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Problem 17.29

The amplifier is operating at the sinusoidal amplifier for which the dissipation isa maximum verify that conversion efficiency is 50%.

+V CC

V i

- VCC

RL

U1

U2

The ideal class B push - pull amplifier in the figure is operating at the sinusoidalamplitude for which the dissipation is a maximum. Verify that the conversion

efficiency is 50 percent.



S olutiont

mi [YY s in!

Tr U1 conducts for positive half cycle & U2 conducts for negativehalf cycle

d.c current obtained from supply = Tm I 2

where L

mm

R

V I !

Output power = L

m

RV 2

2

Power supplied by dcCC CC I V V .!

Tm

dc

I I

2!

Collector dissipation L

m

L

mCC C

R

V

R

V V P

2

2 2

!T



For dissipation to be max

02

0 !! L

m

L

CC

m

C

RV

RV

d V

d P T

TCC

m

V V

2!

Output power = L

CC

RV

21.4

2

2

¹¹ º ¸

©©ª¨

T

Power from supply = L

CC

R

V 2

24

T

%501004

24

2

2

2

2

!v!

L

CC

L

CC

RV

RV

T

TL



Problem 17.3 0 .

In the circuit shown the base - emitter voltage may be assumed to

remain constant at the cut-in value V K for all values of forward bias.The biasing voltage is idealized by two batteries of voltage kV K,

where 0 < k e 1 . Assume that Yi = V s sin [ t.

+V CC

Vi

kV K

kV K

- VCC

Vo

t V V si [s in!



(i) For k = 0, 0.5 & 1 plot output as a sanction of time(Vs= 1 V) (VK = .6V)

(ii) What happens to distortion when V s is increased

(iii) What happens if k exceeds unity

(iv) What happens if k > 1 & a resister is added between two emitters.

(v) What is the class of operation.

0s in 0 !V V kV t V s KK[

1s in0 ! k V t V V s K[

on ±ve half cycle

1s in0 ! k V t V V s K[



Cut in angle

01s in !k V t V s K[

sV

V k t K[ 1s in !

¼

½

»¬

-

«!! s

incu t V

V k t Q K[ 1s in 1

0!k

5.0!k

1!k

01 88.3616.

s in !¹ º ¸

©ª¨!incu t U

01 45.1713.

s in !¹ º ¸

©ª¨!incu t U

01 00s in !!incu t U



Vo

Vo

08.36

05.17

K = 0

K = .5

K = 1



( b) as V s is increased, amplitude of current harmonics increases

The distortion also increases.

(c ) For k > 1

then the output voltage is not zero and is given by

0!oV V kV KK

KV k V o 1!(d) if a resistor is placed between the emitters, then the value of output can be reduced to zero by changing the value of Resister.

(e) it is a class AB operation.



Problem 16.4 0

The gyrator shown is used to simulate an inductance.

(a) Show that Z R R I V i

i 21!

( b) For any resistance between . 1 to k ; what is the range of

inductances possible.

R 1

R 1

I i

Vi+

-

Vi

R 3

R 3

V1 2R

2

Z

Vo

-

+

i V V 21 !



Thus

11

1

RV V

RV V

I oi i i !

111

2 RV

RV

RV V oi i i !

1 R

V I o

i !

at 2 nd op AM P

1

22V

R

Z V o !

i i V R

Z R I 2.

2 21 !

Z

R R

I

V

i

i 21!



When

Z is a capacitor µC¶

Then

2121

1RC R j

C

j

R R Z [

[!!

Thus21 RC R L !

largest possible inductance for R lying between . 1 & 1 0 k ;

0 1.min ! L

100max ! L

for < = 1 F



Problem 16.41

An alternate form of the gyrator is shown. Show that V i/Ii isinductive. Assume ideal O p - Amps.

Find the input impedance of the circuit

Vi +

-

I i

C

R 1

R 2

+

-

Vo



1 RV V

I oi i !

01

2

! R

V V

S C

V oi i

01

2

1 ! R

I R

S C

V i i

i i I R R

S C V 2

1!

S C R

R

I

V

i

i

.2

1!

C j R R

[2

1!C R

R j

[2

1! ¼½»

¬-«!

C R R

j 2

21

[[



Problem 13.6

1000/1100/11102 33

s s s s

sT v!

The low - frequency return ratio of a feedback amplifier is given by

(a)Is the closed - loop amplifier stable?

( b)What are G.M & P.M

S olution

Magnitude plot

? A32log20102log20 3 !v? A330 10.20!

dB54!



(i) due to constant a st.line at - 54 dB

(ii) due to term s 3, a str. line of slope 60 dB passing through [ = 1

(iii) after [ = 1 slope will be 40 dB

after [ = 1 00 slope will be 2 0 dB

after [ = 1 000 slope will be 0 dB

10![at

dB j T 14206054log20 ![100![

dB sT 264014log20 !!1000![

dB sT 462026log20 !!



Problem 13.7

The return ratio is given by

2

765

5.6

10

1

10

1

10

1

101

¼½

»¬-

«¼½

»¬-

«¼½

»¬-

«

¼½»

¬-«

! s s s

sT

sT o

(a)Determine the largest value of T o for which the amplifier is stable

( b) What value of T o results in GM = 1 0dB?

(c) What value of T o results in PM = 45 0



S olution

(a) From Bode plot

48log20 10 !oT

4.2log 10 !oT

18.251!oT

( b) 1048log2010 !oT

38log2010 !oT

9.1log 10 !oT

43.79!oT

(c) For a phase margin of 45 0

dBT o 30!



Problem 13.1 0

Assume that after pole - zero compensator is added

¹ º ¸©

ª¨

v¹ º ¸©

ª¨

v!

6

4

104011

102 s

A s sT

[

Determine [ A , so that 045! M J

S olutionFor dominant pole

1010 2[

[ e A

1002

[[ eA

100

1040 6veA[

5104 ve A[



sT !J

¹ º

¸©ª

¨

v¹ º

¸©ª

¨

v!

7

1

5

1

104tan

104tan

[[

5104 v![

6

10

4 v![7104 v![

8104 v![

00057.4557.45 !!J

09.897.528.84 !!J

04.1344542.89 !!J

1.1742.849.89 !!J

61040 v!G [

51 104 v![and

problem slide

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