"Cecil Moore" wrote in message
...
Richard Harrison wrote:
If we have a Class C amplifier feeding power to the same antenna and
enjoying a conjugate match, we can have a source that takes less than
50% of the available energy. So, the transmitting antenna system can be
more efficient than the receiving antenna system, it seems to me.


Didn't see the original post...

I think the phrase "available energy" may draw discussion, but won't start
it. I suspect it is not the thing to focus on. However...

With a "conjugate match" the source dissipates 50% of the power and the load
the other 50%. This can't be changed. Draw the schematic to see. Rs = RL
therefore Ps = PL. (I'll use a big "L" so it looks like an "L")


------------------------
Then I go to something I have for a long time wondered about; the following
situation:
A (tube) amplifier is in conjugate match conditions. It is dissipating 10
watts in its plate. This is the limit of its plate dissipation. Model this
as a voltage source (Vs) with source resistance (Rs) and a load with load
resistance (RL). The conjugate match has removed all the reactance.

There is also 10 Watts to the load.

Now, assuming you can, increase the (plate) supply voltage by, say 20%.

This raises the source voltage (Vs) [may be the fatal flaw]. If the plate
resistance (Rs) stays the same (I don't know if it will - more flaw-fodder),
then the plate dissipation will also increase. SO... how about adjusting
the match so the plate sees a higher RL to get back to 10 watts plate
dissipation. RL must now increase 40% to do this. The source current must
stay the same if the source resistance stays the same and keep the same
source dissipation.

Now, what we have done is to increase RL and Load voltage (VL), and
therefore load power, but kept 10 watts at the plate. If I did the math
correctly the load now dissipates 40% more power. If, however, we
re-adjust the match back to conjugate we WILL get more power and 50% will be
dissipated in the stage...until it blows. SO...

I think this tells us that if the stage is dissipation limited, but not
breakdown limited, we can non-conjugate match for a higher power than
'before' by raising the supply voltage. It sorta looks like we are getting
more with a non-conjugate match. You don't have the 50-50 division of power
also. It is now 41.7% tube, 58.3% load

Snake oil? The math is correct, but there may be some practical limit not
considered.

--
Steve N, K,9;d, c. i My email has no u's.

Then there's the solid state power amplifier standard output resistance
formula.
Rs = Vcc^2 / (2*Po)
The implication should be obvious...