John --

> Target voltage is 9VDC at 1A.

Whatever. I thought Olin was using 5V in his example.

The OP was looking for +/- 9VDC at a few tens of mA. We've already
established that his capacitors had far too low a voltage rating and were
breaking down internally.

> OK.  Using Olin's formula.  8.3 mCoulombs / 5.6V = 1.482mF or 1400uF.
> So throw in a 2000uF capacitor for good measure.

Now you're doing it basically correctly.

Olin's point was that if you fix the amount of capacitance per output amp,
that basically tells you what the ripple voltage is going to be, regardless
of the actual input and output voltages.

> Let's do the same exercise with your 5V supply.  5V + 1.2V + 2.0V = 8.2V
> which is surprisingly close to the 8VAC value used in the 10A to 30A
> S100 power supplies at the beginning of the computer hobby revolution.

I guess it's kind of a rule of thumb to arbitrarily select the RMS voltage
of the transformer to be roughly the same as the raw DC level required, and
accept a ripple value that's equal to the difference between the peak
transformer output and the minimum raw DC level.

There's nothing that says that this is the only, or even the best, choice.
However, given this assumption, the rest of your calculation looks fine,
except for using the wrong ripple voltage as you noted in your follow-up.

-- Dave Tweed

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