I bought a li-ion battery off of ebay. The listing claims that the battery has a protection circuit, but a week later, the battery didn't cut out or anything. With a multimeter, it is reading 10V when the battery is listed at 10.8-12.6V. Does this mean the battery doesn't have protection or should the cutoff be lower?

  • \$\begingroup\$ 1. It comes from China on eBay, so who knows. 2. I see nothing indicating it has built in protection. \$\endgroup\$ – Asmyldof May 21 '16 at 0:40

The battery is most likely a 3S Li-ion pack, i.e. 3 cells/packs in series. Protection circuits for single cell Li-ion normally have overdischarge protection set somewhere in the range 2.5V-3.2V per cell, which translates to 7.5V-9.6V for a 3S pack. So this is the range that you should test to ensure that the undervoltage protection correctly triggers.

The listed output voltage 10.8V-12.6V is probably meant to be a conservative estimate of the usable voltage range, since there will be little accessible capacity left below 10.8V (except at very small currents).

Edit To respond to a comment on the question, if you do a search on the model number "DC 12680 protection" you will find other listings (e.g. on Amazon) where it claims to include protection circuitry.

You will also find pages that unmask the typical overinflated capacity claims on these packs. For example, here it is discharged at 1A down to 8.5V and tests at only 25Wh. Notice that this is 3.7V * 6.8Ah = 25Wh, not 11.1 * 6.8Ah = 75Wh. The industry standard is to quote capacity at the cell nominal voltage 3.7V not the pack voltage (here 11.1V). Many unscrupulous sellers exploit this ambiguity to mislead the buyer into thinking that the pack has much higher capacity, here 75Wh vs. 25Wh. Caveat emptor.

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    \$\begingroup\$ I would generally advise Li-Ion to cut out at or above 3.0V per cell. A pack for general use with a protection board letting it fall to 2.5V per cell is bad design. From 3V to 2.5V you get about 2% remaining energy content at 1C and increase the risk of premature hardening by 400%. At 3.6V however (10.8 / 3), you still have 18 reasonably safe percent left till you're at 3V with 1C. Other words, for three cells 10.8V as nearly empty is quite conservative, but cutting out below 9V equates to no protection for a compound 3 cell polymer pack. \$\endgroup\$ – Asmyldof May 21 '16 at 0:46
  • \$\begingroup\$ @Asmyldof The listed UVP range was not meant to be advice for any particular design but, rather, to reflect the range that one may encounter in various devices, which is what is relevant for the OP's question \$\endgroup\$ – Bill Dubuque May 21 '16 at 0:51
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    \$\begingroup\$ @Asmyldof One thing to keep in mind is that designers often lower the undervoltage limit for packs designed for high current loads, in order to counteract the high voltage sag caused by cell internal resistance. Failure to do so would result in much less usable capacity in such cases. In practice I've encountered limits across the whole spectrum that I mentioned, and the extremes were not "bad design". Rather, they were designs that were highly optimized for specific applications \$\endgroup\$ – Bill Dubuque May 21 '16 at 1:46
  • \$\begingroup\$ A hard limit at 2.5V is bad design. Period. If you design for more than 2C you use Feed Forwarded design concepts that push down limits based only on currents within the normal ranges. If you can connect a 1k resistor and measure any voltage below 3V per cell, it's just bad design. There's dozen upon dozens design and app notes freely available using feed forward limiting. \$\endgroup\$ – Asmyldof May 21 '16 at 20:02
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    \$\begingroup\$ @Asmyldof The protection circuit is supposed to be set at a voltage you don't want to ever get to. The user product should be shutting off at around 3.0V, the protection circuit is there in case that fails for some reason and so needs to be set to a lower voltage. The protection circuit is a fail safe and should never be triggered under normal operating conditions. \$\endgroup\$ – Andrew Apr 19 '17 at 8:14

Without actually looking in the pack to determine if there is protection circuitry, it would be very difficult to test for the circuitry. The main reason being that in order to do a voltage or current test, you would need to test the extremes( Highest voltage, lowest voltage, max current draw, temp, etc.) and if the battery is not protected you will most likely end up destroying the battery.

Ex: You decide to test if it has under voltage protection, so you start to drain the battery and observe the voltage.

outcome a) The battery has protective circuitry, so as the voltage reaches a low level, around 2.5V per cell(can vary quite a bit, determined by the manufacturer), the output shuts off and you suddenly read 0V.

outcome b) There is no protective circuitry, so the battery keeps draining and the voltage reaches a range where you can determine that there is no protective circuitry and the voltage is still high enough that the cells have not completely died, around 2V-2.5V, depends on chemistry.

outcome c) There is no protective circuitry, and the voltage on the cells becomes low enough that they start plating lithium and become useless.

It then becomes a game of trying to achieve outcome b while avoiding outcome c. If testing is the only option to determine if it has protective circuitry, I would recommend draining the battery to about 2.4V per cell or so, and if it gets that low with out shutting off then there is no circuitry, at least none of any value. You need to be very careful not to go much lower, because all it takes is one cell dropping below 2V to make the entire battery useless. I would not recommend test over voltage or current, as these can easily lead to the battery "rapidly disassembling".

TLDR; It can be done by very carefully lowering the voltage, however it should only be done as a last resort as cell damage can easily occur if not careful.

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    \$\begingroup\$ Above is quite misleading. Li-ion cells don't "completely die" if once drained a bit below the recommended minimum voltage, nor do they "plate lithium" (that is due to overcharging). Rather, copper current collectors can start dissolving if the cell remains at very low voltages (esp. < 2.0V) for long periods. But a slight undervoltage due to a quick test as above is unlikely to cause any serious degradation. \$\endgroup\$ – Bill Dubuque Apr 23 '17 at 4:08
  • \$\begingroup\$ @BillDubuque: so I can safely drain the battery dry as long as I immediately recharge it afterwards? \$\endgroup\$ – SF. Apr 23 '17 at 11:22
  • \$\begingroup\$ @SF No, you should never do that for Li-ion batteries. \$\endgroup\$ – Bill Dubuque Apr 23 '17 at 19:20
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    \$\begingroup\$ This is not optimal answer, but the only one that tries to provide a guide how to answer the titular question instead of trying to help with that particular single battery pack. \$\endgroup\$ – SF. Apr 26 '17 at 6:51

re: "...battery has a protection circuit..." What kind of protection? Over charge protection (max current &/or above the cutoff V), over discharge protection (max current &/or below the cutoff V), max current protection or all three? If you want to know for certain which type/s of protections your battery provides, your battery will need to be monitored (V & A) while charging & discharging it.

If the li-ion cells are 3.7 Vnom, then the max charge should be 4.2 vdc per cell an 12.6 vdc for 3 cells that are connected in series. If the charging current cuts-off before the in circuit charging V exceeds 12.6V, then your battery has over charge protection. If the discharge current cuts-off at or before the battery reaches typical cut-off voltage levels (~3v per cell), then your battery has over discharge protection.


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