From: Keith R. Williams
Subject: Re: Smallest component to step down 240VAC to 18VDC?
Date: Wed, 9 Oct 2002 13:51:47 -0400
References: <email@example.com> <firstname.lastname@example.org> <email@example.com> <firstname.lastname@example.org>
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In article <email@example.com>, a?n?g?e?
> On Wed, 9 Oct 2002 18:41:21 +0200, "fred bartoli"
> >Just because :
> >- your laptop will be *DIRECTLY* connected to mains without isolation. Just
> >a good way to kill yourself.
> >- 120V (mains) * sqrt(2) (rectification + cap) = 170V; then (170V-16V) *
> >4.5A = 690W to keep you hot in case your laptop couldn't succeed in making
> >you cold for the rest of your brand new eternal life...
> I don't quite understand this isolation part.
> The 230VAC comes in, the diode rectifier would change it to 230VDC
Not quite; 320V (230 * sqrt(2)).
> Then spread across 10 (okay 6 might had been too little) regulators,
> each of them drops 7V, and puts out 16V to the laptop.
Even if it were 230V that it would take 30 regulators ( (230-16)/7 ) if
each was 7V. But since it's 320V, that's *43* regulators.
> If the laptop
> draws the full 4.5A, then the regulators would be burning something
> like 25W, which isn't too much right?
Well, the power dissipation would be (230V-16V)*4.5A or over 1300W (and
you think Athlons are hot ;-)! You could dry your hair while you
design your web pages!
Remember, these regulators are floating at 320V. A failure could put
the full 320V across your laptop, or yourself. I agree with the rest.
This is a *VERY* bad idea. Don't even *think* of doing this.