Lead acid batteries get damaged by over-discharge somewhat like lithium polymer batteries.

I need a circuit that will cut off the load from the battery when it drops below about 10v. That is the easy part.

The hard part is having a circuit that consumes a miniscule current and wait until the battery's voltage exceeds around 13v, (which would indicate that it is being charged,) at which point it would re-engage the load.

The best idea I have is to use a low-quiescent current voltage regulator, specifically the TS2950CT50 +5v which has a quiescent current of around 75uA. I would use a mid-range PIC16F MCU that has an internal voltage reference (and either ADC or comparator) that goes into standby mode, periodically checking if the voltage exceeds 13.5v. The MCU could consume as little as a couple of uA.

While 75uA isn't too bad, I was wondering how I could achieve a lower standby current.

Another idea which I'm not certain about at all, is operating a zener diode below its breakdown voltage, relying on a slight increase of leakage current to trigger a MOSFET circuit. Leakage for zener diodes seem to be 1uA per volt, so that would in theory only consume 10uA. But since zeners don't have a great accuracy at the best of times, let alone over different temperatures, with aging, EMI or what have you, could I use this reliably?

At a push, it would be permissible to have the over-discharge protection deactivate when the battery's voltage reaches anywhere between 12.5 and 13.5v.

Because I'm a masochist, I'd prefer not to use an IC which is dedicated to this purpose (not that I've found one). What other solutions are there?


  • 1
    \$\begingroup\$ What is the load current you need to switch? Also I don't think there are other solutions to masochism. \$\endgroup\$
    – Andy aka
    Commented Jun 23, 2015 at 8:56
  • 1
    \$\begingroup\$ since I don't have time to modify and rewrite, here's an answer I wrote some time ago, including some maths to two-transistor Hysteresis (leave out the diode, the OP there had it and I kept it, never heard why it was there): electronics.stackexchange.com/a/132634/53769 maybe you can do something with that. Other than avenues like that, there's chips. Your uC plan is a good one. Single uA solutions with discrete components always have the risk of picking up noise. You mitigate that a little by using an NPN as first element and doing the maths. \$\endgroup\$
    – Asmyldof
    Commented Jun 23, 2015 at 9:01
  • \$\begingroup\$ A real masochist would design this using only discrete components ;-) I'd cheat a little bit and maybe use some CMOS logic or maybe a few comparators. Uhm what you actually need is a window comparator ! Something like this: edn.com/electronics-products/electronic-product-reviews/other/… perhaps ? (the circuit at the top of the page I mean) \$\endgroup\$ Commented Jun 23, 2015 at 9:03
  • 1
    \$\begingroup\$ What is the battery capacity? The bigger it is, the less you care about 100uA or so. The easiest way to design a solution is to use a micropower comparator and voltage reference. The comparator can switch a P-channel MOSFET to connect or disconnect the load. If you insist on not using IC's it will be difficult to achieve the voltage precision and low quiescent current you want. \$\endgroup\$
    – user57037
    Commented Jun 23, 2015 at 19:28
  • 1
    \$\begingroup\$ Perhaps a zener diode could be used to keep a self-opening relay closed while the voltage is above 10v, not sure how you would prevent hysteresis from charging until you got to 13v, maybe disconnect the relay somehow while charging? \$\endgroup\$ Commented Jun 23, 2015 at 23:39

2 Answers 2


If you decide to use a dedicated chip to detect low voltage, you could use a S-1011 Series voltage detector from Seiko.

  • \$\begingroup\$ +1 for the incredible 600nA consumption and 1.5% accuracy with this IC. It would be great if you could give a circuit diagram for its usage in this scenario? Specifically to detect when there's a supply voltage above 13v? \$\endgroup\$
    – Jodes
    Commented Jun 24, 2015 at 8:25

My comments are turning into an answer, so I will post here, but I don't have it fully flushed out yet and will hopefully edit into a proper answer soon

I think you can use a 10v zener diode and a relay to get a "off" state leakage of only a few nA depending on the zener (and assuming 10v/25ºc), as that's typical of the component

Then if we were to choose a relay, such as this one, it has a turn-on voltage of 8.4, so we should aim to keep the voltage applied to it under 8.4 if the battery is under 13v (so it won't activate) - yes, this is a maximum and the particular piece may vary

I think after the 10v zener, if we were to place another diode, say a 12v zener (yes, not 13v, bear with me), and then the relay, it would only turn on when the voltage is above 12v (plus whatever the 10v zener drops in reverse)

Next to keep the relay open even after the voltage drops under 12v and the 2nd zener stops letting current through, a wire from the relay's output (which turns on at 12v) and then connect a transistor to it that would activate from that pin, and have it draw power from after the 10v diode, it would work, as the relay + transistor would keep itself open even after the 12v diode shut off.

Then, to protect the transistor and make it turn off, we would need to limit the input current to the transistor such that it's not enough to keep the relay closed, and the relay needs current to come from the 10v diode that shuts off - really we should only need a couple ma to saturate the transistor, as the coil itself only needs 30ma to stay open, so if we sent say 3ma to the transistor, we would use (12v/3ma -> 4k), say a 4.7k resistor, and the transistor would need a gain of 15x or so

have yet to draw it in-circuit, but I think it would work, and you would need 5 components: a relay, two zener didoes (10v and 12v), a transistor, and a resistor.

for your 50A, you will need a beefier relay, but the principles would be the same (or you could use this circuit to drive a larger relay)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.