JWG wrote:
In short, how does a 5-watt CB produce 12W PEP in the SSB mode?

Once the carrier and one sideband are filtered out of the signal to
produce a SSB signal, and this is fed to the CB's RF amplifier (which
would generate 4-5W of AM), why wouldn't the result just be 4-5W of
SSB?

It is often described as the power being focused into one sideband,
and the increased bandwidth efficiency is clear, but: if the power
input to the final amplifier is the same as with AM, where do the
"extra watts" come from? Or is the amplification somehow applied in a
different way to the SSB signal? Not sure how to conceptualize this.

Thanks!


A 4 watt AM signal sends out 4 watts when there is no modulation (A0)
To fully modulate an AM signal to 100%, you mix 50% more to it, in this
case, your audio. This creates 2 identical mirror image sidebands, each
being 1 watt. Added together, this would show 6 watts on an RMS (average)
reading meter. Ohms law being what it is, and power being related to
ohms law, when you increase the voltage 50% to a fixed resistance (your
antenna), the current also increases by 50%, and now your PEAK TO PEAK
power is now 16 watts. The allowed increase from 4 to 12 comes from the
fact that the FCC allows the power of the selected sideband to be the
amount of peak power that 80% AM modulation would put out. That is part
of the increase. The other part of the increase is that the receiver is
more sensitive when it only has to listen to half the bandwidth of an AM
signal.

All told, sideband is a much more efficient use of power and radio spectrum.