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Sorry if this has already been asked, I tried searching but might not have the right vocabulary.

I'm starting to use lithium-ion batteries to power projects, and I've wound up with some 18650 cells. Not wanting to use unprotected cells, I hooked them up to some common and cheap TP4056 charge/protection boards. This lets the batteries charge up to 4.2V easily, but I've read that the undervoltage protection on these boards is really more of a failsafe than a way to keep the battery healthy; it doesn't cut off until 2.5V, which I'm worried might damage the cells pretty quickly.

It looks like there are affordable over/undervoltage protection ICs for li-ion batteries, but the highest undervoltage cutoff I can find is still only 2.8V on something like TI's BQ29700.

So here's my question - if I want to err on the side of caution when draining the batteries, can I use a 'low voltage supervisor' IC hooked up to an N-ch MOSFET as a cheap/simple extra layer of protection to cutoff at something like 3-3.5V? Would I be better off just using an opamp? Is the voltage supervisor's 3uA enough to continue draining the battery appreciably?

Here's what I'm thinking of specifically - it's just a high-side N-ch switch with its gate attached to the voltage supervisor's "reset" pin, which is pulled low during undervoltage in the MIC2776L model:

MIC2776 circuit

And here are the non-discrete parts; I tried to pick a MOSFET with low on-resistance:

Adjustable low-voltage supervisor - MIC2776L: http://ww1.microchip.com/downloads/en/DeviceDoc/mic2776.pdf

N-Ch MOSFET - PMV20EN: https://assets.nexperia.com/documents/data-sheet/PMV20EN.pdf

Thanks! And sorry if I'm missing something, I'm pretty new to this.

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  • \$\begingroup\$ You will have 3uA in the voltage divider as well as another 3uA going into the IC. That is 6uA total. Unless the battery is very small, that should be reasonable. However, this may not work well depending on how your system works. When reset is asserted, the FET turns off, then the load is disconnected from the battery, which will cause the battery voltage to recover a bit, and that may cause reset to de-assert and turn the FET back on. And, of course, once the load is reconnected, the voltage will sag, and reset will assert again. And so on, over and over. If the load is very low ... \$\endgroup\$
    – user57037
    Mar 13, 2018 at 7:50
  • \$\begingroup\$ then there may be enough hysteresis to avoid this problem. You may not know until you try it. \$\endgroup\$
    – user57037
    Mar 13, 2018 at 7:51
  • \$\begingroup\$ Thanks - I am looking at fairly low-current applications which is why I'm worried about the default 2.5V cutoff, but it does seem like hysteresis could be an issue. Would I maybe be better off with something like a schmitt trigger circuit? \$\endgroup\$
    – Will
    Mar 13, 2018 at 17:04
  • \$\begingroup\$ There are several ways to approach this. Does your design have a microprocessor and monitor the battery voltage with an ADC? And does it also have at least one GPIO output that could be used as an output? \$\endgroup\$
    – user57037
    Mar 13, 2018 at 18:02
  • \$\begingroup\$ Well, most projects I power will have some sort of MCU core, but that would be part of the load on the battery. Are you thinking of something like placing a pull-down resistor on the MOSFET gate and just having the MCU not pull the gate high under low-voltage conditions if it reads a low voltage? I guess there'd need to be a 'reset/start' button to pull the gate high initially, but that could work. \$\endgroup\$
    – Will
    Mar 13, 2018 at 22:35

2 Answers 2

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This is the basic idea. If you push and hold the button, the regulator will be forcibly enabled. Then the output of the regulator will come up and pull up the enable pin so that the regulator stays on when the user releases the button.

The GPIO pin must not go high. When the time comes to shut down the regulator, you can assert the GPIO pin high, and it should disable the regulator. You may need to add a diode and capacitor to the base of the regulator to make sure it is driven low long enough to fully turn off the regulator. I didn't show those. I can add them if you want.

schematic

simulate this circuit – Schematic created using CircuitLab

This is just one way to do it. Not shown is the sense circuit used by the processor to monitor battery voltage. That may need to be controlled by a switch also, so that it doesn't drain the battery when the regulator is disabled.

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  • \$\begingroup\$ Thanks, that makes sense - it looks like a nice solution for most of the projects I'll be using these batteries with. \$\endgroup\$
    – Will
    Mar 14, 2018 at 6:50
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if I want to err on the side of caution when draining the batteries, can I use a 'low voltage supervisor' IC hooked up to an N-ch MOSFET as a cheap/simple extra layer of protection to cutoff at something like 3-3.5V?

Yes. Here is a sample circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

The S-80830CL is a voltage monitoring circuit with CMOS output. When the voltage between the V+ and V- pins goes above 3.15 volts, the output goes high. It will stay high until the voltage falls below 3.0 volts. At that point, the output goes low. The output will not go high again until the voltage is again at 3.15 volts. If 3.0 volts is not the cut-off voltage you are looking for, there are also other members of the S-808xxCL series with cutoff voltages in 100 mV increments.

When the output of the voltage monitor is high, M1 will conduct, but when the battery voltage falls below 3.0 V, the voltage monitor output goes low, and M1 will turn off.

R1 is necessary to protect the output MOSFETs of the S-80830CL.

Would I be better off just using an opamp?

You will still need some sort of voltage reference, so your circuit would not be any simpler. An advantage of using an opamp (or better, a comparator) is that you could add a trimmer potentiometer, and adjust the cutoff voltage if you wished to do that.

Is the voltage supervisor's 3uA enough to continue draining the battery appreciably?

If you leave the battery connected to the supervisor circuit for months without recharging, the small current drawn by the supervisor will be added to the battery's own self-discharge current, and the battery will be drained slightly faster. For long term storage, it would be best to periodically recharge the battery to some mid-level charge (say 3.7 volts) periodically (at least every 6 months, but 3 months would probably be safer). If you need the full 4.2 volts charge, recharge to that level prior to bringing the battery back on active duty.

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