This is an interesting question, and there are two schools of thought. One
says that you get maximum power when you have a conjugate match, and the
other says you get maximum power when feeding an optimum load impedance.
The first school of thought (which includes Tom Rauch and Professor
Belrose) have measured the output impedance of a tubed SSB transmitter and
got an answer of about 50 ohms
The second school of thought (Warren Bruene) argue that if you get maximum
output power when conjugately matched, then efficiency can never be
greater than 50%, and Class C stages work at much higher efficiencies. To
which the first school suggests a non-dissipative resistance in Class C,
but that one passes me by.
If the PA stage is about 50% efficient, then you would expect to get
something like a conjugate match. Look at the curves for the 6146, and you
see the plate resistance is about 2000 ohms, which is roughly the correct
load impedance, efficiency is about 50% and the correct load is something
like a conjugate match..
Consider a transistor PA. It has a very low collector resistance, but
wants a load given by Vcc squared divided by 2 Pout. Here we start getting
to the point where other resistances start to play a part - emitter
ballasting, bond wire, even the connecting tabs. This modifies the
practical case.
In general, I think most linear PA stages do look something like a
conjugate match because they are 50% efficient, give or take a bit.
Finally. Suppose a 12 volt 1 amp power supply. For load change between 0
and 1 amp, the voltage sags by 1 millivolt, so the internal impedance is 1
milliohm. If conjugately matched, it would try to deliver 12,000 amps! So
maximum power in that case is not when conjugately matched - just as the
electricity supply network isn't. If you try, you blow fuses - we had a
chap in college who did just that. Except it was on 1200 amp phase of the
415 volt three phase supply!
73
Peter G3RZP
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