# How is it possible a dead old Lithium battery becomes charged after opened?

I opened a lithuim battery salvaged from my old laptop battery pack. I checked million times that it is 0V and it was by the time i was opening it. For some reason i would love to know, it is now MAGICALLY charged up to 3.99V and that scared me so i left it aside keeping it cool. It has no protection circuit. So is oxygen reacting with Lithium metal and creating such voltage? Or is it really charged?

Update:

I'm charging the battery that was spot welded in parallel to the battery #1. Just for tests and observing its temp. The battery #2 also holds 0V (which also assures that bettery #1 was definitely 0V) so if it stays like that I'll open it and see if it suddenly charges as well up to 4V and what is the reason. Also I sealed up bat#1 Here's bat#2:

I did a 100R load on the battery #1 and here's the data:

-T=0                        3.96V

-T=10 min              3.91V

-T=20 min              3.91V

-T=30 min              3.90V

-T=1 hr                    3.87V

-T=1 hr 30 min       3.86V

-T=2 hr                    3.84V

-T=4 hr                    3.78V

-T=6 hr 30 min       3.72V

-T=9 hr                    3.67V

-T=11 hr 30 min    3.61V

It took a long time so i stopped at 3.6V

Graphed by T. Stewart using data above Using Ic=CdV/dt $C=V/R*dt/dV~$ with units $~= [F]=[V/Ω*[s/V]]$ for R=at 10k load 11h

Next test 100 Ohm load to 3.0V (faster test) taking two measurements; with load and without. This will measure C2 and ESR. Voltage readings were unstable at first ( double layer effect) but averaging 20 F rising to 40 Farads.

Note that a fresh 18650 cell ranges from 5000 Farads to 10kF so this dead battery will have much higher ESR and C1 <1% of it's initial capacity (unknown).

However it still useful for Volatile Memory backup.
Momentary short circuit current will also measure ESR.

There is a protection current interrupt device under the positive terminal. Inside, the battery is still charged. See the structure in this article.

It's not magic, there is a dual layer capacitance ( weak memory ) but main capacitance from aging may have reduced from 10kF to << 10% with ESR > 100x norm.

Considering 18650 normal energy storage up to 10 Joules full, you have <=1% of this energy in storage, is my guestimate.

Remember Voltage is just potential, and Resistance is futile if its too high... ;)

Do a 10K R load test and measure dV/dt=Ic/C from 4 to 3.6V
If you can then do 100R load test from similar conditions down to say 3.1V

Record the results with drop and changes then compute C again. This difference can be modelled crudely by R1+C1//R2+C2

Then present your model to us, if you can.

simulate this circuit – Schematic created using CircuitLab

• I did a 100R load test on it as you suggested. But i stopped at 3.6V. Check the update and the data i recorded. 10k R load would take forever. Do you think it is worth the time for knowledge sake? I can do the test Commented Mar 30, 2018 at 23:32
• Should i charge battery #1? Or shouldn't? Commented Mar 30, 2018 at 23:33
• No i said 1st 10k then 100 ohm or something in between Commented Mar 30, 2018 at 23:37
• Oh i didn't think it would matter. But how it would affect the results if i reversed the values? Commented Mar 30, 2018 at 23:41

Consider that most computer battery packs are actually LION cells packaged with a controlling circuit. Such circuits can disconnect a discharged LION cell to prevent damage to the cell. Note that LION cells can be permanently damaged if deeply discharged. This is likely the primary reason that LION cells are not sold loose such as other secondary (for example NICD) or primary (for example alkaline) cells are.

just feel wired why you cann't charge it up to 4.2V or 4.35V which is the normal 18650 voltage. if batteries cann't charge up to 4.2V about 20% battery capacity are drained.

maybe you can use dripple way of charging to see it can goes to 4.2V?