Sure, Tom.
The efficiency of an amplifier is defined as its output power divided by
the DC input to the stage, or:
e = Po / Pi
If you get 40W out and have 60% efficiency, the input power is Po / e = 40
/ 0.6 = 67W. So the tubes dissipate the difference between input and
output power, or 67W - 40W = 27W.
The tube is just a bit beyond its rating at 40W output.
73,
Ed N3CMI
"Tom Elmore"
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Sent by: To
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Subject
10/07/2004 12:28 [AMRadio] Tube Plate Dissipation
PM Ratings
Please respond to
Discussion of AM
Radio
<[EMAIL PROTECTED]
qth.net>
Can someone please clarify for me the plate dissipation ratings on
vacuum tubes. I am in the process of restoring a Stancor 60N that uses a
HK24 tube and the dissipation rating is about 25 watts. Isn't the 25 watt
rating Ep*Ip and is this a maximum value that should never be exceeded? I
believe this transmitter will put out 30 to 40 watts of carrier and if it
were say 60 efficient wouldn't that bump up the plate dissipation to around
65 watts exceeding the dissipation ratings.
Thank You
Tom Elmore KA1NVZ
Anchorage Alaska
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