Bench power supply help

What about using a lm138 instead?

I am familiar with these, they will handle 5A, and the voltage range you are looking for.

Here is a simple diagram and formula for creating a circuit around it.

I am sorry Lupe but the LM138 won't work.  It's max input voltage is 40v.  30vac once rectified and filtered will give over 42vdc.  Also he needs 5A at 12V so the LM138 would have to drop 42-12=30 volts at 5 amps which is 150 watts.  More than the LM138 can handle.  The requirements call for 2 channels at 12 volts, 5 amps.  This is 60 watts which is what each winding can deliver 30 volts at 2 amps = 60 watts.  With a high efficiency switching power supply, you raise the available current as you lower the voltage so that the output power is close to the input power and little is lost in heat.  Only problem is that a switching power supply is not so easy to build from components.  A solution would be to buy a low cost switching power supply module.

You can look at this switching regulator design collection from Linear Technology http://cds.linear.com/docs/en/application-note/an30fa.pdf

But Yves said it's not easy to build one from parts.

Hi there Anej,

take a look at this device. Ready available, great characteristics, handful components. It would cost a lot less then buying all you kneed for a switch mode.

Linear Technology - Product Page

These parts have exactly the same problem as Yves described for the LM138.

The OP will need to use a switcher or a custom designed linear supply that can deal with the 150W of internal dissipation.

MK

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