I’ve been looking for a simple low-voltage cutoff circuit for a 12v SLA battery, but many of the ones I find have reviews saying that the protection circuit itself drains the battery slowly as well. Is this just inherent in the design, where it has to draw a little to measure the voltage, or are there low-voltage cutoffs that don’t draw anything until the battery is recharged?

  • sweafa
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    1 year ago

    i would probably do some circuit with a TL431 and FETs, but this also drains some small amount of battery current.

  • Susan_B_Good
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    1 year ago

    You want hysteresis and an energy gap - which means putting energy into the system. You could use a latching relay to minimise insertion loss, however the loss in a conducting MOSFET can be pretty minimal.

    SLAs self-discharge, of course.

    If the load versus time is predictable, you could use a latching relay and delay voltage checking until the time window for potential cut-off. Or make it entirely period based and not test at all.

    It may be that you never need to sense voltage, if your time period between recharging is small enough.

    Many operate on that basis - the time interval between recharging may be out of their control (in our case, once, we only had mains electricity between 2am and 4 am each day…) and they provide themselves with enough battery capacity to last that time interval, with a reserve. So no low voltage cut off necessary. So no testing necessary.

  • Saigonauticon@voltage.vn
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    1 year ago

    No, it’s not possible to do it without current draw. You can do it with really, really low current draw though.

    Ignoring MOSFET stages for the moment, I could design a system that could do this, with a power consumption of under 0.1 uA when in the low-voltage cutoff state.

    I’d use a TPL5110 and an Attiny10 to do that.

    Alternatively, if ~50 uA is OK (it really should be), then I’d just use the Attiny10 on watchdog timer, and save the cost of the TPL5110.

    If I absolutely did not want to use the SLA to power that system (as an academic exercise), I’d use a separate CR2032 coin cell. That ought to last 3-5 years. Or if there’s ambient light, a calculator solar cell and a supercapacitor would make it self-powered. I could design a system that could last overnight on just a few hours of ambient light during the day. Modern microcontrollers are a marvel!

    The amount of power drawn by a reasonably designed system should be many orders of magnitude less than the self-discharge of the battery. So not worth worrying about unless it’s very poorly designed for some reason.