Detect power supply voltage threshold, while consuming very low current

Question

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?

Thanks

Answer 1

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

Answer 2

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)