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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? You must be talking about solid-state tx when you mention SWR protection. I don't know of any tube tx that have such protection. Any catastrophic failures in tube tx with pi-network output circuits due to poor impedance match at the output without retuning to match the the network to the load simply leaves the tx detuned away from resonance. The result is excessive plate current that would be reduced to normal by resonating the tank circuit. No operator in his right mind would allow the tx to be operated with the tank not tuned for the resonant dip in plate current. 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 :You said two tx operating on two different frequencies, which means the RF signals from the two tx are not phase coherent. In this condition the signal from each tx does enter the other. On the other hand, the wave reflected from a mismatched termination is phase coherent with the source wave, resulting in the addition of the reflected wave to the source wave when either the antenna tuner or the pi-network is adjusted to deliver all available power at the desired grid drive. W2DU |
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