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RE: more on output impedance

"In this case, the amp has an unknown impedance but we can easily find out what it is by simply connecting it to a variable load. We then find out at what impedance it makes its maximum power and then continue to reduce the load until that value is cut in half. At that point, half of the power is being dissipated in the output circuit itself (causing it to heat up) and the other half in the load. At this point the load must be equal to that of the output circuit."

OK, I need to think about this a bit. May I check that I have your procedure correct? You first adjust the load impedance until the maximum power dissipation in the load is achieved. Let's say that the load impedance for this is Z1. Then, you reduce the load impedance further until you reach a point where the power dissipated in the load is one half of what it was at the maximum. This second, lower, load impedance is, say, Z2.

And you then define Z2 as the output impedance of the amplifier?

Thanks,
Chris

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