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.
Or am I missing something?
Mike
n2mg@contesting.com
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