Download Module – 6 Unit – 6 Power Amplifiers

Module – 6
Unit – 6
Power Amplifiers
Review Questions:
1. In what way the design features of power transistors different from small signal
transistors?
2. What is the basis for the classification of power amplifiers? Mention different
types of power amplifiers?
3. Draw the circuit for commonly used class A – amplifier. If the amplifier draws
10W of dc power, what is the maximum ac power available to the load?
4. Draw the circuit for a push-pull amplifier and discuss its working.
5. Derive an expression for the efficiency of class B – power amplifiers.
6. What is harmonic distortion? How does it arise in Class B-operation? And, how
can it be corrected in push-pull circuit?
7. What do you understand by cross-over distortion? How can it be eliminated in
Class B-operation?
8. What reasons will you assign for higher conversion efficiency of Class B-amplifier
as compared to Class A –amplifier?
9. Draw a circuit for Class C- amplifier and discuss its working?
10. Among all the power amplifiers, Class C-amplifier has the maximum efficiency
but its use is restricted. Give reasons.
Problems:
6.1 Calculate maximum ac output power in the amplifier shown in fig. (Assume V BE = 0)
+20V
50Ω
1k
vi
vo
B
1k
k
50Ω
100Ω
Solution:The ac power in class A-operation, P0 is given by the relation,
VCEQ . I CQ
P0
2
Where VCEQ and ICQ are voltage across collector – emitter of transistor at operating point
and collector current respectively.
First we need the value of ICQ. Now in fig above, the voltage between base and ground
(point B and ground, see fig.) VBB, is 10 V.
Then,
I CQ
VBB
IE
VBE
RE
10V
100
VBB
RE
100 mA
VCEQ can be obtained by summing voltage (dc voltages, capacitors taken open)
VCC = VCEQ + IE(RC + RE)
Or, VCEQ = VCC – IE(RC + RE)
= 20 – 100mA (50 + 100)Ω
= 20 -15
Or, VCEQ = 5V
Therefore, maximum ac power, PO,
P0
VCEQ . I CQ
or P0
2
250 mW
5 100 mA
2
6.2 Calculate maximum ac output power and efficiency of the amplifier shown in fig. V BE
may be assumed negligibly small.
+10V (VCC)
10Ω
vi
~
10Ω
RE
-10V (VEE)
Solution:The operating point current and voltages in the circuit are:
I CQ
VEE
RE
IE
10V
10
1A
And,
VCEQ = VCC = 10V
Therefore, maximum ac output power is,
P0(max)
VCEQ . I CQ
2
10 1
2
5W
To calculate the efficiency, η , the dc power drawn by collector-emitter circuit is,
PDC
VCC
VEE I CQ
(10 10) 1
20W
Therefore efficiency,
P0(max)
PDC
or
25%
5W
100
20W
6.3 Find out the value of resistor R2 to provide trickle current for distortion free output in
the push pull amplifier shown in fig. VBE for each transistor is 0.7V.
+30V
300Ω
R1
vo
vi
R2
R2
300Ω
R1
16Ω
Solution:Trickle current which flows through resistors R2 and produces a voltage drop of 0.7 V
across base – emitter junction over comes cross – over distortion in push – pull
amplifier. For analysis purposes, it is sufficient to consider only half of the circuit for
reasons of symmetry, and VCC of half (= VCC/2 = 30/2 = 15V) is to be taken for one
transistor.
The current through resistors R1 and R2 is,
I
15V
R1 R2
15V
300
R2
But,
I X R2 = 0.7V (desired voltage)
or , I
0.7 V / R2
............( B)
Combining Eqs (A) and (B),
0.7V
R2
15V
300
R2
or , R2
14.7
....................( A)
6.4 Calculate maximum ac output power and the minimum power rating of the
transistors in the push-pull amplifier shown in fig.
+40V
500Ω
vo
vi
8Ω
500Ω
Solution:The maximumac power (output). P0(max) as per the discussion on the topic is,
VCEQ
P0(max)
ic ( sat )
2
Where ic(sat) is maximum (saturated) collector current.
Now,
VCEQ
1
VCC
2
1
2
40V
20V
And, ic(sat) is expressed as,
ic ( sat )
VCEQ
rC
rE
20V
0 8
2.5 A
Here, rC is effective ac resistance seen by the collector and rE is effective resistance
seen by the emitter.
Therefore,
P0(max)
VCEQ
iC ( sat )
2
20
2.5
2
25 W
The maximum power rating, PD(max) is one-fifth of maximum ac power. That is,
1
. P0(max)
5
or , PD (max) 5W
PD (max)
25W
5
6.5 In fig. a basic Class C-amplifier is shown. It uses supply voltage of + 20V and load
resistance of 100Ω. The operating frequency is 3MHZ and VCE(sat) = 0.3 V. Calculate and
efficiency. If peak current is 500 mA, find the conduction angle also.
+20V(VCC)
100pf
3μH
vi
100Ω
Solution:The peak voltage, Vp, as was discussed is,
Vp = VCC – VCE(sat) = 20 – 0.3
Or, Vp = 19.7V
The ac power P0, is
P0
2
V p2
1.97
2 RL
2 100
or , P0
1.69W
And, dc power drawn by the circuit is,
Pdc
VCC
I dc
Where,
I dc
P0
Vp
1.69W
19.7 V
0.0857 A
Therefore,
Pdc = 20 X 0.0857
or, Pdc = 1.714 W
And the efficiency, η, is
P0
Pdc
1.69W
1.714W
100
98.5%
Now, we proceed to find out the conductance angle θ.
For the frequency of 3MHZ, the period of the wave, T, is
T
1
3 106
0.33 s
And transistor’s on- time is,
t
P0 T
I p Vp
1.69W 0.33 10 6
500 10 3 19.7 V
56.6 10 9 s
56.6 ns
or , t
or , t
And, the conduction angle, θ, is
t
T
or ,
360
61.7
56.6 10 9
0.33 10 6
360