Date: Sun, 30 Nov 2014 11:31:31 -0800 From: Darren Pilgrim <list_freebsd@bluerosetech.com> To: Perry Hutchison <perryh@pluto.rain.com> Cc: freebsd-questions@freebsd.org Subject: Re: OT: UPS for FreeBSD Message-ID: <547B7093.1020305@bluerosetech.com> In-Reply-To: <547a52e9.tCBMi6xWobou5Fcd%perryh@pluto.rain.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>
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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. From where did you get your figures?
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