Re: Explain terms for Elecraft DL1 Power Calculation

Posted by Alan Bloom on Jan 22, 2012; 6:57pm
URL: http://elecraft.85.s1.nabble.com/Explain-terms-for-Elecraft-DL1-Power-Calculation-tp7213825p7213937.html

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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