Here is the schematic:
The primary also must have a reactance of > 4 x 19R at 7MHz, so the inductance must > 1.7uH. The output ferrite is chosen to be a Binocular design with Al = 385nH, so for >1.7uH we need n = sqrt(L x 1000/AL) or sqrt(1.7x1000/385) = 2t, so the secondary must be 3t to give a ratio (1:1.5) near the required 1:1.6.
The output passes through a low pass filter to remove harmonics. Seems simple, but will it work?
Critical Parts List
Input transformer: 10t:10t, #24 wire, Farnell torroid 180009 (type 61, TN22/14/7mm, AL = 125)
RFC: 10t:10t, #22 wire, Farnell torroid 180009 (type 61, TN22/14/7mm, Al = 125)
Output transformer: 2t:3t, #20 wire, Fair-Rite Binocular Ferrite BN-61-202, Al = 385
LPF: 1.3uH, 3t, #20 wire, Farnell torrid 180008 (type 61 TN13/7/5.4mm, Al = 125) MOSFETs: IRF510, part number 844-IRF510PBF from Mouser Electronics
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