On Tue, 06 Nov 2007 04:32:37 -0800, Denny wrote:

The whole point of the magnetron's
success in war was its robustness in the face of catastrophic
mismatches. The PFN might flame out or the thyratron burst, but the
magnetron would survive.

A good thing too, or we would all be eating cold leftovers without
that magic 'radar range'...
Jeez, I still have the original Frigidaire that I bought in the late
60's when I worked for GM - got an employees special deal just in time
for xmas, the first family in our neighborhood to have a radar range,
the neighbors marveled... The kids whipped that puppy day and night...
It has outlived three houses... It has been relegated to my R&D shop
where it happily heats epoxy, paint, putty, and cups of tea with
aplomb... I cannot even begin to guess how many tens of thousands of
times it has been cycled... Dozens of times a day for a quarter
century when the kids were still home...

denny / k8do

Hi All,

May I come in with some comments from a somewhat different perspective concerning where the reflected power
goes, or where it doesn't go?

Because I'm not particularly adept concerning solid-state tx, this discussion will involve only tube tx with
pi-network output circuits. We'll assume that when the tx has been adjusted to deliver all its available power
into a resistive load, the output resistance appearing at the output terminals of the tx is equal to the load
resistance. For those who may disagree I refer them either to Chapter 19 in Reflections 2, or the May/Jun
issue of QEX, where I report the results of measurements that prove the statement to be true.

Consequently, we'll begin by adjusting the tx to deliver all its available power into a 50-ohm resistive load,
a real 50-ohm physical resistor. Next, we'll change the load to be the input of a 50-ohm lossless transmission
line terminated with a 3:1 mismatched load. What impedance will the tx see? Until we specify the actual load
terminating the line and the electrical length of the line, the tx may see any impedance appearing somewhere
on the 3:1 SWR circle of the Smith Chart. What will happen when the tx sees this mismatched load? First,
because the 3:1 SWR results from a voltage reflection coefficient of magnitude rho of 0.5 (and a power
reflection coefficient of 0.25), this mismatch will cause the tx to reduce its output to deliver only 75
percent of the power delivered into the matched 50-ohm load.

Will the reflected power enter the tx and cause heating? Or will it cause cooling? The answer is that the
reflected power does not enter the tx. So what does happen?

By changing the length of the transmission line at will, we can cause any impedance found on the 3:1 SWR
circle on the Smith Chart to appear at the input terminals of the line. Let's assume the impedance found there
is 50 ohms x 3 equals 150 ohms. This impedance is a lighter load than the original 50-ohm load, so the plate
current will be less than with the 50-ohm load. However, the reflected power did not enter the tx, it only
caused the 150 plus j0.0-ohm impedance to appear at the line input.

Now the important point: The tx would have responded in exactly the same manner if its new load had been a
physical resistor of 150 ohms instead of a 'virtual' resistance of 150 ohms resulting from voltage/current
relationship appearing at the input of the line.

Let's now consider if the line-input impedance is 50/3 ohms, which equals 16.667 ohms. In this case the new
load is lower than the original 50-ohm load, resulting in overloading the tx, causing the plate current to
rise above the rated level, thus causing overheating. But the overheating was caused only by the increase in
plate current resulting from the lowered load resistance, not by reflected power entering the tx. Again, the
same condition would have occurred if a physical resistor of 16.667 ohms had loaded the tx instead of the same
value of virtual resistance appearing at the line input.

Now a third case important to the issue, that in which the impedance appearing at the line input is 50 plus
j57.738 ohms, which also appears on the 3:1 SWR circle. In this case the 57.738 ohms of inductance simply
detunes the pi-network away from resonance, causing plate current to rise, increasing the temperature, but
still reducing the output power to 75 percent of that with the 50-ohm load. And again, precisely the same
result would occur if the new load had been a physical resistance of 50 ohms in series with a physical
57.738-ohm inductor. (It should be obvious that a simple readjustment of the plate capacitor of the pi-network
would return the network to resonance, with the result that the operation of the tx would be identical to that
when loaded with the 50-ohm physical resistor.

The purpose of this commentary is to show that reflected power does not enter the tx, and that the tx cannot
distinguish between a 'virtual' load impedance appearing at the input of a mismatched transmission line and a
physical load comprising a resistor in series with an inductor-the results will be identical in either case.

Walt, W2DU