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S.55Electronic Circuit Analysis (April/May-2012, Set-4) JNTU-Hyderabad

( JNTU-Hyderabad ) B.Tech. II-Year II-Sem.

S e t - 4S o l u t i o n s

Code No: R09220402/R09

II B.Tech. II Semester Examinations

April/May - 2012ELECTRONIC CIRCUIT ANALYSIS

(Common to ICE, E.COMP.E, ETM, EIE, ECE)

Time: 3 Hours Max. Marks: 75

Answer any FIVE questions

All questions carry equal marks

- - -

1. (a) Define f and fT and also establish the relationship between f and fT. (Unit-III, Topic No. 3.5)

(b) Derive the expression for the CE short-circuit current gain as a function of frequency. [7+8]

(Unit-III, Topic No. 3.5)

2. For the circuit shown in figure, show that,

(a) (AVS

)max

=f

i o r f

h

h h h hif R

L= and R

S= 0

(b) Ri=

i o r f

o

h h h h

hif R

L=.[8+7] (Unit-I, Topic No. 1.3)

hr

V2

V1VS+

+

+

RS hi

V2

IL

ZL

I2 2

hfI1h

o

2'1'

1

hr

V2

V1VS+

+

+

RS hi

V2

IL

ZL

I2 2

hfI1h

o

2'1'

1

Figure

3. (a) Sketch the circuit of a CS amplifier. Derive the expression for the voltage gain at low frequencies. What is the

maximum value of voltage gain? (Unit-IV, Topic No. 4.5)

(b) The FET shown in figure has the following parameters,

IDSS

= 5.6 mA and VP= 4 V. If V

i= 10 V find V

o. [8+7] (Unit-IV, Topic No. 4.3)

+

+24 V

4.7 k

10 kVi

+

Vo

12 V

+

+24 V

4.7 k

10 kVi

+

Vo

12 V

Figure


2/11

S.56 Spectrum ALL-IN-ONE Journal for Engineering Students, 2013

B.Tech. II-Year II-Sem. ( JNTU-Hyderabad)

4. (a) Draw the equivalent circuit of a single tuned capacitive coupled amplifier and derive the expression for gain at

resonance. (Unit-VIII, Topic No. 8.3)

(b) Draw the circuit diagram for tuned RF amplifier and explain its working. [7+8] (Unit-VIII, Topic No. 8.1)

5. (a) Derive an expression for the output power of a class A large-signal amplifier in terms of Vmax

, Vmin

, Imax

and Imin

.

(Unit-VII, Topic No. 7.2)

(b) For a particular power amplifier, the optimum load impedance is 180 . Calculate the turns ratio required tomatch an 8 load to this transistor. If the amplifier takes a mean collector current of 2 A from a 15 V supply anddelivers an A.C load power of 2.5 W to the transformer coupled-load, calculate the efficiency and the collector

dissipation (neglecting the losses). [7+8] (Unit-VII, Topic No. 7.2)

6. (a) Discuss about the types of negative feedback amplifiers giving the effect of each type of feedback on the

parameters of the amplifier. (Unit-V, Topic No. 5.2)

(b) What sort of feedback is employed in a CE amplifier with unbypassed emitter resistor? Discuss its analysis in

detail. [7+8] (Unit-V, Topic No. 5.5)

7. Design a RC phase-shift oscillator to operate at a frequency of 5 kHz. Use a MOSFET with = 55 and rd= 5.5 k. The

phase-shift network not load the amplifier.

(a) Find the minimum value of the drain-circuit resistance for which the circuit will oscillate.

(b) Choose reasonable values of R and find C. [8+7] (Unit-VI, Topic No. 6.2)

8. (a) Write the expressions for overall voltage gain nV(A ) ,

nLf and

n

Hf when n-non-identical amplifier stages are

cascaded. (Unit-II, Topic No. 2.1)

(b) Compute f H

and fL

a 2-stage amplifier if1H

f = 3 kHz and2H

f = 4 kHz,1L

f = 300 Hz and2L

f = 600 Hz. [15]

(Unit-II, Topic No. 2.1)


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Q1. (a) Define f and fT and also establish the relationship between f and fT.

Answer : April/May-12, Set-4, Q1(a) M[7]

For answer refer Unit-III, Q21, Topics: cut-off frequency,f, Current Gain Bandwidth Product,fT.

(b) Derive the expression for the CE short-circuit current gain as a function of frequency.

Answer : April/May-12, Set-4, Q1(b) M[8]

For answer refer Unit-III, Q17.

Q2. (a) For the circuit shown in figure, show that,

(a) (AVS

)max

= f

i o r f

h

hh -h hif R

L= and R

S= 0

(b) Ri=

i o r f

o

h h -h h

hR

L= .

hrV

2V

1V

S

+

+

+

RS

hi

V2

IL

ZL

I2

2

hfI

1

ho

2'1'

1

hrV

2V

1V

S

+

+

+

RS

hi

V2

IL

ZL

I2

2

hfI

1

ho

2'1'

1

Figure

Answer : April/May-12, Set-4, Q2 M[8+7]

Given that,

hrV

2V

1V

S

+

+

+

RS

hi

V2

IL

ZL

I2

2

hfI

1

ho

2'1'

1

hrV

2V

1V

S

+

+

+

RS

hi

V2

IL

ZL

I2

2

hfI

1

ho

2'1'

1

Figure

(a) (Avs

)max

=f

i o r f

h

h h h h

if RL = and RS = 0

To prove that (AVs

)max

=froi

f

hhhh

h

at R

L= andR

S= 0, consider the overall voltage gain expression

including source.

The expression for voltage gain including source from figure is given as,

AVS

=V i

i s

A R

Z R+... (1)

SOLUTIONS TO APRIL/MAY-2012, SET-4, QP


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By substitutingRS

= 0 in equation (1), we get,

AVS

=V i

i

A R

Z

[Q Ri=Zi]

=V i

i

A R

R

=AV

... (2)

Since,AV

=)( rfoiLi

Lf

hhhhRh

Rh

+

... (3)

By substituting equation (3) in equation (2), we get,

AVS

=)( rfoiLi

Lf

hhhhRh

Rh

+

=

+

)( rfoiL

iL

Lf

hhhhR

hR

Rh

=

+

)( rfoiL

i

f

hhhhR

h

h... (4)

SubstitutingRL

= in equation (4),

AVS =froi

i

f

hhhhh

h

+

= 01

Q

=froi

f

hhhh

h

+

0

max( )f

VS

i o r f

hA

h h h h

=

(b) Ri= i o r f

o

h h h h

h

if R

L=

The expression for input impedance from figure is

given as,

Ri= h

i

1

f r

oL

h h

hR

+... (5)

SubstitutingRL

= in equation (5), we get,

Ri= h

i

1

f r

o

h h

h +

= hi

0

f r

o

h h

h +

= hi

f r

o

h h

h

i o f r

i

o

h h h hR

h

=

Q3. (a) Sketch the circuit of a CS amplifier.Derive the expression for the voltagegain at low frequencies. What is themaximum value of voltage gain?


A common source amplifier is as shown in figure (1).

+

Vi V0

+

RG

S

RS C

S

D

RD

VDD

G

+

Vi V0

+

RG

S

RS C

S

D

RD

VDD

G

Figure (1): Common Source Amplifier

The small signal equivalent circuit of common source

amplifier is shown in figure (2).

+

Vi Vgs

G

RG

+ Vgs

RD

+

V0

D id

S

+

Vi Vgs

G

RG

+ Vgs

RD

+

V0

D id

S

Figure (2)

Voltage Gain

The capacitor Cs

is used as a bypass for source

current. Because ofCs,R

sis removed.

Applying KVL to source-drain loop, we get,

D

o

R

V+

Dd

gs

Rr

V

+

= 0

Vo =dD

D

rR

R

+

Vgs


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IfVgs

= Vi

o D

V

i D d

V RA

V R r

= =

+The maximum value of voltage gain is nearly equal to

maximum theoretical gain (i.e.,) from FET amplifier circuits.

(b) The FET shown in figure has thefollowing parameters,

IDSS

= 5.6 mA and VP

= 4 V. If Vi= 10 V

find Vo.

+

+24 V

4.7 k

10 kVi

+

Vo

12 V

+

+24 V

4.7 k

10 kVi

+

Vo

12 V

Figure


Given that,

For a Field Effect Transistor (FET),

IDSS

= 5.6 mA

VP

= 4 V

Vi

= 10 V

Vo

= ?

+

+24 V

4.7 k

10 kVi

+

Vo or VDS

12 V

VDD

Rd

Id

D

S

G

RsVGS

VSS

+

+24 V

4.7 k

10 kVi

+

Vo or VDS

12 V

VDD

Rd

Id

D

S

G

RsVGS

VSS

Figure

From figure,

VDD

= + 24 V

VSS = 12 V

Rd

= 4.7 k

RS

= 10 k

Vi

= VGS

= 10 V

From figure, VGS

is expressed as,

VGS

= ID

RS

ID

=S

GS

R

V

= 31010

10

= 1 103

1mADI =

The output voltage Vo

can be obtained by applying

KVL (Kirchoffs Voltage Law) at the output terminal,

i.e., VDS

Vo= V

DDI

D(R

S+R

D) V

SS

= 24 ( 1) 103(4.7 103+ 10 103) ( 12)

= 24 + 103(4.7 + 10) 103 + 12

= 36 + 103+3 (14.7)

= 36 + (1) (14.7)

VDS

= 50.7 V

50.7VoV =

Q4. (a) Draw the equivalent circuit of a singletuned capacitive coupled amplifier andderive the expression for gain atresonance.


For answer refer Unit-VIII, Q6.

(b) Draw the circuit diagram for tuned RFamplifier and explain its working.


Tuned RF Amplifier

Amplifier which are used to select and amplify a

particular range of high frequencies or radio frequencies are

known as tuned RF amplifier.


6/11



The circuit diagram for tuned RF amplifier is as shown

in figure,

L OutputR1

Cin

L

R2 REInput

CE

+VCC

C L OutputR1

Cin

L

R2 REInput

CE

+VCC

L OutputR1

Cin

L

R2 REInput

CE

+VCC

C

Figure: Tuned RF Amplifier

It consists of a parallel tuned circuit at the collector

end and two resistorsR1

andR2

connected in parallel with

parallel tuned circuit and emitter resistance respectively. The

input and output capacitors are used as a tuning dial which

tune to a desired frequency by rejecting all the unwanted

frequencies.

Tunning can be accomplished by varying the

capacitance and inductance values in the parallel tunedcircuit. The impedance of the tuned circuit varies with

varying frequency. At resonant frequency, the circuit offers

a very high impedance whereas low impedance is achieved

at other frequencies. If the frequency of signal applied to

the tuned circuit is identical to the frequency of LC circuit,

larger amplification of signal takes place in the amplifiers. In

contrast to this, when various signals with different

frequencies are applied, the circuit selects signals with

frequencies lying between the range of resonant frequency

and rejects all the other frequencies.

Applications

It has wide range of applications in various areas.

For example, in electronic devices such as television

receivers, radio receivers and radar systems.

Q5. (a) Derive an expression for the outputpower of a class A large-signal amplifierin terms of V

max, V

min, I

maxand I

min.


For answer refer Unit-VII, Q2.

(b) For a particular power amplifier, theoptimum load impedance is 180 .Calculate the turns ratio required to

match an 8 load to this transistor. Ifthe amplifier takes a mean collectorcurrent of 2 A from a 15 V supply anddelivers an A.C load power of 2.5 W tothe transformer coupled-load, calculatethe efficiency and the collectordissipation (neglecting the losses).


Given that,

For a power amplifier,

Optimum load impedance, 'LR = 180

Load resistance,RL

= 8Mean collector current,I

CQ= 2 A

Power supply, VCC

= 15 V

A.C load power, Pac

= 2.5 W

(i) Turns ratio, n = ?

(ii) Efficiency, = ?(iii) Power dissipation, P

D= ?

(i) The expression for turns ratio of a transformer is given as,

'

2

L

L

R

Rn =

... (1)

By substituting given values in equation (1),

We get ,

n2 =180

8

n2 = 0.044

n = 044.0

21.0=n

(ii) The efficiency of a power amplifier is expressed as,

DS

ac

P

P= ... (2)

Where,

D.Cinput power PD.C

is given as,

PDC

=ICQ

. VCC

PDC

= 2 15 [Q ICQ= 2 A, VCC= 15 A]

30 WDCP = ... (3)


7/11



On substituting PDC

and Pac

in equation (2), we get,

=30

5.2= 0.083

%3.8=

(iii) Collector power dissipation, PD

is obtained as,

PD

= PDC

Pac

= 30 2.5

= 27.5

27.5 WDP =

Q6. (a) Discuss about the types of negative feedback amplifiers giving the effect of each type of

feedback on the parameters of the amplifier.Answer : April/May-12, Set-4, Q6(a) M[7]

Types of Negative Feedback Amplifier

For answer refer Unit-V, Q5.

Effect of Each Type of Feedback on the Parameters of the Amplifier

The effect of each type of feedback on the parameters of the amplifier is illustrated in table (1),

Types of Parameters

Negative Feedback Improved Desensitises Bandwidth Non Linear Noise Overall

Amplifier Characteristics Distortion Transmission Gain

1. Voltage- Voltage AV Increases Decreases Decreases Decreasesseries amplifier

feedback

amplifier

2. Current Transcon-

-series ductance Gm

Increases Decreases Decreases Decreases

feedback amplifier

amplifier

3. Current Current

shunt amplifier Ai


feedback

amplifier

4. Voltage- Transresis-

shunt -tance Rm


feedback amplifier

amplifier

(b) What sort of feedback is employed in a CE amplifier with unbypassed emitter resistor? Discussits analysis in detail.


The type of feedback which is employed in common emitter amplifier with unbypassed emitter resistor is current-

series feedback.

For remaining answer refer Unit-V, Q25.


8/11



Q7. Design a RC phase-shift oscillator to operate at a frequency of 5 kHz. Use a MOSFET with = 55and r

d= 5.5 k. The phase-shift network not load the amplifier.

(a) Find the minimum value of the drain-circuit resistance for which the circuit will oscillate.

(b) Choose reasonable values of R and find C.

Answer : April/May-12, Set-4, Q7 M[8+7]

Given that,

For a phase shift oscillator using MOSFET,

Frequency of oscillation,fr= 5 kHz

Amplification factor,= 55

Drain resistance, rd= 5.5 k

(a) Minimum value ofRD

for which circuit oscillates = ?

(b) For the oscillator to work satisfactorily, reasonable values ofR = ?, C= ?

The circuit arrangement of a phase shift oscillator using MOSFET is as shown in figure,

G

R RVi

+

R

Vo

C C C

CS

VDD

RD

D

S

RS

G

R RVi

+

R

Vo

C C C

CS

VDD

RD

D

S

RS

Figure

(a) Minimum Value of RD

In the phase shift oscillator, the FET is required to have a gain greater than 29 i.e.,A > 29 for obtaining |AB| = 1

Consider |A| = 40

Then, the expression for open loop gain of phase shift oscillator using MOSFET is given as,

|A| =dD

Ddm

rR

Rrg

+

|A| =

dD

D

rR

R

+

[Q = gmrd]

|A|RD

+ |A|rd

=RD

( |A|) RD

= |A|rd


9/11



RD

=A

rA d

=4055

105.5403

=15

102203

= k6.14DR

(b) R and C Values

For the proper and satisfactory operation of the circuit,

Consider, C= 0.01F

Since, the frequency of oscillation of a phase shift oscillator using MOS is expressed as,

fr =62

1

RC

R =62

1

rCf... (1)

On substituting values offr

and Cin equation (1), we get,

R =61051001.014.32

136

= 310769.0

1

= 1300

= k3.1R

Q8. (a) Write the expressions for overall voltage gain )(A ''V ,''

Lf and''

Hf when n-non-identical amplifier

stages are cascaded.

Answer : April/May-12, Set-4, Q8(a)

Overall Voltage Gain )(nvA

The expression for overall voltage gain ( )( )nVA , when n-non-identical amplifier stages are cascaded is given as,

( )nVA =Av1Av2 ... (1)

It can be observed from equation (1) that, over all voltage gain is the product of voltage gain of each amplifier stage.

Over All Lower 3dB Frequency (n)Lf

The lower 3 dB frequency )(nLf can be expressed as,

2)(

f

f nL=

2

1

f

fL+

2

2

f

fL+ ...

...22)(

21++= LL

nL fff ... (2)


10/11



Over All Upper 3dB Frequency ( )(n)HfThe upper 3 dB frequency )(n

Hf can be expressed as,

...11

1

22

)(

21

++=

HH

nH

ff

f

... (3)

(b) Compute fH

and fL

a 2-stage amplifier if1H

f = 3 kHz and2H

f = 4 kH;1L

f = 300 Hz and2L

f = 600

Hz.

Answer : April/May-12, Set-4, Q8(b)

Given that,

For a multistage amplifier,

n = 2

1Hf = 3 kHz

2Hf = 4 kHz

1L

f = 300 Hz

2L

f = 600 Hz

fH

= ?

fL

= ?

The expression for lower frequencyfL

of a multistage amplifier is given as,

fL

= )(nLf 12

1

n ... (1)

Where,

)(nLf =

22

21 LLff +

= 22 )600()300( + = 36000090000+

)(nLf = 670.82 Hz. ... (2)

On substituting )(nLf and n values in equation (1), we get,

fL

= 670.82 1122 1

= 670.82 414.0 = 670.82 0.64

= 429.3 Hz

Hz429= Lf

The higher frequency (fH

) can be expressed as,

fH

=

121

)(

n

nHf

... (3)


11/11



Where,

)(n

Hf =22

21

11

1

HH ff+

=

2323)104(

1

)103(

1

1

+

=

16000000

1

9000000

1

1

+

= 41015.4

1

)(nHf = 2409.6 Hz

On substituting values of )(nHf and n in equation (3), we get,

fH

=

( )

1

2 1

nH

n

f

=

12

6.2409

21

=

41.0

2404

= 5863.4 Hz

kHz8.5= Hf