Posted by
Jack Smith-6 on
Jan 22, 2012; 7:21pm
URL: http://elecraft.85.s1.nabble.com/Explain-terms-for-Elecraft-DL1-Power-Calculation-tp7213825p7214002.html
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.htmAfter 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
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