Richard Harrison wrote:

Richard Clark wrote:
"The Thevenin model only requires an IMPEDANCE."

Yes, but "To secure maximum power output from a generator whose emf and
whose internal impedance are constant, the load must have an impedance
equal to the conjugate of the generator`s internal impedance." (page 43,
"Transmission Lines, Antennas, and Wave Guides", King, Mimno, and Wing)

Anything about dissipationless resistances or negative resistances?
--
73, Cecil http://www.qsl.net/w5dxp



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"Richard Clark" wrote in message
...
On Mon, 24 May 2004 16:40:44 -0500, "Steve Nosko"
wrote:
This is an interesting twist, Tam. I think if this were the case,
then
there would be MORE power dissipated in the Tx

Hi Steve,

And why would that be?

73's
Richard Clark, KB7QHC

Reconstructing the long ago comment not having the original comment, Steve
comments further:

I think he had the power returning to the PA device and being dissipated
there.
Steve

Yea! See. That's what I inferred and I said he was implying.

Man! This he said...he said...he said gets confusing who said what...
Steve


"Tam/WB2TT" wrote in message
...

"Steve Nosko" wrote in message
...

"Tam/WB2TT" wrote in message
...

"Richard Fry" wrote in message
.........................
...............................
Concept below

However this is not an accurate model of a transmitter.

For an example, take an old Heathkit DX-100 generating a measured
180
watts
of CW RF into a matched 50 ohm load. To do this, it does NOT also
dissipate
180 watts of RF into some "virtual" internal RF load in the DX-100.
In
fact, the PAs and power supply in the DX-100 could not produce a
total
RF
output power of 360 watts without exceeding their ratings.

The dissipation in the PA is essentially related only the DC to RF
conversion efficiency of the PA, which in this case probably is
about
75%,
max (Class C). So a PA input power of about 240 watts DC is
required
to
produce 180 watts of RF output power. The other 60 watts of plate
input
power is converted to heat by the PA tube anodes.

The entire RF output generated by the PA stage is applied virtually
100%
to
the output connector. How much of that is absorbed by the load
connected
there is a function of load SWR and system losses.

- RF
There is a Motorola ap note that agrees with what Richard is saying.
To
paraphrase it, if the the DX100 had an output impedance of 50 Ohms,
then
the
overall efficiency would be 37.5%.

Unfortunately I can't read all the digressions in the thread. I
skim
by
author...

This is an interesting twist, Tam. I think if this were the case,
then
there would be MORE power dissipated in the Tx than Mr. Fry is saying -
making the situation worse. By that, I mean, getting further from what
is
going on. I think this goes in the wrong direction. I believe the flaw
is
believing that the Rs=RL must exist for the transmitter.

That is what I am saying. The efficiency goes from 75% to 37.5%; so,
there
is more power dissipated in the TX.

Tam/WB2TT
--
Steve N, K,9;d, c. i My email has no u's.

On Wed, 26 May 2004 09:05:30 -0500, (Richard
Harrison) wrote:

Richard Clark wrote:
"The Thevenin model only requires an IMPEDANCE."

Yes, but "To secure maximum power output from a generator whose emf and
whose internal impedance are constant, the load must have an impedance
equal to the conjugate of the generator`s internal impedance." (page 43,
"Transmission Lines, Antennas, and Wave Guides", King, Mimno, and Wing)


Hi Richard,

Not if it is strictly resistive. This is the rhetorical dance of
conditions. Too many arguments turn on a resistive load, and then
demand efficiency in terms of a resistive Thevenin model. That just
doesn't pass.

73's
Richard Clark, KB7QHC

Richard Clark wrote:

(Richard Harrison) wrote:
Yes, but "To secure maximum power output from a generator whose emf and
whose internal impedance are constant, the load must have an impedance
equal to the conjugate of the generator`s internal impedance." (page 43,
"Transmission Lines, Antennas, and Wave Guides", King, Mimno, and Wing)

Not if it is strictly resistive.

Of course, it works for the purely resistive.
R+j0 is the conjugate of R-j0
--
73, Cecil http://www.qsl.net/w5dxp



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Ahhh! Finally some meat... SORRY,... LONG POST WITH MUCH techincal MEAT IN
IT...


"Richard Clark" wrote in message
...
On Tue, 25 May 2004 02:59:52 GMT, "Henry Kolesnik"
wrote:

Richard
...(what is) the "mechanism" in
the transmitter (such that) it can't
dissipate a reflection because there's some kind of one way
....checkvalve...reflects.
tnx

Hi Hank,
...
It can also reflect (some or all of) what it does not absorb, ...
is determined by the ratio of its Z to the line/load
at the antenna terminal. (Of the transmitter, I believe you mean)

Fine point: I think it must reflect ALL it does not absorb.
Important point: However, I believe the case is that it absorbs little
of this energy.

In this case (bear with me), the source Z is
NOW found at the load's mismatch, transformed through the line. This
means matching works both ways (often the singlemost ignored aspect of
obvious reciprocity in any of these arguments). Such circuit analysis
is called superposition.

Yes, no argument. Matching does, er um, CAN work both ways on the line,
is true. And the TX is, indeed, a "load" for the reflected energy (or wave
if you prefer).
I say CAN because the match does work both ways, but that is not to
imply that the impedances which exist on the line are matched - nor that one
way IS matched while the other is NOT. If there is a mis-match in one
direction, there will be another mismatch going the other way. This may be a
difficult to understand considering the "this situation must be matched"
mind-set we are in at this point, but I do not believe it is pixie dust,
however choose to skip it for now. See my conclusion at the end....
I think "superposition" is ther wrong term for this. If I could spell
reciprocity, I'd say that it is a better word.
Digression, though minor: Superposition refers typically, to the
application of two signals to a circuit and the resulting response being the
sum of the individual responses. no?


Now some meat I'd like to pursue...

...a 500 Ohm Source...you will never launch much power into a 50 Ohm
system because it would immediately hit a reflective interface at the
antenna connector.

There's a question I see here as to wheather the power was "launched" in
the first place and this may be part of the assumption (or set thereof)
which (my gut feel says) leads the logic astray.


This means that a 500 Ohm source, when confronted
by power going towards it from a 50 Ohm system will reflect most of
that power (but how did we get this power into the system in the first
place - Karma?) This, by design, will never happen in any transistor
ham rig built in the last quarter century.

I think this is true...BUT let's ask this same question of a 5.0 ohm
source?? How would you carry this through Richard C? I submit that THIS
source CAN launch significant power down the line.



...snip non technical stuff

The answer is the transmitter source Z is 50 Ohms at rated power.

OK, so now I see you are of the "Zs=50 ohms" camp.

If
a watt of power is chooglin' down the line toward it, that 50 Ohms is
going to dissipate into a watt worth of calories. This can be argued
with wave mechanics, or lumped circuit equivalents - doesn't matter
because it's all the same calories.

All ok with me _IF_ Zs=50. I'm not convinced however. I have to think
more about the modern broad-band transistor PA to have an strongly arguable
opinion on this concept.


Modern rigs can tolerate this...snip...

=== WARNING to the logic impaired, the following is a supposition ===

Now, lets simply accept those answers that require the magic pixie
dust

Uncalled for... I can make it work (so far though I am open) with the
electrical/electronic principles I think are true and work everywhere else.

of total reflection from the transmitter. Fine, the mismatched
antenna reflects some power, this power returns to the transmitter,
the transmitter simple routes it ALL back to the antenna,
round-and-round until the antenna finally radiates it. If this were
true, what do we need tuners for?

I think the answer here is that if the transmitter can take the strange
impedance it sees, then it isn't needed. Like Hanks 30L1 that, however,
couldn't take it very well.
Furthermore, there is an implicit assumption here in your reasoning
which I believe is key and leading to some error. There is an unspoken
assumption that the power which makes it to the antenna after all the "
round-and-round until the antenna finally radiates it" is the FULL output
and I believe it AIN'T. It is that little dribble that made it into the
line in the first place. a.k.a. the Tx is not generating the full output.
I think this assumption is causing much trouble.

Unless, of course, I have been smoking the wrong brand.

Comments R.C. ?

Steve

Cecil Moore wrote:

Jim Kelley wrote:
Electromagnetic waves can cancel, but rocks can't.

However, two EM waves have to exist before they can cancel.

And that makes rocks like waves?

If they exist, they posses both energy and momentum.

Bet ya can't prove it without first transfering it to something.

73, Jim AC6XG

Cec and Co., I keep telling you it's a waste of bandwidth talking about
'conjugate matches'. In the transmitter/transmission line context conjugate
matches do not exist because nobody knows what the internal impedance of the
transmitter actually is.

There's never any need to know. Of what possible use would it be if you knew
it.?

Except provide a topic of conversation for this newsgroup. ;o)

Jim Kelley wrote:

Cecil Moore wrote:
However, two EM waves have to exist before they can cancel.

And that makes rocks like waves?

That makes real waves tangible like real rocks. The wave particles
are just smaller. OTOH, "People who live in glass houses shouldn't
throw stones." is an intangible.

If they exist, they posses both energy and momentum.

Bet ya can't prove it without first transfering it to something.

_Optics_, by Hecht is good enough for me. "It is possible
to compute the resulting (momentum) force via Electromagnetic
Theory, whereupon Newton's Second Law suggests that the *wave
itself carries momentum*. (all emphasis his, not mine) ... As
Maxwell showed, the *radiation pressure* equals the energy density
of the EM wave. ... When the surface under illumination is perfectly
reflecting, the beam that entered with a velocity of +c will emerge
with a velocity of -c. This corresponds to twice the change in
momentum that occurs on absorption, ..."

It's obvious that the energy in the TV ghosting wave makes a round-
trip to the match-point and back to the RCVR. That's obviously a
change in the direction of momentum of the reflected wave. It is
twice the change in momentum than if it encountered a circulator/
load and was dissipated.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Tam/WB2TT wrote:
You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance?

The question is a little misleading because the direction of the flow of
current changes every half cycle and is transverse, or orthogonal to the
direction of wave propagation. In a transmission line, the current
flows through Z0, ostensibly, which is essentially the impedance from
one conductor to the other at every point along the transmission line.
Other than that, superposed forward and reflected waves behave just as
you described, naturally.

73, Jim AC6XG