The original poster was using a sensor connected to a P3. For me, the big use for this is to view the modulation pattern of my transmission. I have other devices (like the KPA500 itself) that tells me the power. And if I really want to see the match on the antenna feed (which for my tribander should be close to 50 ohms), I will use my W2 and sensor between the KAT500 and the antenna feed line. I doubt there will be much distortion in the modulation pattern with a mismatch, though, Just don’t want to surpass the KAT500's 10:1 maximum rating.
OK, so truth be told, I use it to make sure my signal is going out. And to see the nice pictures it produces… 73, Jack, W6FB > On Apr 19, 2023, at 2:42 PM, Al Lorona <alor...@sbcglobal.net> wrote: > > Yes, if you're willing to do the math in your head, or can set up your fancy > wattmeter to do the math for you, this'll work. > > But the main reason why I don't like the idea of putting the wattmeter (or > sensor, actually) in a non-50-ohm location is the additional uncertainty due > to the mismatch(es). In the original poster's situation, there would be a > mismatch at the interface between the input of the sensor and the KAT-500's > output, and another mismatch at the interface between the output of the > sensor and the input of the feedline. > > How large would the additional uncertainty be? It depends, but assuming that > the wattmeter sensor has a (very good) match of 1.2:1, and the antenna has > 4:1 -- which is not out of the question for many antennas-- the additional > uncertainty is about ±1.0 dB. > > This means that if the actual forward power were 500 W, the wattmeter may > read (after doing the math mentioned above) as much as 1 dB lower, which is > 397 W, or 1 dB higher, which is 630 W. That's a whole heck-of-a-lot of error > there that many hams wouldn't tolerate. > > Nobody had raised the issue of uncertainty and that's why I wanted to point > it out. Y'all can have the last word on this. > > R, > > W6LX/4 > > ____________________________________________________ > I'm fairly sure that (forward power) - (reverse power) gives the correct > nett output power, before cable and antenna losses. I'd need to review > the maths to be sure. Most reflected power ends up re-re-reflected, as > additional forward power. A high SWR will giver reflected power almost > as high as forward power. > > -- > David Wooley > > > ______________________________________________________________ > 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 > Message delivered to jackbrin...@me.com ______________________________________________________________ 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 Message delivered to arch...@mail-archive.com
