1. 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.
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.

2. Class B Output Stage Q1 and Q2 form two unbiased emitter followers
Q1 conducts only when the input is positive
Q2 conducts only 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
(We are temporarily ignoring the need to FWD
Bias the BE junctions for conduction to occur, which causes “crossover distortion” in the output.)

3. Class B Current Waveforms
Iout
time
IC1
Ideal waveforms shown.
-ignoring
“crossover distortion”
IC2
time
time

4. Class B Efficiency IC1 Average power drawn from the positive supply:^
Vo/RL p 2p
Phase, q = wt
A sin(q)
Waveform of + supply
current has avg. value
given by peak/pi
By symmetry, power drawn from +ve and –ve supplies will be the same.
Total suply power, therefore:

5. The average load power will be
Efficiency
Max efficiency occurs for Vo = VCC
and equals pi/4 = 0.75 =75% ^
(In actual practice max. value is limited to
Vcc – VCE sat  VCC)
Max. avg.load power is found by subst. Vo = VCC
into equation for PL above
and equals (1/2 ) (VCC)2/RL

6. 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/heatsinks
For Class B quiescent power dissipation = 0
(it was max. under quiescent conditions for Class A)
When an input signal is applied the avg. power
Dissipated in the Class B stage is
Subst. For Ps and PL from eqns. on previous page
(eqn 9.19)
To find maximum differentiate
Subst this value in eqn. (9.19) yields
Thus
(From symmetry half the total PD is dissipated in each
transistor.)

7. We can find the efficiency at the point of max power
dissipation by subst.
into the eqn. for Class B efficiency to get h = 50%
Plotting eqn. (9.19), which gives avg. power
dissipated vs. output signal amplitude, shows that
power dissipation decreases after it it reaches
a maximum while operating at a higher signal
amplitude. However, at higher signal amplitudes
there is greater nonlinear distortion as a result of
approaching saturation in the transistors.

8. Þ Example It is required to design a class B output stage to
deliver an average power of 20W to an 8 ohm load.
The power supply voltage VCC is to be 5 volts
greater than the peak output voltage.
Determine:
the supply voltage required.
the peak current drawn from each supply,
the total supply power,
the power conversion efficiency,
the maximum power that each transistor can dissipate
safely.
Solution: Þ
Thus =

9. Efficiency / Power Dissipation
Example (Cont’d)
Efficiency / Power Dissipation
Peak efficiency of the class B output stage is %, 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.

10. Cross-Over Distortion
A small base-emitter voltage is needed to turn on a transistor
Q1 actually only conducts when vin > 0.5 V
Q2 actually only conducts when vin < -0.5 V
When 0.5 > vin > -0.5, nothing conducts and the output is zero.
i.e. the input-output relationship is not at all linear.

11. Actual Input-Output Curve

12. Effect of Cross-Over Distortion

13. 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.

14. 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.