2. Power amplifiers EL= IxRL
To deliver a large power in to a load, Large voltage and current amplitudes are required.
This requires reducing the base collector concentration, to avoid punch through. This results in lower β.

3. Characteristics of Power Transistors.
Lower current gain.
Heat sinks to be used to dissipate the heat.
When there is no input signal, large current flows through the transistor, resulting maximum power dissipation.

4. Use of Transformers. Advantages. Better Impedance matching.
Less DC power dissipation under quiescent conditions.
Disadvantages.
Frequency Limitations.

6. In class-A amplifier, D.C power is wasted even in the absence of signal. So efficiency is small.
Class-A Power Amplifier.
Direct Coupled
Transformer Coupled.
a) Direct Coupled.
Graphical representation of class A amplifier

7. Applying Kirchhoff s voltage law to the circuit shown in the Fig. 4
Applying Kirchhoff s voltage law to the circuit shown in the Fig. 4.11, we get Vcc – IcRl = Vce 1
The slope of the load line = -1/RL
Y intercept = Vcc /RL
D.C. Operation
The collector supply voltage VCc and resistance Rb decides the d.c. base-bias current IBq. The expression is obtained applying KVL to the B-E loop and with
ICQ = β IBQ
VCEQ = VCC – ICQ RL

8. (Vmax – Vmin)(Imax – Imin)
D.C. POWER INPUT.
PDC = VCC ICQ
A.C. POWER OUTPUT.
VPP = Vmax – Vmin
IPP = Imax – Imin
RMS of Pac = 8
Vm x Im
Pac=
(Vmax – Vmin)(Imax – Imin)
Efficiency
The efficiency of an amplifier represents the amount of a.c. power delivered or transferred to the load, from the d.c. source i.e. accepting the d.c. power input. The generalized expression for an efficiency of an amplifier is,

9. %𝜂= ( 𝑉 𝑚𝑎𝑥 − 𝑉 𝑚𝑖𝑛 )( 𝐼 𝑚𝑎𝑥 − 𝐼 𝑚𝑖𝑛 ) 8 𝑉 𝐶𝐶 𝐼 𝐶𝑄 𝑋 100
%𝜂= ( 𝑉 𝑚𝑎𝑥 − 𝑉 𝑚𝑖𝑛 )( 𝐼 𝑚𝑎𝑥 − 𝐼 𝑚𝑖𝑛 ) 8 𝑉 𝐶𝐶 𝐼 𝐶𝑄 𝑋 100
Maximum voltage and current swings
%𝜂= ( 𝑉 𝐶𝐶 −0)(2 𝐼 𝐶𝑄 −0) 8 𝑉 𝐶𝐶 𝐼 𝐶𝑄 𝑋 100= 2 𝑉 𝐶𝐶 𝐼 𝐶𝑄 8 𝑉 𝐶𝐶 𝐼 𝐶𝑄 =25%

10. Class A Output Stage - Recap
Class A output stage is a simple linear current amplifier.
It is also very inefficient, typical maximum efficiency between 10 and 20 %.
Only suitable for low power applications.
High power requires much better efficiency.

11. Why is class A so inefficient ?
Single transistor can only conduct in one direction.
D.C. bias current is needed to cope with negative going signals.
75 % (or more) of the supplied power is dissipated by d.c.
Solution : Eliminate the bias current.

12. Transformer-coupled transistor circuit.

13. TRANSFORMER COUPLED POWER AMPLIFIER.
Reflected Impedance from Secondary to Primary of theTransformer 2 RL X
Pacmax =

14. Family of collector curves for power transistor.

15. Transformer-coupled class B push-pull amplifier.
Ic1 and Ic2 flow in opposite directions in the primary of the transformer. So their magnetizing effects cancel and hence no magnetic saturation problems.

16. Combined collector curves for push-pull operation.

17. Transistor characteristics and operating points:
(a) input characteristic of a silicon transistor; (b) bias operating points on a load line.
(b)

18. η=

19. FIGURE 14-16 A class AB push-pull amplifier.

20. Class B Output Stage Q1 and Q2 form two unbiased emitter followers
Q1 only conducts when the input is positive
Q2 only conducts when the input is negative
Conduction angle is, therefore, 180°
When the input is zero, neither conducts
i.e. the quiescent power dissipation is zero

21. Class B Current Waveforms
Iout
time
IC1
time
IC2
time

22. Class B Efficiency Average power drawn from the positive supply: IC1
Phase, q
A/RL p 2p
A sin(q)

23. Load power: Efficiency:
By symmetry, power drawn from +ve and –ve supplies will be the same. Total power, therefore:
Load power:
Efficiency:

24. Power Dissipation
To select appropriate output transistors, the maximum power dissipation must be calculated.
Just need to find the maximum value of PD to select transistors/heat sinks

25. Peak Output Amplitude, A [V]
E.g. VS = 15 V, RL = 100 W
1.5 PL PS PD 1
Power [W]
0.5 5 10 15
Peak Output Amplitude, A [V]

26. Maximum Power Dissipation
PD is a quadratic function of A,
maximum when:
Substitute A in above equation.

27. Efficiency / Power Dissipation
Peak efficiency of the class B output stage is 78.5 %, much higher than class A.
Unlike class A, power dissipation varies with output amplitude.
Remember, there are two output devices so the power dissipation is shared between them.

28. Design Example
Design a class B amplifier which will deliver up to 25 W into a 4 ohm load.
Supply voltages must be larger than Amax so choose Vs = 15V.

29. Each of the two output transistors must be able to safely dissipate up to 5.7 Watts. Using a TIP120 & TIP 125:
But, with qJC = 1.92 °C/W
i.e. Either two heatsinks rated at less than 20°C/W are required or a single heatsink rated at less than 10°C/W.

30. Suggested heatsink
Dimensions, 50mm x 50mm x 9.5mm
Accommodates two devices
Rating 6.5°C/W
Cost 60p inc VAT

31. Cross-Over Distortion
A small base-emitter voltage is needed to turn on a transistor
Q1 actually only conducts when vin > 0.7 V
Q2 actually only conducts when vin < -0.7 V
When 0.7 > vin > -0.7, nothing conducts and the output is zero.
i.e. the input-output relationship is not at all linear.

32. Actual Input-Output Curve
vout
-VBE
vin
+VBE

33. Effect of Cross-Over Distortion

34. Class B Summary
A class B output stage can be far more efficient than a class A stage (78.5 % maximum efficiency compared with 25 %).
It also requires twice as many output transistors…
…and it isn’t very linear; cross-over distortion can be significant.

35. DISTORTION IN POWER AMPLIFIERS.
The possible distortions in any amplifier are
Amplitude
Frequency (Not significant in Power Amplifiers)
Phase (Cannot be detected by ear.)
Because of the non linear dynamic characteristics of transistors, waveform distortion occurs. This gives raise to Harmonic Distortion.
The second harmonic has the highest amplitude.

36. Elimination of Even Harmonic Distortion. (Push-Pull)
The amplitude of second harmonic is the largest. As the order of harmonic increases, it’s amplitude decreases.
ic1= a1cos𝝎t + a2cos 2𝝎t + a3cos 3𝝎t + ………………
ic2 = a1cos(𝝎t + π) + a2cos2(𝝎t + π) + a3cos3(𝝎t + π) + ………….
= -a1cos 𝝎t +a2cos2𝝎t –a3cos3𝝎t + ………………………
i0 = ic1 – ic2
So even harmonics are eliminated and the % harmonic distortion is only due to odd harmonics given by,

37. If B1 is the Amplitude of fundamental Frequency
Bn is Amplitude of nth Harmonic Component
% Harmonic Distortion due to nth harmonic
component
Total Harmonic Distortion

39. Power Transistor Derating Curve
Power transistors dissipate a lot of power in heat. This can be destructive to the amplifier as well as to surrounding components.

40. With Diode Compensation Dual Supply
With Driver & Diode Compensation
Single Supply

41. Quasi Complimentary With NPN Matched Pair.
A Practical Circuit.

42. Thermal Analogy of Power Transistors.
θ=Thermal Resistance.
Pd = Heat due to Power Dissipation.

43. Heat Flow with Heat Sink.

44. The thermal resistance as per the above diagram

45. Problem:
Determine the power handling capacity for a 60 W power transistor rated at 25 "C if de-rating is required above 25 "C at a case temperature of 100 "C. The de-rating factor is 0.25 W/°C.
Thus due to de-rating at 100 deg.C, the power handling capacity decreases to W.

49. 𝜂 > 98%