ECA

CHAPTER 7 Analysis of first- and second-order

transient circuits

1

7.5 Application Examples

2

Application Example 7.12 • The Xenon flash has the following specifications;

Determine values for , , and . And find recharge time, the flash

bulb’s voltage, current, power, and total energy during the flash.

7.5 Application Example

Voltage required for flash: min: 50 V/max: 70 V

Equivalent resistance:

Time constant during flash time: 1 ms

The resistor must dissipate no more than 100 mW 1R

SV FC 1R

Xenon bulb need the voltage at least 50 V,

We can select 60 V for . SV

80BR

During flash time, the time constant is FBF CR

Given , we find 80 ms, 1 BF R F 5.12 FC

At the beginning of the charge time,

the capacitor voltage is zero → Both the current and power in are max. 1R

→ Voltage across the is 60-V 1R

0.13600

11

2

max RR

VP S

R k 36 1R

3

Application Example 7.12


The recharge time is the time required for the capacitor to charge up

to at least 50 V t

CF eKKtv 21)(The capacitor voltage

Using the conditions 60)( ,0)0( CFCF vv

V 6060)( 1 FCRtCF etv

At , chargett V 50)( tvCF

F 5.12 ,k 361 FCRusing

ms 806 charget

4



Since 80BR

mA 750)(

)( 1000t

B

BB e

R

tvti

The power is W45)()()( 2000t

BBB etitvtp

Finally the total energy consumed by the bulb during flash is

dtedttptw t

BB

0

2000

045 )()(

mJ 5.22

Consider in the flash time again,

V 60)( 1000tB etv

The bulb and capacitor voltage are the same

during flash using 60)0( Bv ms 1 , F

5

Application Example 7.13 • 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.

• 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 sec.


When in pos.1 1

0in

1)(

T

dtVL

ti

Then the switch moves from pos.1 to pos.2

at time

The peak inductor current is L

TVip

1in

1T

Figure 7.44 (a) Pulse generator with switch in position 1. Inductor is

energized.

6



While in pos.1 the resistor voltage is zero.

At , 1Tt

Figure 7.44 (b) Switch in position 2.

As energy is drain from the inductor, the voltage and

current decay toward zero.

The inductor current flows into the resistor producing the voltage

RTtiTtvo )()( 11

We know that the form of the voltage /)( 1)(Tt

o Ketv

And , according to the spec , 500)( 1 TvoRL)(10 1

5

500)(Tt

o etv

L

TVRRip

1in500 For , 1T A 5 ms, 1 1 piT

At the end of the 2 ms period, that is, ms 22 1 Tt

the voltage is or essentially zero. 100500 e

7



The complete waveform for the voltage is

Consider the rating of the various components

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!

Fortunately, the resistor power is pulsed rather than continuous;

thus, a lower power rated resistor will work fine, perhaps 500 W.

8

Application Example 7.14 • The SCR 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

←

Once the voltage across the SCR reaches 5 V.

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 , is 0 V and the capacitor begins to charge toward

the 6 V source voltage.

Find the resistor value such that at .

At , the SCR fires and begins discharging the capacitor.

Find the time required for to drop from 5 V to 0.2 V.

Finally, plot for the three cycles.

0t )(tvC

V 5)( tvC s 1ts 1t

)(tvC

)(tvC

Figure 7.46 Heart

pacemaker

equivalent

circuit.

9



s 1tFor

V 66)( / RCtC etv

RCt 034.01

RCt 792.12

s 1758.112 RCtt s 569.0 RC

k 569R

A voltage of 0.2 V occurs at

and 5 V at

We desire that s 112 tt

s 1tFor

s 1t

s 1tRCteKKtv /)1(21)(

Just after the SCR fires

at , is still 5 V,

at , s 1t

t

)(tv

IRtvC 6)(

521 KK IRK 61and

RCtc eIRIRtv /)1()1(6)(

10



Let be the time beyond 1 s necessary for to drop to 0.2 V T )(tvC

2.0)1(6)1( / RCTc eIRIRTv

Using the values, s 11.0T

The output waveform is

11

Application Example 7.15 • and have been chosen 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.


0)( A, 1)0( LL ii

L

Reti t

L A, 1)(

SV R

Since

and when switch is open the is infinite

and the time constant L/R is zero R

At , 0t the peak inductor voltage is negative infinity!

→ disrupt the inductor current instantaneously

→ the peak switch voltage must be infinity

→ This phenomenon called inductive kick

We look for to reduce this excessive voltage.

The capacitor voltage can not change suddenly

→ Therefore let’s put an RC network!

tLL e

dt

tdiLtv

)()(

12



We need the characteristic equation for the series RLC circuit

011

2 2200

2

LCs

L

Rsss

CLC 4

1220

10

1110

4

60

10

111022

R

L

R

nF 10C 199R

Assume that the switching speed has a critically

damped system : ,1 rad/s 1060

Solving these equation, and

Focus on the peak voltage.

Although the switch is open, the capacitor voltage would not change

instantaneously.

So, the peak resistor voltage is V 199LRi

This is tremendous improvement over the first scenario! 13



A plot of the switch voltage is,

Note that the switch voltage is controlled at a 199-V peak value

and the system is critically damped; that is little or no overshoot!

This RC network is called snubber and is the solution of controlling

inductive kick.

14

Application Example 7.16 • In the electronic system such as TVs, radios, and computers, the ac

waveform is converted to a quasi-dc voltage by an inexpensive ac-dc

converter. And unregulated dc is converted to a higher quality dc output

by a switching dc-dc converter called the boost converter.

Let us develop an equation relating the output voltage to the switching

characteristic


Boost converter with switch setting for time interval ont

The inductor is storing the energy

The capacitor is discharging

and the is exponentially decrease. oV

15



Boost converter with switch setting for time interval offt

The current flows into the capacitor

and the resistor.

The capacitor is recharging

Note that the energy added to the inductor during must go to the

capacitor and the load during . offtont

The inductor energy is )(2

1)( 2 tLit

The inductor current must also be

the same at the end of each

switching cycle. 0I

16



)()0()0(1

)0()(1

)( offonin

onin

0 in0ononon

titL

Vit

L

VidtV

Lidttv

Lti LLL

t

L

t

LL

)()(1

)()(1

)( ononoffoffon

on 0

offon

on

tidtVVL

titvL

ti L

tt

t inL

tt

t LL )( onoff titL

VV oin

So, offon )( tVVtV inoin ))(( ontTVV ino offon ttT

Solving for oV

DV

TtV

TtTV

tT

TVV ininin

onino

1

1

)/1(

1

/)(

1

onon

where TtD on /

Since is always a positive fraction,

is always bigger than oVinV

D

+ +

where

=

17

Application Example 7.17 • When Sw-1 is close, the capacitor bank is charging

When Sw-2 is close, the capacitor discharges and the current causes the

foil at the end of the gun to explode.


Experimental schematic for a railgun. Railgun discharge circuit

The differential equation for the circuit is

0)()()(

busbus

bus2

2

CL

ti

dt

tdi

L

R

dt

tid

Let us use the characteristic equation to describe the current waveform.

Resistor of foil is negligible

18



Using the values, the characteristic eq. is

0103.58105.37 1042 ss

And the roots are 4

21 10)7475.18(, jss → The system is underdamped!

The damped resonant frequency is

Therefore,

and the period the waveform is

krad/s 740d

kHz 1182/ ddf

s 5.81

df

T

An actual plot of the current is

19

7.6 Design Examples

20

Design Example 7.18 • Design an efficient electric heater. It operates from a 24V and creates

heat by driving a 1 Ω resistive heating element. For the temperature

range, the power absorbed by the heating element should be between

100 and 400 W. There are two techniques; (a) the voltage divider and (b)

a switch inductor circuit.

7.6 Design Examples

(a) the voltage divider circuit

adjhe

he

RR

RVV so

The heating element voltage is

2adjhe

he2

he

2

)( RR

RV

R

VP s

oo

The power can be expressed as

, 2.01400

)1)(24( 2

hemax,

he2

minadj, RP

RVR

o

s 4.11100

)1)(24( 2

hemin,

he2

maxadj, RP

RVR

o

s

By substituting the max. and min. values of the power

we can determine the range of resistance

In the worst case, the efficiency of the power is only 50% when , 1adj R

21

Design Example 7.18

7.6 Design Examples

tL

VdtV

Ldttv

Lti s

sLL 1

)(1

)(

1tL

VI s

peak

/')'( tpeakL eIti

Assume the inductor current is zero and the switch

has just moved to pos.1.

The inductor current increases until the switch

moves at time at which point the peak

current is 1tt

This inductor current will discharge

exponentially.

(b) a switch inductor circuit.

where is zero when the switch

move to pos.2 and

't

he/ RL

Repeated switching cycles will transfer

power to the heating element 22

Design Example 7.18

7.6 Design Examples

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(50%).

23

Design Example 7.19 • There are disturbances in dc power supply. The disturbances can be

modeled with a switch. We can reduce disturbance using decoupling

capacitor. Find the equation for in terms of

7.6 Design Examples

DC ' , , , , , tVVVVR oSoSS and

Disturbances model in

the source voltage

Reducing the disturbances

with decoupling capacitor.

24

Design Example 7.19

7.6 Design Examples

)1()(/ DSCRt

SSLSSo eVRIVVtv

To determine the time constant’s equivalent resistance,

reduce all independent source to zero, and view from the capacitor.

It is easy to see that the time constant is DSCR

/21)( t

o eKKtv The form of the voltage of is DC

sLso RIVKKv 21)0(

sLsso RIVVKv 1)(

0t

t

When ,

When ,

0t t

25

Design Example 7.19

7.6 Design Examples

s

o

soos

s

V

V

VVVV

Vf

1ln

1

/1

1ln

1

ln

1

Let us isolate the logarithm term and express it as

Note that for very small this term is very large

We find that the price for excellent decoupling is very large capacitance.

oV

os

ss

D

VV

VR

tC

ln

'

DSCRtSSLSSoSLSo eVRIVVVRIVtv

/')'(

DSCRtSoS eVVV

/'

At exactly , the output voltage is its original value,

Substituting this condition into Eq.(1)

'tt oo VV )0(

Which can be reduced to the expression

26

Design Example 7.19

7.6 Design Examples

A plot of the function, versus fs

o

V

V

Finally, as an example

V 2.0 ms, 5.0' V, 1 , 20 V, 5 oSSS VtVRV

The required capacitance is

μF 0.112

ln

'

os

ss

D

VV

VR

tC

27

Design Example 7.20 • The inductor is the ignition coil. The switch is the keyed ignition switch. To

start the motor we close the switch, thereby discharging the capacitor

through the inductor.

Assume that optimum starter operation requires an overdamped for

that reaches at least 1 A within 100ms after switching and remains above 1

A for between 1 and 1.5 s.

Let us find the value for the capacitor, plot the response and verify design

7.6 Design Examples

Before the switch is moved

A 0)0()0( LL ii

V 12)0()0( cc vv

After switching (1) 012

LCs

L

Rs

When the roots are

(2) 0)())(( 21212

21 ssssssssss

21 , ss

28

Design Example 7.20

7.6 Design Examples

Comparing Eq.(1) and Eq.(2) ,2021 ssL

R21

1ss

LC

Since the network is overdamped tstsL eKeKti 21

21)(

0)0( 21 KKiL 12 KK Just after switching

Since , the resistor voltage is zero and 0)0( Li V 12)0()0(

LC vv

L

12)0()0( 1211

KsKs

dt

diLv L

L

121

s

60

sK

29.414

60

s

60

121

sK

If we choose 17 ,3 21 ss

mF 98)17)(3)(2.0(

1

s

1

21

Ls

C

29

Design Example 7.20

7.6 Design Examples

Hence,

It satisfy first specification (at 100 ms, 2.39A),

but has fallen down to 0.16 A at

So, we have to slow down the response

A 29.4)( 173 ttL eeti

s 1.1t

21 ss Since ,

It is better to reduce than reduce 1s 2s

If we choose 19 ,1 21 ss

mF 2639)(0.2)(1)(1

11

21

sLs

C

33318

6060

121 .

ssK

A 333)( 19ttL ee.ti s 5.1tIt 1.11 A at .

So this capacitor satisfy all specification

30

Design Example 7.21 • A defibrillator is a device that is used to stop heart fibrillations by

delivering an electric shock to the heart.

Let us find the necessary values for the inductor and capacitor.

7.6 Design Examples

Lown defibrillator waveform Lown defibrillator simplified circuit

Since the waveform is underdamped, and

teKtvt

oo

sin)( 1

0)0( ov

where ,2L

Ro ,1 2 o LC

o

1

Since the period of the waveform is 10 ms

ms 10T

31

Design Example 7.21

7.6 Design Examples

rad/s 20012 2

oT

Thus,

At two instant of time the sine function is equal to +1, -1

12250

3000

)4/3(

)4/( )2/(

)4/3(

1

)4/(

1

T

T

T

o

o o

o

o

eeK

eK

tv

tv

So, 0.497o

Using , 50R mH 3.50L

2222 )200( oo

22 )0.497(1

)200( LC

F 0.31 C

32

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.

− The form of a first-order differential equation with a constant

forcing function is

and the solution is

where is referred to as the steady-state solution and is

called the time constant.

Atx

dt

tdx

)()(

/2)( teKAtx

A

33

Summary

− The function decays to a value that is less than 1% of its

initial value after a period of . Therefore, the time constant, ,

determines the time required for the circuit to reach steady

state.

− The time constant for an RC circuit is and for an RL circuit

is , where is the Thevenin equivalent resistance

looking into the circuit at the terminals of the storage element

(i.e., capacitor or inductor).

/te

5

CRTh

ThRThRL

34

Summary

− 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.

− 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.

35

Summary

• Second-Order Circuits

− The voltage or current in an RLC transient circuit can be

described by a constant coefficient differential equation of the

form

where f(t) is the network forcing function.

− The characteristic equation for a second-order circuit is

where is the damping ratio and is the undamped natural

frequency.

)()()(

2)( 2

002

2

tftxdt

tdx

dt

txd

02 200

2 ss

0

36

Summary

− If the two roots of the characteristic equation are

− real and unequal, then and the network response is

overdamped

− real and equal, then and the network response is

critically damped

− complex conjugates, then and the network response

is underdamped

− The three types of damping together with the corresponding

network response are as follows:

1. Overdamped:

2. Critically damped:

3. Underdamped:

where and

1

1

1

)( )1(

2)1(

1

200

200 tt

c eKeKtx

)sincos()( 21 tAtAetx ddt

c

)( 00

21

tt

c teBeBtx

20 1 d0

37

Summary

− Two initial conditions are required to derive the two unknown

coefficients in the network response equations.

38

Documents

ECA