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7/28/2019 Jntu Hyd 2 2ece Eca Set 4
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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
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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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S.57Electronic Circuit Analysis (April/May-2012, Set-4) JNTU-Hyderabad
( JNTU-Hyderabad ) B.Tech. II-Year II-Sem.
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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S.58 Spectrum ALL-IN-ONE Journal for Engineering Students, 2013
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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?
Answer : April/May-12, Set-4, Q3(a) M[8]
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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S.59Electronic Circuit Analysis (April/May-2012, Set-4) JNTU-Hyderabad
( JNTU-Hyderabad ) B.Tech. II-Year II-Sem.
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
Answer : April/May-12, Set-4, Q3(b) M[7]
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.
Answer : April/May-12, Set-4, Q4(a) M[7]
For answer refer Unit-VIII, Q6.
(b) Draw the circuit diagram for tuned RFamplifier and explain its working.
Answer : April/May-12, Set-4, Q4(b) M[8]
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.
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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.
Answer : April/May-12, Set-4, Q5(a) M[7]
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).
Answer : April/May-12, Set-4, Q5(b) M[8]
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)
7/28/2019 Jntu Hyd 2 2ece Eca Set 4
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S.61Electronic Circuit Analysis (April/May-2012, Set-4) JNTU-Hyderabad
( JNTU-Hyderabad ) B.Tech. II-Year II-Sem.
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
Increases Decreases Decreases Decreases
feedback
amplifier
4. Voltage- Transresis-
shunt -tance Rm
Increases Decreases Decreases Decreases
feedback amplifier
amplifier
(b) What sort of feedback is employed in a CE amplifier with unbypassed emitter resistor? Discussits analysis in detail.
Answer : April/May-12, Set-4, Q6(b) M[8]
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.
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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
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S.63Electronic Circuit Analysis (April/May-2012, Set-4) JNTU-Hyderabad
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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)
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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)
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S.65Electronic Circuit Analysis (April/May-2012, Set-4) JNTU-Hyderabad
( JNTU-Hyderabad ) B.Tech. II-Year II-Sem.
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