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Date: Sun, 30 Nov 2014 11:31:31 -0800
From: Darren Pilgrim <list_freebsd@bluerosetech.com>
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To: Perry Hutchison <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?