Random question

Sometimes learning this new stuff I start feeling dumb.

But then I remember there are places like this with you smart people and I feel better!

So here is my question.

If you have 2 batteries. Lets say we are using 18650 because... why wouldn't we be.

1 is charged at 4.2v (nice and full)
1 is lower (not dead) but sitting around 3.4 because it goes used in a flashlight or something.

Now. If I connect these 2 batteries in parallel, my understanding is that the 4.2 is going to try to charge the 3.4 battery.

What kind of current are we talking there? Is there a formula?

Now lets say we have a nice big pack of 79 cells that are all already in parallel and we decide to throw mr 3.4 in as the 80th cell. What kind of current is going to be flowing to that battery?

Are we talking like full speed ahead multiple amps at a time? or is there a nice formula to give an idea on how much current would flow.

Thanks Smart People!!!!

Yes, full speed amps. I=V/R is your formula. So, for Current, let's take voltage of 4.2V and divide it by almost no resistance. Then, that means there's going to be about 4.2Amps of current going from full cell to less full cell. Not a good situation. Best case, you get really warm cells. Worst case, one or both could fail horribly after enough times of doing this.

If you want to charge the lesser with a full charged one, you need to put a resistor between them to limit the current going into the lesser cell. I don't remember where, but I saw someone make a dirty charger like this and soldered a resistor on the positive end (though, in reality, it probably doesn't make a difference which end) and put this on a cell holder that we are well familiar with. That way they can charge less cells easily. This was actually done for cells that are too low to trigger the battery charger and this brought the "dead" cell up to workable voltage.

Definitely full speed ahead and multiple amps. The fully charged cell isn't just trying to charge the other one, it is doing it for sure. Although is has nothing to do with charging as such. Charging is usually a controlled method, this situation isn't though. There is a potential difference between the cells (4.2V-3.4V=0.8V) and therefore there will be a current between them to equalize this difference.
18650s are capable of short circuit currents in the region of 50A. How much will be actually flowing between the cells depends on how big the potential difference is and what the resistance of the circuit is. It is Ohm's Law, I = U / R. At least I think it applies here. Since 0.8V isn't that big of a difference and if we assume an resistance of the cell of 0,25 Ohm and neglect the resistance of the parallel circuit then this is 0.8V / 0.25 Ohm = 3.2A. If the cell has lower resistance, which is likely, then the current will be much higher up until you have practically a short circuit where the short circuit current of the cell will come into play. I will happily accept a correction here, I was never good at this sort of calculations. The principle is right though, the current scales with the difference in voltage between the cells.
If you do this in a pack with not 2, but 80 cells in parallel, then this will be essentially the same, but worse because of the scale.

Just put a high power resistor between should limit the current going into the battery.

Remembered who did the video. It was Daromer:

Korishan said:

Remembered who did the video. It was Daromer:

Click to expand...

oh man that just came out a couple days ago!! been a little out of it with hurricane irma messing up my area. still without house internet. having to tether. was without power for a week.

Thanks for the link!!!

Yeah, that's the dirty way of jump starting a dead battery. Or actually the attempt to make the dirty way less dirty. Obviously this would otherwise be much worse than you example because we have a potential difference of almost (or even exactly) 4.2V and almost no resistance. That is short circuit current territory. Current will drop soon, but it is still bad enough.

About that video.....

If we say each battery has an internal resistance of 50mOhm, and we neglect wires and such:
A = 3.3V SOC, IR=0.050 Ohm
B = 4.0V SOC, IR=0.050 Ohm
Then the current flow would be (V_a-V_b)/(R_a+R_b) = 0.7V/0.100Ohm = 7A

Bump charing with an inserted resistor would give you an artificial high "IR", say we insert a 1 Ohm:
A = 3.3V SOC, IR=0.050 Ohm
+ 1 Ohm
B = 4.0V SOC, IR=0.050 Ohm
Then the current flow would be (V_a-V_b)/(R_a+R_b+R_extra) = 0.7V/1.100Ohm = 0.63A
The heat lost to resistance would be U*I = I^2*R = U^2/R
So back-of-the-envolope would be 0.44W out of a total of 2.5W as R-losses so 18%

So this means that you dont even loose that much charge to heat if you resistance-limit equalization, essentially you loose delta_V/V_max = 0.7/4=18%
This drops as the voltage delta lowers, assuming linear voltage-changes, the average loss to heat would be ie. 9%

Yeap bump charge with a resistor will not stress the heck out of the cells that much and the lost energy is the least of your issues. note that you rarely never need to go with 1 cell at 3.3 and another at 4. its generally the ones that is closer to 0V if so

Korishan: yes it was me

so in my scenario of a charged 4.2 and a lower charged 3.4 the numbers come out to be around 1ohm resistor to keep things around 1 amp.

since the battery is not "dead dead' i am thinking 1 amp is still acceptable since this is what we are charging the batteries at anyway. Am I on the right track for my scenario?

Yes you are. Not sure why you want to do it?

That is correct. However, what is your use case for that sort of thing?

DarkRaven said:

That is correct. However, what is your use case for that sort of thing?

Click to expand...

fortunately no real case at this point, just curiosity