Jack, Interesting. I didn't realize it was a Schottky diode. But it makes sense. Using the rule that the current increases by e (2.718) for every 26 mV change in forward voltage: If you assume that the forward voltage drop is 0.3V at (let's say) 1 mA forward current, then at 1 uA (1V into a 1M load), the voltage should drop by a factor of 0.026V x ln(1000) = 0.18V. 0.3 - 0.18 = 0.12V, pretty close.
With a silicon diode the answer would be something like 0.6 - 0.18V = 0.42V. That's why Schottky-diode detectors have better linearity. Alan N1AL On Sun, 2012-01-22 at 14:21 -0500, Jack Smith wrote: > 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 > ______________________________________________________________ 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
