"Richard Harrison" wrote in message
...
Steve Nosko wrote:
"Then there`s the solid-state power amplifier standard output resistance
formula.
Rs = Vcc^2 / (2*Po)
The implication should be obvious."

It looks like Ohm`s law to me, P=Vsq / R.

Well... I don't call THAT ohm's law, but rather, oh, I suppose, the
power formula, but that's symmantics. It is a transposition of (whatever
you call) that formula.



The implication of (2*Po) is that 50% of the power is in the source and
50% of the power is in the load.

Again, I don't recall teh derivation, but it works. I don't believe it
related to a mathematical constraint that the power be equally split. Can't
speculate further withoug working it out.
This is the "maximum output" load. Don't know off-hand what the limiting
factor is, but this is what they desigend for and I don't thing you could
get more out without killing the part or its lifetime. Don't recall anyone
blowing parts with the wrong load...pretty robust parts. You just couldn't
get any more out.


If so, it`s a Class-A amplifier

They are class C. VHF FM PAs.


formula, but the semiconductors could be biased to cut-off (Class-B) to
reduce dissipation in the transistors when they are idle. The best
collector load resistance is often not that which produces maximum
output, but that which produces maximum "undistorted output".

Best regards, Richard Harrison, KB5WZI