Jim Candela jcandela at prodigy.net
Sun May 7 13:32:22 EDT 2006

```Rick asks:

Jim, correct me on my math if I am heading the wrong
direction.
Since I don't have a true RMS voltmeter can't I take
the square root of
the power dissipated in a resistor across the output
to find the actual
RMS voltage?

E^2=PR, so E = sq root (PR)

Yes, but how do you measure P?

Another idea:
If you have an RF ammeter, this will measure rms
current regardless of ac waveform or even dc, then
I=E/R so E = IR. Pick a known resistor that will work
with the voltage and ammeter you have, and do the
calculation. The ammeter must be the thermoucouple
type however to get the correct answer.

One drawback of the "quasi-sine-square" wave from a
Ferro transformer is that certain power supplies are
designed to charge the capacitors to the peak value of
the AC voltage, and the designers expect the peak to
be 1.414 times the RMS value. With a square wave the
peak and RMS are the same, and with a Ferro you might
have a value of 1.2 or so. Certain equipment like a
SSB linear with a full wave voltage doubler P/S will
be way low on peak power when powered from a Ferro
versus normal sine wave power.

Some of the power inverters on the market that
transorm say 12 volts DC to 115 vac do so with a peak
either side of the 60 HZ half wave pulse to in effect
maintain a workable RMS at 115, and yet peak to about
150 to satisfy capacitor input power supply needs.
This can be done quite efficently, and is pretty
clever.

Regards,
Jim
JKO

```