Background: I have a laptop with a bad battery. Upon opening it up and testing the six 18650 cells that make up the battery, I found one to be completely discharged while the other five were still at an acceptable voltage. I partially charged this cell in isolation to bring its voltage back up to the average of the others in an attempt to get the battery working again. Which it did. But the battery isn't lasting nearly as long as it used to, so I am assuming this one cell is damaged.

Now my actual question is whether I could e.g. replace the cell with a 2000mAh cell when the others are 2600mAh, or if that's a bad idea. We're talking about a laptop battery here with a relatively low current draw and fairly advanced charge controller. Though I admit that I know nothing of these things aside from what I learned in school up to 10th grade in the 90's.

I salvaged a number of these cells from other laptop batteries but I have no idea whether they're all the same capacity, so I just want to make sure that I can safely do this, or whether I'd have to bite the bullet and spend the $50 it'd cost me for 6 brand new high quality cells (which may as well all be the nice 3000mAh ones while I'm at it).

  • \$\begingroup\$ Bad idea. Oh and that's 2.6 Ah (2600 mAh) not 2600 Ah. \$\endgroup\$
    – user16324
    Sep 28, 2020 at 13:43
  • \$\begingroup\$ That's what I mean LOL. Care to explain? I assume the smaller cell would discharge faster and just get damaged again? Though I also assumed that since the rate of discharge is fairly low, the cells would be able to equalize before this happens. \$\endgroup\$ Sep 28, 2020 at 13:47
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    \$\begingroup\$ You can keep "messing" with used cells like that forever. The only "proper" way is to determine the capacity of each cell individually (you need special equipment for that) and then combine cells that all have a similar capacity. But even then you will suffer many more frustrations than simply buying a completely new battery. Yes it is expensive but laptop batteries wear out, it is a fact of life. \$\endgroup\$ Sep 28, 2020 at 14:37
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    \$\begingroup\$ Well when I tried, I replaced all the cells with matched ones. However I omitted to patch in a temporary power supply while removing/replacing cells. The proprietary BMS lost the contents of its CMOS memory and the laptop never recognised it as a battery again... \$\endgroup\$
    – user16324
    Sep 28, 2020 at 14:43
  • \$\begingroup\$ I had an Asus netbook that did that with the not recognising the battery after I had it in storage for 2 years. I managed to get it working via some weird method I found on a support forum. I think I had to uninstall the battery driver, then do something in the BIOS or something. It wasn't an obvious process. But I remember it not being recognised, but after a lot of "random" poking, I got it recognised and charging again. Couple years later I threw it away and salvaged the cells. They're used in LED flashlights now. \$\endgroup\$ Sep 29, 2020 at 7:09

4 Answers 4


Don't mix different batteries!

(unless you know very, very well what you are doing)

The battery management system is probably going to keep them from exploding, but they will not work as intended and will probably fail quickly.

If you have a balancing charger/analyser and a lot of patience, you may be able to measure the capacity of all the cells that you have and match six of the available that are within 5..10% capacity difference.

p.s. don't be tempted to replace the cells with much more powerful ones. The battery management may not like them much. More powerful ones are also more picky about charging conditions.

10% or 20% over is probably OK.


The cells in your current battery are all the same age. They've all been put through the same load in the same environment. Barring some obvious external trauma, the likelihood that another one of the cells wears out soon is fairly high. You'll probably have better (and possibly cheaper) results replacing the whole thing at once than replacing the cells ship-of-Theseus style over several months.


I'd suggest a single slightly LARGER capacity replacement cell.
A smaller mAh one will again "bottom out" first and if the BMs is not doing a per cell protection it will again die.

A larger cell will tend to have more charge than the others IF all are fully charged to start.

Testing cell capacity is a very good idea and not hard - only annoying without a proper tester. Charge all fully. Discharge each with the same value load. You could get fancy and use an LM317 + 1 resistor per cell and get constant current. But a say 10 Ohm resistor (2 Watts or higher) will give you 420 mA at full charge (4.2V) and 300 mA at "as flat as you should allow" = 3V. You can monitor battery voltages with a meter occasionally (set an alarm) or if you have some sort of logging (a $5 Arduino would do it) log voltage against time. You can do all at once still connected with 6 dischargers or one at a time or disconnect the cells.

Constant current load:

If you use an LM317 plus resistor Vin min will be around 3V - as you need 1.25V for the resistor and maybe 1.7V headroom.

LM317 - 45c each in 10's here

Datasheet here
Dropout voltage shown in fig 4.
Typically 1.75V at 500 mA and falls with increasing temperature.


simulate this circuit – Schematic created using CircuitLab

In this case Iout is shorted to ground.
The LM317 will dissipate V x I watts = (Vbattery - 1,25V) x Iout
At say 4V and 400 mA that's P = V x I = (4-1.25) x 0.4 = 1.1 Watts. A modest heatsink will be a good idea. For the TO220 part Tja = 60 C/W so 66 C rise at 1.1 W (only when battery is fully charged) or nearly 100C in typical ambient. Tjmax is allowed to be 125C so no heatsink in open air is "notionally OK". Notionally.

  • \$\begingroup\$ Nice writeup. Thanks. \$\endgroup\$ Sep 28, 2020 at 14:46

There is a protection circuitry on the batteries, once it senses a failure or interference it transmit to the computer not to charge or use the batt, there is nothing you can do to reset it. Replacing any of the cell is a waste of time.

  • \$\begingroup\$ I managed to get the battery to charge (and maintain charge) by simply charging the one cell in isolation AFTER my battery reported that it had failed. I do believe I managed to "reset" it. \$\endgroup\$ Sep 28, 2020 at 14:50
  • \$\begingroup\$ If it works normally in the laptop then you proved me wrong. \$\endgroup\$
    – Moty
    Sep 28, 2020 at 14:54
  • \$\begingroup\$ Well, "normally" as in it works for 20 minutes where it used to work for 45. I'm probably just extremely lucky. \$\endgroup\$ Sep 28, 2020 at 14:58

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