I have a 1.25 to 28 V variable power supply I'm working on. here's the schematic (from here):

enter image description here

What I'd like to do is put in a fan (12v 0.16A) so it runs when i turn it on, however i don't know where to insert it so that it doesn't blow. I was also thinking of putting in a small bulb as an indicator light - when i turn it up, the light gets brighter. I have quite a few different rated lights, so I'm good for that, but still- don't know how to set it up.

I'll be using a cheap multimeter as a voltmeter - i don't have a standalone voltmeter on it's own and they are quite expensive actually for something so simple.

  • 2
    \$\begingroup\$ You want to make sure the fan doesn't blow? ;-) \$\endgroup\$
    – joeforker
    Apr 4, 2012 at 12:29

4 Answers 4


You can get a simple LCD panel voltmeter from Futurlec, they have various models. The PM128E for $7.90 looks like it would do what you want. I'd suggest using this instead of trying to use the brightness of a bulb to show the output.

For your fan, probably you should just have a separate voltage regulator to supply power to it, since it needs a constant voltage, and not an adjustable one.

  • \$\begingroup\$ +1. bulb brightness is neither a consistent nor a linear function of voltage. \$\endgroup\$
    – Jason S
    Feb 6, 2010 at 14:11

You can also get a fan that runs off 115VAC and put it before the transformer T1.


The unregulated voltage (across the 1000 μF capacitor) is about sqrt(2) * 25V, or 35V.

So you could wire three 12V fans in series. Lots of cooling!

Or you could put a (35-12)/0.16, or 150 ohm resistor in series with the fan. This resistor will dissipate 150*(0.16)^2, or 3.8 watts. You would need a 5 or 10 watt resistor.

Alternatively, you could connect a three-terminal 12V regulator (e.g., LM7812). It's not too hard to use, just read the datasheet. The regulator will still dissipate the 3.8 watts, so it should be heatsinked, and in the airflow of the fan.

For a cheap meter, you can buy little Chinese DMMs for < $10 at the right places.



This power supply doesn't need a fan. It just needs a decent heatsink.


  • 3
    \$\begingroup\$ Worst case dissipation is 1.5A, 1.25V output. The dissipation is (35-1.25)*1.5 or 34 watts. The heatsink will need to be 0.8 K/W or less. That's a pretty big heatsink. (eg, Digi-Key 345-1046-ND) With a fan, you can use a much smaller heatsink. \$\endgroup\$
    – markrages
    Apr 4, 2012 at 23:40

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