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
