Hi Chris,
Yes, right. But my _expression_ is the same. 0,707 = 1/sqrt(2), so U^2/R
= P =(û/sqrt(2))^2/R =û^2/2/R =û^2/2*R. Measuring 100 Vp at 50 Ohm is
100 W :-)
73, Stefan
Am 09.09.2010 10:11, schrieb Chris:
Hi Stefan,
I thought the power was the RMS
voltage squared (i.e. peak voltage X 0.707) divided by the resistance.
Chris, G4AYT.
-----
Original Message -----
Sent:
Tuesday, September 07, 2010 9:31 PM
Subject:
Re: LF: 137.5kHz WSPR Tuesday evening - reports appreciated
Roger,
For measuring the output power take a 50 Ohm dummy load (50, not 55
Ohms or so) and measure the peak voltage with an oscilloscope. Type
this into the excel file and calculate the output power (P=(û)^2/(2*R)).
73, Stefan
|
|