[AMRadio] Modulation power required

Gary Schafer garyschafer at comcast.net
Thu Feb 14 13:02:51 EST 2008



Since the discussion the other day about reducing drive when using a linear
amp seemed to have some interest maybe this will be of interest to some


For a plate modulated final it is common thought that if you have a 100 watt
input final amp that you need 50 watts of audio to modulate it 100%. But
what does that really mean? You look at the modulator tube specs in the
handbooks and they show how much power they will produce into a certain
load. Start doing the math to figure out what that load is and things start
to get a little fuzzy sometimes.


Take a final that has 1000 volts and draws 100 mA. That is 100 watts input
and common knowledge says that we need 50 watts of audio for 100%

To find the load impedance we divide plate voltage by plate current so we
have 1000/.1 = 10000 ohms that the modulator is going to see.

We know that the peak modulator voltage must match the final plate voltage
for 100% modulation so let’s do it that way first.


With 1000 volts audio applied to the final from the modulation transformer
(this is peak voltage) it is working into 10000 ohms. To find the power
required we do E squared /R. So we have 1000x1000=1,000000 divide by R which
is 10000 ohms. That gives us 100 watts. Wait a minute!! Isn’t it only
supposed to take 50 watts to modulate that 100 watt final?


Here’s the thing that lots of people overlook: The 50 watts that is often
quoted is AVERAGE audio power and that is talking about using a sin wave
single tone. The peak power in a 50 watt sin wave signal is 2 times the
average so all this works out.


The point is that we must remember when we are figuring modulation voltage
and power across the plate resistance that we are dealing with peak voltages
and currents. The average power of a sin wave is going to be ½ that amount.


When we say that a transmitters power output increases by 50% with
modulation we are talking about average power of the modulation. A
transmitter with 75 watts carrier (same as above with 75% efficiency) out
will contain 37.5 watts total side band power so the total transmitter power
out will increase to 112.5 watts with full modulation. But remember that
this is with the transmitter modulated with a single tone sine wave audio

Average voice power is going to be quite a bit less than that as average
voice is anywhere from ¼ to something less than that.


Now we get to PEP, peak envelope power. Remember our peak modulating voltage
above shows that we need that voltage to be the same as the DC plate voltage
for 100% modulation. That effectively doubles the plate voltage at the
instant the audio sin wave is at its peak (peak audio voltage plus DC plate
voltage). When plate voltage is doubled so does current get doubled. 

This makes for 4 times the power at the instant of the audio signal peak
than what we had with just the carrier. 

Just as the peak modulation voltage is only there for the duration of the
audio peak, so is the peak plate voltage there for that duration and so is
the peak power there for the duration of the audio peak. 


That is where the PEP (peak envelope power) figure comes from. It is the
power that is being generated during the peak (or crest) of the modulation
signal (envelope).


The PEP that we are concerned about is the output PEP of the transmitter and
you can work it backwards if you observe the output power of the

Our same transmitter above that produced 75 watts carrier out and had a
total side band power of 37.5 watts average will be used. You can not add
powers together when figuring PEP; you must add the voltages first then
calculate the power just as we did with the peak input power of the final in
the above.


But here is the catch! The Peak Envelope Power is the COMPOSITE of all the
signals involved. That means the two individual side bands and the carrier.
Just as these 3 components make up the familiar modulation envelope that we
are used to looking at on the scope.


75 watts into 50 ohms has a voltage of square root of PxR.  So the square
root of 75x50= 61.2 volts.

Now we find the voltage in EACH of the side bands:

The total side band power is 37.5 watts so each side band will contain 18.75

18.75 watts into 50 ohms = 30.6 volts.

Now we add all the voltages together:  30.6 + 30.6 + 61.2 = 122.4 total

P = E squared /R.

To find the PEP, 122.4 squared = 14982/50 ohms = 299.6 or rounded to 300
watts.  4 times the carrier power.


We are using RMS voltages here rather than peak voltages. Shouldn’t we be
using peak voltages as we did in the plate circuit? 

Well, PEP is defined as the AVERAGE power over at least one audio envelope
cycle. So we need to use the average power of the carrier at least. Average
power is derived from RMS voltage so we use the RMS voltage of the carrier.

With the side band part which represents the envelope we used the RMS
voltage of each individual side band. If we found the PEAK voltage of the
composite side band power (which makes up the envelope) it would work out
the same. For example:

37.5 watts (total side band power) into 50 ohms is the square root of
37.5x50 = 43.3 volts RMS. The peak voltage of that is 43.3x1.414 = 61.2
volts peak. 

Notice that this peak voltage is the same as the sum of the two 30.6 RMS
voltages of each side band. Add that to the 61.2 volts for the carrier and
we again have 122.4 total volts which gives us the 300 watts PEP.


Note that AVERAGE power is what you get when you multiply RMS volts times
RMS current or if you are figuring it with resistance, E squared /R = P.
Where E is RMS voltage.


Some want to call this RMS power which does not exist. It is a wrong
statement as there is only average power.



Gary  K4FMX

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