Perry Hutchison perryh at pluto.rain.com
Sat Nov 29 23:13:20 UTC 2014

```Ralf Mardorf <ralf.mardorf at rocketmail.com> wrote:
> On Sat, 29 Nov 2014 00:49:07 -0800
> perryh at 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

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.
```