Batteries in parallel, charging and tricks

Thread Starter

ArakelTheDragon

Joined Nov 18, 2016
1,366
Hi guys! Happy new year!

Batteries used: 4x18650 (the same).

I want to know will 4 batteries in paralel balance each other's voltage when they are left unused for a while? Meaning will the voltage become equal on all batteries or will the more charged ones remain more charged, while the less charged ones remain less charged. Also how long will it take?

EXAMPLE:
Bat0 == "4VDC", Bat1 == "3.5VDC", Bat2 == "4.2VDC", Bat3 == "3.8VDC" /* Parallel connection. */

after how much time or never?

All batteries have the same voltage.
 

ian field

Joined Oct 27, 2012
6,536
Hi guys! Happy new year!

Batteries used: 4x18650 (the same).

I want to know will 4 batteries in paralel balance each other's voltage when they are left unused for a while? Meaning will the voltage become equal on all batteries or will the more charged ones remain more charged, while the less charged ones remain less charged. Also how long will it take?

EXAMPLE:
Bat0 == "4VDC", Bat1 == "3.5VDC", Bat2 == "4.2VDC", Bat3 == "3.8VDC" /* Parallel connection. */

after how much time or never?

All batteries have the same voltage.
Put it this way - the best way of balancing series cells is to re configure them into a parallel bank and charge them as one cell.
 

oz93666

Joined Sep 7, 2010
742
Sure they will ... any type of rechargeable battery ...or any capacitors if connected in parallel will quickly approach the same voltage.

How quickly is determined by internal resistance and voltage difference , your cells have very low int resistance , will be within about 0.01 V of each other with in about half an hour
 

MrAl

Joined Jun 17, 2014
13,707
Hi,

Just a note here.
When batteries are fully charged they get up to some target voltage level like 4.2 volts.
After they are taken off charge, the voltage will drop slightly.
When placed in parallel, the one with the higher voltage ends up charging the one with the lower voltage.
The current is given by the voltage difference divided by the sum of the internal resistances:
i=(v2-v1)/(R1+R2)

Since R1 and R2 are typically very small (0.05 Ohms) this means we could have this:
i=(v2-v1)/0.1

and since v2-v1 is just the voltage difference vd we have:
i=vd/0.1=10*vd

so the current could be ten times the voltage difference. Depending on how big vd is, the charge current could be high or low. If the cells are good, vd will probably be low, but if one of the cells is not that good it could be high.

Another point is charging in parallel.
When charging in parallel we would like to charge two cells at 2x the normal charge rate. But if we do that and one cell is already charged or has high internal resistance, then all the current goes into one cell. That would mean that one cell would be taking two times the safe charge current level. That's something to keep an eye on also and maybe just not charge the two cells at too high of a current.
 

Thread Starter

ArakelTheDragon

Joined Nov 18, 2016
1,366
Thanks for the replies!

In my case I have 4 batteries charged together at half the rated current. What I needed to know was will I need an equalization charge for each battery separately also.

Happy new year!
 

ian field

Joined Oct 27, 2012
6,536
Thanks for the replies!

In my case I have 4 batteries charged together at half the rated current. What I needed to know was will I need an equalization charge for each battery separately also.

Happy new year!
You need to balance cells charged in series - there's probably an off the shelf module.

A quick and dirty cheat is to use individual USB charger PCBs for each cell - but you have to use that many USB wall warts with floating outputs, or do all the cells one at a time.
 
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