>>What difference does it make that the anode supply is unregulated when it
>>comes to peak power readings?
>
>? peak power is related to anode supply volts. When the anode begins to
>draw heavy current, the charged filter C initially delivers a bit more
>volts for a mS or so..
Two problems here:
1.) I'll agree that the peak power HANDLING capability of the tube can be
related to its anode voltage. But to me the peak of a particular signal is
based on the level of the input signal and the gain of the tube. If you mean
to imply that the tube has less gain with less anode voltage, I'd agree, but
how much less? Is it enough to make a difference? If we hold the gain of the
tube constant then all bets are off. If the anode supply drops then following
Ohms law, current increases and power remains the same. In the real world, I
would bet that given the few hundred volt sag in the anode supply that the gain
of the tube doesn't change all that much to be noticeable. But if a tube has a
gain of 10 dB and you input 100 Watts, it will output 1000 Watts. If the
amplifier has an efficiency of 50% this means you are dissipating 2000 Watts
(input power). Input power is the product of plate current and plate voltage.
If the input power remains constant then as plate voltage dips, plate current
increases. So I have a technical problem with your argument.
2.) Let me requote you for clarity:
>? peak power is related to anode supply volts. When the anode begins to
>draw heavy current, the charged filter C initially delivers a bit more
>volts for a mS or so..
OK, second problem. You say the charged filter C initially delivers a bit more
volts for a mS or so.
HOW?? The filter caps help hold the anode voltage closer to the no-load value
but they certainly can't add MORE voltage to it can they??? If the HV is at
5KV under not load, the voltage on the filter-C is 5 KV (no load), so how can
the filter caps INCREASE the voltage? I fail to understand what you are
saying. Perhaps I am reading wrong.
>
>> The CW signal is going to pull the anode
>>supply down to a specific level. The SSB signal will pull that anode
>>supply down to the same level at its peak.
>
>? HV sag depends on duty cycle. SSB duty cycle is typically 15%, except
>on Channel 6. . CW duty cycle is about 30%.
OK. So then. If the SSB duty cycle is lower, then that means that the voltage
on the anode is less which means the average voltage on the anode will be
higher. Therefore, by your argument, the higher the voltage on the anode, the
more power. Then SSB should see HIGHER peak power than CW because the sag is
less. This is not the case according to the original poster.
Even allowing for your argument, still however, at the PEAK which is what we
are talking about, the SSB and CW powers will be identical. The sag will be
the same. And power out will be the same.
>
>> I fail to see how regulating
>>that supply would make a difference in PEP levels between CW and SSB.
>>
>? Because CW vs.SSB HV sag is different with an unreg. HV supply.
Average sag might be. But again, we are talking about peak levels. At the
peaks, all is the same.
You also made one other comment in a previous message that I question:
>>Peak and average power for a CW signal is identical.
>>
>? ... only if the anode-supply is regulated. For a garden-variety
>amplfier with an unregulated anode-supply, an oscilloscope indicates more
>RF peak volts at the beginning of a dash than it does at the end. On my
>SB-220, the difference is more than 5%, and so is the HV sag.
Are you sure this difference that you see from the beginning of a dash to the
end isn't the typical overshoot one sees on a normal CW signal? If the
transceiver doesn't put out a true square wave, but a ramped wave with a little
overshoot, this will show up on the output of the amplifier and consequently in
your RF voltage, HV sag, etc.
73,
Jon
KE9NA
--------------------------------------------------------------------------
The Second Amendment is NOT about duck hunting!
Jon Ogden
jono@enteract.com
www.qsl.net/ke9na
"A life lived in fear is a life half lived."
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