Is open circuit voltage of the supply important for Lithium battery as long as we guarantee that the maximum allowed voltage will not be exceeded when the supply is connected to the battery?

Suppose we have a normally discharged (say 3.1V at that moment) single cell (3.6V) lithium battery and a high internal resistance voltage source, say 12V. If we guarantee that the terminal voltage will drop to some value under 4.2V, is it safe to connect the battery to the source? Would the initial terminal voltage give any harm to the battery, or the damage due to the excessive voltage is only related to the electric field inside the battery, meaning that the damage state is a function of "state of charge" and the current?

Of course, the terminal voltage MUST be monitored in order not to exceed 4.2V limit while the battery is getting close to the fully charged state. However, this safety precaution is beyond the scope of this question.

  • \$\begingroup\$ "Maximum voltage AT the battery (1 cell) under ANY current is also Vmax" of the cell.. \$\endgroup\$
    – Natsu Kage
    Commented Feb 1, 2020 at 3:34
  • 1
    \$\begingroup\$ If you connect a 12V supply with a large series resistor to a discharged lithium ion battery, there is no problem as long as the current is within spec for the battery. Current is (12V - VBATT) / Rseries. However, as the battery charges, the cell voltage will increase, and you must make sure that the supply is somehow disconnected once the cell reaches its terminating voltage or voltage/current condition. You can't just leave the supply attached to the cell. Because of this critical need to remember to disconnect the supply later, I consider this to be a dangerous practice. \$\endgroup\$
    – user57037
    Commented Feb 1, 2020 at 6:44

1 Answer 1


If we guarantee that the terminal voltage will drop to some value under 4.2V, is it safe to connect the battery to the source?


But how can you guarantee that the battery voltage will not exceed 4.2 V? Electrically a Lithium-ion battery looks like a large capacitor, and just like a capacitor if you push even a tiny current into it the voltage will continue to rise as it accumulates charge. Any amount of current that charges the battery at all will eventually overcharge it.

Therefore you need an external method of preventing the voltage from rising above 4.2 V. A crude way of doing it is to simply measure the cell voltage (either with a comparator circuit or a multimeter), and electronically or manually cut the charging current when it reaches 4.2 V. But this will only get to around 70~80% of capacity, depending on the charge rate.

A more sophisticated charging circuit would automatically reduce the charge current if voltage tries to go above 4.2 V. This is called CVCC (Constant Voltage Constant Current) charging. It isn't good for the cell long-term though, because it keeps it at maximum stress. Most Li-ion chargers shut off completely when current drops to ~10% of the constant charging current, which gets to ~99% of maximum capacity.

Many chargers also have a timeout and/or capacity limit as a backup in case the CVCC cycle fails to complete, and a lower charge rate if the cell voltage is very low. Finally the cell temperature can be monitored to (hopefully) cut the charge off before it explodes if something goes wrong. As well as all that, a PCM (Protection Circuit Module) may be built into the battery or on the charger, to cut it off if the voltage goes too high or low.

These extra safety functions are required for consumer products that may be left unattended in situations where an exploding battery would be disastrous. If you need to charge a Li-ion cell and all you have is a constant current then if you constantly monitor the cell voltage, and if you don't try to charge an over-discharged or damaged cell, and if you keep it away from flammable material and have a fire extinguisher nearby then it should be OK. But you must guarantee that the battery limits won't be exceeded.

  • \$\begingroup\$ This answer is potentially misleading to beginners. It is not enough to say that you must prevent the cell voltage from rising above 4.2V. You actually must disconnect the cell from the supply and cause charge current to drop to zero. Floating the cell at 4.2V is not safe. \$\endgroup\$
    – user57037
    Commented Feb 1, 2020 at 6:46
  • 1
    \$\begingroup\$ @mkeith that's why I said that 'floating' "isn't good for the cell long-term though, because it keeps it at maximum stress.". For the OP's proposed 'just charge from 12V through a resistor and hope that terminal voltage will drop to some value under 4.2V' disconnecting is the only option anyway. \$\endgroup\$ Commented Feb 1, 2020 at 7:32
  • \$\begingroup\$ Fair enough I guess. \$\endgroup\$
    – user57037
    Commented Feb 1, 2020 at 7:48

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