Ya lost me there Larry. P = E^^2 / R. R is 50 ohms, P is 10 watts, E works out to sqrt(500) or 22.36 volts. But are we talking RMS, peak, or peak-to-peak?
Let's see, if my number is an RMS and you're number is peak, then 22.36 is about 22.4. Multiply by 1.41 and I get 31.6 Pretty close to your number. Or am I way off base? - Keith - -----Original Message----- From: Larry Phipps [mailto:[EMAIL PROTECTED] It's a standard diode peak detector I believe, so it would indicate 31.7 VDC (minus the drop) at 10 W for a 50 ohm load. I expect the drop is about 0.3 V. Darwin, Keith wrote: > I never completed the RF probe with the K2 (didn't need it). I assume > it simply converts RF AC voltage into DC voltage (probably RMS) so you > can measure it with a DVM, right? So at 10 watts into a dummy load, I > should see 22.4 volts? Does the diode drop some voltage so I'll > actually see 21.7? _______________________________________________ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
