Electrical – Why is the maximum efficiency of a push-pull amplifier 78.5%

amplifierpush-pull

One of the practice questions for my upcoming test is this:
enter image description here

After some googling, it seems like the correct answer is A. But why?

Best Answer

It is good to review this.
@Barry gives the good answer. Am attempting a more graphic solution using LTSPICE. The assumption is that a sine wave is the signal source, whose peak voltage just grazes the DC power supply that powers the Class-B output device(s).
This is unrealistic, because any transistor has some overhead voltage drop: at least its saturation voltage. The answer assumes it is zero, and the transistor works linearly all the way through the sine wave peak.

In this simulation, a sine wave current source of one amp is fed to a load resistor of two ohms. The peak voltage at this load resistor reaches 2V. So a 2V DC power supply is required to power this Class-B simulation. It can only do the top half of the sine wave - only one quarter cycle is scanned from a 1Hz source.

LTspice circuit
Below is a plot of power vs. time. One trace shows power dissipated in R1, while the other shows power dissipated in I1:
Power in I1, Power in R1
The result of the two LTspice measurements shows the average power during the quarter-second run:

p_source: AVG((2-v(n002))*i(i1))=0.273242 FROM 0 TO 0.25
p_r1: AVG(v(n002)*v(n002)/2)=0.999993 FROM 0 TO 0.25

Efficiency for your question seems to be defined as p_r1/(p_r1 + p_source)
That is: 1/(1 + 0.273242) = 78.54%

Note that for every watt dissipated into the load resistor, 0.273W is dissipated in the pull-up transistor, but that only lasts for half a cycle. During the next half cycle, this top transistor (it might be a PNP) dissipates nothing.

During the bottom half cycle (while the PNP dissipates nothing), the NPN would take over, and dissipate a similar 0.273W. So the PNP and NPN share power dissipation, each taking half of the 0.273W.
Don't forget, this is a unrealistically ideal case.

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