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[AMPS] Peak reading vs True peak

To: <amps@contesting.com>
Subject: [AMPS] Peak reading vs True peak
From: measures@vc.net (Rich Measures)
Date: Wed, 23 Jun 1999 13:35:02 -0700

? ? ?  WARNING   This reply is too long.  

>>>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. :-)
>>
?  hardly amusing 

>>> 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?  

?   The bottom-line is that less anode-supply volts makes less suds. 

  
>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???

?  Depends on the amplifier.  Depends on the microphone gain setting.  
>
> > > 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?  

?  Confuse not RL with output. 

>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?  
>
?  You gotta measure how much for the particular amplifier.  

>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.
>
?  A measurement with a calibrated 'scope and HV multiplier probe is 
needed. 

>I'll grant you there *may* be a difference, but by how much?
>
?  Unless one measures the potentials with the appropriate test 
equipment, it's guesswork.  
>
>>>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?  

?  In radios that have an independant Drive control, and use a regulated 
Collector or Drain supply, it is possible. 

> So the transceiver generates a perfect square wave? 

? to my eye, the envelope shape is rectangular, but only if No ALC is 
going on.  At the output of the amplifier, the envelope is a tilted 
rectangle.  
>> 
>>>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??
>
?  One either either applies arithmetic to the measurements, on one turns 
down the Drive control until transceiver ALC stops mucking up the 
envelope. .  
>
-  cheers, Jon


Rich...

R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures  


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