Sizing transformer for linear DC supply

DVDdoug:
The biggest concern I have is with the power dissipation in the regulator. I'm pretty sure you can overheat the thing (depending on the heatsink) at less than 3 Amps when there's more than a couple of volts dropped across the regulator.

Which is why I want to not oversize the transformer, in terms of output voltage. I will have heatsinking, but no point in pushing that limit, or wasting power.

It's been a long time since I've calcuated ripple. The voltage regulator will knock-out the ripple, as long as the ripple doesn't exceed the point where the voltage regulator drops-out. So, I usually just use a capacitor that's 1000uF or more, one that's available, and one that will fit physically. (A higher voltage into the regulator will tend to minimize ripple problems, but more it's more power/heat in the regulator.)

I'll keep digging on the ripple question.

I'm thinking 15W 12V secondary, but I could be way off. I'm figuring ~2V drop on the regulator, and 1.7 for the bridge, giving 9.7V, and then just adding some headroom.

12VAC is plenty for a 6VDC supply. I think you're forgetting that the peak AC voltage is about 1.4 times the RMS. At 12VAC your capacitor will charge-up to about 17V minus the diode drops. (I'll usually use a 12V transformer for a 12V power supply.) And at no-load, the transformer will probably put-out more than 12VAC. If you can find a 9V transformer, that should be enough.

No, I'm not forgetting about it. I honestly don't know how to treat it. Hence, asking for info. :slight_smile: I understand the notion that as each peak occurs in the rectified voltage, the cap charges up a bit more. And so ... eventually, it would reach peak V. Except that there's a constant discharge as well. We can, as thought expermiment, imagine a load across the capacitor producing an RC time constant less than 1/60th of a second. And maybe, in a practical circuit such as I'm contemplating, this can be simply discounted.

Peak on 6VAC is ~8.5 -- not enough. Peak on 9VAC is ~12.73, so I'd be dropping 5V across the regulator, if the VDC does indeed approach the peak. Going back to my initial back-of-the-napkin 9.7V * 1.414 = 6.86, that get me to this brutishly expensive transformer (for my purposes, anyway). I'm selecting for current output, not VA rating. Moving to 10V out, the price improves considerably, but then I'd be dropping 6V across the regulator.

If it really comes to a $46 transformer to do this well, I might as well buy a new Pila charger -- they're ~$45. Yeah, I've looked at Digikey for a 6V wall-wart supply, which can be had for about $17. Just thought it might be a nice little project to do one evening.