Question Uneven charged batteries

[Bakuku]

Hi,

I have 8x 18650 cells but 7 of those are charged to 3.6v and one is 4.2v. Thats because the 4.2v one came from another pack. My question is how can I make them all evenly charged?

 

[Noah]

Run their charge down all of the way and then recharge them together (4 at a time is probably optimal)

 

[Bakuku]

Run their charge down all of the way and then recharge them together (4 at a time is probably optimal)

Thing is I have an uneven amount of same charged batteries. 1 and 7

I was thinking about draining the 4.2v one but doing that unprotected doesnt seem wise

 

[Noah]

Could you run down the other one until it is 3V6 as well? I feel like that'd be the easiest way to do it.

 

[Luke]

I was thinking about draining the 4.2v one but doing that unprotected doesnt seem wise

Seems legit to me. Those protection circuits are protecting against over-charging and over-discharging. Obviously you don't have to worry about over-charging. Over-discharging will likely not be an issue either, just make sure whatever load you put on it doesn't pull a crazy amount of current (i.e. don't simply connect the positive and negative terminal with a single wire).

Somebody back me up here, I haven't done this before.

 

[cheese]

Yeah, I would say run down the 4.2v to the same voltage as the others, something like a few kOhm resistor between + and - should do the trick. If the resistor gets hot, take it off and use a higher wattage resistor (can handle more current) or higher ohm resistor (decrease current).

 

[Bakuku]

Yeah, I would say run down the 4.2v to the same voltage as the others, something like a few kOhm resistor between + and - should do the trick. If the resistor gets hot, take it off and use a higher wattage resistor (can handle more current) or higher ohm resistor (decrease current).

Just to make sure... I take a Resistor (10kOhm) and connect it to the + and - side of the battery "Shorting the battery" ?

Edit: I'm using Panasonic NCR18650(B) batteries

 

[cheese]

Since it's a resistor, it's a load instead of a short, you can calculate the current with I=V/R

 

[Luke]

Yeah, I would say run down the 4.2v to the same voltage as the others, something like a few kOhm resistor between + and - should do the trick. If the resistor gets hot, take it off and use a higher wattage resistor (can handle more current) or higher ohm resistor (decrease current).

Since it's a resistor, it's a load instead of a short, you can calculate the current with I=V/R

This

 

[Bakuku]

Since it's a resistor, it's a load instead of a short, you can calculate the current with I=V/R

Engrish prease

 

[Aurelio]

Different batteries (even if they are the same model, but came from different packs) should never be used together. Even if you charged them to the same level, you can't be sure that they would discharge in the same way

 

[Bakuku]

Different batteries (even if they are the same model, but came from different packs) should never be used together. Even if you charged them to the same level, you can't be sure that they would discharge in the same way

The cells were all bought seperately.
4.2v was just removed from a pack I had made.

 

[Luke]

Since it's a resistor, it's a load instead of a short, you can calculate the current with I=V/R

Engrish prease

K, here's the long explanation

It's a resistor; power is being dissipated across it, it's not acting as a "short" which would be just a wire from one end to the other.

Batteries have a max discharge current, which is what I was warning against. The equation I = V/R (where I is current, V is voltage, and R is resistance) can be used to select a resistor that will ensure that you don't end up with too high of a current (discharge current). For example, your battery has V = 4.2, and lets say you want to drain it with a resistor R = 1kohm. Plugging this into I = V/R, we get I = 4.2/1000 = 4.2mA, which is no where NEAR the max discharge current. In fact it will take a longer time to drain because of such a low current. If you wanted to use a 100ohm resister, then I = 4.2/100 = 42mA, still well within the realm of reason, but it will get you down to 3.6V quicker (and naturally, monitor the whole process with a multimeter).

EDIT:
Okay maybe I'll share some more, as to why you wouldn't want to go any lower than around a 100ohm resistor...
Another thing to consider is the amount of power your resistor is capable of dissipating, probably what you have is 1/4 watt resistor. Power can be calculated with the equation P = V^2 / R. So if you're at 4.2V and your Resistor is 100ohms, then P =0.1764 W, which is just barely under 0.25W. If you were to use, say, a 50ohm resistor, then P would equal 0.3528 W, and then you're potentially burning a resistor.

 

[Bakuku]

K, here's the long explanation

It's a resistor; power is being dissipated across it, it's not acting as a "short" which would be just a wire from one end to the other.

Batteries have a max discharge current, which is what I was warning against. The equation I = V/R (where I is current, V is voltage, and R is resistance) can be used to select a resistor that will ensure that you don't end up with too high of a current (discharge current). For example, your battery has V = 4.2, and lets say you want to drain it with a resistor R = 1kohm. Plugging this into I = V/R, we get I = 4.2/1000 = 4.2mA, which is no where NEAR the max discharge current. In fact it will take a longer time to drain because of such a low current. If you wanted to use a 100ohm resister, then I = 4.2/100 = 42mA, still well within the realm of reason, but it will get you down to 3.6V quicker (and naturally, monitor the whole process with a multimeter).

EDIT:
Okay maybe I'll share some more, as to why you wouldn't want to go any lower than around a 100ohm resistor...
Another thing to consider is the amount of power your resistor is capable of dissipating, probably what you have is 1/4 watt resistor. Power can be calculated with the equation P = V^2 / R. So if you're at 4.2V and your Resistor is 100ohms, then P =0.1764 W, which is just barely under 0.25W. If you were to use, say, a 50ohm resistor, then P would equal 0.3528 W, and then you're potentially burning a resistor.

So just a resistor between + and - and regulary check the voltage? When desired voltage is met cut of connections.
This can be done without a protection circuit? The panasonic NCR don't have one built in like some other 18650 cells do.

 

[Luke]

That's what I think. And it wouldn't hurt to check the current, at least in the beginning.
(Just to clarify, and you probably already know this but on the off chance you don't, when measuring current, disconnect one of the ends of the resistor and put one probe to the disconnected end of the resistor, and one to the disconnected end of the battery, do NOT put it across the battery like you would if you were measuring voltage)