>>>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.
? this is not what I said.
> 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? 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 anode supply
>drops then following Ohms law, current increases and power remains the
>same.
? no.
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???
? 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.
>
>If the HV is at 5KV under not load, the voltage on the filter-C is 5 KV (no
load),
? The charge on the filter C decreases under load.
> so how can the filter caps INCREASE the voltage? I fail to
>understand what you are saying. Perhaps I am reading wrong.
>>
? yes
>>> 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
? hardly. As the load on an unregulated anode supply decreases, volts
out increases.
>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?
>>
>>> 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.
>
? It takes a calibrated 'scope and x1000 probe to see what be.
>
>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?
? 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. .
>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. .
>
- cheers, Jon
Rich...
R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures
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