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Old September 29th 06, 04:16 PM posted to rec.radio.cb
Frank Gilliland Frank Gilliland is offline
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First recorded activity by RadioBanter: Jul 2006
Posts: 432
Default Look at what Griffey thinks his amp will do...

On 29 Sep 2006 08:01:56 -0700, "Telstar Electronics"
wrote in
.com:

Frank Gilliland wrote:
You claim 516 watts PEP and 55% efficiency. And I'll be super-nice and
say that's at 14 volts and you aren't going into saturation. So that
means:

516 watts / 14 volts = 36.86 amps
36.86 amps / 55% efficiency = 67 amps
67 amps / 2 transistors = 33.5 amps Ic on each transistor.


Amplifier efficiency isn't based on peak power (PEP) you putz... see
http://www.rf-amplifiers.com/index.php?topic=dc_input



Once again you didn't understand what I wrote. I wasn't calculating
amplifier efficiency; I calculated peak collector current -based on-
your very own amplifier efficiency claim. Would you rather I use the
transistor's collector efficiency of 35%?


Are you really this stupid?... or are you just putting us on?



I have thought of asking the same question of you, but I haven't
because you would ignore it just like all the other tough questions.
That, and I already know the answer.