Problem Slide4

8/3/2019 Problem Slide4

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Problem 16.40

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

Ii

Vi

+

-

Vi

R 3

R 3

V12R

2

Z

Vo

-

+

i V V 21!



Thus

11

1

R

V V

R

V V I oi i

i

!

111

2

R

V

R

V

R

V V oi i i

!

1 R

V I o

i

!

at 2nd op AMP

1

22

V

R

Z V o !

i i V R

Z R I 2.

22

1!

Z

R R

I

V

i

i 21!



When

Z is a capacitor µC¶

Then

21

21

1RCR j

C

j

R R Z [

[

!!

Thus21

RCR L !

largest possible inductance for R lying between .1 & 10 k ;

01.min

! L

100max

! L

for < = 1 F



Problem 16.40

The gyrator shown is used to simulate an inductance.

(a) Show that Z R R

I V

i

i 21!

(b) Assuming resistance range between 0.1 to10k ;, what is

the range of inductances possible.

R 1

R 1

Ii

Vi

+

-

Vi

R 3

R 3

zVi2R

2

Z

Vo

-

+

Vi



iov

R

Z v 2

2 2

!

iv

R

Z

2

!

i

i

ii I R

R

Z v

R

vv

!

¼

½

»¬

-

«

@

1

2

1

1

2

iiI

R R

Z

R Rv !¼

½

»¬-

«

2111

11

Z

R R

I

v

i

i 21! Hence proved

If resistance lies between 0.1 & 10K ; then the range of inductance

= 0.01 to 100. Ans.



Problem 16.41

An alternate form of the gyrator is shown. Show that Vi/Ii is

inductive. Assume ideal Op-Amps.

Find the input impedance of the circuit

Vi

+

-

Ii

C

R 1

R 2

+

-

Vo



1 R

V V I oi

i

!

01

2

!

R

V V

SC

V oi i

01

2

1 ! R

I R

SC

V i i

i i I R

R SCV

2

1!

SC 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

2

1

[[



Problem 17.29

The amplifier is operating at the sinusoidal amplifier for which the dissipation is

a maximum verify that conversion efficiency is 50%.

+VCC

Vi

-VCC

RL

U1

U2

The ideal class B push-pull amplifier in the figure is operating at the sinusoidal

amplitude for which the dissipation is a maximum. Verify that the conversion

efficiency is 50 percent.



Solution

t mi [YY sin!

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

d.c current obtained from supply =T

m I 2

where

L

mm

R

V I !

Output power = L

m

R

V

2

2

Power supplied bydc

CC CC

I V V .!

T

mdc

I I

2!

Collector dissipation

L

m

L

mCC C

R

V

R

V V P

2

22

!

T



For dissipation to be max

0

2

0 !! L

m

L

CC

m

C

R

V

R

V

d V

d P

T

T

CC m

V V

2!

Output power = L

CC

R

V

2

1.

42

2

¹¹ º

¸©©ª

¨T

Power from supply =

L

CC

R

V 2

24

T

%501004

2

4

2

2

2

2

!v!

L

CC

L

CC

R

V

R

V

T

TL



The ideal class B push pull Amplifier is operating at the

sinusoidal amplitude for which the dissipation is maximum.Verify that the conversion efficiency is 50%.

Problem 17.29

+VCC

Vi

-VCC

R LQ2

Q1



t V V mi

[sin!

Q1 conducts for +Ve half cycles & Q2 conducts for ±Ve half cycles.

d.c. current obtained from supply =T

m I 2

L

m

m R

V I !

Input power = CC

L

mV

R

V .

2

T

Output power = L

m

R

V

2

2

Power supplied bydcCC CC I V V .!

T

m

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 maximum:-

0!m

C

dV

dP 02

! L

m

L

CC

R

V

R

V

T

T

CC

m

V V

2!

Output power L

CC

R

V

2

1.42

2

¹¹ º ¸©©

ª¨T

Power from supply =

L

CC

R

V

2

24

T

%5010042

4100

4

2

4

2

2

2

2

2

2

2

2

!vv!v!CC

L

L

CC

L

CC

L

CC

V

R

R

V

R

V

R

V

T

T

T

TL Ans.



Problem 17.30.

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

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

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

+VCC

Vi

kVK

kVK

-VCC

Vo

t V V si [sin!



(i) For k = 0, 0.5 & 1 plot output as a sanction of time

(Vs= 1V) (VK = .6V)

(ii)What happens to distortion when Vs 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.

0sin0! V V kV t V s KK

[

1sin0

! k V t V V s K[

on ±ve half cycle

1sin0

! k V t V V s K[



Cut in angle

01sin ! k V t V s K[

sV

V k t

K[ 1sin !

¼½

»¬-

«!!

s

incut V

V k t Q

K[ 1sin

1

0!k

5.0!k

1!k

0188.36

1

6.sin !¹

º

¸©ª

¨!

incut U

0145.17

1

3.sin !¹

º

¸©ª

¨!

incut U

0100sin !!

incut U



Vo

Vo

08.36

05.17

K = 0

K = .5

K = 1



(b) as Vs 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

K

V 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.

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