There's a new revision of the DL1 manual (Rev C) that should appear on the Elecraft web site soon. It has this new formula:
P(watts) = (V + 0.25)2 /25 where V= volts and 0.25 is the RF voltage drop across the diode. Ron AC7AC -----Original Message----- Hi all, Thanks for the tips. I might have taken the instructions too literally and not taken into account the receiver attenuating the signal. However, I did confirm that AGC is turned off on the receiving radio, so that can be ruled out. What is hard for me to understand is that the signal is load and clear at 10 Watts, but drops off significantly at 11 Watts when the high power relay kicks in. In my mind, there shouldn't be that much of a difference from 10 to 11 watts, one way or the other, to kick in any attenuating circuits or AGC. Measuring the actual power will be my next step: I'm relying on using a DL1 and taking the voltage measurements and converting it into power. Now this has me worried that I'm using the wrong formula since I've seen at least two floating around on the reflector. Here's what I'm using: Power (Watts) = (2x((DC Volts + 0.15)^2))/50 Or: add 0.15 to the measured voltage, squaring the sum, multiplying the result by 2, and then dividing the final product by 50 Can you all verify that this is the correct Power formula for the DL1? I'm using the DL1 up until about 25 watts, then I'm switching over to a 100 Watt dummy load. 73 N5BCN - Brian ______________________________________________________________ Elecraft mailing list Home: http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/mmfaq.htm Post: mailto:Elecraft@mailman.qth.net This list hosted by: http://www.qsl.net Please help support this email list: http://www.qsl.net/donate.html
