Re: Explain terms for Elecraft DL1 Power Calculation
Posted by
Don Wilhelm-4 on
Jan 22, 2012; 8:14pm
URL: http://elecraft.85.s1.nabble.com/Explain-terms-for-Elecraft-DL1-Power-Calculation-tp7213825p7214098.html
Alan,
Actually the correct formula is even easier. If you note, you are
squaring the 2 x .707 term, and the overall result of that is 2. So the
formula becomes:
P (watts) = 2(Volts + diode drop)^2/50
or even easier
P (watts) = (Volts + diode drop)^2/25
Substitute whatever you think is correct for the diode drop - see Jack
Smith's recent post for details.
The formula above was supposed to make it into the DL2 manual at one
time, but I see it has not been inserted.
73,
Don W3FPR
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
>
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