Date: Sun, 30 Nov 2014 21:03:05 +0000 From: RW <rwmaillists@googlemail.com> To: freebsd-questions@freebsd.org Subject: Re: OT: UPS for FreeBSD Message-ID: <20141130210305.2d62bbb1@gumby.homeunix.com> In-Reply-To: <547B7093.1020305@bluerosetech.com> 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> <547a52e9.tCBMi6xWobou5Fcd%perryh@pluto.rain.com> <547B7093.1020305@bluerosetech.com>
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On Sun, 30 Nov 2014 11:31:31 -0800 Darren Pilgrim wrote: > On 11/29/2014 3:12 PM, Perry Hutchison wrote: > > 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. > > An unloaded, ideal, single-phase FWR produces Vdc = 0.900Vrms. The > factor is 2sqrt(2)/pi, to be precise. A half-wave rectifier produces > half that. That's the mean voltage, which is of limited relevance. > From where did you get your figures? He's referring to a rectifier charging a smoothing capacitor; the voltage will rise to sqrt(2)*Vrms. He's added on an unnecessary factor of two - possibly because he's mixing it up with a bridge-rectifier in a centre-tap configuration.
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