High Wind alert!

"Richard Harrison" wrote in message
...
Steve Nosko wrote:
"I am missing just what it is that gets to the "Loss-less resistance"
conclusion."
If part of the source resistance were not lossless, efficiency would be
limited to 50%.

I disagree.
DC or RF:
Real source = XX volts
Real source resistance = 10 ohms
Real load 50 ohms

Efficiency is 50% DONE. (and I have no "loss-less resistance" anywhere)

I can't figure out where you are going nor what the hole is in the
technology that needs this extra stuff... are there formulas for it?

Fact is, Class-C amplification frequently is 2/3 or 66.6% efficient.
Twice the power is delivered to the load as is lost in the source.

Yea. We both know all this. I'm trying to get to the WHY part. The
issue is WHERE or WHAT is this loss-less resistance? Where is it? Why is
is even needed?



If we deliver 1000 watts into 50 ohms, we have 224 volts and 4.47 amps
as a load.

Our source is identical,, 224 volts and 4.47 amps. Its volts to amps is
the same ratio, but its loss is not the same power as the power
delivered to the load because part of the source resistance is lossless

I stop right here and say: It is because the source resistanace is
*less than* the load resistance. Simple as that! I can not come to any
other conclusion.


because it is the product of interrupted energy delivery, not energy
conversion into heat.

My eyes glaze over here. "interrupted energy delivery" -- can't
get a grip on this.

With 2/3 efficiency, when we have 1000 watts into the load, we have 500
watts lost in the source.

Yep. I believe the true issue is WHY does this happen? What is the
cause of this effect?
"Loss-less resistance" or Rs RL?
I say the latter, I think you say the former.


500 watts lost means only 1/2 the dissipative resistance as we have
resistance in the load, and thus the source resistance consists of 25
ohms of dissipative resistance and 25 ohms of non-dissipative

Why is this needed? I think I see. You are saying that because we
have a "match" We are at the "conjugate match" condition and therefore
*MUST* be at Rs=RL. Here's where I disagree. I believe this is absolutely
NOT the case as I say above. It's as simple as the fact that the amplifire
IS NOT internal resistance limited, but either dissipation limited or
perhaps breakdown voltage limited or cathode emission limited. NOW I THINK
I GET IT (your tack)!. You believe that the source resistance MUST equal
the load resistance.. Well I say Nope!
DC or RF:
Real source = XX volts
Real source resistance = 10 ohms
Real load 50 ohms

Efficiency is 50% DONE. What was that guy's name Ocam? of the Ocam's
razor fame:"Take the simple answer", or something to that effect.



resistance. The total source resistance is 50 ohms which matches the
load resistance.

I beileve this is where your error of reasoning is. I can not accept
that there is some other resistance which is not
quantifiable/observable/measureable. The source resistance must be less
than the load R for Eff 50%. I believe you are making this much more
complicated that is really is.



A match allows maximum power transfer.

I believe this is what is leading you astray. I believe you are
assuming that the maximum power of the maximum power theorem, and the
maximum power out of the real transmitter, are the same maximum power.
This is where I believe the error in reasoning is that requires the
loss-less resistance. They are not the same maximum power.


The typical DC power supply is operated with Rs RL. The power output
is limited by factors other than the maximum power transfer theorem (its
output resistance) suggest. The power supply is not limited by its internal
resistanse, but usually its ability to dissipate what little amount of heat
(relative to the load power) that is can.

I'll stick my neck out (because I believe I have a firm
understanding of what physical laws can not be violated) and say that my
conclusion is that the tube amplifier we commonly think of, when "tuned up"
absolutely can not be operating in an Rs=RL configuration. This is for the
very reasons you have stated. Because the power dissipated in the tube is
less than in the load, Rs does not = RL. This is is, of course, looking at
the load R presented to the tube by its output tank circuit. I believe that
this (the RL.-RS stuff) is the natural law which can not be violated.

There is indeed SOME load which the tube likes to see in order to get
whatever power it can provide to come out. And it AIN'T Rs.


The less than 360 degrees of energy supplied to the load makes 25 ohms
of dissipationless resistance as a part of our 50-ohm source in our
example.

Keep working on the idea and eventually you may get the model.
Best regards, Richard Harrison, KB5WZI


I 'spose it's briefly is an interesting idea for discussion, but sorry
Richard, I find it hard to beieve it actually has any merit. My (I think
generally accepted) model works fine, it can explain all the phenomona(sp)
so far observed and... I work in the field and have never met anyone with
this idea and I have yet to see anything in the (uh, oh, here it comes. nose
in the air Engineer snob talk) profession which suggests this lossless stuff
is there, nor is there anything that remains unexplained which needs some
other effect.

I guess we close disagreeing.

If it works for you...

Interesting journey into some serious examination of principles, however.
--
Steve N, K,9;d, c. i My email has no u's.