>
>I'll risk sticking my neck out again into the battle
>(I learned something last time)
>
>
>>>With half of the total P being dissipated in the
>>>generator, and half of the power being delivered
>>>to the load R, the efficiency is obviously 50%.
>
>
>Is it being suggested, in essence, that if an amplifier delivers (as per
>Bird wattmeter) 1500W RF to the coax, that due to the power transfer
>rule, the amp's tubes must be generating at least 3KW RF? My HV and
>plate current suggest otherwise.
? To get a handle on this, one has to look at instantaneous values of e,
i, and p. For example a 3-500z typically runs quite linearly at 1.2 peak
amperes of anode current, so the peak world is a whole nuther ballgame,
Mike. .
Rich...
R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures
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