To: | "[email protected]" <[email protected]>, [email protected] |
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Subject: | LF: Re: I: Fw: For today the FETs survived... |
From: | Andy Talbot <[email protected]> |
Date: | Sun, 28 May 2017 20:18:12 +0100 |
Cc: | [email protected] |
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First thing I noticed is that your turns ratio on the output transformer doesn't look right. You quote " ... with primary winding of 15 turns and secondary of 12 turns..." 180V DC in a half bridge is 180V peak-peak square wave. The fundamental sine part of that is 4/pi * 180 = 229V pk-pk so is 229V /[2.SQRT(2)] = 81V RMS To a good approximation RMS(fund) from a half bridge is Vrms(fund) = 0.45VDC For 500 Watts out, Rload = 81 ^ 2 / 500 = 13 ohms So to match to 50 ohms you need a turns ratio of SQRT(50/13) = 1.9:1 so call it 2:1 Keeping 12 turns on the secondary means you need 6 turns on the primary When operating at reduced voltage, the power out will vary exactly as the square of the voltage. Recalculating from first principles for a 12V supply: 12V DC = 12V pk-pk = 12 / [2.SQRT(2)] * 4/pi = 5.4V RMS (fundamental) in 13 ohms should give 5.4^2/13 = 2.2 Watts check using ratio of voltages, squared : (12V/180V) ^ 2 * 500W = 2.2 Watts which is the same as above. QED Your 15:12 ratio result sin a load impedance of (15/12)^2 * 50 = 78 ohms At 40V DC == 18V RMS(fund) that will give 18^2/78 = 4.1 watts which is actually LESS that you are seeing - the 2* discrepancy is odd, but the low power is in the area of what you measured.. Andy G4JNT On 28 May 2017 at 19:34, [email protected] <[email protected]> wrote: Hi Chris, |
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