W6WRT wrote:
> Divide the plate voltage (under load) by the plate current
> in amps, then divide
> that by a factor of about 1.6. I say "about" because the
> exact factor depends on
> how the tube is biased. The closer the tube is biased
> toward class A, the
> smaller the factor and the closer toward class B the
> higher the factor. Most
> modern amps run class AB2 and will be somewhere between
> 1.5 and 1.7.
>
> Any recent ARRL handbook will have a further explanation.
Bill, that information is exact! But you stopped short
of actually answering the question.
I think you meant to also add on . . .
- Jim W4ENE
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