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: [email protected] [mailto:[email protected]] 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 ______________________________________________________________ Elecraft mailing list Home: http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/mmfaq.htm Post: mailto:[email protected] This list hosted by: http://www.qsl.net Please help support this email list: http://www.qsl.net/donate.html
