On Sun, 25 Mar 2007 19:14:07 -0800, Richard Clark wrote:

On Sun, 25 Mar 2007 22:20:25 GMT, Walter Maxwell
wrote:

Richard, assume a mismatched load has produced both voltage and current refleftions on the line that result in
a particular reactive impedance at the line input. The line input is connected to the output of the
transceiver that has a pi-network output coupling circuit.

I call this condition 1. It exhibits a mismatch and it exhibits the
probability of the reflected energy being absorbed by the source to
the degree of the phase relationships.

Richard, although it exhibits a mismatch, and thus detunes the source, the probability of the reflected energy
being absorbed by the source is zero. The additional power dissipated in the source is due to lowered
impedance of the network resulting from off-resonance operation, thus increasing the plate current. The
reflected energy does not enter the network, but only results in a decrease in the power delivered relative to
that when the reactance in the load is cancelled by correct retuning of the source network.

When the network has been adjusted to deliver all
the available power into the line the output source impedance is the complex conjugate of the line input
impedance.

I call this condition 2. It exhibits just what you describe:

In this condition the reflected voltage and current waves are totally re-reflected back into the
line, while adding in phase to the voltage and current waves from the source, respectively. Consequently, the
reflected waves do not pass rearward through the network to be incident on the plate. Only if the network is
mistuned, such as being connected to the reactive input of the line without being retuned to resonance, in
which case the excessive plate current due to being mistuned will result in an inordinate amount of heating of
the plate.

It seems to me that in your initial post in the original thread (that
was largely ignored for comment) you made mention of injecting a
signal from an external source into the mouth of the dragon for the
purposes of measuring the source Z. Am I wrong?

Yes, you are wrong here, because I made no mention of the 'mouth of a dragon'. That comment must have come
from another poster, twarn't I.

From March 14:
"2. The amplifier is now powered down and the load resistance RL is
measured across the input terminals of the resonant pi-network tank
circuit (from plate to ground) with an HP-4815 Vector Impedance
Meter."

Richard, try this on for size and then determine whether you believe RF understands the function of the
phasing in impedance matching:

...

Now, when adjusting the output network of a tube-type transceiver to deliver all the available power into a
line having reflections, the adjustment of the network accomplishes the same function as the stub on the line
in the above discussion. Consequently, this is the reason why the reflected power is totally re-reflected at
the output terminals of the network, and is never seen at the plate of the amp to cause heating.

If you follow my separation of arguments above, you will find it
demands that the source MUST have the intervention of an outside agent
to resolve its probability of facing destructive energy. In the arts,
this is called Deus ex Machina. In the world of science, we would
have to say that the source is extremely non-linear when its internal
state of Z (a distinctly non-Thevenin characteristic) changes to
follow the load.

In short, of course no energy finds its way in, we twisted knobs to
make that a self-fulfilling prophecy. Without this intervention
reflected energies present the real probability of destruction by
heat.

Twisting the knobs is what adjusts the output impedance of the network to the complex conjugate of the load.
But without this intervention the real probability of destruction by heat is not resulting from the reflected
energy reaching the plate--it is only that the reflected energy detuned the network, causing the plate current
to rise because the network is off resonance.

This is the concept I believe Richard Fry is not appreciating. If I'm wrong on this I hope he'll straighten me
out.

I will let him speak for himself.

OK, Richard F, speak up.

73's
Richard Clark, KB7QHC