Hi Peter,
A little circuit analysis never hurts anyone who is correct.
> >You've made the common mistake of looking at "pictures in books"
> >and not reading all the text explaining the pictures.
>
> There are at least as many very eminent engineers and academics who
> disagree with Tom as there are those who agree. Positions are sufficiently
> entrenched that I don't believe either side will convince the other.
> Personally, I can't swallow 'non dissipative' resistance. But I think that
> all this thread will produce is acrimony and discord.
That's almost any conversation on this reflector. Or haven't you
noticed? <grin>
In any event, what you prefer to call the impedance in the model
that, by rule of the model, has no defined resistor properties like
"dissipation" is up to you. Call it what you like. Fred would do
nicely, if he is non-dissipative.
It is the unfortunate result of poor planning that we call non-
dissipative things R in ohms (the impedance of a transmission line
or free-space) while we also call things that dissipate power R in
ohms.
At least the mechanical engineers were smart enough to use
different terms for different effects, so as not to confuse those who
scan the text and leap to conclusions.
It can not, no matter whatever you name that thing in the model, be
used to discuss or define dissipation or any form of energy
conversion. The rules of the model are clear about that.
Anyone who has even a very basic and even the most crude
understanding of a PA and how it functions almost certainly
realizes the PA can not be modeled as a resistance in series with
a generator, or in parallel with one, so far as efficiency is
concerned.
If you misuse a model none of the results make sense when other
models are substituted, and you can not interchange the Norton
model with the Thevinin and get the same results.
A model is either correct, or it is not a good model. In the case of
power transfer to the linear time-invariant load it is a good model. If
you try to use the model to describe actions inside the PA, it is
totally useless.
For those who "think" it is a good PA model, what happens when
we shut the drive off? The efficiency reaches zero. What happens
when we design a Thevenin model with an efficiency of 90% and
change the model to a Norton model? The efficiency is now 10%.
If you mismatch a class A, AB, B C, or D PA in EITHER direction
from an optimum load value, efficiency drops. Does it do that in the
model? Of course not!
It simply doesn't work if you use the model for anything more than
it was intended to describe, power transfer. Dissipation is nowhere
to be found in the model Rich used and it is not allowed by its rules.
73, Tom W8JI
w8ji@contesting.com
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