circuitry ch07 first-order circuits

10

Chapter 7 First-Order Circuits

Circuit elements Passive elements : resistors, capacitors and inductors Active element : OP Amp (operational amplifier)

Analysis of RC and RL circuits by applying Kirchhoffs laws RC and RL circuits are first-order circuit A first-order circuit is characterized by a first-order

differential equation Excitation of first-order circuits Source-free circuit : initial conditions on inductor and capacitor Independent sources : dc, sinusoidal and exponential sources

1

7.2 The Source-Free RC Circuit

A source-free RC circuit occurs when dc source is suddenly removed

The resistor and capacitor may be equivalent circuit of combinations of them

We want to know the circuit response, i.e. the capacitor voltage v(t). Assuming, v(0) = V0

The energy stored is

Applying KCL:

20CV2

10w =)(

0RCv

dtdv0

Rv

dtdvC

0ii RC

=+=+=+

or

22

The Source-Free RC Circuit

This is a first-order differential equation

Integrating both sidesdt

RC1

vdv =

RCt0

0

RCt

eVtvVA0v

AetvRCt

Avln

AlnRCtv ln

/

/

)(

)( ,conditions initial theFrom)(

constant nintegratio theis A where

===

==

+=

An exponential decay of the initial voltage

3

The Source-Free RC Circuit

The natural response of a circuit refers to the behavior (in terms of voltages and currents) of the circuit itself, with no external sources of excitation

Natural response depends only on the nature of the circuit alone, nothing related to the external sources

The time constant ( = RC)of a circuit is the time required for the response to decay by a factor of 1/e or 36.8% of its initial value

/

/

)(

.)(

t0

01

0RC

0

eVtvand

V3680eVeVv

=

===

34

Steady state v.s. transient state

v(t) is less than 1% of V0 after 5; in other words,It takes 5 for the circuit to reach its final state or steady statewhen no changes occur

The state before steady state is called transient state The smaller the time constant, the faster the response

0.0067450.0183240.0497930.1353420.36788v(t)/V0t

5

Energy in RC circuit

)( toequal is which,)( , tas that Notice

ere wh)(

)(

is time toupresistor by the absorbedenergy The

)(

isresistor thein dissipatedpower The

)()(current resistor The

/

//

/

/

0wCV21w

RCe1CV21

0t

eR2

VdteR

Vpdttw

t

eR

Vvitp

eRV

Rtvti

C2

0R

t220

t22

0t

0

t22

0t

0R

t22

0R

t0R

==

===

==

==

46

Another perspective of time constant

The time constant is the initial rate of decay, or the time taken for v/V0 to decay from unity to zero, assuming constant rate of decay

This initial slope interpretation is used to find from the response curve

Draw the tangent to the curve at t = 0 The tangent intercepts with the time axis at t =

1e1

Vv

dtd

0t

t

0t0

==

=

=

/ /)( t0eVtv=

7

Work with a Source-free RC Circuit

First find: The initial voltage v(0) = V0 across the capacitor The time constant

The capacitor voltage is then

The time constant is identical regardless of what output is When a circuit contains a single capacitor and several resistors

or dependent sources, the Thevenin equivalent can be found at the terminals of the capacitor

/)()()( tC e0vtvtv==

58

Example 7.1

In the figure, let vC(0) = 15V. FindvC, vx, and ix, for t > 0.

Sol: The equivalent resistance

The time constant Thus

20 5 420 5eq

R = = +4(0.1) 0.4eqR C s = = =

/ / 0.4

2.5

2.5 2.5

2.5

(0) 15 ,15

12 0.6(15 ) 912 8

0.7512

t t

tc

t tx

txx

v v e e Vv v e V

v v e e V

vi e A

= == == = =+= =

9

Practice Problem 7.1

For the circuit, let vC(0) = 30V. Determine vC, vx, and io, for t 0.

Ans: 30e-0.25tV, 10e-0.25tV, -2.5e-0.25tV

610

Example 7.2

The switch in the circuit has beenclosed for a long time, and it is opened at t = 0. Find v(t) for t 0 and the initial energy stored in the capacitor.

Sol: Using voltage division

vC(0) is the same as vC(0-)

9( ) (20) 15 , 09 3c

v t V t= =

712

7.3 The Source-Free RL Circuit

Analysis goal is to find the circuit response Select inductor current as the response in order

to take advantage of that: inductor current cannot change instantaneously

Assuming i(0) = I0, the energy stored in the inductor is

This is a first-order differential equation

0iLR

dtid0Ri

dtidL

0vv RL

=+=+=+

or

: KVLApplying

20LI2

10w =)(

13

The Source-Free RL Circuit

Rearranging

Integrating both sidesdt

LR

iid =

where)(

)(

)( )(

//

)(

RLeIeIti

LRt

Itiln

0LRtIlntiln

0t

LR

Iti

iln

dtLR

iid

t0

LRt0

0

00

t

0

ti

I0

===

=

+==

=

An exponential decay of the initial current

814

Natural response of RL circuit

The natural response of the RL circuit is an exponential decay of the initial current

/0

2 2 /0

2 2 /00 0

2 2 / 2 2 /0 0

0

20

( )

( )

1 1 (1 )2 2

1Note that as , ( ) , which is equal to (0)2

tR

tR

t t tR

tt t

R L

v t iR I Rep v i I Re

w t pdt I Re dt

I Re LI e

t w LI w

= == =

= =

= =

15

Key to Work with a Source-free RL Circuit

First find: The initial current i(0) = I0 through the inductor The time constant of the circuit

The inductor current is then

Once iL is determined, inductor voltage vL, resistor voltage vRand resistor current iR can be obtained

When a circuit contains a single inductor and several resistors or dependent sources, the Thevenin equivalent can be found at the terminals of the inductor

/)()()( tL e0ititi==

916

Example 7.3

Assuming that i(0)=10A, calculate i(t) and ix(t) in the circuit.

Sol: Method 1:

1 2 1 2

2 1 1 2 1

1 0 1

0

0

12( ) 1 02

56 2 3 06

3 , 31Hence 3eq Th

Applying KVL

i i i i

i i i i i

i A i i AvR Ri

+ = =

= == = =

= = =

17

Example 7.3

Method 2:/ (2 / 3)

3The timeconstant is 2

Thus, thecurrent through the inductor is( ) (0) 10 0

eq

t t

L sR

i t i e e A t

= =

= = >

11 2

2 1 1 2 1

11

1

1

11 2( ) 02

52 6 2 3 06

2 03

2Rearranging terms3

diApply KVL for loop i idt

for loop i i i i i

diSubstituting idt

di dti

+ =

= =

+ =

=

10

18

Example 7.3

Practice Problem 7.3 Let i(0)=5A, find i(t) and vx(t) in the

circuit.

Ans: 5e-53t A, -15e-53t V

( )

(0)0

(2 /3)

2ln3

( ) (0)

ti t

ii t

i t i e

= = (2 /3)

(2 /3) (2 /3)

(2 /3)

10 02 100.5(10)( )3 3

( ) 1.667 , 02

t t

t t

tx

e A tdiv L e e Vdtvi t e A t

= >= = =

= = >

19

Example 7.4

The switch in the circuit has been closed for a long time. At t = 0, the switch is opened. Calculate i(t) for t > 0.

Sol:

1

1

Combine the 4 and 124 12resistors 34 12

40Hence, 82 3

Usingcurrent division12( ) 6 , 0

12 4(0) (0 ) 6

i A

i t i A t

i i A

= += =+

= =

11

20

Example 7.4

Combining the resistors, we have

The time constant is

Thus,

Practice Problem 7.4 For the circuit, find i(t) for t > 0.

Ans: 2e-2t A

816||)412( =+=eqRs

41

82 ===

eqRL

A6)0()( 4/ tt eeiti ==

21

Example 7.5

In the circuit, find io, vo, and i for all time, assuming that the switch was open for a long time.

Sol: For t < 0

For t > 0A)(

V)()( A,)(

20i

6ti3tv232

10ti 0

====+=

A)()(

s

||

/ ttTh

Th

e2e0iti

1RL

263R

======

12

22

Example 7.5

Since the inductor is in the parallel with the 6 and 3

0

0

0 0

( ) 2( 2 ) 4 , 0

2( ) , 06 3

0 , 0 6 , 0( ) , ( )2 , 0 4 , 0

32 , 0

( )2 , 0

t tL

tL

t t

t

div t v L e e V tdt

vi t e A t

ThusA t V t

i t v te A t e V t

A ti t

e A t

= = = = >

= = >

13

24

7.4 Singularity Functions ()

Singularity functions are functions that either are discountinuous or have discountinuous derivatives.

Singularity functions (also called switching functions) serve good approximations to the switching signals in the circuits with switching operations

These functions are used to describe the sudden application of an independent dc voltage or current source

Three most widely used are unit step, unit impulse, and unit ramp functions

25

Unit step functions

The unit step function u(t) is 0 for negative values of t and 1 for positive values of t.

>

14

26

Unit step voltage source

Step function is used to represent an abrupt change in voltage or current, like the changes that occur in the circuits of control system or digital computers

)()( aslly mathmatica drepresente be canIt ,,

)( like is a voltage If

00

00

0

ttuVtvttVtt0

tv

=

>

15

28

Unit impulse function

Unit impulse function (t) is zero everywhere except at t = 0, where it is undefined

Unit impulse function is the derivative of the unit step function

Unit impulse function is also known as delta function The unit area is known as the strength

of the impulse function

0, 0( ) ( ) , 0

0, 0

tdt u t Undefined tdt

t

unit area

0

0( ) ( ) 1t dt t dt + = =

29

If an impulse function has a strength other than unity, the area of the impulse is equal to its strength

Integrating the function results in the value of the function exactly at the point where the impulse occurs

Useful feature known as sampling or sifting property

Sampling or sifting property

0

0

0

0

0 0

0 0

0 0 0

( ) ( ) where

( ) ( )

( ) ( ) ( )

b

a

t

t

t

t

f t t t dt a t b

f t t t dt

f t t t dt f t

+

+

<

=+=+

or

For

)(or )(

0

( )

0

00

ln( )

ln( ( ) ) ln( ) 0 or ln

tv t

s V

ss s

s

tv VRC

t v V tv t V V VRC V V RC

= = + =

23

44

Step Response of an RC Circuit

Taking the exponential of both sides

Thus,

This is known as the complete response (or total response) of the RC circuit

0/

0

, 0( )

( ) , 0ts s

V tv t

V V V e t

assuming Vs > V0

45

Step Response of an RC Circuit

If assuming the capacitor is initially uncharged, then V0 = 0

The above can be alternatively written as

The capacitor current is obtained by i(t) = Cdv/dt

/

0, 0( )

(1 ), 0ts

tv t

V e t

/( ) (1 ) ( )tsv t V e u t=

/

/

( ) , 0

or ( ) ( )

ts

ts

dv Ci t C V e tdtVi t e u tR

= = >

=

24

46

Complete response

Complete response = nature response + forced response

Originally Alternatively

Natural response vn is the source-free response, dying out gradually with the transient part of the forced response

Forced response vf is the response induced by external force, existing continually to form the steady-state component

stored energy independent source

)( , where // tsft

0n

fn

e1VveVv

vvv ==

+=0( ) ( ) , 0

t

s sv t V V V e t= + >

0

V0

Vs

vn

vfvtotal

47

Complete response

Complete response = transient response + steady-state response

The transient response is the circuits temporary response that will die out with time

The steady-state response is the behavior of the circuit a long time after an external excitation is applied

Temporary part Permanent part

ssst

s0t

sst

VveVVvvvv

==+=

, )( where /

0

0

s

vt

vssvtotal

V

V

25

48

Decomposition of complete response

The first decomposition of the complete response is in terms of the source of the responses; the latter is the permanency of the responses

Under certain conditions, the natural and transient responses are the same; so as forced and steady-state responses

Either way, complete response can be written as

where v(0) is the initial voltage at t = 0+ and v() is the final or steady-state value

/( ) ( ) [ (0) ( )] tv t v v v e = +

49

Solving RC circuit with response decomposition

To find the step response of an RC circuit requires three things The initial capacitor voltage v(0) The final capacitor voltage v() The time constant

If the switch changes position at time t = t0, there is a time delay in the response

where v(t0) is the initial value at t = t0+

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

t tv t v v t v e = +

26

50

Example 7.10

The switch in the figure has been in position A for a long time. At t = 0, the switch moves to B. Determine v(t) for t > 0 and calculate its value at t = 1s and 4s.

Sol: By voltage division

3 3

5(0 ) (24) 155 3

Since the capacitor voltage cannot change(0) (0 ) (0 ) 15

The time constant 4 10 0.5 10 2Th

v V

v v v VR C s

+

= =+

= = == = =

51

Example 7.10

Since the capacitor acts like an open-circuit to dc at steady state

Practice Problem 7.10 Find v(t) for t > 0. Assume the switch has

been open for a long time and is closed at t = 0. calculate at v(t) t = 0.5.

Ans: -5+15e-2tV, 0.5182V

/

/ 2 0.5

0.5

2

( ) ( ) [ (0) ( )]30 (15 30) (30 15 )1 (1) 30 15 20.9024 (4) 30 15 27.97

t

t t

v t v v v ee e V

At t s v e VAt t s v e V

= + = + = = = == = =

27

52

Example 7.11

In the figure, the switch has been closed for a long time and is opened at t = 0. Find i and v for all time.

Sol: By definition of the unit step function

Since the capacitor voltage cannot change

0, 030 ( )

30, 0For 0

10 , 110

tu t

tt

vv V i A

0, by using voltage division

/

(3/ 5) 0.6

0.6 0.6 0.6

20( ) (30) 2020 10

10 20 2010 2030 3

20 1 53 4 3

( ) ( ) [ (0) ( )]20 (10 20) (20 10 )

201 0.5 0.25( 0.6)( 10) (1 )

Th

Th

t

t t

t t t

v V

R

R C s

v t v v v ee e V

v dvi Cdt

e e e A

= =+= = =

= = == + = + =

= += + = +

28

54

Example 7.11

Practice Problem 7.11 The switch is closed at t = 0. Find

i(t) and v(t) for all time. Note that u(-t) = 1 for t < 0 and 0 for t > 0. Also u(-t) = 1- u(t).

Ans:

0.6

0.6

10 , 0(20 10 ) , 0

1 , 0(1 ) , 0

t

t

V tv

e V tA t

ie A t

55

7.6 Step Response of an RL Circuit

Consider the RL circuit, which may be replaced by the circuit applying unit step function

Our goal is to find the inductor current Decomposing the answer into transient and

steady-state responses, i.e. i = it + iss Since transient response is a decaying

exponential; that is

After long time, the inductor short-circuited

/ , , is a constantttLi Ae AR

= =

sss

ViR

=

29

56

Step Response of an RL Circuit

The complete response is Let I0 be the initial inductor current Since the current through the inductor cannot change

instantaneously, thus

At t = 0,

Therefore,

0(0 ) (0 )i i I+ = =

00 0 or s

V VI A A IR R

= + =

/0( ) ( )

ts sV Vi t I eR R

= +

/t sVi AeR

= +

57


The response may be written as

where i(0) and i() are the initial and final values of i To find the step response of an RL circuit requires: The initial inductor current i(0) at t = 0 The finial inductor current i() The time constant

If the switching happens at time t = t0 instead of t = 0, then

/( ) ( ) [ (0) ( )] ti t i i i e = +

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

t ti t i i t i e = +

30

58


If I0 = 0, then

The voltage across the inductor

/

/

0, 0( )

(1 ), 0

( ) (1 ) ( )

ts

ts

ti t V e t

RVor i t e u tR

=

/

/

( ) , , 0

( ) ( )

ts

ts

di L Lv t L V e tdt R R

v t V e u t

= = = >=

59

Example 7.12

Find i(t) in the circuit in the figure for t > 0. Assuming that the switch has been closed for a long time.

Sol:10(0 ) 52

(0) (0 ) (0 ) 50,

10( ) 22 3

2 3 51

135 15

Th

Th

i A

i i i AWhen t the switch is open

i A

R

L sR

+

= == = =

> = =+= + =

= = =

31

60

Example 7.12

/ 15

15

15 15

( ) ( ) [ (0) ( )] 2 (5 2)2 3 , 0:0

10 5

15 [10 15 ] [ (3)( 15) ] 103

t t

t

t t

Thusi t i i i e e

e A tCheckfor t KVL must be satisfied

dii Ldt

dii L e edt

= + = + = + >

>= +

+ = + + =

61


The switch has been closed for a long time. It opens at t = 0. Find i(t) for t > 0.

Ans: (2 + e-10t) A, t > 0

32

62

Example 7.13

At t = 0, switch S1 in the figure is closed, and switch S2 is closed 4s later. Find i(t) for t > 0. Calculate i for t = 2s and t = 5s.

Sol: For t < 0 , For 0 t < 4

00i0i0i === + )()()(

( )4t0Ae14

e404ei0iiti

50105

RL

1064RA464

40i

t2

t2

tTh

Th

4

s2215

322

5RL

3226

624624R

72721130

6vi

11180v

6v

2v10

4v40

i4e144i4i

Th

8

Th

===

=+=+=

===

==+

==

||

A.)(

V

KCLUsing,)(findTo)()()(

33

64

Example 7.13

( )4)272.24(272.2

)()4()()()4(4667.1

/)4(

>+=+=

teeiiiti

t

t

2

1.4667( 4)

4

1.4667

0, 0( ) 4(1 ), 0 4

2.727 1.273 , 42

(2)

T

4(1 ) 3.935

(5) 2.727 1.273 3.02

o sum up,

t

t

ti t e t

e tAt ti e AAt ti e A

= + =

= ==

= + =

65


Switch S1 is closed at t = 0, and switch S2 is closed at t = 2s. Calculate i(t) for all time. Find i(1) and i(3).

Ans:9

5( 2)

0, 0( ) 2(1 ), 0 2

3.6 1.6 , 2(1) 1.9997 , (3) 3.589

t

t

ti t e t

e ti A i A

= = =

34

66

7.7 First-Order Op Amp Circuits

An op amp circuit containing a storage element will exhibit first-order behavior

Differentiators and integrators are examples of first-order op amp circuits

For practical reasons, inductors are hardly used in op amp circuits; only RC type circuits are considered

The following three examples illustrate the location of the capacitor will influence the circuit behaviors; i.e. located in the input, output, or the feedback loop

67

Example 7.14 (source-free op amp circuit)

For the op amp circuit in the figure, find vo for t > 0, given that v(0)= 3V. Let Rf = 80 k, R1 = 20 kand C = 5 F.

Sol: Method 1:

1

1

11

/ 100 1

00

3 6 10 100

0Applying at 1 ,

0

( ) , , ( ) 3Applying at 2

0

80 10 5 10 ( 30 ) 12 , 0

t t

ff

t t

v dvKCL node CR dt

dv vv vdt CR

v t V e R C v t eKCL node

dv v dvC or v R Cdt R dt

v e e V t

=

= + == = =

= =

= = >

Q

35

68

Example 7.14

Method 2:

0

/0 0 0 0

10 10

Applying at 23 0 (0 ) 0, (0 ) 12

20,000 80000Applying to the input loop20,000(1) 0 20

2010.1. ,

( ) ( ) [ (0) ( )]

0 (12 0) 12 , 0

o

eq

eq

t

t t

KCL nodev v V

KVLv v kV

vThen R k

and R C Thus

v t v v v ee e V t

++

+ = =

= == =

= == + = + = >

69


For the op amp circuit in the figure, find vo for t > 0, if v(0) = 4V. Assume that Rf = 50 k, R1 = 10 k and C = 10 F.

Ans: -4e-2t V, t > 0

36

70

Example 7.15

Determine v(t) and vo(t) in the circuit. Sol:

/

3 6

1

0 1

1 0

( ) ( ) [ (0) ( )] , 050 10 10 0.05

0, (0) 0200, 3 2

20 1050( ) (1 ) 3.5 2 720

So that ( ) 2 7 5

tv t v v v e tRC

For t v

For t v V

v v V

v v vv V

= + >

= = =< => = =+

= + = = =

= =

71

Example 7.15

Practice Problem 7.15 Find v(t) and vo(t) in the op amp circuit.

Ans: 40(1-e-10t) mV, 40(e-10t -1) mV

20 20

200 1

Substituting( ) 5 [0 ( 5)] 5( 1) , 0

We obtain ( ) ( ) ( ) 7 5 , 0

t t

t

v t e e V tv t v t v t e V t

= + = >

= = >

37

72

Example 7.16

Find the step response vo(t) for t > 0 in the op amp circuit. Let vi = 2u(t) V, R1 = 20 k,

R2 = R3 = 10 k, C = 2 F. Sol:

1

3 3

2 3 2 3 1

2 32 3 0

2 3

To find Thevenin voltage

To find Thevenin resistance

( 0)

fab i

fTh ab i

Th

RV v

R

RR RV V vR R R R R

R RR R R RR R

=

= = + +

= = =+

73

Example 7.16

3

2 3 1

2 3

2 3/

03 6

1000

10 50 2 ( ) 2.5 ( )20 20

5

( ) 2.5(1 ) ( )

where 5 10 2 10 0.01Thus the step response

( ) 2.5( 1) ( )

fTh i

Th

t

Th

t

RRV v u t u tR R RR RR k

R Rv t e u t

R C

v t e u t V

= = = += = +=

= = =

=

38

74


Obtain the step response vo(t) in the op amp circuit. Let vi = 2u(t) V, R1 = 20 k, Rf = 40 k, R2 = R3 = 10 k, C = 2 F.

Ans: 6(1 - e-50t)u(t) V

75

7.9 Transient Analysis with PSpice

Transient response is the temporary response of the circuit thatsoon disappears

PSpice can be applied to obtain the transient response of a circuit with storage elements

If necessary, dc PSpice analysis is first carried out to determine the initial conditions

The initial conditions are used in the transient PSpice analysis to obtain the transient responses

It is recommended that during dc analysis, all capacitors are open-circuited while all inductors are short-circuited

39

76

Example 7.17

Use Pspice to find the response i(t) for t > 0 in the circuit.

77


For the circuit, use Pspice to find v(t) for t > 0.

40

78

Example 7.18

In the circuit, use Pspice to find v(t) for t > 0.

Circuit for t > 0 Simplified circuit

( ) 10 18 tv t e V=

79

Example 7.18

Method 1 (dc transient analysis)

dc analysis

transient analysis

41

80

Example 7.18

Method 2 (simulate directly)

81

Practice problem 7.18

The switch was open for a long time but closed at t = 0. If i(0) = 10A, find i(t) for t > 0 by hand and also by Pspice.

Ans: i(t) = 6 + 4e-5t

42

82

7.9 Applications

Devices making use of RC or RL circuits include filtering in dc power supplies, smoothing circuits in digital communications, differentiators, integrators, delay circuits, and relay circuits

The short or long time constants of the RC or RL circuits are applied in the devices

In the following, delay circuits, photoflash unit, relay circuits and automobile ignition circuit are introduced

83

7.9.1 Delay Circuits

An RC circuit can be used to provide various time delays

When the switch is closed, the capacitor voltage increase gradually toward 110V at a rate determined by the circuits time constant,(R1+R2)C

The lamp will not emit light until the voltage across it exceed 70V

When the voltage level is reached, the lamp fires and the capacitor discharge through it

43

84

Delay Circuits

Due to the low resistance of the lamp when on, the capacitor voltage drops fast and the lamp turns off

Therefore, the lamp will turn on and off repeatedly Adjusting R2, we can introduce either short or long time delay The warning blinkers commonly found on road construction

sites are the application examples

85

Example 7.19

Consider the circuit in the figure, and assume that R1 = 1.5 M, 0 < R2 < 2.5 k. (a) Calculate the extreme limits of the time constant of the circuit. (b) How long does it take for the first time after the switch is closed? Let R2 assume its largest value.

Sol: (a)

26 6

1 2

26 6

1 2

The samllest value for is 0

( ) (1.5 10 0) 0.1 10 0.15The largest value for is 2.5

( ) (1.5 2.5) 10 0.1 10 0.4

RR R C s

R MR R C s

= + = + =

= + = + =

44

86

Example 7.19

(b)

0 0

0 0

/ /

/ /

/ /

0

(0) 0, ( ) 0

( ) ( ) [ (0) ( )] 110[1 ]0.4 , 70

770 110[1 ] 111

4 1111 4

Taking the natural logarithm of both sides11ln 0.4ln 2.75 04

c ct t

c c c c

c

t t

t t

v vv t v v v e ewhere s the lamp glows when v V

e e

or e e

t

= == + =

= == =

= =

= = = .4046s

87

Example 7.19

Practice Problem 7.19 The RC circuit is designed to

operate an alarm which activateswhen the current through it exceeds

120A. If 0 R 6k, find the range of the time delay that the variable resistor can create.

Ans: Between 47.23ms and 124ms

0 00

0

0

(0) ( )A more general formula for finding is ln( ) ( )

The lamp will fire repeatedly every seconds if and only if ( ) ( )

v vt tv t v

tv t v

=

45

88

7.9.2 Photoflash Unit

This application exploits the ability of the capacitor to oppose any abrupt change in voltage

The charging time is approximately tcharge = 5R1C(when the switch is in position 1)

The discharging time is approximately tdischarge = 5R2C (when the switch is in position 2)

89

Photoflash Unit

This circuit provides a short-duration, high-current pulse

11

sVIR

=

22

sVIR

=

46

90

Example 7.20

An electron flashing has a current-limiting 6k resistor and 2000F electrolytic capacitor to 240V. If the lamp resistance is 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, and (e) the average power dissipated by the lamp.

Sol: (a) The peak charging current is

(b)

(c) The peak discharging current is

(d) The energy stored is

1 31

240 406 10

sVI mAR

= = =3 6

charge 15 5 6 10 2000 10 60 1 minutet R C s= = = =

22

240 2012

sVI AR

= = =2 6 21 1 2000 10 240 57.6

2 2sW CV J= = =

91

Example 7.20

(e)

Practice problem 7.20 The flash unit of a camera has a 2mF capacitor charged to 80V.

(a) How much charge is on the capacitor?(b) What is the energy stored in the capacitor?(c) If the flash fires in 0.8ms, what is the average current through the flashtube?(d) How much power is delivered to the flashtube?(e) After a picture has been taken, the capacitor needs to be recharged by a power unit that supplies as maximum of 5mA. How much time does it take to charge the capacitor?

Ans: (a) 0.16C, (b) 6.4J, (c) 200A, (d) 8kW, (e) 32s

6discharge 2

discharge

5 5 12 2000 10 0.1257.6 4800.12

t R C sWp W

t

= = == = =

47

92

7.9.3 Relay Circuits

A magnetically controlled switch is called a relay

The coil current gradually increases and produces a magnetic field

Eventually the magnetic field is sufficiently strong to pull the movable contact in the other circuit and close switch

At this time the relay is said to be pulled in The time interval td between the closure of

switch S1 and S2 is called the relay delay time

93

Example 7.21

The coil of a certain relay is operated by a 12V battery. If the coil has a resistance of 150 and an inductance of 30mH and the current needed to pull in is 50mA, calculate the relay delay time.

Sol:

/

3

/

The current through the coil is given by( ) ( ) [ (0) ( )]

12where (0) 0, ( ) 80 ,150

30 10and 0.2150

Thus ( ) 80(1 )If ( ) 50 , then

t

t

d

i t i i i e

i i mA

L msR

i t e mAi t mA

= + = = =

= = ==

=

48

94

Example 7.21

Practice problem 7.21 A relay has a resistance of 200 and an inductance of 500mH. The relay

contacts close when the current through the coil reaches 350mA. What time elapses between the application of 110V to the coil and contact closure?

Ans: 2.529ms

/ /

/ /

550 80(1 ) 18

3 88 3

By taking the natural logarithm on both sides, we get8 8ln 0.2ln 0.19623 3

d d

d d

t t

t t

d

e e

or e e

t ms ms

= =

= =

= = =

95

7.9.4 Automobile Ignition Circuit

The ability of inductors to oppose rapid change in current makes them useful for arc or spark generation

To form a spark requires thousands of volts between the air gap

How can such a high voltage be obtained from the car battery? Since the voltage across the inductor is v = Ldi/dt , we can make di/dt

large by creating a large change in current through the inductor

When the ignition switch is closed, the final current value isi = Vs / R

The time taken to charge is charge 5 5LtR

= =

49

96

Automobile Ignition Circuit

At steady state, di/dt = 0, therefore, inductor voltage v = 0 When the switch suddenly opens, a large voltage is developed

across the inductor causing a spark The induced large voltage is due to the rapidly collapsing field The spark continues until the energy stored in the inductor is

dissipated in the spark discharge

97

Example 7.22

A solenoid with resistance 4 and inductance 6mH is used in an automobile ignition circuit. If the battery supplies 12V, determine: the final current through the solenoid when the switch is closed, the energy stored in the coil, and the voltage across the air gap, assuming that the switch takes 1s to open.

Sol:

2 3 2

36

The final current through the coil is12 34

The energy stored in the coil is1 1 6 10 3 272 2

The voltage across the gap is36 10 18

1 10

sVI AR

W LI mJ

IV L kVt

= = =

= = =

= = =

50

98

Example 7.22

The spark coil of an automobile ignition system has a 20mH inductance and a 5 resistance. With a supply voltage of 12V, calculate: the time needed for the coil to fully charge, the energy stored in the coil, and the voltage developed at the spark gap if the switch opens in 2s.

Ans: 20ms, 57.6mJ, and 24kV

99

Assignments

7.3, 7.7, 7.16, 7.19, 7.26, 7.39, 7.43, 7.49, 7.56, 7.66, 7.73

circuitry ch07 first-order circuits

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