Re: +/- 18VDC PS basic questions


"Pooh Bear" <rabbitsfriendsandrelations@xxxxxxxxxxx> wrote in message
news:436322F0.5BB578F4@xxxxxxxxxxxxxx
>
>
>
> The other winding is likely 18-0-18 V AC and used for the split (
bipolar )
> supply. The centre tap connects to the 'ground' of the supply.

There are two windings, one gives 5VAC single-tapped or two leads, the
second has four leads with varying voltages depending on which leads are
measured. No center tap. Too bad.
>
> Note that 18V AC rectified will be about 24V DC. A +/- 24V DC supply will
toast
> most op-amps.You need voltage regulators to reduce it to typically +/-
15V. The
> regulators also remove most of the supply ripple.
>
Ah, now this is useful information. I have been poring over schematics for
power supplies and nowhere is it evident (and I obviously lack the training
to know) that you get higher DC voltage out of a rectifier than the AC
voltage in. Is there a "rule-of-thumb"? I see that above you note about
1.5X multiplier for the DC out of a rectifier. The particular circuit I'm
looking at shows a 32VAC CT transformer which is rectified, filtered,
regulated down to 18VDC, and filtered again. For my particular app, I don't
necessarily WANT a regulated 18VDC, I am okay with letting it float as the
op-amps are protected by more downstream regulators. 32VAC would give
~48VDC? You'd need one helluva beefy regulator (most of them that I've seen
can handle up to 30VDC) and heatsink to drop 30VDC!!! If the voltage is
filtered between the rectifier and regulator, why would you need to filter
it again after the rectifier? Could one assume that you'd need smaller
filter caps as you work your way downstream?

In this regard, I have a 12.6VDC CT x-former (which I thought would not be
enough voltage). I really only need 15VDC, so can I expect ~18VDC out of my
rectifier if I use 12.6VAC in?

While I'm at it with this enlightened audience, I have another basic
question: After the rectifier I have a 3300uF 50V electrolytic "filter cap"
and a 1uF disc "bypass cap". It is my understanding (and, hey, I'm not
batting a thousand here so bear with me) that the filter cap stores up
charge and compensates for voltage drops. Is this right? What does the
bypass cap do?

Thanks in advance for any and all replies.

Dave

> Graham
>


.

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