Date: Mon, 1 Dec 2014 15:28:48 +0000 From: RW <rwmaillists@googlemail.com> To: freebsd-questions@freebsd.org Subject: Re: OT: UPS for FreeBSD Message-ID: <20141201152848.33a54703@gumby.homeunix.com> In-Reply-To: <547BF99D.3070801@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> <20141130210305.2d62bbb1@gumby.homeunix.com> <547BF99D.3070801@bluerosetech.com>
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On Sun, 30 Nov 2014 21:16:13 -0800 Darren Pilgrim wrote: > On 11/30/2014 1:03 PM, RW wrote: > > On Sun, 30 Nov 2014 11:31:31 -0800 Darren Pilgrim wrote: > >> 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. > > The mean voltage is the output DC voltage. Defining the DC component of a pure rectified sine wave as the output DC voltage isn't really useful, unless the design smooths to that value. > >> 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. > > No, the voltage across the capacitor will rise to 0.900*Vrms. Look > at it this way: > > The output of the diodes is a time-varying (AC) signal with a DC > offset. The cap provides a low-impedance path for the AC signal, > turning the voltage cycle into a current cycle that accumulates a > charge in the cap. Once the cap is fully charged, it fully negates > the AC component and you're left with just the DC offset of > 0.900*Vrms. When you connect a smoothing capacitor it charges to the peak rectified voltage; when the rectified voltage move off-peak, the capacitor can't discharge through the diodes. Most linear regulators operate with a smoothing capacitor kept close to the peak voltage to minimise the headroom needed by the regulator, which in turn reduces the heat generated. This is the most substantial reason why a square wave is problematic, it can't have both the correct RMS and peak voltage. In practice a computer power-supply would have active circuitry between the rectifier and the capacitor for power-factor correction. In that case the voltage on the capacitor is arbitrary. > > 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. > > > A bridge rectifier doesn't use a center-tapped transformer. You > wouldn't see 2*Vpeak in a two-diode rectifier either. The voltage > between the center tap and either diode tap is at most Vpeak. > Bridge rectifiers are commonly used with a grounded centre-tap to provide positive and negative rails. In that case the smoothed voltage on the bridge output is ~ 2*sqrt(2)*Vrms, but only because Vrms is specified per secondary winding.
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