Bench power supply help

Why not us an ATX power supply? Most will push 10A or more at 12V. Plus, you'd have 5V & 3.3V outputs, as well.

Bob

Why not us an ATX power supply? Most will push 10A or more at 12V. Plus, you'd have 5V & 3.3V outputs, as well.

Bob

The OP said he wanted it to be adjustable but it's along time since we heard from him.

PC supplies are cheap but a bit brutal for test use without additional parts for current limiting and load protection.

MK

Ti has this on the bench too to play with. Looks good to me:

http://www.ti.com/lit/ds/symlink/lm2679.pdf

Just follow the design advises in the document, also about the capacitors.

The OP seems to have lost interest but he did say his power source would be  a transformer with nominal 30V windings.

When you are designing a system using  a mains transformer/rectifier/capacitor as the raw supply you can calculate the peak no load voltage on the capacitor as follows:

Vpkdc = Vnomac * 1.1 * 1.1 * 1.414 (the first 1.1 is for 10% mains tolerance, the second is 10% transformer regulation (check the spec of the actual part) and the 1.414 is the peak to RMS ratio of the AC)

So for a 30V transformer we will see a maximum voltage on the smoothing capacitor of 51.32V (it might be a bit less due to the voltage drop in the rectifier).

( 30 * 1.1 * 1.1 * 1.414)

Good engineering practice would be to allow a margin on this worst case of at least 5V so the switcher needs to be specced to cope with 56V but the 2679 has an absolute maximum voltage specced at 45V so it won't do for this application.

To finish the job you should also calculate the least voltage you might see which is a little harder:

Vmindc =     (Vnomac * 0.9 *1.0 * 1.414) - ripple = 38.18 - ripple.

At current I with a smoothing capacitor C the ripple will be I * t/C where t is the time during each half cycle (assuming full wave rectification) where the capacitor is discharging - rough rule of thumb makes t half the half cycle time so at 50Hz mains its 1/200 seconds (0.005).

So the ripple will be (5 * .005)/4700e-6 = 5.3V.

So the min voltage on the smoothing capacitor at full load = 32.9V.

Mostly we have confirmed that the OP would have an easier time if he bought a more suitable transformer !

MK

You are completely right Michael overlooked (read badly) the maximum input VDC level of the chip.

Indeed 2x24VAC from the Transformer would have been easier - now he either has to unwind the transformer secondary windings or maybe try this chip http://rohmfs.rohm.com/en/products/databook/datasheet/ic/power/switching_regulator/bd9611muv-e.pdf