Re: Explain terms for Elecraft DL1 Power Calculation

Posted by Kevin Stover on Jan 22, 2012; 6:17pm
URL: http://elecraft.85.s1.nabble.com/Explain-terms-for-Elecraft-DL1-Power-Calculation-tp7213825p7213841.html

Volts X 1.414 gives you peak volts.
I'm not quite sure why only half the diode voltage drop (.15) is added.
I've only ever seen it all added.
All that squared and divided by the impedance (50Ohms) gives you power.


On 1/22/2012 12:05 PM, 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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--
R. Kevin Stover
AC0H
ARRL
FISTS #11993
SKCC #215
NAQCC #3441

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