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Efficiency and maximum power transfer
"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? 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? RF |
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