Re: Measuring power
Charles W. Powell
Lloyd,toggle quoted messageShow quoted text
I don’t know about anyone else, but I am measuring power on mine, key down, using a 50 ohm dummy load, and connecting my high impedance oscilloscope leads across the load. The scope is a a Siglent 200 mHz scope, so it is measuring far away from the rolloff. Using the peak-to-peak voltage, I make my calculation by deriving the RMS voltage, then using RMS-squared/resistance. So with about 13.4 DC volts input, my peak output voltage on the HT-20 was 52 volts. ((52 * 0.707/2)^2)/50 = 6.8 watts into 50 ohms. I get approximately the same readings with my MFJ tuner’s power readout and also my QRP-o-meter, although I don’t expect those to be nearly as accurate.
Caveats: I might be entirely wrong in my calculations and I might be missing some subtlety in making this calculation.
Chas - NK8O