[Elecraft] Explain terms for Elecraft DL1 Power Calculation

[Ron D'Eau Claire]

It does use a 1N511 Schottky diode which as a very low voltage drop -
something like 0.15 V at a mA or two of forward current. 

Ron AC7AC

-----Original Message-----
From: elecraft-bounces at mailman.qth.net
[mailto:elecraft-bounces at mailman.qth.net] On Behalf Of Alan Bloom
Sent: Sunday, January 22, 2012 10:57 AM
To: Chip Stratton
Cc: Elecraft Reflector
Subject: Re: [Elecraft] Explain terms for Elecraft DL1 Power Calculation

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