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SYSU - 2010 Spring Lecture 05 - 1

Chapter 5:First Order Circuit Analysis

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Topics To Cover

Special Supply Functions

DC, Step, Ramp, and Pulse

First Order Circuit Source-free RC and RL circuits

Step response

Textbook: Chapter 7

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Common Supply Functions

Voltage or Current Sources: DC

Step

Rectangular Pulse

Ramp Triangular Pulse

Impulse

Singularity functions: functions thateither are discontinuous or have

discontinuous derivatives

sVtv =)(

0>sV

0

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4/36SYSU - 2010 Spring Lecture 05 - 4

Special Supply FunctionsStep

Unit Step Function(corresponds to a switch)

>

0Time advanceto

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Rectangular Pulse Function

Unit Rectangular Pulse Function

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Ramp Function

Unit Ramp Function Integration of unit step function u(t)

== t

ttudutr )()()(

0

By inspection: vc(0-) = 0

Since the capacitor voltage

cannot change instantaneously,

vc(0+) = v

c(0-) = 0

For t>0, apply KVL:

Rs

CVs

vc(t)

+

-

ic

First-order ordinary

differential equation

0= ccss viRV

0= cc

ss vdtdvCRV

)1()(

t

sc eVtv

=

CRs=

t > 0

)1()( CR

t

scseVtv

=

,0)0(useandIntegrate =cv

or

The solution is

CRdt

vVdv

scs

c =

( ) ( )CR

tVvV

s

scs = lnln CRt

s

sc s

eV

Vv

=

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ELE 1110A - 2009 Autumn Lecture 05 - 16

The Response

0.37Vs

)1()( t

sc eVtv

=

Rs

Vs

C

vs=Vsu(t)

vc

(t)

+

-

ic

Initial capacitor voltage = vc

(0+) = 0

Final capacitor voltage = vc() = Vs

for t > 0

s

t

sc

c ReVdt

tdv

Cti /

)(

)(

==Vs/Rs

0.37Vs/Rs

for t > 0

SYSU - 2010 S rin

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Lecture 05 - 17

RC step responseCase 2

for t < 0

Rs

Vs vc(t)

+

-

Cic

= Vo at t = 0

for t > 0

Rs

C

vs=V0+ (Vs V0) u(t)

vc(t)

+

-

ic

Vs

V0 t = 0

Vo Cic

vc(t)

+

-

0)0( VtvS = 0

By inspection: vc(0-) = V0

Since the capacitor voltage

cannot change instantaneously,

vc

(0+) = vc

(0-) = V0

For t>0, apply KVL:

Rs

C

Vsvc(t)

+

-

ic

0= ccss viRV

0= cc

ss vdt

dvCRV

CRs=

t

ssc eVVVtv

+= )()( 0

V0 = initial capacitor voltage

Vs = final capacitor voltage

The solution is

,)0(useandIntegrate oc Vv = ( ) ( )CR

tVVvV

s

oscs = lnln

CR

t

so

sc seVV

Vv =

CRdt

vVdv

scs

c =

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Lecture 05 - 19

The Response

Vs

t

Rs

C

vs=Vsu(t)

vc(t)

+

-

ic

Initial capacitor voltage = V0Final capacitor voltage = Vs

s

tt

VeeV )1(0 +=

V0

t

ssc eVVVtv+= )()( 0

0.37(Vs V0)

sVV )1(0 += 10

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Lecture 05 - 20

Step ResponseSummary

>+

0

Rs

vs=V0+ (Vs V0) u(t)

vL(t)

+

-

Vs

V0 t = 0

LiL

Vo

iLvL(t)

+

-

Rs

L

Rs

Vs vL(t)

+

-

iLL

iL(t) =I0 = V0/RsAt t = 0, vL(t) = 0

RL Step Response

Let Is = Vs/Rs

SYSU - 2010 Spring

R t t 0

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ELE 1110A - 2009 Autumn Lecture 05 - 24

Response at t > 0

By inspection:IL(0-) =I0 = V0/Rs

Since the inductor current cannot

change instantaneously,IL(0+) =

IL

(0-) =I0

For t>0, apply KVL:

0= LLss viRV

0=dt

diLiRV LLss

sRL /=

t

ssL eIIIti

+= )()( 0

I0 = initial inductor current

Is = final inductor current

Rs

Vs vL(t)

+

-

iLL

Is = Vs/Rs

L

Vi

L

R

dt

di sL

sL =+

L

tR

sL

tR

L

ss

eLVei

dtd =

Ae

RVei L

tR

s

sL

tR

L

ss

+=

L

tR

sL

s

AeIti +=)(0)0( IiL =

SYSU - 2010 Spring

St R f I d t S

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Lecture 05 - 25

Step Response of InductorSummary

Steady-state

response, i.e.

the final current Transient response,

i.e. the current

waveform frominitial to final

(2.4)

Initial current in the inductor

Time constant

Rs

vs=V0+ (Vs V0) u(t)

vL(t)

+

-

Vs

V0 t = 0

LiL

>+

0

Vs

0.37(Vs V0)

for t > 0

t

ssL eIIIti

+= )()( 0

Rs

Vs vL(t)

+

-

iLL s

RL /=

SYSU - 2010 Spring

E i

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Lecture 05 - 27

Exercise

Find i(t) for t > 0. The switch had been closed for a long timeand before it opens at t = 0.

SYSU - 2010 Spring

Application: Photoflash Unit

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Lecture 05 - 28

Application: Photoflash UnitR1: current-limiting resistor

(Large)

Resistance of Flash lamp

(small)

The battery (dc voltage supply) has a limited output current.

Directly connecting it to the flash lamp may exceed the current

limit and may damage the battery.

Solution

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Lecture 05 - 29

First, use the batter to charge a capacitor through a larger resistor

R1 (switch to position 1). The maximum current drawn from the

battery is VS/R1.

The charging time is about five times the time constant R1C:

CRt 1charge 5=

(This time constant is large and it is why sometimes you need towait a while before you press the flash button of a camera)

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Lecture 05 - 30

To flash the lamp, switch is thrown to position 2.The capacitor is discharged. The low resistance R2 of the

photolamppermits a high discharging current with peak

I2=VS/R2 in a short duration. The discharging time is:

CRt 2discharge 5=

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Lecture 05 - 31

Capacitor voltage and current waveforms

(-ve: flows out of

the +ve terminal)

SYSU - 2010 Spring

Example 5 2

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Lecture 05 - 32

Example 5.2

Vs=9V, R1 = 6k, C = 2000F, Lamp resistance R2 = 12. Find:(a) The peak charging current

(b) The time required for the capacitor to fully charge

(c) The peak discharging current(d) The total energy stored in the capacitor

(e) The average power dissipated by the lamp

SYSU - 2010 Spring

( ) Th k h i

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Lecture 05 - 33

(a) The peak charging current

(b) The time required for the capacitor to fully charge

(c) The peak discharging current

(d) The total energy stored in the capacitor

(e) The average power dissipated by the lamp

mAR

V

I

s

5.16000

9

1

1

===

sCRt 60)102000(600055

6

1charge ===

AR

VI s 75.0

12

9

2

2 ===

mJCVEs

819)102000(

2

1

2

1 262 ===

sCRt 12.0)102000(1255

6

2discharge

=== W

m

t

Ep 675.0

12.0

81

dischargeaverage

===

SYSU - 2010 Spring

Application: Automobile Ignition Circuit

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Lecture 05 - 34

Application: Automobile Ignition Circuit

By creating a large voltage (thousands of volts) between the

electrodes, a spark is formed across the air gap, thereby

igniting the fuel.

How can we create such a high voltage from the car battery,

which supplies only 12V?

SYSU - 2010 Spring

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Lecture 05 - 35

Let R = 4, Vs = 12V, L=6mH and assume the switch takes1s to open.

The steady state current through

the inductor before switch opens is:A

R

VI s 3

4

12===

The voltage across the air gap when switch suddenly opens is:

kVt

ILV 18

101

3106

6

3 =

=

=

(Large enough!)

SYSU - 2010 Spring

Summary

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Lecture 05 - 36

Summary

1st order circuit consists of one energy storageelement (capacitor or inductor)

The circuit equation corresponds to a 1st order

differential equation

To solve the differential equation, need to determine

the initial conditions of the circuit, i.e. capacitor =open circuit; inductor = short circuit given long

enough time before switching

t

ssL

eIIIti

+= )()(0

t

sosc eVVVtv

+= )()(Time constant

For capacitor:

For inductor: sR

L=

CRs=

SYSU - 2010 Spring

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