Alan: A couple years ago, I measured the DC output for a 1N5711 Schottky diode at 2, 10 and 21 MHz with RF input from essentially 0 to 1.8V peak (half-wave rectification). The diode operated into a high Z load (an HP digital multi-meter) and the RF was from an HP signal generator.
A plot of this is at http://www.cliftonlaboratories.com/diodes_for_rf_probes.htm After reading your question about the diode drop today, I ran a linear regression fit of the average of the 2, 10 and 21 MHz data, over the range starting a bit above the knee, at 0.3V peak up to the 1.8V peak. Below this point, the diode is in square law range and a simple fixed voltage drop assumption fails. Over this linear range, the Y intercept is -0.11V, with an R^2 > 0.999 and standard error of 0.00275V. Into a high Z load, therefore, 0.11V is a suitable diode offset. The actual diode offset, leaving aside unit-to-unit variations, will be a function of temperature and load impedance amongst other things. Given the other inaccuracies in the process, 0.1V is a good estimate. Jack K8ZOA 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 > > 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 > ______________________________________________________________ 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
