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Efficiency and maximum power transfer
"Richard Fry" wrote in message ... "Walter Maxwell" wrote The source resistance appearing at the output of either a Class B or C amplifier is R = E/I, where E is the peak voltage at the output terminals and I is the peak current at the output. Or RMS values can also be used.Since E/I is simply a ratio, R is also a ratio. And we know that a ratio cannot dissipate power, or turn electrical energy into heat, thus the output resistance R is non-dissipative. I have made many measurements that prove this. It is also the reason why reflected power does not dissipate in the tubes, because it never reaches the tubes. The reflected power simply causes a mismatch to the source, causing the source to deliver less power than it would if there were no mismatch. __________ If the source resistance of a tuned r-f PA stage was truly non-dissipative, and the tx simply supplied less power into poor matches, how would that explain the catastrophic failures to the output circuit components often seen when high power transmitters operate without suitable SWR protection into highly mismatched loads? if the load impedance seen at the transmitter terminals is outside the range that it was designed for you end up with either arcing from excessive voltage or meltdown from high currents. remember, the pa output as long as it is connected to a linear load and you consider only the sinusoidal steady state condition can be completely replaced by a lumped load impedance, hence no reflections necessary to figure it out... same load, same result, hence the reflections have no effect on the internals of the pa. Another reality is that r-f power from two co-sited, tuned transmitters on two frequencies in the same band can be present in each others output stage due to antenna coupling, which causes r-f intermodulation between them. The non-linear (mixing) process occurs at the active PA stage. If reflected (reverse) r-f energy never reaches the PA stage as you assert, then how could this IM generation occur? non-coherent input would of course pass through the matching network going into a pa, there is nothing that precludes that. that current would cause mixing in the non-linear tube or transistor and therefore im generation. but that incoming rf is not reflected, it is the incident wave, with nothing coherent to interfer with why would it not pass into the pa? Now, have fun with this one... two transmitters on exactly the same frequency feeding a common load through equal lengths of coax... what happens if you change the phase relationship between them? make it simpler, remove the coax and connect both pa outputs togther with a single load, now no reflections to worry about... does it matter? do the two pa's feed all their power into each other? where do the reflections go?? |
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