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RE: output impedance

The technique is quite solid when no current is involved.

I have to assume though that if there is a resistance that is used as a load that the voltage can only be there due to the fact that current is flowing. So power is involved.

Now if that is the case, you can't have power with no load, right? So you would start with some very high impedance well outside the circuit's range, and decrease the impedance until a peak in power is observed. Now we know the maximum power the circuit can produce. To sort out the output impedance, it will be equal to the load impedance when equal amounts of power are dissipated by the load as is by the output circuit. This point is of course when the load impedance equals that of the internal impedance that is dissipating the other half of the power.

This is how it was done in the old days. If you were educated in audio after the 1970s, its likely that the method you proposed would have been taught. You might want to look at this link to see what is going on.

The Voltage Paradigm has a number of 'charged terms' which don't quite mean the same thing as output impedance. I can give you a wonderful example of how this can affect your viewpoint.

If you were taught this method, you were probably also taught that applying loop negative feedback will lower the output impedance. But have you ever considered that such would violate Kirchoff's Law?


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