Battery not fully charging

[pears_91]

Hi,

to give update to the thread.., it has taken a while to arrive and for me to get around to it.. but I received the charger and charged up individually each set of cells.

As saneagle suggested above, I did have to find a way to connect the charger and I chose to solder it to the lead strip on the battery pack. Around half of the cell groups needed charging until they reach fully charged, the other half were already full.

So in summary , all of the cells that I replaced were nominal voltage 3.6v (as supplied) and at the time I did this all the other cells in the pack were fully charged 4.1v

The charger has brought the all up to 4.1v by charging individually so now the pack is pretty well balanced.. according to my volt meter there's about 0.05v max different between the cell groups. So I will get much better range out of the battery now I hope .

 

[pears_91]

I am now wondering,,, I am not entirely sure that I didn't replace 1 cell,, that may have been grouped with an existing pre charged cell..

So I connect a cell at 4.1v and one at 3.6v in parallel.

I'm quite sure I didn't do this, because when I read them with volt meter all cell groups measure either 4.1 or 3.6 as a pair..

I think if I did replace just half of a pair it would measure 3.85v as a cell group..

But in the event that I did do that, am I right in thinking that the cells in parallel would not balance.. and upon charging, it would bring the group up to 4.1v.. one of the cells would be 3.85 the other would be 4.35v for example,, giving average 4.1v?

As I say I'm quite sure I didn't do that just because the voltage of the pairs would have been different had I done that, but I just had a moment of slight worry.

 

[saneagle]

I am now wondering,,, I am not entirely sure that I didn't replace 1 cell,, that may have been grouped with an existing pre charged cell..

So I connect a cell at 4.1v and one at 3.6v in parallel.

I'm quite sure I didn't do this, because when I read them with volt meter all cell groups measure either 4.1 or 3.6 as a pair..

I think if I did replace just half of a pair it would measure 3.85v as a cell group..

But in the event that I did do that, am I right in thinking that the cells in parallel would not balance.. and upon charging, it would bring the group up to 4.1v.. one of the cells would be 3.85 the other would be 4.35v for example,, giving average 4.1v?

As I say I'm quite sure I didn't do that just because the voltage of the pairs would have been different had I done that, but I just had a moment of slight worry.

If you put any cells in parallel, they immediately become the same voltage, so if you did put one at a lower voltage to the others into the group, they would equalise at a slightly lower voltage . The charging rate of the lower one should just about be in it's safe range since the difference was about 0.5v and its internal resistance to DC is about 0.1 Ohm, so current (I) =V/R = 5 amps; however, that would decrease fairly rapidly as the cell charges up because the voltage difference becomes less as it charges.