"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.

Finally someone who is getting closer to the flaw in the way many
mis-interpret the maximum power transfer theorm concept.
--
Steve N, K,9;d, c. i My email has no u's.

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

Cecil Moore wrote:

snip

For what it is worth, I believe that the first homo sapien
originated about a quarter of a million years ago and was
a female with dominant genes.
--
73, Cecil http://www.qsl.net/w5dxp


My wife has me convinced that ALL women have the dominant genes!!!!

Deacon Dave

The first ghost is the most visible and the offspring become less and less
visible. Ghosts of ghosts are hard to see. If you're a whiteman you
haven't seen any skinwalkers, aka Navajo ghosts..

--
73
Hank WD5JFR
"Cecil Moore" wrote in message
...
Henry Kolesnik wrote:
SWR ghosts are usually smear because the transmission line is short and
the
displacement fo the image is small compared to object reflection shost
which
have a greater image displacement becasue the reflected signal travels
over
a greater distance.

If reflected waves don't make a round trip back toward the source
and get re-reflected, how can there be ghosts? :-)
--
73, Cecil http://www.qsl.net/w5dxp



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If the anser were at the library I'd already have it and wouldn't be here.

--
73
Hank WD5JFR
"Dave" wrote in message
...

"Cecil Moore" wrote in message
...
Dave wrote:
of course it is against guidelines. you were starting a new thread
without
changing the subject. in effect hijacking the thread for your own
discussion.

Of course, it is not against the guidelines as the subject was
"Reflected
Power". I simply continued to talk about reflected power, the subject of
the thread. That's a broad subject. Perhaps the originator erred in
choosing
that broad of a subject but he chose that subject nevertheless. All of
my
postings to this thread have been about reflected power, including this
one,
in defense of my postings about reflected power.

then obviously you just struck on the convenient subject line and ignored
the original question, subverting the whole thread for your own twisted
use... including the argument that you were not taking over the
discussion.
this obviously just raises the noise floor and does nothing to help the
original poster who by now is likely heading for his local library if he
is
smart.

Richard Clark wrote:
Strange belief for a creation of barely 5000 years old.

Wow, religious dogma in both your fields of religion and electronics?

What has this to do with Reflected Power?

You asked - I answered.



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Jim Kelley wrote:
... the exchange becomes degraded to ascii mud wrestling.

That's exactly why you are in my email kill file. Please find
someone else with which to enjoy that favorite sadistic sport
of yours.
--
73, Cecil http://www.qsl.net/w5dxp



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Thanks, I knew this but what I don't know is why a final doesn't dissipate
the reflected wave but just reflects 100% I assume. Transistor and tube
finals dissipate a bunch producing the RF but what is the mirror, check
valve or diode that keeps it reflecting

--
73
Hank WD5JFR
"Jim Kelley" wrote in message
...
Henry Kolesnik wrote:

I know that any power not dissipated by an antenna is reflected back to
the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an
open
is required to reflect power and I'm searching for which it is, an open
or a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or
analogy
and some math wouldn't hurt.
tnx
Hank WD5JFR

Hi Hank,

Here's a link which talks about (and illustrates) the physics of waves
at a boundary.

http://www.physicsclassroom.com/Class/waves/U10L3a.html

It shows what happens at the two extremes for transverse waves along a
string - not unlike electromagnetic waves in a transmission line in some
ways. In one instance, the string is fastened directly to the boundary
- a rod in this case. This is analogous to a short across a
transmission line. It will be seen that a reflection occurs, and that
the wave becomes inverted upon reflecting.

In the other case, the string is fastened to a ring which can slide
freely up and down the rod. This case is analogous to an unterminated,
or open transmission line. It can be seen that this too causes a
reflection, only this time the wave is reflected back without a phase
reversal. The amplitude of the reflected wave in both these cases
equals the amplitude of the incident wave.

Now imagine that some friction between the sliding ring and the rod can
be added in varying amounts. This friction would be proportional to the
conductance in an electrical circuit, and would be infinite at one
extreme and zero at the other. The greater the friction, the greater
the conductance (and the lower the electrical resistance). As we begin
to increase the friction (conductance) from zero, the amplitude of the
inverted, reflected wave begins to decrease. The decrease in amplitude
continues with increasing friction until the amplitude of the reflected
wave becomes zero. It could be said that this value is equal to the Z0
of the transmission line.

As the amount of friction is increased still further, a small reflection
once again begins to appear. Only now the phase is opposite from what
it was before. Further increases in friction produce further increases
in reflection amplitude until the amplitude of the reflected wave once
again equals the amplitude of the incident wave.

What we noticed in the exercise is that there is some value of friction
(conductance or resistance) for which no reflection occurs. The exact
value depends on the medium through which the wave is propagating. All
other values produced a reflection.

There is no perfect explanation, and this certainly isn't a perfect
analogy but I hope that it will at least help give you a little more of
a feel for the idea.

73, Jim AC6XG

Henry Kolesnik wrote:
Thanks, I knew this but what I don't know is why a final doesn't dissipate
the reflected wave but just reflects 100% I assume. Transistor and tube
finals dissipate a bunch producing the RF but what is the mirror, check
valve or diode that keeps it reflecting

There is none! The definition is the problem. It is simply a copout.
Sources can dissipate reflected energy. The amount is unknown.
--
73, Cecil http://www.qsl.net/w5dxp



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On Mon, 24 May 2004 17:01:44 -0500, Cecil Moore
wrote:

Richard Clark wrote:
Strange belief for a creation of barely 5000 years old.
Wow, religious dogma in both your fields of religion and electronics?
Nice to see you accede.
What has this to do with Reflected Power?
You asked - I answered.
Did you? What was the original question? [accredited stock response]