> If one were to build a class "A" RF tube amplifier capable
> of producing an
> AM signal of 375 watts carrier and 1500 watts PEP, what
> would be the plate
> dissipation required by the tube? (I know how stupid and
> inefficient this
> is, but I really would like an answer)
A linear amplifier is also sometimes called "efficiency
modulation".
What happens is this:
100% modulated AM requires PEP to be four times the carrier
power. This occurs in a plate modulated class-C low mu
triode because the power output follows the square of the
anode voltage. When the anode voltage is doubled on
positiove peaks, power output is quadrupled. Double the RF
output voltage is 4 times the PEP.
With a linear stage driven by a 100% modulated AM signal,
the drive voltage is only doubled. This creates a little
problem since efficiency is not constant with varying drive
power. Normally the carrier efficiency is half or less of
the peak power. You double the drive voltage, and the
efficiency also doubles.
So at carrier with a linear PROPERLY tuned, you might have
25% or less efficiency with carrier. This means 75% of the
plate input power is heating the anode. This means anode
heat is about 3 times the carrier power. With 375 watts
carrier you would have 1125 watts anode heating and 1500
watts dc anode input.
At peak power it might be 3000 watts plate input, 1500 watts
output, and 1500 watts dissipation. But that is intermittent
dissipation.
> As background, I have an AL-1200 (3CX1200A7) which starts
> to flattop beyond
> 375 watts of carrier at 100% modulation and this amp is
> class "AB2", so I
> know class "A" would require significantly more
> dissipation than 1200 watts.
Dissipation has nothing to do with flat topping. Emission
does. You may not have the loading control open far enough.
You have to load for the PEAK power, not anything less.
My AL1200 will easily do 1500 watts PEP on AM with 375 watts
of carrier. Be prepared for a lot of heat in the tube anode.
73 Tom
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