From: John Woodgate
Subject: Re: PSU's: Exceeding transformer rated current
Date: Wed, 8 Jan 2003 11:13:10 +0000
Organization: JMWA Electronics Consultancy
Reply-To: John Woodgate
NNTP-Posting-Date: Wed, 8 Jan 2003 13:14:20 +0000 (UTC)
X-Newsreader: Turnpike (32) Version 4.01 <5Z8C9wtxbnpWyFnyfFzqmVF739>
I read in sci.electronics.design that Richard wrote (in ) about 'PSU's:
Exceeding transformer rated current', on Tue, 7 Jan 2003:
>Where the transformer supplies an ac current, ie say without any rectification,
>transformer rated at 10v rms @ 1A supplying it's max rating has 0.5 watts
>in the internal resistance (I^2 x R; 1A x 0.5R) . Connect as half wave
>the available voltage for R load is 4v when dissipating 10 watts in R load and
>current is 2.4A. Dissipation in R internal is 2.4A ^2 x 0.5R = 2.88 watts. So,
>are supplying 10watts in both instances to a load, but greater copper losses
>have used half-wave rectification. And that's assuming the 2.4A rms current,
>would be 6.7892504A peak would not melt the wire.
Ah, well, in that case, your original question meant 'can I get 20 W
*into the load* from a 10 VA transformer, using a half-wave rectifier?'.
The answer is that you can't even get 10 W, *because the rectifier
circuit is nowhere near 100% efficient*.
The question of whether you could get 20 W *out of the transformer* is
more reasonable, but the answer is still 'no', and again you can't even
get 10 W out.
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
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