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Total 3 documents matching your query.

1. Re: LF: CLASS D E (score: 1): Author: "Andy Talbot" <[email protected]>; Date: Mon, 5 Nov 2007 12:26:01 +0000; Its not difficult to measure RF voltage at this frequency. A scope calibration at DC will extrapolate accurately. Then you can measure I using a current transformer / resistor. Getting within a few p; /rsgb_lf_group-archives/html/rsgb_lf_group/2007-11/msg00390.html (11,356 bytes)
2. LF: CLASS D E (score: 1): Author: "hamilton mal" <[email protected]>; Date: Mon, 05 Nov 2007 11:40:46 +0100; Hi all Tnx to all for the information, but some of the figures do not work out. How are the % figures measured. One simple method is to calculate the DC input PWR in watts (I x V) then using an RF cu; /rsgb_lf_group-archives/html/rsgb_lf_group/2007-11/msg00452.html (8,690 bytes)
3. Re: LF: CLASS D E (score: 1): Author: John GM4SLV <[email protected]>; Date: Mon, 5 Nov 2007 12:13:59 +0000; Again...with my TX At 50W output :- Dc supply = 18.7v at 3.4A Efficiency = 78% The math's is trivial..... 18.7V x 3.4A = 63.58W DC input 50W RF output (as measured on a homebrew power meter calibrate; /rsgb_lf_group-archives/html/rsgb_lf_group/2007-11/msg00483.html (11,038 bytes)

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