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I am running one of my projects from two 2000 mAh Lithium Ion cells wired in parallel

I decided to let the battery run until it died, just once, to see how long it would last. It lasted 25.9 hours, and when I checked the voltage on the cells, they had gone down to 2.5 V.

I've read in many places that Li-Ions should be 3.7 V when full and 3.2 V when empty, but I've never seen anything about 2.5 V or anything lower than 3 V for that matter. I have heard and seen people talk about "over-draining" a Li-Ion cell, and that when it goes below 3 V a microchip disconnects the battery to protect it from discharging too far.

In this case, my battery still works, and it is charging right now, I don't plan to run it down that low again, but if it were to happen again, is it a big problem? Could this affect the longevity/performance of the cells?

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    \$\begingroup\$ Your project really should include a low battery cutoff if you are powering it from Li-Ion... \$\endgroup\$
    – Ecnerwal
    Commented Feb 25, 2016 at 3:14
  • \$\begingroup\$ @Ecnerwal It does! The computer powered off but I can't power off the LED on the step up board. That was the only thing that kept running. \$\endgroup\$ Commented Feb 25, 2016 at 5:28
  • \$\begingroup\$ @PatrickCook You could cut it off. \$\endgroup\$
    – Passerby
    Commented Feb 25, 2016 at 5:30
  • \$\begingroup\$ Are you using a "naked" battery, or does yout battery include some kind of BMS circuit? \$\endgroup\$ Commented Apr 26, 2022 at 13:55
  • \$\begingroup\$ ZDNet posted article claiming that cell phone manufacturers were intentionally damaging non-replaceable batteries after several years of phone use. Article quickly disappeared from ZDNet. \$\endgroup\$ Commented May 2, 2022 at 14:00

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Yes, lithium-ion cells undergo unwanted chemical reactions when discharged below 3 V, causing their internal resistance to be permanently and significantly raised. Their capacity will suffer as well, meaning that they won't accept the same amount of charge anymore. When such an over-discharged cell is "brought back to life", it will likely become chemically unstable, creating a risk of a short circuit developing inside the cell.

Even worse, assuming that you measured 2.5 V at no load, your cells have dropped even lower when they were being discharged and have subsequently rebounded to 2.5 V after the load was removed.

Li-ion cells have a maximum voltage of 4.2 V or less, I am not sure where you got the 4.7 V figure from but it's a recipe for fireworks. OP has since edited the question, to a still incorrect 3.7 V. 3.7 V is the nominal voltage (average voltage during a complete constant current discharge), while 4.2 V is the maximum voltage. These figures will vary slightly from cell to cell.

I would completely discharge the cells and get rid of them, 2 Ah 18650s are cheap and not worth the risk of them blowing up.

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  • \$\begingroup\$ Yep -- for Li-Ion batteries there are three important protections: OCP (over-current protection), UVP (under-voltage protection) and OVP (over-voltage protection). OCP applies in both directions, charge and discharge, and the value at which it trips (especially charge) varies with temperature -- it's a bad idea to charge a Li-Ion battery at a high charge rate when it's cold (generally <10C or so, but varies on brand). \$\endgroup\$ Commented Feb 25, 2016 at 4:50
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    \$\begingroup\$ Thanks for the answer, but I just realized I was measuring the wrong voltage... The board I'm using to step up the voltage to 5v has an enable pinout that reads 2.5v... My battery was actually somewhere around 3.3v. I charged it again and discharged it to 3.3v and found that the mAh was still ~3990mAh, so everything seems to be ok. \$\endgroup\$ Commented Feb 25, 2016 at 5:39
  • \$\begingroup\$ This is not the best answer; for older cells it's mostly correct, but for most newer cells is it wrong. The best answer is the one Michal Leon where he says you need to read the specs of the cell in question. \$\endgroup\$
    – KyferEz
    Commented Feb 19 at 19:31
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And that when it goes below 3V a microchip disconnects the battery to protect it from discharging too far.

That's true, for batteries that have built in protection circuits. Not all batteries do. Most Li-Ion batteries are raw cells that do not. The ones that do will be slightly longer than the raw cells.

enter image description here

There is also circuits for multiple cells, and in various variations of externally visible or not. You can purchase the cells with the protection built in or purchase the circuits by themselves. Not just for 18650, all form factors of Lithium cells can have them.

You obviously have a non-protected cell, and because you didn't add a low voltage lockout, drained it beyond the safe limits.

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Yes, depleting a rechargeable cell under certain voltage level is harmful to it. The discharge voltage level depends on the cell chemistry.

The minimum discharge voltage varies between various sites, datasheets, etc. but 3.0 V - 2.7 V is an empirical value. If discharged under this voltage, the cell may be permanently damaged.

To get the precise value of min discharge voltage, consult the datasheet of your cell.

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This requires an update in 2020:

For most modern Li-ion cells, 2.5 V is the discharge limit. Older batteries were usually rated at 2.75 V or 3.0 V, but as I've said, that's not the case in 2020. However, to be completely sure, you do need to consult the cell's manual, as the parameters vary wildly.

For example, a typical Sanyo cell will have safe discharge current at around 1C to 2 C, while a Sony power tool cell will be allowed to give 10 to 20 A. That's a 4 to 8 times difference.

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    \$\begingroup\$ 10 to 20A? What capacity are you assuming here for the power tool cell? Or do you mean 10C to 20C? \$\endgroup\$
    – nvd
    Commented Mar 10, 2022 at 0:23
  • \$\begingroup\$ @nvd fast-discharge batteries have lower capacity, usually 1.3..1.8 Ah, while modern "standard" cells reach almost 4Ah. \$\endgroup\$ Commented Mar 11, 2022 at 1:03
  • \$\begingroup\$ @nvd The answer used the proper units when they said 10 to 20 A as in amps. He's stating the continuous discharge amperage, not the C rating. It is similar, but 10C for a 1AHr battery would be 10Amps whereas 10C for a 2AHr battery would be 20Amps. The battery C is related to the allowable discharge amps, but not the same. \$\endgroup\$
    – KyferEz
    Commented Feb 19 at 19:28
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Battery manufacturers in 2022 still firmly say that the cutoff voltage should be no lower than 2.7 V to avoid degrading the cell. Their specifications for mAh capacity are based on the minimum cutoff voltage so discharging below that will not add any additional capacity, it will only damage the cells. Under specified "standard" loads, the cell voltage will rise to about 3.2 V after the load is removed at the specified cutoff voltage. Much smaller loads the cutoff voltage should be higher, at 3.0 to 3.2 volts.

Despite this, very many 'protection' boards and circuits now use a "DWO1" or similar 'protection' IC that has a cutoff voltage around 2.4 V. It's actually becoming difficult to find such 'protection' boards with a correct cutoff voltage of 2.7-3 V.

There's a lot of trolls around trying to convince people that 2.4 - 2.5 volts is OK, while draining the cells that low actually significantly reduces their service life... or destroy the cell fairly quickly if it's used with small loads.

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    \$\begingroup\$ You are confused about the role of an over-discharge protection circuit. "Do not use over discharge protection as signal for when to charge batteries, it might wear the batteries down faster." (source). For redundancy and to reach max. battery life, your device has to track voltage and shut down before the over-discharge protection will. That would be at the 2.7 V under load that you mention. \$\endgroup\$
    – tanius
    Commented Jun 30, 2022 at 18:00
  • \$\begingroup\$ the overdischarge circuit IS the device's function to prevent overdischarge. How is it a "protection board" with "overdischarge protection" if you need an overdischarge protection board after you install it? That makes no sense. \$\endgroup\$
    – user309900
    Commented Jul 1, 2022 at 19:06
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    \$\begingroup\$ Device and battery can often be separated, that's why overdischarge protection circuit on the battery is added as a last resort protection against battery damage, but not to maximize battery lifetime. Just as lithium chargers have to stop at 4.2 V before the battery's overcharge protection will kick in. Please have a look here at "But my over-discharge protection do first trip at 2.3 volt, is it faulty?". \$\endgroup\$
    – tanius
    Commented Jul 2, 2022 at 8:23
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    \$\begingroup\$ I'm not talking about batteries with internal protection circuits, but about bms boards that are intended to act as the battery protection in devices. \$\endgroup\$
    – user309900
    Commented Jul 3, 2022 at 11:00
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    \$\begingroup\$ Final comment: You quote battery manufacturers but miss out on quoting them in full: they have two different cutoff values, one "voltage for terminate discharging" (maximizing service life) and one lower "voltage of over-discharging protection" (ensuring battery survival and safety; 2.0-2.5 V depending on battery and load). (source, quoting Samsung specs). If I understand it right, BMS uses the first, while protection circuits are for removable batteries and use the second. \$\endgroup\$
    – tanius
    Commented Jul 6, 2022 at 7:40
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I just wanted to add to this thread some info I found. I tried to find out just what happens to cells when the are discharged below 2.5V. I have since learned thru experience that their rate of self discharge goes way up. If you just charge them up and measure the capacity of the pack, they will seem to work fine with capacity remaining stable for 4-5 cycles. But if you let them sit on the shelf for a few days or a week they will self discharge. This makes them useless for any normal battery application. Hope this is helpful.

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