First-Order CircuitsPart II

Instructor: Chia-Ming TsaiElectronics Engineering

National Chiao Tung UniversityHsinchu, Taiwan, R.O.C.

Contents• Introduction

• The Source-Free RC Circuit

• The Source-Free RL Circuit

• Singularity Functions

• Step Response of an RC Circuit

• Step Response of an RL Circuit

• First-Order Op Amp Circuits

• Applications

Singularity Functions

• To aid the understanding of transient analysis

• To serve as good approximations to the switching signal in circuits

• Singularity functions are functions that either are discontinuous or have discontinuous derivatives.

• Three most widely used types are introduced– Unit step function– Unit impulse function– Unit ramp function

Unit Step Function

0 ,1

0 ,0

t

ttu

Unit Step Function (Cont’d)

0

00 ,1

,0

tt

ttttu

0

00 ,1

,0

tt

ttttu

Unit Step Function (Cont’d)

00

00

0

,

,0

ttuVtv

ttV

tttv

00

00

0

,

,0

ttuIti

ttI

ttti

Unit Impulse Function

0 ,0

0 ,Undefined

0 ,0

t

t

t

tudt

dt

10

0

dtt

Unit Impulse Function (Cont’d)

Unit Impulse Function (Cont’d)

00

00

000

0

,Consider

tfdttttf

dttttf

dttttfdttttf

b ta

b

a

b

a

b

a

b

a

(Sampled at t0)

Unit Ramp Function

0 ,

0 ,0

tt

ttr

ttudttutrt

Unit Ramp Function (Cont’d)

00

00 ,

,0

tttt

ttttr

00

00 ,

,0

tttt

ttttr

Summary

dt

tdrtu

dt

tdut

dttutr

dtttu

t

t

Example 1

5210

510210

tutu

tututv

=

Example 1 (Cont’d)

5210 tututv 5210 tt

dt

tdv

Step Response of an RC Circuit

RC

Vv

dt

dv

t

tuRC

V

RC

v

dt

dvR

tuVv

dt

dvC

R

tuVvi

dt

dvCi

ii

Vvv

S

S

S

SRC

RC

0,For

)(or

0)(

)( and

0

gives KCL Applying

)0()0( Assume 0

=

iC

iR

Cont’d

0 ,

0 ,)(

0 ,)(

where,

ln

ln

0

0

0

0

0

0

)(

||0

teVVV

tVtv

teVVVtv

RC τeVVVv

RC

t

VV

VvRC

tVv

RC

dt

Vv

dv

tSS

tSS

tSS

S

S

ttv

VS

S

Forced Response (V0=0)

)()(

),( 1)(

0 , 1

0 ,0)(

,0 i.e.

initially, unchargedcapacitor Assume

0

tueR

V

dt

dvCti

RCτ tueVtv

teV

ttv

V

tS

tS

tS

Step Response (I)• Complete Response

= Natural Response (vn) + Forced Response (vf)

(stored energy) (independent source)

tSf

tn

fn

tS

t

tSS

eVv

eVv

vv

eVeV

eVVVv

1

where

1

0

0

0

Step Response (II)• Complete Response

= Transient Response + Steady-State Response

(temporary part) (permanent part)

where

0)( and

0

0

Sss

tSt

t

tss

tSS

Vv

eVVv

v

vv

eVVVv

Short-Cut Method• Three items required to describe the response

– The initial capacitor voltage v(0) (or v(t0) )

– The final capacitor voltage v()– The time constant = RC

0)()()()(

or

)()0()()(

0tt

t

evtvvtv

evvvtv

v(0)

v()

Example 1

s 2105.0104

104 and V 30)(

V 152435

5)0()0(

33Th

3Th

CR

Rv

vv

0 V, 1530

)3015(30

)()0()()(

5.0

5.0

te

e

evvvtv

t

t

t

Example 2

t < 0 t > 0

10)0( v t

Th

evvvtv

Rv

)()0()()(

32020||10 ,20)(

Step Response of an RL Circuit

tSS

S

S

St

Sf

tn

fn

eR

VI

R

Vti

R

VIA

IR

VAi

R

VAei

R

Vi

RLτAei

iii

0

0

0

)(

)0(

,

iL

iR

=

Forced Response (I0=0)

)()(

),( 1)(

0 , 1

0 ,0)(

,0)0( i.e.

current,inductor initial no Assume

0

tueVdt

diLtv

R

Lτ tue

R

Vti

teR

Vt

ti

iI

tS

tS

tS

Short-Cut Method• Three items required to describe the response

– The initial inductor current i(0) (or i(t0) )

– The final inductor current i()– The time constant = L/R

0)()()()(

or

)()0()()(

0tt

t

eitiiti

eiiiti

i(0)

i()

Example 1

0 A, 32)25(2

)()0()()(

s 15

1

5

31

532 and

A 232

10)(

A 52

10)0()0(

1515

Th

Th

tee

eiiiti

R

L

R

i

ii

tt

t

Find i(t).

Example 2

A )1(4)40(4

)()0()()(

s 2

1

10

5

1064 and

A 464

40)(

40 )2(

0)0( ,0 (1)

.considered are intervals timeThree

22

Th

Th

tt

t

ee

eiiiti

R

L

R

i

t

it

Find i(t).

Example 2 (Cont’d)

)4(

15

22

/)4(

Th

Th

Th

8

273.111

30

)()4()()(

22

15

322

5

727.211

30

322

20)(

3

22

4121

16

20

4)1(4)4(

4 (3)

t

t

S

S

t

e

eiiiti

R

L

R

Vi

R

V

ei

t

Example: OP AMP Circuits

tt

t

o

o

o

o

ee

evvvtv

CR

vR

vv

v

v

v

v

1010

/

6Th

Th

12)012(0

)()4()()(

1.010520000

200001

200000120000

(b).circuit in KVLapply , find To

12)( and

12)0(

080000

)0(0

20000

3

2, nodeat KCLapply ,)0( find To

(b)

(a)

Applications: Delay Circuit

t

v70 V

I II

I: II:

= (R1+R2)C = RC

R

Applications: Relay Circuit

closed

, If

2

0

S

Vv

v+

_

Applications: Ignition Circuit

• Two steps to work– S is closed to build the inductor current

– Open S to force the inductor current to pass through the air gap

S

RVi S