That's a reasonable alternative, but provides less power, since it only
rectifies half the available cycles, so q/t is lower, though not halved. Ripple
is likely to increase, though. In any case, it's as I often say, "Where there's
a will, there's mourners..."
If he really wants to do this, a little calculation will tell him how much power
he needs, a little more calculation will tell him which of the methods I
previously mentioned will be needed in order to reach his goal, and a meter on
the transformer when it's loaded with an appropriate load resistor, will tell
him what the AC voltage he has to work with under full load is. Different
manufacturers interpret that differently, though I suspect that valve heaters
are pretty similar.
If he has a 6.3 volt output at full load, he'll certainly be able to do the job
with a schottky bridge and a series resistor, and may not even have to replace
the classic regulators with a low-dropout replacement. Moreover, since it's
unlikely he'll be operating at the full load current, or anywhere near it, he'll
probably have on the order of 5-10% headroom without objectionable ripple if his
filters are large enough. I've got one system with 800K microfarads at 10
working volts as its filter, with just such a low-voltage transformer. The low
voltage is high enough to feed the regulators, yet low enough that heat isn't a
problem.
Dick
----- Original Message -----
From: "Tony Duell" <ard_at_p850ug1.demon.co.uk>
To: <classiccmp_at_classiccmp.org>
Sent: Wednesday, August 29, 2001 4:56 PM
Subject: Re: S-100 Power Supplies: thanks!
> >
> > It's 6.3 volts * SQRT(2) less two diode losses. If one uses a SCHOTTKY
bridge
>
> Alternatively, if you use a centre-tapped 12.6V winding (or maybe 2 6.3V
> windings in series) you can use the 2-diode 'biphase' rectifier circuit
> and have than output of 6.3 * sqrt(2) - _one_ diode drop.
>
a resonable alternative, though that wasn't on his list.
>
> -tony
>
>
Received on Wed Aug 29 2001 - 19:02:19 BST