I've always been reading my AM modulated signal on a Bird wattmeter and trying to set the carrier to 1/4 of the peak envelope power (PEP).

Recently, I was told by a CBer that you only can see it that way on a scope, and that with a PEP meter, you should be setting the carrier to 1/2 the PEP. Is this correct? Have I been doing it wrong?


2 Answers 2


Say the unmodulated carrier amplitude (voltage) is 1. At 100% modulation, the envelope will vary between amplitude 0 and 2. So at peak, the amplitude is twice an unmodulated carrier, but because power is proportional to the square of amplitude, power is $2^2=4$ times the carrier power.

So I think your CBer friend has it backwards.

If you are looking at the signal on a scope, you are likely looking at amplitude, and peak amplitude should be twice that of an unmodulated carrier, assuming 100% modulation.

If you are monitoring the signal with a peak power meter, peak power should be four times that of an unmodulated carrier.


It depends...

With a 100% steady modulated carrier, the PEP will be 4 times the unmodulated carrier power. The average power under this condition will be 1.5 times the unmodulated carrier power with 2/3 of this contained in the carrier and 1/3 in the sidebands (1/6 each in the USB and LSB sidebands).

The real issue is - are you obtaining 100% modulation? For example, a plate modulated 6146 final (or nearly any pentode or tetrode final) will only reach a PEP that is 2 times the unmodulated carrier because the positive peaks are about 50% of the negative peaks. In this case your friend would be correct.

Finally, keep in mind that if you use a scope instead of a PEP meter, you will likely be viewing the amplitude of the voltage - not power - at the point of insertion of the scope. This will work for comparing relative voltages of the carrier to the modulated signal since both are sampled at the same impedance. However, recall that the voltages observed will be relative to the square root of the powers present.


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