)(t

180)(3

20)( t

dt

td

27)( tpt

h eKt 3

20

2)(

t

hp eKttt 3

20

227)()()(

5,3227)0( 22 KK

tet 3

20

527)(

][6

5

2

9

6

)()( 3

20

mAek

tti

t

EMLAB

18Problem-Solving Strategy

Step 1. We assume a solution for the variable x(t) of the form

Step 2. Assuming that the original circuit has reached steady state before a switch was thrown. Solve for the voltage across the capacitor, υc(0-) or the current through the Inductor iL(0-), prior to switch action.

Step 3. Recall that voltage across a capacitor and the current flowing through an in-ductor cannot change in zero time after the switch is changed. Solve for the initial value of the variable x(0+) using υc(0+) = υc(0-), iL(0+) = iL(0-).

/21)( teKKtx

Step 4. Assuming that steady state has been reached after the switches are thrown,draw the equivalent circuit, valid for t >5τ by replacing the capacitor by an open cir-cuit or the inductor by a short circuit.

)()(5

xtxt

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19

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 circuitcontaining a voltage source, resistor, and a storage element (i.e., capacitor or in-ductor) by forming a simple Thévenin equivalent circuit at the terminals of the storage element. This Thévenin equivalent circuit is obtained by looking into the circuit from the terminals of the storage element.

ThTh R

LCR ,

Step 6. Using the results of steps 3, 4, and 5, we can evaluate the constants.

)()0(),(

)(,)0(

21

121

xxKxK

KxKKx

/)]()0([)()( texxxtx

EMLAB

20Example 7.4

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>0.

/21)( teKKt

][3

8

31

61

31

3636

4

24)0( AiL

Initial condition (before the switch action)

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21

Right after the switch action : (t = 0+)

Applying Kirchhoff current law to node υ1,

][3

20)0(0

12

)0(8

6

)0(

4

24)0(1

111 V

][3

52)0(24)0( 1 V

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22

Steady state solution :

][24)( V ][2123

12

121

61

41

1

ThR

][22

4s

R

L

Th

][3

2024

3

52)()0(],[24)(

)(,)0(

21

121

VKVK

KKK

][3

2024)]()0([)()( 2/ Veet

tt

EMLAB

23Pulse response

)0(1

)0(0)(

t

ttu

EMLAB

24Unit step function

)(1

)(0)(

0

00 tt

ttttu

)]()([)( TttuAt

EMLAB

25Example 7.6Let us determine the expression for the voltage υo(t). Since the source is zero for all negative time, the initial conditions for the network are zero

)]3.0()([9)( ttut

0)0()0( tt CC

0)0(84

8)0(

t

kk

kt Co

4)(846

8)(

t

kkk

kto

)()1(4

)()0()()(

4.0

1

tue

etttt

t

oooo

kkkkRTh 4)84(||6

)3.0()1(4)( 4.0

3.0

2

tuet

t

o

)3.0()1(4)()1(4

)()()(

4.0

3.0

4.0

21

tuetue

ttttt

oo

9

0.3

EMLAB

267.3 Second order transient circuit

Integrating both sides,

)()()(1

00

tidt

dCtidxx

LR SL

t

t

)()()(

10

0

tdt

diLtdxxi

CiR SC

t

t

dt

di

CLCdt

d

RCdt

d S112

2

dt

d

LLC

i

dt

di

L

R

dt

id S12

2

EMLAB

27Solution to 2nd order differential equation :

Homogeneous differential equation :

20201 ,2 aa

Trial solution :

Characteristic equation :

ζ : damping ratioω0 : natural frequency

2

)()()a

AtxAtfex p

02 2002

2

ccc x

dt

dx

dt

xd

0212

2

ccc xa

dt

dxa

dt

xd

)(212

2

tfxadt

dxa

dt

xd

)()()( txtxtx cp

stc Ketx )(

0)2(2 200

2200

2 stststst KessKesKeKes

02 200

2 ss

Standard form :

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28

Characteristic equation :

Homogeneous solution :

Initial conditiontsts

cp eKeKa

Atxtxtx 21

212

)()()(

212

)0( KKa

Ax

22110

)(KsKs

dt

tdx

t

02 200

2 ss

12002,1 s

tstsc eKeKtx 21

21)(

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(1) |ζ|>1 : over damped

(2) |ζ|=1 : critically damped

(3) |ζ|<1 : under-damped

tsts eKeKA

tx 21212

0

)(

2120

)0( KKA

x 2211)0( KsKs

dt

dx

tAtAeA

tx ddt

sincos)( 212

0

0

120

)0( KA

x 210)0( KK

dt

dxd

tetKKA

tx 0212

0

)(

12

0

)0( KA

x 210)0( KK

dt

dx

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30

t

t

ttt

eBtBtx

BtBtxe

txetxee

txtdt

dxt

dt

xd

0

0

00

0

)()(

)(

0])([0])([1

0)()(2)(

21

21

2002

2

Critically damped: ζ=1

Derivation for the critically damped case :

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31

)(ts

Typical unit step responses with ζ changed

(1) |ζ|>1 : over damped

(2) |ζ|=1 : critically damped

(3) |ζ|<1 : under-damped

02 2002

2

ccc x

dt

dx

dt

xd

02

2

LC

i

dt

di

L

R

dt

id

L

CR

LC 2,

10

(ζ is proportional to R.)

EMLAB

32Example 7.7

( R=2, C=1/5F, L=5 H)

0)0()(1

0 dt

dCidxx

LR L

t

01

2

2

LCdt

d

RCdt

d

Let us assume that the initial conditions on the storage elements are iL(0) = -1 A and υC(0) = 4 V. Let us find the node voltage v(t) and the inductor current.

05.22

2

dt

d

dt

d

stKet )(

015.22 ss

5.022

5.15.2

2

45.25.2 2

ors

Applying Kirchhoff current law to node υ(t),

Trial solution :

Characteristic equation :

EMLAB

33

tt eKeKt 5.02

21)( General solution :

Initial condition for υ(0)

4)0( 21 KK

Initial conditions for 0tdt

d

210

5.02)(

KKdt

t

t

are used to determine unknown coefficients.

0)( 0 dt

dCti

R L

Initial conditions can be extracted from KCL on node υ

5)(1

0 tiCRCdt

dL

2,2 21 KK

EMLAB

34

tt eet 5.02 22)(

EMLAB

35Example 7.8

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 υC(0) = -4 V. Let us determine the expression for both the current and the capacitor voltage.

0)0()(1

0 dt

diLdxxi

CiR C

t

02

2

LC

i

dt

di

L

R

dt

id 02562

2

idt

di

dt

id

02562 ss

jj

s 432

86

2

10066 2

stKeti )(

Applying Kirchhoff voltage law to the loop,

Trial solution :

Characteristic equation :

EMLAB

36

)4sin4cos()( 213 tKtKeti t

Initial condition for i(0) : 4)0( 1 Ki

Initial conditions for 0tdt

di

014640)()(1

00

dt

di

dt

diLtdxxi

CiR C

t

t

20dt

di

2043)4cos44sin4()4sin4cos(3 21213

213

0

KKtKtKetKtKedt

di tt

t

2,4 21 KK

)4sin24cos4()( 3 tteti t

EMLAB

37Example 7.9Determine the current i(t) and the voltage υ(t) in the figure.

0)()(1

00

dt

diLtdxxi

CiR C

t

t

AiV

HLFCRR

LC 2

1)0(,1)0(

2,8

1,8,10 21

Using KVL : 0)(1 tiRdt

diL C

2

)(Rdt

dCti

Using KCL :

01

2

211

22

2

LCR

RR

dt

d

L

R

CRdt

d096

2

2

dt

d

dt

d

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38

0962 ss

stKet )(

3s

Trial solution :

Characteristic equation :

tetKKt 321 )()(

Initial condition for υ(0)

11)0()0( KC

Initial conditions for 0tdt

d

tt eKeKdt

d 31

32 3

5.0)0(20

iRdt

dC

t

3185.0

)0()0(

20

CRC

i

dt

d

t

33 120

KKdt

d

t

62 K

tett 3)61()( tetRdt

dCti 3

2 2

3

2

1)(

EMLAB

39Example 7.11

Let us determine the output voltage υ(t) for t>0.

HLFCRR 2,4

1,2,10 21

24)(1 tiRdt

diL C

2

)(Rdt

dCti

LCLCR

RR

dt

d

L

R

CRdt

d 241

2

211

22

2

48127

2

2

dt

d

dt

d

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40

481272

2

dt

d

dt

d

Particular solution : Atp )(

48121272

2

Adt

d

dt

dp

pp

4A

Homogeneous solution :

sth Ket )(

4301272 orsss

tthp eKeKttt 4

23

14)()()(

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41

Initial conditions :

][212)0()0(21

2 VRR

RCC

][112

)0()0(21

ARR

ii LL

0)0()0(

12020

CRC

i

dt

d

Rdt

dC

tt

043,24)0( 210

21

KKdt

dKK

t

6,8 21 KK

tt eet 43 684)(

Before the switch action:

Using KCL :

EMLAB

42Application example 7.13Consider 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.

1

0

1)(

T

indtVL

tiSwitch in position 1 :L

TVi in

p1

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43

Switch in position 2 :

0)()(

0)()(

tiL

R

dt

tditRi

dt

tdiL

)(

0

1

)(Tt

L

R

eiti

)(

0

1

)()(Tt

L

R

KeRtit

500)( 110 K

L

RTVT in 500

10

10053

11

KT

L

RTVin

][5],[11 AImsT p

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44

Consider the simple RL circuit, shown in the figure, which forms the basis for essen-tially 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

Before the switch action :

L

Vti

L

R

dt

tdiVtiRR

dt

tdiL S

SSL )()(

)()()(

][1)0( AR

Vi

S

S

After the switch action :

tL

R

SS

tL

RS e

RRVKe

R

Vti

11)(

LR

SR

),( SLSL RRRRR

S

S

LSS

LS

LSSS

S

SS

R

V

RRR

RV

RRRVK

R

VK

R

Vi

)(

11)0(

)(

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45

Inductor voltage :

tL

R

S

LSL e

R

RV

dt

diLt 1)(

Switch voltage :

t

L

R

SS

LL eV

R

RtiRt )()(switch

1

S

L

R

R0t

)(switch t

SS

L VR

R

(inductive kick)

LS RR

L

• At the peak inductor voltage is negative infinity! This voltage level is caused by the attempt to disrupt the inductor current instantaneously.

• The peak switch voltage jumps up to positive infinity.• This phenomenon is called inductive kick, and it is the nemesis of power supply

designers.

EMLAB

46Circuit design to avoid inductive kick

Before the switch action :

t

CSCLS dxxiC

VtiRRdt

tdiL

0)(

1,)()(

)(

R

Vi S )0(

After the switch action :

02

2

LC

i

dt

di

L

R

dt

id

iRout

SR

LR

C

S

t

LS VdxxiC

tiRRdt

tdiL 0 )(

1)()(

)(

EMLAB

47

01

2 2200

2 LC

sL

Rsss

]/[10,1 60 srad Set arbitrarily

][199],[10 SL RRRnFC

020 2,

1 L

R

LC ][2001010010122

],[10101010010

11

660

86122

0

LR

nFL

C

tetKKti610

21 )()(

][1)0( 1 AKi

SLSt

ViRRdt

diL

)0()(0

Initial conditions :

44

0

10199)12001(10)0()(1

iRRVLdt

diLSS

t

tt etKKeKdt

di 66 10621

102 )10)((

16

20

10 KKdt

di

4442 10991019910100 K

EMLAB

48

tetti6105 )109.91()(

tLout ettiRt

6105 )109.91(199)()(

][199)0( Vout

EMLAB

49Application example 7.16

][24 V

• DC voltage sources or power supplies are fed from AC outlets on walls.

• The ac waveform is converted to a quasi-dc voltage by an inexpensive ac–dc con-verter 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 sev-eral versions of dc–dc converters.

• We will focus on a topology called the boost converter.

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50

The boost converter with switch set-tings for time intervals (a) ton and (b) toff.

)(1

)( 000 ononin

t

in tiIL

tVIdtV

Lti

on State #1 :

State #2 :

0)(

)()(1

)()(1

)(

ItitL

VV

tidtVVL

tidttL

ti

onoffoin

on

tt

t oinon

tt

t L

offon

on

offon

on

(I0 is the initial current at the be-ginning of each switching cycle)

00 IIL

tVt

L

VV oninoff

oin

Assuming that Vo is nearly constant,

EMLAB

51

0)(0

oninoffoinonin

offoin tVtVV

L

tVt

L

VVin

off

onoff Vt

ttV

0

)(1

)( 000 ononin

t

in tiIL

tVIdtV

Lti

on )()(1

)( on

tt

t oin tidtVVL

tioffon

on

EMLAB

52

inin

onoff

onin

off

onoff VD

V

ttt

Vt

ttV

1

1

1

10

onoff

on

tt

tD cycle)Duty (

EMLAB

53

)(2

)(1

21)( onon ttsttsino eKeKVt

Rdt

dCtiVt

dt

diL oo

ino

)(,)(

LC

Vt

LCdt

d

RCdt

d ino

oo )(11

2

2

State #2 :

)(1

)(1

onoontt

o tRC

tiCdt

d

on

State #1 :

0)()(

R

t

dt

tdC oo

0)(1)(

tRCdt

tdo

o RC

t

CC et )0()(

21)( KKVt inono

RC

t

oono

on

et )0()(

ont offt0

2211 sKsKdt

d

ontt

o

)(to

EMLAB

54Application example 7.20 : automobile ignition system

• The voltage source represents the standard 12-V battery.• The inductor is the ignition coil. The inductor’s internal resistance is

modeled by the resistor.• The switch is the keyed ignition switch.

Optimum starter operation requires an over-damped 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.

Requirement:

EMLAB

55

100 ms

1~1.5s

02

2

LC

i

dt

di

L

R

dt

id

0)0()(1

0 iR

dt

diLdxxi

C C

t

Characteristic equation :

0)())((2 21212

21200

2 ssssssssssss

stKeti )( 012

LCs

L

Rs

The values of s1 and s2 should be set to satisfy the requirements.

Requirement:

EMLAB

56

2121

1, ss

LCss

L

R

Comparing the two expressions, we see that

1221 0)0( KKKKiL

0)0()0(0

tC dt

diLRi

)()( 2112211 ssKsKsKtdt

diL

60125)0()0(1

0

C

t

RiLdt

di

60)( 121 ssK

2021 L

Rss

Derived constraints:

Derived constraints:

tstsL eKeKti 21

21)(

EMLAB

57

(1) Let us arbitrarily choose s1=3 and s2=17, which satisfies the condition.

29.460

121

ssK ][98

1

21

mFsLs

C ][)(29.4)( 173 Aeeti ttL

Design :

0.5s

Fail

(2) Let us arbitrarily choose s1=1 and s2=19, which satisfies the condition.

33.360

121

ssK ][263

1

21

mFsLs

C

][)(33.3)( 19 Aeeti ttL

1.5s

Success

EMLAB

58Application example 7.21 : Defibrillator

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. Its key feature, as shown in the figure, is its voltage waveform. A simplified circuit dia-gram that is capable of producing the Lown waveform. Let us find the necessary values for the inductor and capacitor.

• The Lown waveform is oscillatory. → Under-damped (ζ <1).• The waveform starts from zero voltage.

LCteKtx dd

t 1,1,sin)( 0

201

0

Lown waveform

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59

• The period = 10 [ms]. 2002

Td

12250

3000

)4/3(

)4/( 2/4/3

1

4/1 0

0

0

T

T

T

o

o eeK

eK

T

T

250

49712ln2

0 T

• From the ratio of voltage peaks

012

LCs

L

Rs

02 200

2 ss

][3.504972

49722 0 mHR

LL

R

20

2220

20 1801)497()200()( dd

Characteristic eq. for a series RLC circuit.

• Natural frequency :

][311120

20

FL

CLC

T

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60