[Boatanchors] 813 grid to filament short / AM PEP

[Gary Schafer]

If we get the math right PEP is easier to understand.

First, there is a difference between peak power and peak envelope power
(PEP).

A 375 watt carrier requires 187.5 watts of AVERAGE power to modulate to 100%
which equals 1500 watts PEP.

Note: Average power is derived from RMS voltage X RMS current or RMS
voltage^2/R. It is not RMS power as many call it.

As we know a 100% modulated transmitter must double the plate
voltage/antenna voltage in order to achieve this.

Ever wonder how that 187.5 watts of audio power makes the plate voltage
double? 

Let's say we have a DC plate voltage of 2500 volts and running 375 watts DC
input for carrier. Our transmitter is 100% efficient. (only for ease of
explanation)
This means that our plate load impedance is 16,666 ohms. (P=E^2/R)
We know that to modulate to 100% our modulator must provide 2500 volts peak
voltage in order to double the plate voltage.

To find the average modulator power required we first need to find the RMS
value of the peak voltage needed. So we multiply 2500 x .707 = 1767.5 volts
RMS. To find average modulator power we use ohms law P = E^2/R or 1767.5 x
1767.5= 3124056 / 16,666 = 187.5 watts average modulator power.

With a constant load impedance of the final when the plate voltage doubles
so does the plate current double. That gives 4 times the power input at the
peak of the audio voltage.

LOOKING AT THE OUTPUT OF THE TRANSMITTER:

375 watts into 50 ohms has a voltage of 137 volts RMS. If the voltage
doubles that gives 274 volts at the 50 ohm antenna. To find the power
related to this voltage we square the voltage and divide by 50 ohms (ohms
law) 274 x 274 =75076 / 50 = 1500 watts at the peak of the modulation
envelope.

The output signal is composed of 3 signals:
The carrier and two identical side bands.

How do we get 137 volts from 187.5 watts of audio?

Since 1/2 of the 187.5 watts of modulation power is in each side band, we
have 93.75 watts in each side band. At 50 ohms that is 68.5 volts for each
side band at the 50 ohm antenna. We add both of these together for 137 volts
RMS of voltage in the side bands. This equals the carrier voltage of 137
volts.
Adding all 3 together we get 274 volts RMS on the antenna at the crest
(peak) of the audio signal. This is equal to 1500 watts Peak Envelope Power.
(E^2/R)

Viewing the composite AM signal on a scope we see the peak to peak voltage.
When modulated 100% we see the scope voltage double. This is the result of
the addition of the 3 signals.

In order to find PEP we add voltages not power.


Per the FCC, The definition of PEP is the AVERAGE POWER contained in 1 cycle
at the PEAK of the modulation envelope.

This is the rule that we have so we should follow it and educate others of
the proper way of operating.
There is enough confusion out there about this so promoting "run what you
want" etc doesn't help anyone understand.

In the days of reading only DC input power to transmitter one could run any
amount of positive peak modulation and still be legal. PEP could have been
many KW under the old rules.



73
Gary  K4FMX


> 
> On the PEP issue, here is a brain exercise. I know there is a
> mathematical
> basis for PEP, and, that mathematically, 375 watts AM is 1500 watts PEP.
> So, lets assume (to simplify the math) that our transmitters are 100%
> efficient. On CW, if we output 1500 watts, PEP is 1500 watts, and our
> 100%
> efficient amplifier draws 1500 watts from the power supply. Makes sense.
> Now, if we go to AM, and we put out 375 watts 100 % modulated, the RF
> puts
> out 375 watts, and the modulator adds 187.5 watts. Total draw from the
> power supply is 562.5 watts, yet produces 1500 watts PEP. The 1500 W PEP
> on
> CW requires 1500 watt dissipation in a dummy load, yet the AM 1500 watt
> PEP
> only requires a 562.5 watt load. Perhaps a single PEP value is not he
> best
> way to specify transmitter power. --73, Mike, WV2ZOW
>