[AMRadio] Power, and the bad manners on the list

[John E. Coleman]

This may not be my place to speak but here goes.

You were perfectly right to ask about the power thing, and the AMRadio
list was the correct place to do it.  I was really flabbergasted at all
that was said. There have been a number of people ask the same or
similar questions and never been chastised about it but you happened to
get the attention of a few very out spoken and rude people some of whom,
I think, have been removed from the list and or no longer able to post
rude messages there.  At any rate the majority of the folks there are
very helpful and knowledgeable.  So if the ADMIN person doesn't invite
you back (I understand Brian was out of pocket for a while) and when he
returned he said he got rid of the rude ones.)  I asked him to send a
letter to the newcomers that had left the list because of the flames and
invite them back.  But just incase he couldn't get around to it; I
thought I would take it upon my self to do so.  I hope you will come
back; there is enough knowledge on the archives of this list to fill a
library.  You some time have to weed through the BS to get to the good
stuff.  For this I am always sorry.  

In case you didn't ever get a real good explanation.  Here Goes:

I apologize first if some of this is too elementary.
But I always say it never hurts to review. 

Power is of course measured as a product of voltage and current.

In AC, RF or any sine wave type of energy, the voltage and current or
not present all the time as with DC.  It is found that 1/2 the power
would be delivered with AC when the peak of the AC is equal to the DC
test.  In other words if a resistor has 100 volts DC across it and the
resistor dissipates 10 Watts in heat then if 100 Volts Peak (that's 100
volts from 0 center or 200 volts Peak to Peak "PTP") of AC is put across
it will only dissipate 5 Watts in heat.  When the AC voltage is raise to
1.414 (square root of 2) the heat was back to 10 watts dissipation. This
is because the current also had to rise as well, to 1.414 of what ever
it was. So that when the two, Voltage and current, are multiplied it
will equal twice as much. This is why we have the term RMS value ROOT
MEAN SQUARE. The peak voltage times .707 (.707 is sqr of .5) equals the
RMS Voltage

Here is a project that you may like.  Using a 6AL5 diode or equivalent
(must be a tube) connect the plate directly to your 50 OHM dummy load
with short wires. And connect a .05 uf @500 volt ceramic capacitor from
the cathode to ground and light up the tube.  Apply RF to the dummy Load
and measure the DC across the capacitor with a sensitive voltmeter like
a VTVM or a 100K ohm per volt or more VOM.  The Capacitor voltage will
be equal to the peak of the RF Voltage.  Let's say it measures 150 volts
DC.  We must have RMS value to calculate power.  
The RMS of 150 V is (.707 X 150) 106.5 Vrms
Power is E*E / R

106.5 * 106.5 = 11342.25

11342.25 / 50ohms = 226.845 WATTS or about 227 WATTS

It is accurate on HF if the leads to the plate of the 6AL5 or short.
The 6AL5 is a multi diode tube but it is only necessary to use one diode
and the heater XFMR that you light up the tube with should be a stand
alone and do not ground one leg of the XFMR let it float because the fil
to cathode insulation may brake down on the tube if you were to ground
on leg of the fil XFMR.

Now to the PEP thing:

When the AM rig is modulated with a SINE WAVE audio to 100 percent, the
envelope is pinched or it reaches 0 volts RF at the trough of the audio
wave. At the peak of the audio wave the RF voltage has doubled, in the
above example that would be 213 Volts RMS RF

DO the math

213 * 213 / 50 = 907.38 Watts PEP

Peak Envelope Power
(Not to be confused with PTP (peak to peak) voltage or current)

Doubling the voltage will quadruple the power if the load is constant.

As Bacon (WA3WDR) points out, the power of both sidebands added is only
1/2 of the carrier, and interesting thing that can be seen through
vector analysis and more voltage to power calculations.

Working the 1500 W PEP output power thing backwards we see that 1/4 of
that can be carrier 375 Watts output if you are to modulate 100 %

375 / Efficiency*100 = DC input power to the Final Amplifier

If Class C / Plate Modulated with Efficiency of 75%

375 /.75 = 500 watts DC input to the Final

DC Plate Voltage might be 1700 Volts @ 394 milliamps of plate current or
some other product 

And, by the way, you will need 250 watts of sine wave audio to modulate
that 100%.
For Voice completed waves, better figure on 300w or more audio.
 
If your running a linear amplifier and want to run 1500 watts PEP

You will need to tune the amp and driver to max output of 1500 watts and
when in operate mode be sure the drive level to the AMP is reduced to
half RF voltage if looking at a scope or 375 watts if looking at a watt
meter.

Good luck and come back.
John, WA5BXO