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A while back when I ordered a 3S Li-ion protection board I didn't think about low voltage protection so I made a small circuit using an opamp. Is this gonna be good or will the magic smoke escape?

circuit

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    \$\begingroup\$ Could you please explain how the circuit works? It might help discovering points yourself :-) \$\endgroup\$
    – Huisman
    May 2, 2020 at 21:55
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    \$\begingroup\$ Can you find a worse op-amp to make it even harder? The 741 is not recommended for new designs. It doesn't work at low voltages. \$\endgroup\$
    – Transistor
    May 2, 2020 at 22:06

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You need latch as use in BMS module. because when you cut battery off from load. the voltage will rise and it connect to load agian then voltage drop and cut off and so on. when it cut-off once you need to hold cut-off state until manual reset or until battery get charge again. Note: Most of BMS module have under voltage protection build-in.

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  • \$\begingroup\$ I see. I think I'll just order a proper BMS and replace the one without LV protection. Thanks. \$\endgroup\$ May 3, 2020 at 7:46
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So it seems you want to use the ua741 in a comparator application. You have the LM317 provide a 7.86V reference voltage to compare your Voltage source, the 3S li-ion battery which is about 11.1V nominally. When the Vcc is above Vref then output is high and MOSFET is on. When Vcc is below Vref, output is low, and MOSFET is off.

The in theory the circuit operation is sound but you’ll find it will have some problems. Issues:

  1. Your Vcc, which will power op amp and is an input, shouldn’t be the same because max common mode is about 8.1V worst case.

  2. Op amp peak output is about 8.1V . Not big deal but not rail to rail. If yo use MOSFET that needs 10V then you’re out of luck.

  3. When Vcc drops, so does the supply for LM317. This is an issue when you want to compare Vcc to 7.86V because the LM317 needs about 3V headroom above its output to maintain regulation. So minimum it can accept is 10.86V. This is a problem because then you can’t compare properly to the reference.

The circuit will won’t cause magic smoke but it won’t work properly.

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  • \$\begingroup\$ I see. Thanks for answering. \$\endgroup\$ May 3, 2020 at 7:42

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