# Class A Output Stage - Recap

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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.

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.)

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

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:

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

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.)

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.

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 =

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.

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.

Actual Input-Output Curve

Effect of Cross-Over Distortion

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.

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.