I have a DC adapter rating 12 V 750 mA. I broke open its case and saw that its rectified but not regulated.

I need 12 V for my design and do not want to buy a transformer. Do you know of any regulator with the least dropout voltage?

I know LM317 is adjustable and its Vdo is 1.2 V(min) but can't use that because then regulated voltage will be 10.8 V and I need 12V regulated.

I haven't tested the adapter yet, will do that tomorrow its offload voltage.


2 Answers 2


What you need is an LDO (Low DropOut) regulator. The Micrel MIC2937 has a dropout of only 370 mV at 750 mA out.

You'll have to check the output voltage of the adapter at maximum current. Connect a 16 \$\Omega\$/10 W resistor and measure the voltage. You can combine resistors you have in your box, keeping in mind that P = V\$^2\$/R. The voltage should be higher than 12.4 V.

The MIC29372 (same datasheet) is a version with adjustable output, which may be of help if the output voltage isn't high enough for 12 V out.

If the voltage is too low, and you really want to use the adapter, you can replace the rectifier diodes with Schottky types. That will give you an extra volt output.

edit 2 (regarding the LM317 hli also proposes)
If it's a 12 V adapter it will deliver about 12 V under load. Saying that the unloaded voltage is high enough for the LM317 is meaningless; you do want to use it, don't you? 12 V, even a bit higher is insufficient for the LM317. Don't design for unreliability and use an LDO.

edit 3 (2012-06-14, re your comment) If the adapter is only going to be used only for powering a relay you don't need a regulator. The relay will only show a small load to the 12 V, so it may well be a few volts higher (unloaded transformers output a higher than nominal voltage). This relay is rated at 12 V, but has a maximum voltage 1.7 \$\times\$ that, so that's 20.4 V, which should be safe. Nominal power is 360 mW, and that will be somewhat higher at 15 V for instance (560 mW). If the voltage at 30 mA load is much higher than the 12 V you could still use a regulator. This will also consume some power, but the sum of regulator and relay power would be less than the relay alone.

Also nice about it is Must operate voltage, which is maximum 75 % of rated voltage. That means it will still work if the adapter's voltage would drop a few volts. The Micrel regulator will drop less than 200 mV at 30 mA, if we go for the adjustable, and choose 10 V out, then we will have proper regulation down to 10.2 V in, and the relay will only consume 250 mW instead of 360 mW. So we win twice.

  • \$\begingroup\$ thanks man i searched on farnell and thats wat i found. Ure the man u answer most of my questions \$\endgroup\$ Jun 13, 2012 at 6:23
  • \$\begingroup\$ @David - Always glad to help. How's the optocoupler thing going? \$\endgroup\$
    – stevenvh
    Jun 13, 2012 at 6:25
  • \$\begingroup\$ Actually thats wat im doing rite now. Shud i post my answer as a new question? Or in the same thread where u explained everything \$\endgroup\$ Jun 13, 2012 at 6:35
  • \$\begingroup\$ @David - Keep things separate. This is power supply, the optocoupler is a different problem, even if it's for the same product. If you want feedback on what you're doing, post it as an update to the original question, and make clear that you want feedback. Success! \$\endgroup\$
    – stevenvh
    Jun 13, 2012 at 6:38
  • 1
    \$\begingroup\$ @miceuz - If it has a DC output it will refer to that, not the transformer's secondary's voltage, which means nothing to an average user. But being unregulated the voltage may be way higher than 12 V unloaded. \$\endgroup\$
    – stevenvh
    Jun 13, 2012 at 6:50

I think you could go with the LM317. Usually, the voltage output of such power supplies is higher that stated when they are not under full load.

Remember, the transformer delivers an AC voltage (a sine wave). When it is rectified, you get sine half-waves. Their peak voltage is (assuming the transformer delivers 12V AC) 12V*sqrt(2)=16.9 volts. When you then add a capacitor, you get 16.9V DC voltage (when there is no load connected, nothing discharges the capacitor). Depending on the load, the voltage drops. (I remember a basic guideline of using 3000µF capacitor per ampere of output, but I don't know how good this advice is).

So you should just try using a LM317, and measure the voltages under the load you expect for your project.

Btw: Wikipedia explains the rectification much better than I can...

  • 1
    \$\begingroup\$ An answer to a comment to my answer: Even if it were 12V AC the 16.97 V peak might still not be sufficient, and surely not "plenty". You've got two 1 V drops across the rectifier bridge, that leaves 14.97 V. Some LM317s only specify Vref for a minimum Vin - Vout of 3 V. That leaves 0 V for ripple, and that's impossible. For this regulator you'll need a 15 V transformer. \$\endgroup\$
    – stevenvh
    Jun 13, 2012 at 13:29
  • \$\begingroup\$ You can evaluate how good the advice of 3mF per Ampère is: at 50 Hz with half wave (10 ms low), 1 mA will be equal to 10 µC: with 3 mF capacitor, it'll be a drop of 10µ/3m = 3.3 mV. Errors excluded :) \$\endgroup\$
    – clabacchio
    Jun 13, 2012 at 13:30
  • \$\begingroup\$ @stevenvh: whoops, I've considered 1 mA instead of 1 A! Well, in that case it would be perfect! Even overkill \$\endgroup\$
    – clabacchio
    Jun 13, 2012 at 13:37
  • \$\begingroup\$ Ignore clabacchio's comment, he's confused, too much drinking ;-). V = I \$\cdot\$ t / C = 1 A \$\cdot\$ 10 ms / 3 mF = 3.3 V. That leaves 11.7V!!! From my first comment you know that you don't have a mV to spare. \$\endgroup\$
    – stevenvh
    Jun 13, 2012 at 13:39
  • \$\begingroup\$ By the way, rectification is free these days? It's not because Wikipedia doesn't mention diode voltage drops that they don't exist. They do! 1 V per diode. Back to the drawing board... \$\endgroup\$
    – stevenvh
    Jun 13, 2012 at 13:52

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