I've got a 18.5 volt battery made up of 5 3.7 volt cells (Li-Ion). For charging the datasheet for the single cells says the battery requires .5 volts more, to counteract the internal resistance, I guess.

My question: Since the nominal voltage is additive per cell, is the additional voltage as well? As in: do I have to charge at 21 (18.5 + 0.5*5) or at 19 (18.5+0.5) Volts?

I believe it is the the former, since in a serial circuit resistance is additive as well, but since I don't want to make a mistake, I decided to simply ask.


  • \$\begingroup\$ This question seems to stem from a misconception. That extra half a volt is due to the fact that a fully charged li-ion cell has a terminal voltage of 4.2V, and has nothing to do with internal resistance at all. Battery voltage changes with the state of charge, so a "3.7V" cell can have anywhere from 3.0V or below (fully discharged) to 4.2V (fully charged LiPo) or 4.35V (fully charged LiFePO₄ cell). \$\endgroup\$
    – Hearth
    Oct 14, 2019 at 17:09
  • \$\begingroup\$ Thanks for the answer, but this isn't true. The maximum and minimum voltages are given, and yes, the maximum voltage is 4.2 volts as well, but under Charge/Discharge Cycle it specifically asks for an input voltage of 4.2 and and input current of 2. My question simply is wether this stacks with more cells or if the charging voltage just has to be 0.5 volts higher than the nominal voltage, no matter what it actually is. I added a Link to the datasheet \$\endgroup\$
    – Mat NX
    Oct 14, 2019 at 17:21
  • \$\begingroup\$ Sorry, but @Hearth is right. It's the chemistry of a Li-ion cell that requires charging at 4.2V for a nominal 3.7V cell, not the cell's internal resistance. I don't see that the "Charge/Discharge Cycle" specification is relevant to this issue. \$\endgroup\$ Oct 14, 2019 at 17:42
  • \$\begingroup\$ Right, sorry. i didn't mean that he was wrong with the resistance thing, I didn't know that. What I meant with incorrect was actually that I wasn't referring to SoC of the battery, but the voltage required to "charge" (connect to power supply) said battery, not the voltages it has at any potential "charge" (fuel). \$\endgroup\$
    – Mat NX
    Oct 14, 2019 at 18:22
  • \$\begingroup\$ You seem to know that, based on your comment. And I added the Charge/Discharge Cycle to signify that I was looking for information on a different value. Sorry if my comment might have sounded a bit ignorant, that wasn't my intention. \$\endgroup\$
    – Mat NX
    Oct 14, 2019 at 18:30

3 Answers 3


Consider that the cells don't know they're in a pack. Your responsibility is to each cell. Each cell requires 4.2V to charge effectively, so that's what you have to provide.

The simplest way to do that is to charge at up to 21V, and hope that the voltage is evenly distributed between the cells. Generally it will be, unless one of the cells is significantly weaker than the others, or damaged. If so, it may exhibit an internal resistance or cell voltage significantly different to the other cells, and therefore see a different voltage to the other cells.

Often these non-ideal effects are compensated for by individual cell electronics, but if you don't have that luxury then you can still provide some protection by simply setting an overall current limit. Without the current limit, you risk destroying cells or even starting a fire.

A sensible current limit for Lithium Ion cells is generally "0.5C - 1.0C", which means half the one-hour capacity to one times the one-hour capacity. Your cells are 2.15Ah, so 1A is a safe current limit, and you can go up to 2.15A if you need to charge quicker.


Since the nominal voltage is additive per cell, is the additional voltage as well? As in: do I have to charge at 21 (18.5 + 0.5*5) or at 19 (18.5+0.5) Volts?

Yes, the 'additional voltage' is additive, but there really isn't additional voltage. The nominal voltage of a lithium ion battery is 4.2V, and that voltage is multiplied by the amount of cells you have in series. 5 cells would be 21V. Lithium ion cells can actually be charged by a lab power supply, you put the supply on constant current, then on constant voltage (which would be the max voltage for the cells in series). (Which is a really fun educational activity when done properly and safely). The curve looks like this:

enter image description here

  • \$\begingroup\$ Charging Li-Ion or Li-Po batteries with a lab power supply is NOT recommended, and is a recipe for DISASTER. \$\endgroup\$
    – romain145
    Oct 14, 2019 at 17:38
  • \$\begingroup\$ @romain145 And why? our battery engineer (20 yrs experience) at my last job did it all the time and trained me how to do it. It's no different than using a charge controller. A lab supply on CC mode is self regulating on voltage (as the battery voltage goes up, the current goes down). Driving a car is dangerous if you don't know what your doing. \$\endgroup\$
    – Voltage Spike
    Oct 14, 2019 at 17:44

The 100mA cut-off current in the datasheet you provided is not a recommendation, it's an obligation. Not respecting this could lead the cells to overheat and/or explode. The proper way to charging Lithium elements is to use a charger compatible with that particular technology, and to follow the cell's datasheet for the charging current and end of charge condition.

As Heath Raftery said, you are responsible for balancing the charge in between the cells. Circuits exist, they are called "cell balancers". in case any one cell charges faster than the others, this circuit will ensure the cell never goes beyond the 4.2V limit. Even if each cell had its own BMS (and I don't think yours does), you shouldn't rely on the BMS to properly charge the cells.

Charge management recommendations by TI: http://www.ti.com/lit/an/slyt334/slyt334.pdf


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