Gary, you may be right regarding my power output measurement, but I thought
I knew how you calculate DC power and AC power of sinusoidial waveforms on
my scope, which, by multiplying with 0.707 you have to reduce first to the
aquivalent of a DC voltage.

What confuses me in your suggestion (printed below) is that even after you
rectify an AC signal and send it through a filter capacitor you STILL
suggest to multiply it times 0.707.
I would have thought that would give you the wrong result??

Where do I go wrong, if at all??

Uwe


in article , JGBOYLES at
wrote on 3/12/04 17:19:

So a 50 resistor serves as my dummy load and a meter connected to
the dummy load indicates around 10 volts RMS. If I did my homework that
would indicate around 2 watt output.

Uwe, The 10 Vrms that your meter indicates may not be accurate at 3.5 or 7
MHZ. It is probably OK at 60 hz. What you can derive from the 10 Vrms is
that
the AC-1 is producing power, just not sure how much.
One thing you might try is a method to remove the frequency dependency of
your measurements. Use a 10:1 voltage divider (10k and a 1.11k). Run this
thru a Germanium or Schottky detector diode and a .01 filter capacitor. You
now have a DC voltage that is proportional to power, and relatively frequency
independent.
To calculate the RMS voltage across the 50 ohm load: Read the DC volts out
of the detector-Vdc. Then Vrms=(Vdc*.707)*10.
Example: You read 2Vdc out of the detector. Vrms=14.14 volts. Power into
the
50 ohm load is then: 14.14^2/50=4 Watts. The diode drop in the dector will
introduce some error at QRP levels, hopefully not too much for what you are
trying to do.
73 Gary N4AST