I have a set of devices using Li-Ion batteries and featuring a charging IC (LTH7). Charging works normally on recent batteries, but does not on used ones.

Knowing that chargers initially try to maintain a constant current (CC phase), I was assuming to have a very low or no charging current for those used batteries, and thus expecting the charger to respond like if no battery were connected: increasing Vbat until the max value (4.2V).

Instead of that, I observe that Vbat remains constant around 0.1V. Any idea why the charging IC does not increase the battery voltage ?

  • \$\begingroup\$ It is doing what it should do, protecting you from harm. ErikR explained it nicely. He did forgot to mention how brilliant the fire ball is when the Li batteries fail. \$\endgroup\$
    – Gil
    May 29, 2021 at 15:11
  • \$\begingroup\$ "I was assuming to have a very low or no charging current" - so you are telling me this assumption is wrong, and that dead batteries can potentially draw high current (if no protection), resulting in overheat ? \$\endgroup\$
    – rclyde
    May 29, 2021 at 17:53

1 Answer 1


It is unsafe to rapidly charge a deeply discharged li-ion battery, and so many li-ion charging chips will use a pre-set trickle charge if the battery is below a certain voltage. In the datasheet this is usually indicated by V_TRIKL and I_TRIKL.

As another safety precaution, most charging ICs will refuse to charge a battery which is lower than a certain voltage. This is called under voltage lockout. I wasn't able to find what that is for the LTH7 but I would suspect it is probably around 2.0V. In any case, I definitely would regard a 0.1V battery voltage to trigger the lockout.

One way that usually works to "revive" deeply discharged cells is to charge them using a constant-current power supply at a very low current. Set your power supply to 3V and the current limit to a low value like 10mA (you need a current limiting power supply). A lot of times the voltage on the cell will rapidly rise to a voltage around 3V at which point you can put it on a normal charger.

  • \$\begingroup\$ The datasheet reads "Vtrickl = 2.9V trickle charge voltage" (涓流充电电压)" and "Vuv = 3.4V VCC undervoltage lockout threshold (欠压锁定阈值)". So we are not in trickle charging but definitely in the lockout. Great answer, thanks! \$\endgroup\$
    – rclyde
    May 29, 2021 at 9:02
  • \$\begingroup\$ you should ideally do this in a fire-proof container \$\endgroup\$
    – user253751
    Apr 3 at 15:04

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