What is the basis for the classification of power amplifiers? Mention ... Among all
the power amplifiers, Class C-amplifier has the maximum efficiency but its use is
...
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
