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At present, my little atom PC is being powered by an 240V AC to 12V DC PSU which connects to a 12V DC-DC pico PSU. Our home is off grid, so all of our power comes from 12V lead acid batteries. I've tried connecting the 12V batteries directly to the pico PSU to improve efficiency and it works as long as the voltage from the batteries is no greater than 13.5V or so. Unfortunately, while they are being charged, the voltage from the batteries can initially be anything up to 14.5V. Is there a simple way to 'cap' the voltage from the batteries at 12V?

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  • \$\begingroup\$ How much current do you need? \$\endgroup\$ Jan 27, 2014 at 13:27
  • \$\begingroup\$ The PC draws a maximum of about 40W, so if my rudimentary understanding of electricity is correct, that would be about 3.5A. \$\endgroup\$
    – magicalex
    Jan 27, 2014 at 13:28
  • \$\begingroup\$ What's the minimum voltage the PicoPsu will work at? A power diode (Sized for 3 Watts) can drop 0.7v, two will drop 1.4v, but then you might have the opposite problem. Also, what model picopsu? \$\endgroup\$
    – Passerby
    Jan 28, 2014 at 1:13
  • \$\begingroup\$ I've never let the batteries get below 12v, Passerby, so I don't really know, but that might be just what I need. Thank you. \$\endgroup\$
    – magicalex
    Jan 28, 2014 at 5:29

3 Answers 3

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If you are up for some DIY then you can make the following LDO circuit

enter image description here

It uses an opamp, a P-mosfer, a zener diode and a few resistors.
The zener provides a reference voltage, in this case about 6.4v.
R2,R3 and trimmer RV1 form a voltage divider, the divider output (in the middle) is about half of the mosfet output voltage. The trimmer adds the ability to set the output voltage from about 11.5v to 13.5v.
The opamp (triangle) compares the reference voltage with the output voltage and drives the mosfet so that it keeps the output level constant (assuming the input is higher than the output voltage).

The excess voltage times the output current will be dissipated as heat on the mosfet so you'll need a small heat-sink.
As an example for 4A and input voltage 14.5v and ouput 12.8v the power dissipation will be (14.5v-12.8v) * 3A = 5.1W.
When the input voltage matches the set voltage or is lower, the dissipated power will be negligible.
Note that you should choose a low ON resistance mosfet so that it drops only a few mV when it's fully on.

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  • \$\begingroup\$ With no feedback compensation, this circuit is sure to oscillate. \$\endgroup\$ Jan 27, 2014 at 23:16
  • \$\begingroup\$ @Madmanguruman Added \$\endgroup\$
    – alexan_e
    Jan 27, 2014 at 23:37
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There are wide input range picoPSUs available, e.g. the picoPSU-120-WI-25 supports up to 25V input voltage. Why don't you use one of these?

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  • \$\begingroup\$ Perhaps this would be better posted as a comment than as an answer. \$\endgroup\$ Jan 27, 2014 at 17:39
  • \$\begingroup\$ I would if I hadn't already bought the one I have. I thought there may be a simpler, less expensive way to drop the voltage. \$\endgroup\$
    – magicalex
    Jan 27, 2014 at 17:49
  • \$\begingroup\$ Given OP's relative lack of knowledge and just wanting something that works, this answer gets a big +1 from me. \$\endgroup\$
    – John U
    Jan 27, 2014 at 18:33
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Yes, you could make your own low dropout regulator to feed this "pico PSU" (whatever that really is).

However, the better solution would be to get or make a proper power supply that can handle the variation in voltage on your battery bus. That really isn't hard. There are plenty of buck switcher chips that can easily to this with a minimum of external parts.

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  • \$\begingroup\$ Thanks for your answer. Unfortunately, I've no idea how to make a low dropout regulator (despite reading up about them), but I'll certainly look into it. \$\endgroup\$
    – magicalex
    Jan 27, 2014 at 17:53
  • \$\begingroup\$ @magi: If you didn't want a answer to get or make a power supply, why did you come here? What can we possibly say that you are willing to do? \$\endgroup\$ Jan 27, 2014 at 21:48
  • \$\begingroup\$ @OlinLathrop chill. Reread what he said. \$\endgroup\$
    – Passerby
    Jan 28, 2014 at 1:06

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