Date: Sat, 29 Nov 2014 15:12:41 -0800 From: perryh@pluto.rain.com (Perry Hutchison) To: ralf.mardorf@rocketmail.com Cc: erichsfreebsdlist@alogt.com, kudzu@tenebras.com, freebsd-questions@freebsd.org Subject: Re: OT: UPS for FreeBSD Message-ID: <547a52e9.tCBMi6xWobou5Fcd%perryh@pluto.rain.com> In-Reply-To: <20141129113018.17759e2a@archlinux> References: <CAHieY7QGp2ELF-R91eu=vSrPsimVmVNJQ4kfucQ56PR7EEZmig@mail.gmail.com> <m57qdq$did$1@ger.gmane.org> <54777AB1.9010800@bluerosetech.com> <m581p1$65m$1@ger.gmane.org> <54779629.302@bluerosetech.com> <alpine.BSF.2.11.1411271433320.60866@wonkity.com> <5478BD4F.7020306@yahoo.com> <5478BEE6.30308@bluerosetech.com> <5478CC08.9090307@yahoo.com> <20141128204722.561f948e@archlinux> <5478F16A.80605@yahoo.com> <CABhTyc9m7fOoeV170dj=foAhmyYWphzc8KD8wBacu5gNRPhT%2BQ@mail.gmail.com> <54791d3a.w/pI0kak03d%2B3nKC%perryh@pluto.rain.com> <CAHu1Y71vVbdx6Yd1VbE7kb_8k9O5UG93RXEaORPU0tULCpMsCQ@mail.gmail.com> <20141129113405.3d1bd1d6@X220.alogt.com> <54798883.saa13h6lE6rPwZCf%perryh@pluto.rain.com> <20141129113018.17759e2a@archlinux>
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Ralf Mardorf <ralf.mardorf@rocketmail.com> wrote: > On Sat, 29 Nov 2014 00:49:07 -0800 > perryh@pluto.rain.com (Perry Hutchison) wrote: > > multitapped > > 120VAC ==> step-down ==> 6VAC ==> full-wave ==> ~15VDC ==> battery1 > > transformer rectifier > > | > > +---> 3VAC ==> full-wave ==> ~7.5VDC ==> battery2 > > rectifier > > Now I'm confused. Being con-fused was in a different branch of the thread :) > The ratio should be around 1.41, the square root of 2 We may be thinking of different "full wave" rectifier designs. Presuming a reasonable amount of capacitance on the output, a half wave (single diode) rectifier should produce a DC output roughly equal to the peak value of the AC input, which is indeed ~1.4 * the RMS AC voltage. A two-diode full-wave rectifier using a center tap will also produce DC of ~1.4 * the RMS voltage of the entire winding (and with the advantage of less ripple). However, a full wave bridge (4 diodes) should produce a DC output roughly equal to the peak-to-peak value of the AC input, or ~2.8 * the RMS voltage. > "-" 1.5V assumed it's a diode bridge rectifier. Yes, at these voltages the diode drop really should not be ignored: 6 VAC(RMS) * 2.8 - 1.5 = 15.3 VDC leaving 3.3 VDC for regulator drop to produce 12 VDC regulated. That may be a bit thin; perhaps we should use the same winding ratio as in an old-time filament transformer: 6.3 VAC => 16 VDC or so. Alternatively, 12.6 VAC-CT (center-tapped) would produce approximately the same result with only two diodes. OTOH 3 VAC(RMS) * 2.8 - 1.5 = 6.9 VDC leaving only 1.9 V for regulator drop to produce 5 VDC regulated. That "3VAC" winding should be more like 3.8 VAC (or 7.6 VAC-CT), producing 9 VDC.
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