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 On Sun, 2012-01-22 at 18:05 +0000, Chip Stratton wrote: > The DL1 50 ohm dummy load is basically composed of two 25 ohm resistors in > series. For power calculations, RF is taken off between the two resistors, > rectified via a diode with a 0.3v drop. The DC is then smoothed with a .01 > uf capacitor. before reaching the test point. > > The power calculation is then given as P(watts)=((Voltsx1.414) + 0.15)^2 / > 50 > > I'm wanting to understand where the terms in this equation come from? I > presume 0.15 is somehow related to the 0.3 Volt drop across the diode, but > how is 0.15 arrived at and why not 0.3? And where does 1.414, obviously the > sqrt of 2 come from - the divided voltage from the dummy load? > > Thanks for any enlightenment! > Chip > AE5KA > ______________________________________________________________ > Elecraft mailing list > Home: http://mailman.qth.net/mailman/listinfo/elecraft > Help: http://mailman.qth.net/mmfaq.htm > Post: mailto:[email protected] > > This list hosted by: http://www.qsl.net > Please help support this email list: http://www.qsl.net/donate.html > ______________________________________________________________ Elecraft mailing list Home: http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/mmfaq.htm Post: mailto:[email protected] This list hosted by: http://www.qsl.net Please help support this email list: http://www.qsl.net/donate.html
