"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...

Steve Nosko wrote:
"I think the phrase "available energy" may draw discussion, but I won`t
start it."

I plead guilty. I knew when I wrote it that the choice was poor but
couldn`t think of a better phrase at the moment.

Terman says on page 76 of his 1955 edition:
"Alternatively, a load impedance may be matched to a source of power in
such a way as to make the power delivered to the load a maximum. (see
footnote) The power delivered to the load under these conditions is
termed the "available power" of the power source."

Walter Maxwell straightened me out on my carelessness on this long ago,
and I repeated anyway. Dang me!

Best regards, Richard Harrison, KB5WZI

"Richard Harrison" wrote in message
...
Steve Nosko wrote:
"I think the phrase "available energy" may draw discussion, but I won`t
start it."

I plead guilty. I knew when I wrote it that the choice was poor but
couldn`t think of a better phrase at the moment.........
Best regards, Richard Harrison, KB5WZI

I think this is what causes much of the discussion here. Terms are not
always as clear cut as things like: voltage or resistance. We know those,
but some of the others are more vague...or just what concept they point to
is not obvious. It is also sometines difficult to determine what the KEY
point or word is of a question or statement.
When I read yours, I made the assumption that "available power" was not
where your question pointed. I went along the; "Can we get a non 50-50
division? and not waste so much in the stage" path.
It is clear that different responders "Hear" different question.

Cryminy crum! I think my brain has forgotten how to hit the keys "on" in
the correct order. I keep getting things like "questino" for
"question"....or is it a keyboard timing issue??? As my fingers fly
along...
--
Steve N, K,9;d, c. i My email has no u's.

Steve Nosko wrote:
"With a "conjugate match" the source dissipates 50% of the power and the
load the other 50%."

This is true only if all the source resistance is the type that converts
electrical energy to heat. There is a non-dissipative resistance.
Switched-off time in the Class-C amplifier is part of its internal
resistance.

You can deliver all the available power of the Class-C amplifier to the
load and dissipate less than 50% in the source.

Best regards, Richard Harrison, KB5WZI

"Richard Harrison" wrote in message
...
Steve Nosko wrote:
"With a "conjugate match" the source dissipates 50% of the power and the
load the other 50%."

This is true only if all the source resistance is the type that converts
electrical energy to heat. There is a non-dissipative resistance.
Switched-off time in the Class-C amplifier is part of its internal
resistance.

You can deliver all the available power of the Class-C amplifier to the
load and dissipate less than 50% in the source.

Best regards, Richard Harrison, KB5WZI


I think this becomes very academic (or perhaps a better word is awkward)
because originally you framed this as a "conjugate match" situation. Now,
you are in terms of time varying parameters. I think the analysis must stay
in one realm or the other. My mental models have trouble switching back and
forth.
I can't speak to tubes, but I do know that to get the most out of a
transistor power amp (transmitter type) up to about 200 MHz you design the
output match to be for a collector resistance of Vcc^2/(2Po). IF I recall,
this can be derived easily if you assume the transistor pulls all the way to
ground and the output tank swings up to 2 x Vcc. I was shown, and
understood it way back then, but can't recall the path to the solution off
the top of my head.
This, then, always started the discussion of whether this "matched" the
transistors output impedance or some other more esoteric concept. Then talk
of "average resistance" came in and eyes would glaze over....

The designer would then go back to the bench, work to optimize the design to
his requirements, test it over temp, etc. and ship it.

"There comes a time to shoot the Engineer and ship the product." is the
title of a famous editorial from the early 70's (IIR). Got it around here
somewhere...
-- 73
Steve N, K,9;d, c. i My email has no u's.

Steve Nosko wrote:
"A (tube) amplifier is in conjugate match conditions. It is dissipating
10 watts in its plate. This is the limit of its plate
dissipation----There is also 10 watts in the load.

Now assuming you can, increase the (plate) supply voltage by , say
20%---(may be the fatal flaw)."

Likely so. If everything remains linear, 20% more voltage increases
power by 1.2 squared, or 1.44 times. If the tube was already dissipating
its maximum sustainable power, expect an early failure due to the
overload.

If you were not already in a maximum power transfer condition, and a
condition which might provoke flashover within the tube, a readjustment
of the match might shift more of the available power to the load and
thus relieve the tube of some of the dissipation.

Best regards, Richard Harrison, KB5WZI

"Richard Harrison" wrote in message
...
Steve Nosko wrote:
"A (tube) amplifier is in conjugate match conditions. ...

Now assuming you can, increase the (plate) supply voltage by , say
20%---(may be the fatal flaw)."

Likely so. If everything remains linear, 20% more voltage increases
power by 1.2 squared, or 1.44 times. If the tube was already dissipating
its maximum sustainable power, expect an early failure due to the
overload.

That's what I went on to say (get more power and 50% will be
dissipated in the stage...until it blows), but there was more to the story
which is relavant.
--
Steve N, K,9;d, c. i My email has no u's.

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.

The implication of (2*Po) is that 50% of the power is in the source and
50% of the power is in the load. If so, it`s a Class-A amplifier
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

"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

"Steve Nosko" wrote
With a "conjugate match" the source dissipates 50% of the power and the
load
the other 50%. This can't be changed.
=============================
Without disagreeing with what you say -

A conjugate match is not relevant in the present discussion because there is
seldom, if ever, a conjugate match between a PA and its antenna system.

The tuning-up process is NOT intended to produce such a match.

Tuning up is just the simple process of adjusting the transmitter load
resistance to be equal to its designed-for load resistance, usually an
arbitrary 50 ohms.

The internal resistance of a transmitter is NOT 50 ohms. It is not a design
feature. It is whatever happens to appear after the designer has met a
series of other requirements. The designer himself does not know what the
internal resistance is unless, out of curiosity, he bothers to measure or
calculate it.
----
Reg, G4FGQ