Confirming PSU spec for steppers.

[irving2008]

So, you have a 50v AC transformer giving 12A, what about the rest of the components?

Here are my suggestions and the calculations needed to ensure it all works reliably...

Bridge rectifier: Buy Bridge Rectifiers Bridge Rectifier Single 400V 35A GBPC4 Vishay GBPC3504-E4/51 online from RS for next day delivery. @ £3.60

This bridge has a forward voltage drop of 1.1v per diode so your output volts will be 50 X 1.4 - 2 x 1.1 = 68v! Also each diode will dissipate 1.1v * 12A = 13.2W so you'll have 26W of heat to dissipate. Assuming air temp of 25degC, and a rectifier temp of 125degC you need a thermal resistance of no more than (125-25)/26 = 3.8degC/W, since the rectifier has a thermal resistance of 1.4degC/W from the diode to the mounting base it'll need to be mounted on a heatsink rated at 2.4degC/W or better. Something like this: Buy Heat Sinks Heatsink 2.4K/W 97x50x25mm ABL Components 333AB0500B online from RS for next day delivery. @ £3.79 and don't forget the thermal compound between rectifier and heatsink and between heatsink and the case...

Smoothing Capacitors:

You need C = 0.1 * A/V for a 5% ripple = 0.1 * 12/70 = .017F = 17,000uF with a voltage rating of at least 100V and a ripple rating of 12A. (for derivation see here)

Doing this in one capacitor is expensive (18000uF, 400v, 12A ripple = £271!!!) So usually you make it up with a bank of capacitors in parallel. The exact value is not critical so 3 x 5000uF @ 100v/4A would do, or whatever you can get cheap on eBay :)

For example, 3 of these: Buy Aluminium Capacitors GP Al electrolytic capacitor,4700uF 100V Epcos B41456B9478M online from RS for next day delivery. would be £33, with the matching mounting clips @£1.70 for a pack of 5. These have screw terminals so you'll need the appropriate solder ring tags.

You'll also need a bleed resistor across each capacitor to discharge it when the power is turned off (for safety). The energy stored is 0.5CV^2 joules. For each capacitor spec'd above, the energy is 0.5 * .005 * 70^2 = 12joules. A watt is a joule per second, so for a 5 second discharge you need to dissipate 12/5 = 2.4W so we're going to need 3 or 5W rated resistors. Assuming the resistor dissipates the same 2.4W when the power supply is turned on the value will be R = v^2/W = 70^2/2.4 = 2000ohm. I'd use 2200ohm (2k2) @ 3W wire ended and solder them directly to the ring tags on the capacitor: Buy Through Hole Fixed Resistors ROX3S metal oxide film resistor,2K2 3W TE Connectivity ROX3SJ2K2 online from RS for next day delivery.. (pack of 10 @ 99p)

Don't forget fuses for primary and secondary sides, push-on tags for the rectifier, screw tags for the capacitors, suitably heavy wire (red & black, at least 1.5mm sq, e.g. 30 x 0.25mm)...