GG

1. 4.11
Q21. A serles fed classA ampllfiler uses a supply
voltage of 10 V and load resistance of 20 W. The
A.C input voltage results in a base current of4
mA peak. Calculate,
() D.C Input power
() A.C output power
(1) %offlclency.
Ans:
Given that,
For a series fed class-A amplifier,
Supply voltage, Vcc 10V
Load resistance, R, =20 2
Base current, , = 4 mA
To find,
(i) D.C input power, Ppc "?
i) A.C output power,P4c?
(ii) Percentage efficiency, n =?
Assuming8
B 50 and
R=1 kQ
3) The D.C input power for a series fed class A amplifier
is obtained as,
()
PDC Vccx Ice (1)
Where,
4) coB./0 (2)
Ix10°
10-0.7
10
15)
9.3 x 10-
B9.3 mA
he
fier By substituting /ßo value in equation (2), we get,
c 5 0 x 9.3 x 1003
urs co465 mA
By substituting lco value in equation (1), we get,
'PDc= 10 x465 x 103
'. Ppc=4.65W
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2. ii) But
-B
50 x
4x 103 : Given/,= 4mA]
I= 200 mA
200x10
dtm 2
em141.42 mA =1,ms
r.m.s
Then, the A.C outputpowerfora series fed class A power
amplifier is obtained as,
PAC12 R,
r.m.s
=
(141.42 x
103)(20) =
0.4 W
PA.C 0.4W
(ii) Efficiency of a series fed class A power amplifier is
obtained as,
%nACx100
Ppc
0.4
x100 8.6
4.65
%n=8.6

3. Q22. A single ended class A amplifier has a
transformer coupled load of 8 2. If the
transformer turns ratio is 10, find the maximum
power output delivered to the load. Take the
zero signal collector current of 500 mA.
Ans:
Given that,
For a single ended transformer coupled class A
amplifier,
Loadresistor,R, =8Q
N 1
Np
Turns ratio of transformer, n i.e.,
10
Collector current value at Qpoint, I- 500 mA
Maximum output powerat load, Plima?
The expression for maximum output power at load is,
VPe
8RL
PL(max) . (1)
=
Where,
-Peak-to-peak load voltage
NsPp
V= xP,
PP
(2)
Np
PP

4. RONIC IRCUIT ANALYSIS IJNTU-HYDERABADI
Where,
P-Output complianceofthe transformer-coupled
amplifier
..3)
P,-2eo
For a transformer coupled amplifier, r, equals the
impedance 2, on primary of the transformer
i.e.,r =2,
But the impedance on primary of the transformer, Z, is
obtained as,
- .(4)
RL
N
[: 2,-R
=
(10)P x 8
2800
8002
On substituting the correspondintg values in equation (3),
we get,
P =2 x 500 x 103x 800
P, =
800 V
.(5)
N
On substituting the values of P, and in equation,
NP
(2), we get,
1X800
10
PP
8 0 V .. (6)
On substituting equation (6) in equation (1), we get,
(80) 6400
Pua -
8x8 64
= 100
Pmax)= 100W
Q23, A transf

5. a23. A transformer-coupled
class-A
amplifie
supplies 2 W of power to speaker. f the supph
voltage is 36 V and Ico is 150 mA, find the
efficiency of the circuit.
Ans:
Given that,
For a transformer coupled class-A amplifier,
Power delivered to the load (speaker), P.,=2*
Supply voltage, Ve 36 V
D.C collector current, Ico 150 mA
Efficiency ofcollector circuit, n
=
?
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6. ignal Amplifiers and uned Amplifiers)
UNIT-4 (Large Signa
The
efficiency of collector circuit (n) for a transformer
Power delivered to the load
c o u p l e d
class-A
amplifier. is given by,
D.C power supplied to the output circuit
Po
nPae
. (1)
The D.C power (Pd) can be calculated as,
PaeVccxco .(2)
By substituting the values ofVccandIcoin equation (2),
we get,
Pa 36 x 150 x
10
P5.4 W
Bysubstituting the values ofP.,and P, in equation (1),
a.c
we get,
2
n 5.4
n0.37
Percentage efficiency is given by.
on= 37%

7. 024, For a particular power amplifier, the óptimum
load Impedance is 180 . Calculate the turns
ratio required to match an 8 2 load to this
transistor. Ifthe amplifiertakes a mean collector
current of 2 A from a 15 V supply and delivers
an A.C load power of 2.5 W to the transformer
coupled-load, calculate the efficiency and the
collector dissipation (neglecting the losses).
Ans
Given that,
For a power amplifier,
Optimum load impedance, R =
180 2
Load resistance,R, =8 2
Mean collector current, I2 A
Power supply, V=15 V
A.C load power, P..=2.5 W
ac
(i) Turns ratio, n =?
(i) Efficiency, n=?
(ii) Powerdissipation, Pp=?
() The expression for turns ratio ofa transformer is given
as,
.(1)
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n
R

8. 4.13
By substituting given values in equation (1), we get,
8
180
n=0.044
n= vo.044
n= 0.21
The efficiency of a power amplifier is expressed as,
(2)
Ppc
D.C input power P,cis given as,
D.C
Ppc cVec
PDC2x 15 I2A, Vec=15A
PDC30 w . (3)
On substituting Pac and P in equation (2), we get,
ac
2.5
-300.083
30
=8.3%|
(ii) Collector power dissipation, Pp is obtained as,
(iii)
P-PP
= 30-2.5
27.5
Pp27.5 W

9. 10. TOper Matcning or uran
Q31. For a class B amplifier providing a 20 V peak
signal to a 162 load operates on a power supplyy
ofVec30v. Determine the input power, output
ofVo
power and circult efflclency.
Ans:
Given that,
For a class B amplifier,
Peak value ofoutput voltage, V= 20 V
Load resistance, R, =
16 2
Supply voltage, Ve 30 V
Inputpower, P(d.c) =?.
Output power,P.(a.c)=?
Circuit efficiency,n=?
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10. RONICCIRCUITANALYSIS{JNTU-HYDERABAD)
Input Power P. (d.c)
Input power of a class B amplifier is given by.
PndoVclae (1)
Where
And =
20
16 A
=1.25 A . (2)
0.3979 .. (3)
On substituting corresponding values in equation (1).
we get,
Pimde)30 x 0.3979
Pad.l1.93 W
Output Power P.ac)
Output power ofa class B power amplifier is expressed as,
-..)
Poa.c)
> Paad
x20x 1.25 Fromequation(2)
= 6.25
Poda.c) 6.25W
Circuit Eficieney (n)
Circuit efficiency of a class B power amplifier is given as
n-ex100
Pin(d.c)
6.25x100
n 11.93
=0.523 x 10
52.3%
52.3%|

11. Amplifiers and TunedAmplifiers
UNIT-4
(Large
SignalAmplife
For
an
deal class B transistor amplifler the
collector
supply voltage Voc and the effective
load
reslstance R =
(N,/N_P R are flxed as the
hase Current
excitation is varled. Show that the
collector
dissipation P. ls zero at no signal, rises
as m
increases
and passes through a maximum
atV2V/n.
a 3 2 .
AnS
Given that,
For an ideal class-B transistor amplifier,
Collector supply voltage = Ve
Effective load resistance- R,-
As and V. be the maximum output current and
maximum output voltage.
Then, the instantaneous voltage and current are given
by,
a n d ,
r.m
Thus, collector power dissipation,
P-V
P.=VEms.ma
From the above expression, it is clearthat 'P,' is directly
POportional to square of maximum output voltage (
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12. When the input signal is made zero i.e.,
V-0, the power dissipation at collector(P,
becomes zero,
ie. P-0.When -0
Maximum collectordissipation (P )can be obtained
as follows,
Powerdissipationby hoth transistor, P Ppc-Pac
2
2V
a)TR, d 2R' dV
TtR
nRR
But, for maximum power dissipation,
2 0
TR R
2
This is the condition to get maximum power dissipation.
P Ptl,
"(max)
4V2 2V2
TRnR'
Hence, the collector dissipation is zero at no signal and
2
maximum at V=
uDENTS-
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13. ELECTROL
4.32
Q33. A push pull amplifler
utillzes a
t r a n s f o r m e r
whose primary
has a total of 160 turns and
whose secondary
has 40 turns. It must be
capable of delivering
40 W to an 8 n load
under maximum power
conditlons.
What is the
minimum possible
value of Vce?
a3
C
Anst
Given that,
For a push pull amplifier,
Transformer primary
turns, 2N,
=
160
Secondary turns, N, 40
Load Resistance, R,
=
82
Maximum powerat load, (Pa)nas40 W
Minimum value ofVo=?
Since, 2N,
=
160
N = 80
-0.5
N
n
80
8 8
R- (0.5)2 0.2532
Under maximum condition,
The expression for maximum power delivered to the
load is,
1Vc
P 2 Ri ..(1)
Equation (1) can be rewritten for cc as,
Vc 2Pac)maxxR
Vac=40 x 2 x 32
.
80 x 32
= 2560
Vv2560
= 50.60 V
Vec =
50.60V
Therefore, the minimum possible value ofVo50.60 V.
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