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It's really tricky to do a frequency domain analysis on a waveform as
distorted as the one appearing on a class C amplifier collector. Unless you're extremely careful and very aware of exactly what you're doing, results are likely to be very wrong. Among other things, you have to rigorously preserve phase information for each frequency. I personally think a time domain approach is much more likely to yield correct results, and this has been done. Search for papers by Sokal and Sokal about class E amplifiers. I believe Frederick Raab did a lot of the mathematical analysis, and published a number of papers on the topic. I'd be very skeptical of any frequency domain analysis, or even general intuitive arguments based solely on frequency domain considerations. You'll probably find more references on class E amplifiers in _Experimental Methods_. Class E amplifiers are essentially class C amplifiers in which the output networks are carefully and rigorously designed (using time domain techniques) to maximize efficiency. Roy Lewallen, W7EL Mike Silva wrote: wrote in message . .. It seems that the formula should be adjusted to account for the conduction angle, e.g. Rl should be smaller by at least a factor of 2 to compensate for the conduction angle. What am I missing? More current is flowing thru the transistor when conducting than would normally be used if class-A ? ... this would average out because it's being pulsed rather than continuous. And it's reasonable to expect higher current in the pulse than Ohm's Law would give, because the pulse has a whole bunch of energy in harmonics, for which the load impedance is lower (the load C reactance being lower). Since the harmonic content of the pulse goes up as the conduction angle gets smaller, it makes sense that the average current through the whole cycle would be somewhat independent of conduction angle. Just don't ask me to prove it! ;-) 73, Mike, KK6GM |
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