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

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

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??

Uwe, Wouldn't be the first time I was wrong.
If you rectify and filter an AC waveform, and the read the DC voltage, this
is Vpeak of the AC. In the example I gave, you read 2 vdc out of the detector
circuit. Multiply this times the X10 divider, and you get 20vdc which is Vpeak
across the 50 ohm load. To get the rms voltage across the load
Vpeak*.707=20*.707=14.14 Vrms. 14.14Vrms across 50 ohms=4 watts.
BTW, the reason I suggested a voltage divider (attenuator) rather than
reading directly with a VOM or scope is to reduce the likelyhood of the scope
probe or meter impedance from detuning the AC-1 causing further errors.
If you are using a scope, measure the Vpeak across the 50 ohm resistor,
multiply by .707 to get Vrms.
73 Gary N4AST

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??

Uwe, Wouldn't be the first time I was wrong.
If you rectify and filter an AC waveform, and the read the DC voltage, this
is Vpeak of the AC. In the example I gave, you read 2 vdc out of the detector
circuit. Multiply this times the X10 divider, and you get 20vdc which is Vpeak
across the 50 ohm load. To get the rms voltage across the load
Vpeak*.707=20*.707=14.14 Vrms. 14.14Vrms across 50 ohms=4 watts.
BTW, the reason I suggested a voltage divider (attenuator) rather than
reading directly with a VOM or scope is to reduce the likelyhood of the scope
probe or meter impedance from detuning the AC-1 causing further errors.
If you are using a scope, measure the Vpeak across the 50 ohm resistor,
multiply by .707 to get Vrms.
73 Gary N4AST

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