Volts X 1.414 gives you peak volts. I'm not quite sure why only half the diode voltage drop (.15) is added. I've only ever seen it all added. All that squared and divided by the impedance (50Ohms) gives you power.
On 1/22/2012 12:05 PM, 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 > -- R. Kevin Stover AC0H ARRL FISTS #11993 SKCC #215 NAQCC #3441 ______________________________________________________________ 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
