Alan, Actually the correct formula is even easier. If you note, you are squaring the 2 x .707 term, and the overall result of that is 2. So the formula becomes:
P (watts) = 2(Volts + diode drop)^2/50 or even easier P (watts) = (Volts + diode drop)^2/25 Substitute whatever you think is correct for the diode drop - see Jack Smith's recent post for details. The formula above was supposed to make it into the DL2 manual at one time, but I see it has not been inserted. 73, Don W3FPR On 1/22/2012 1:57 PM, Alan Bloom wrote: > I think it can be made clearer by re-casting the equation from: > > P(watts)=((Voltsx1.414) + 0.15)^2 / 50 > > to > > P(watts) = [2 x 0.707 x (Volts + 0.106)]^2 / 50 > > A diode rectifier is a peak detector. So you multiply by 0.707 to > convert from peak to RMS and then multiply by 2 to correct for the 2:1 > voltage divider. I'm not sure why only 0.1V is added to compensate for > the diode voltage drop. It's true that the voltage drop is less with a > high load impedance, but 0.1V seems too small. > > Alan N1AL > ______________________________________________________________ Elecraft mailing list Home: http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/mmfaq.htm Post: mailto:Elecraft@mailman.qth.net This list hosted by: http://www.qsl.net Please help support this email list: http://www.qsl.net/donate.html
