In the design of a class AB amplifier, there is part of the understanding that I miss. It is well-known that the maximum power transfer is achieved when the load impedance is the complex conjugate of the source impedance, $$Z_L=Z^*_S.$$

It is also shown in many textbooks that using the conjugate match, the maximum attainable power efficiency is 50% (and also on Wikipedia).

I am well aware that the conjugate matching is not a good choice for large-signal amplifier design, however, if I was to load my amplifier with the conjugate match, does the maximum power theorem imply that I will only be able to get 50% efficiency out of my amplifier, no matter the biasing?

In simulations, I tried loading my class AB amplifier with the conjugate of the small-signal output impedance, and I achieved a drain efficiency of 55%, with the amplifier driven pretty heavily into saturation.

How is this possible?


1 Answer 1


You'll find that the Wikipedia article you mention does make a distinction between maximum power transfer and maximum efficiency. It further states that

"The theorem results in maximum power transfer across the circuit, and not maximum efficiency. If the resistance of the load is made larger than the resistance of the source then efficiency is higher, since a higher percentage of the source power is transferred to the load, but the magnitude of the load power is lower since the total circuit resistance increases.[2]"

In the case of the Class AB amplifier this is indeed the case, the source impedance should be very small and the load mostly larger (if you take an audio amp for instance). For reference, a Class B amplifier has a maximum theoretical efficiency of π/4 (≈ 78.5%), so in your simulation (that you did not present the details of) you should be pretty much be on the conservative side, and there's probably room for improvement.

Getting back to your simulation it would be usefull to know how you established the conjugate of the source impedance? I'm not aware of any simulators that will automaticall do that for you (as far as I know). This may be a source of your error that provides you with an underestimated efficiency perhaps.


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