http://elecraft.85.s1.nabble.com/Explain-terms-for-Elecraft-DL1-Power-Calculation-tp7213825p7214243.html
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
0.18V. 0.3 - 0.18 = 0.12V, pretty close.
0.42V. That's why Schottky-diode detectors have better linearity.
> 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:
[hidden email]
> >>
> >> 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:
[hidden email]
> >
> > 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:
[hidden email]
>
> This list hosted by:
http://www.qsl.net> Please help support this email list:
http://www.qsl.net/donate.html>