On Tue, 10 May 2011 15:34:36 -0700 (PDT), Richard Fry
wrote:

On May 10, 4:37*pm, walt wrote:
This is for Richard Fry,

Richard, because the output source impedance of a p-network following
a tube RF power amp is non-dissipative, no reflected wave enters it to

Here is what I believe you missed: When the 111.11w arrives at the
mismatch, 11.111w is reflected, leaving 100w dissipated in the load.

Walt,

Let us start at T=0. The transmitter is keyed, and instantaneously
delivers 100W to the input of a lossless transmission line. That 100W
travels to the end of the transmission line and encounters a load
connected there having a 2:1 mismatch to the Zo of the transmission
line.

That load will dissipate about 88.889W, and reflect about 11.111W back
to the source.

Likewise, the load will dissipate only 88.889% of the RE-reflected
power from the source for the same reason it dissipated only 88.889%
of the power in the original incident wave.

So even if the RE-reflection from the source is lossless, and phase-
coherent with the original incident power at the load, how does this
yield the 100W of load dissipation that you posit?

RF

73's
Richard Clark, KB7QHC

On Tue, 10 May 2011 15:34:36 -0700 (PDT), Richard Fry
wrote:

So even if the RE-reflection from the source is lossless, and phase-
coherent with the original incident power at the load, how does this
yield the 100W of load dissipation that you posit?

The PI network between the tube and the output connector transforms
the tube Hi-Z to 50 Ohms (or any other suitable Z); however, the same
PI network's input Z (from the perspective of the returning, original
reflection) is NOT 50 Ohms, it is the complex conjugate (whatever 2:1
brings to the table) of the load answering phase-coherency, hence it
reflects the reflection (by the same mechanism of mismatch) to the
dissipative load the antenna, and thus into the æther.

The transmission line is bounded at either end by identical mismatches
in conjugation. There is a "ringing" in the line, which at only one
end there is dissipation (neglecting loss and damage to the tube for
exceeding maximum tolerances, of course).

73's
Richard Clark, KB7QHC

On May 10, 6:48*pm, Richard Clark wrote:
On Tue, 10 May 2011 15:34:36 -0700 (PDT), Richard Fry
wrote:

So even if the RE-reflection from the source is lossless, and phase-
coherent with the original incident power at the load, how does this
yield the 100W of load dissipation that you posit?

The PI network between the tube and the output connector transforms
the tube Hi-Z to 50 Ohms (or any other suitable Z); however, the same
PI network's input Z (from the perspective of the returning, original
reflection) is NOT 50 Ohms, it is the complex conjugate (whatever 2:1
brings to the table) of the load answering phase-coherency, hence it
reflects the reflection (by the same mechanism of mismatch) to the
dissipative load the antenna, and thus into the æther.

The transmission line is bounded at either end by identical mismatches
in conjugation. *There is a "ringing" in the line, which at only one
end there is dissipation (neglecting loss and damage to the tube for
exceeding maximum tolerances, of course).

73's
Richard Clark, KB7QHC


For Richard F.

Where does the 88.889w of power delivered to the load come from? I
calculate it as 100w. 100/111.11 = 0.9000. Since 11.111w of power
reflected at a 2:1 mismatch with a 100 source and lossless line is
indisputable, then the mismatched load will dissipate 90% of the
forward power. 90% of 111.11w is 100w. So please explain the
significance of 88.889w. I can't see it as relevant.

Walt

On May 10, 7:22*pm, walt wrote:
On May 10, 6:48*pm, Richard Clark wrote:









On Tue, 10 May 2011 15:34:36 -0700 (PDT), Richard Fry
wrote:

So even if the RE-reflection from the source is lossless, and phase-
coherent with the original incident power at the load, how does this
yield the 100W of load dissipation that you posit?

The PI network between the tube and the output connector transforms
the tube Hi-Z to 50 Ohms (or any other suitable Z); however, the same
PI network's input Z (from the perspective of the returning, original
reflection) is NOT 50 Ohms, it is the complex conjugate (whatever 2:1
brings to the table) of the load answering phase-coherency, hence it
reflects the reflection (by the same mechanism of mismatch) to the
dissipative load the antenna, and thus into the æther.

The transmission line is bounded at either end by identical mismatches
in conjugation. *There is a "ringing" in the line, which at only one
end there is dissipation (neglecting loss and damage to the tube for
exceeding maximum tolerances, of course).

73's
Richard Clark, KB7QHC

For Richard F.

Where does the 88.889w of power delivered to the load come from? I
calculate it as 100w. 100/111.11 = 0.9000. Since 11.111w of power
reflected at a 2:1 mismatch with a 100 source and lossless line is
indisputable, then the mismatched load will dissipate 90% of the
forward power. 90% of 111.11w is 100w. So please explain the
significance of 88.889w. I can't see it as relevant.

Walt

Richard, the point I missed until rereading your post, is that there
is total re-reflection of the reflected wave at either the output of
the correctly-adjusted pi-network, or at the output of the antenna
tuner, not the same mismatch as between the line and the load.

Walt

On May 10, 5:48*pm, Richard Clark wrote:
The transmission line is bounded at either end by identical mismatches
in conjugation. *There is a "ringing" in the line, which at only one
end there is dissipation (neglecting loss and damage to the tube for
exceeding maximum tolerances, of course).

If, as you have posted, dissipation occurs only at one end of the
transmission line -- which presumably for the most useful benefit
would be at the end of the line opposite the source, e.g., the load --
and the incident power generated and delivered to that line by the
final r-f stage in the transmitter is a constant, then how could
"damage to the tube" occur?

Just to note that as a field engineer for RCA Broadcast for many years
I have had to troubleshoot and repair failures that occurred in
transmission lines, transmitters, and output networks that resulted
from reflections -- either from within the transmission line alone, or
together with the load connected at its far end.

RF

On May 10, 6:22*pm, walt wrote:
Where does the 88.889w of power delivered to the load come from?
I calculate it as 100w. 100/111.11 = 0.9000.

Yes, Walt, 100/111.11 = 0.9.

But a transmitter generating 100W does not deliver 111.11 watts in the
original incident wave arriving at a load VSWR of 2:1 w.r.t a lossless
transmission line used to convey that power. It delivers only 100W.

Of that 100W, 88.889W is absorbed by the load, and 11.111W is
reflected back to the source.

THAT is where the 88.889W comes from.

Also note that 0,9 does not equal 0.8889.

RF

On Tue, 10 May 2011 16:28:23 -0700 (PDT), Richard Fry
wrote:

If, as you have posted, dissipation occurs only at one end of the
transmission line -- which presumably for the most useful benefit
would be at the end of the line opposite the source, e.g., the load --
and the incident power generated and delivered to that line by the
final r-f stage in the transmitter is a constant, then how could
"damage to the tube" occur?

Hi Richard,

Note how you were careful to specify "phase coherent."

This is dangerous to speak of power in this way, energy would be
better, but I will proceed with caution thrown to the winds:

When the reflection hits the transmitter port, it sees the conjugate
of the load. As the load was 2:1 (and trusting everyone's math with
my own rounding) this means that the transmitter port also presents a
2:1 mismatch at this near end of the coax. Some (10%) of the power in
the reflection is re-reflected to the load. Some (90%) of the power
in the reflection re-enters the transmitter to add to the next cycle
of power going out. In this scenario, that amounts to 9W going into
and being coherently impressed upon the tube. The transmitter, then,
becomes a buck-boosted 109W source conjugately matched into a 2:1
mismatched load.

9 additional watts at the plate isn't much, but I dare say there are
many problems with transmitter tolerances even if you could tune into
a short, or an open. Yes, somewhere the SWR meter says everything is
hunky-dory, while the smoke is poring out.

Another way of putting it is saying the ringing line is holding 10W
until it is dissipated. So far, only the antenna qualifies as
dissipation.

Through the magic of Integral Calculus, all the bits and pieces
ringing around the system sum up to the whole numbers we started with.

No doubt this will provoke examinations, re-examinations,
cross-examinations, innuendo, rants, and raves (or possibly disdainful
silence) - something for everyone to enjoy. ;-O

73's
Richard Clark, KB7QHC

On May 10, 7:28*pm, Richard Fry wrote:
On May 10, 5:48*pm, Richard Clark wrote:

The transmission line is bounded at either end by identical mismatches
in conjugation. *There is a "ringing" in the line, which at only one
end there is dissipation (neglecting loss and damage to the tube for
exceeding maximum tolerances, of course).

If, as you have posted, dissipation occurs only at one end of the
transmission line -- which presumably for the most useful benefit
would be at the end of the line opposite the source, e.g., the load --
and the incident power generated and delivered to that line by the
final r-f stage in the transmitter is a constant, then how could
"damage to the tube" occur?

Just to note that as a field engineer for RCA Broadcast for many years
I have had to troubleshoot and repair failures that occurred in
transmission lines, transmitters, and output networks that resulted
from reflections -- either from within the transmission line alone, or
together with the load connected at its far end.

RF


Yes Richard, I'm familiar with your work at RCA Broadcast Div.
However, in my posts I'm concerned only with tube rigs used in the
Amateur Service. In these rigs reflected power doesn't cause
overheating, or other damage to the tubes. But I'll qualify that
statement--if the pi-network is originally resonated into, say, a
dummy load, and is then switched to a mismatched line without
retuning, the reactance appearing at the input of the mismatched line
detunes the network and the result is the same is if the network was
left off resonance initially. As you well know, a mis-tuned pi-network
results in excessive plate current, which, if high enough will damage
the tube. Consequently, it's the mis-tuning that causes the damage,
not the reflections per se.

Incidentally Richard, I was also an electrical engineer with RCA, from
1949 thru 1980, first at the RCA Labs in Princeton and beginning in
1958 with Astro.

Walt

On May 10, 7:00*pm, Richard Clark wrote:
Another way of putting it is saying the ringing line is holding 10W
until it is dissipated. *So far, only the antenna qualifies as
dissipation.

Apparently you have never had to repair the components of a
transmitter PA, an output network, or a transmission line that arced
over and/or melted down due to load reflections (coherent, or not).

RF

On Tue, 10 May 2011 17:14:56 -0700 (PDT), Richard Fry
wrote:

On May 10, 7:00*pm, Richard Clark wrote:
Another way of putting it is saying the ringing line is holding 10W
until it is dissipated. *So far, only the antenna qualifies as
dissipation.

Apparently you have never had to repair the components of a
transmitter PA, an output network, or a transmission line that arced
over and/or melted down due to load reflections (coherent, or not).

Not more than a quarter million Watts, fur shure. Does it show?

Seriously, if that was your only rant, based on that snippet, like
Wim, you have just indicted your own witness (experience?).

RF

73's
Richard Clark, KB7QHC