>>1.) I'll agree that the peak power HANDLING capability of the tube can be
>>related to its anode voltage.
>
>? this is not what I said.
I know. I was intentionally putting words into your mouth. :-)
>
>> 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?
>
>? depends on HV sag.
>
>> Is it enough to make a difference?
Again, does the gain drop enough to make any difference in output power? Of
course perhaps I should have asked how much difference you have measured in
peak output powers between SSB and CW. Is it watts? 10s of watts? 100s of
watts???
If we hold
>>the gain of the tube constant then all bets are off.
>
>? To do this, all supply potentials to the tube must be regulated.
If the gain vs. supply voltage varies substantially then we cannot assume it to
be constant. However, my gut tells me the variation is minimal and therefore
we can approximate the gain to be constant. If the variation in gain of a tube
is not minimal (less than a few tenths of a dB) then please provide what that
number is.
>
>>If the anode supply
>>drops then following Ohms law, current increases and power remains the
>>same.
>
>? no.
>
No? Huh? If gain is approximated to be constant, it has to be the case as I
explained earlier. I left response below:
>>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.
>>
>
>? With light loads, the filter C essentially charges to the peak v. from
>the transformer. The filter C maintains the charge. until increased load
>consumes it.
>
Agreed. The way you wrote it, it sounded like you were saying that as the caps
begin to discharge the increase the anode voltage.
>>
>>OK. So then. If the SSB duty cycle is lower, then that means that the
>>voltage on the anode is less
>
>? hardly. As the load on an unregulated anode supply decreases, volts
>out increases.
Typo here by me. I meant to say,
>> "If the SSB duty cycle is lower, then that means that the voltage on the
>> anode is greater >>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.
>>
>? even though the duty cycle of the SSB and CW loads on the unreg.
>supply are different?
Again:
Let's say peak power out of the rig is 100 Watts. The rig will deliver 100
Watts upon keydown on CW or on instantenous voice peaks on SSB. If the gain of
the tube in question under FULL load (after voltage sag) is 10, then the peak
output power will be 1000 Watts.
If instantenous peak power out of the transceiver is identical for SSB or CW,
then it matters not what the duty cycle of the loads are on the unregulated
supply. They are reaching the same value at the peak.
I agree that the filter cap action of the HV supply will prevent the SSB peak
from sagging the HV as much as for CW. But by how much?
Also again, I point out that if your argument is correct, then peak SSB power
on a meter should be higher than CW peak power. The original poster claimed
just the opposite.
I'll grant you there *may* be a difference, but by how much?
>>Average sag might be. But again, we are talking about peak levels. At
>>the peaks, all is the same.
>>
>? It takes a calibrated 'scope and x1000 probe to see what be.
I'll take your word for that.
>>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?
>
>? With ALC action, there is sure to be overshoot. However, with the
>drive control backed off a bit, the ALC threshold is not reached and a
>series of dashes exhibits flat tops.
Are you saying that there is no overshoot of any kind without ALC? So the
transceiver generates a perfect square wave?
>
>>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.
>>
>? Quite true. However, with an oscilloscope and HV multiplier probe,
>one can see what's what, wherever. Never leave home without one.
Well, if there is overshoot on the CW wave, how do you know that the overshoot
you see on the output of the amp is not due to the input signal??
cheers, Rich.
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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