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I made a nightlight and was very happy with the dim yet usable output and runtime (months), but my assumptions about battery safety were wrong. I expected the LDO 3.3 V regulator to stop output when the source voltage dipped below 3.3 V, and that it would not draw much current in that case. I must have been wrong, since the 18650 lithium ion cell was irredeemably dead after a couple months. The circuit is below.

Nightlight 18650 LED circuit

(Note: the output would be more efficient with two LEDs and a changed resistance to reduce the current as needed. At tiny currents like this, even a white LED does not have the expected large voltage drop.)

How could I better protect the battery from over-discharge? Could I use an additional regulator (3 V), a comparator, and a MOSFET so the comparator switches off the MOSFET if the battery's output voltage becomes less than 3 V? Even if it works, finding low power versions of these chips might be hard (or expensive). Is there a better way to cause the circuit to turn off instead of draining the battery? I assume a dedicated battery protection IC would be best, but most seem to come in a package that I won't be able to solder at home.

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    \$\begingroup\$ You can find a battery protection module that is solderable. \$\endgroup\$
    – user20574
    Commented Feb 7, 2020 at 13:22
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    \$\begingroup\$ Why not just find an LDO with an acceptable drop-out voltage? \$\endgroup\$
    – Ron Beyer
    Commented Feb 7, 2020 at 13:23
  • \$\begingroup\$ @RonBeyer I don't care about the drop-out as long as it's positive. Is it atypical that the battery was drained below ~3.4 V in this case? I thought the regulator would stop drawing current, but it seems not to have. \$\endgroup\$
    – piojo
    Commented Feb 7, 2020 at 13:25
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    \$\begingroup\$ The 2950 you use doesn't drop out (stop regulating) until the input voltage gets very low, something like 300mV. If you choose an LDO with a drop out voltage closer to 3.4V it won't continue to discharge the battery below that value because it turns voltage regulation off. A battery protection circuit would be better, but if you use the right parts it doesn't need to be complicated. \$\endgroup\$
    – Ron Beyer
    Commented Feb 7, 2020 at 13:28
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    \$\begingroup\$ Some LDO's use that metric for shutdown, yes, but looking at the datasheet for the 2950, it doesn't look to be the case for this chip. Take a look at page 12, the "Shutdown Threshold Voltage", that says that the regulator won't be off (at room temp) until about 1.3V. Page 8 has the dropout characteristics for a 5V output which you may be able to translate to 3.3V. \$\endgroup\$
    – Ron Beyer
    Commented Feb 7, 2020 at 13:48

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