If I wire five 3.6V 3500mAh cells in series I get an 18V 3500mAh battery (pretty easy) now if I parallel this with another 18V battery (five 3.6V in series):
1. what is the difference if I ONLY tie the top (18V) AND bottom (ground) together vs tying each of the 4 taps together also? For that matter can I do each of these cases?
2. In question 1 are both considered 5s2p?
3. Which instance of question 1 will give me 3500mA*3 or 10500mA?
Thanks
Basic question??
Re: Basic question??
They both will give you 3500mA * 3 = 10500mA
The problem with the 3 serie strings in parallel is balancing. If you place 3 cells in parallel, you can see that entity as one cell with triple current capacity.
A charger is connected to all the batteries in serie. If one of them decreases in capacity due to whatever reason, it's voltage will be lower compared to the others. So, if your charger charges up to a specific voltage, the other cells will charge up to a higher voltage.
That's what balancing is 4. The voltage on all cells of the serie string is equalised, usually by slowly discharging the cells that are at a higher voltage.
If you have 3 cells in parallel, and one is having a lower capacity, the other one will compensate and make the voltage difference less significant.
So, if you have the cells in serie first, you have triple the number of individual cells to measure their voltage and see if they are equal.
The problem with the 3 serie strings in parallel is balancing. If you place 3 cells in parallel, you can see that entity as one cell with triple current capacity.
A charger is connected to all the batteries in serie. If one of them decreases in capacity due to whatever reason, it's voltage will be lower compared to the others. So, if your charger charges up to a specific voltage, the other cells will charge up to a higher voltage.
That's what balancing is 4. The voltage on all cells of the serie string is equalised, usually by slowly discharging the cells that are at a higher voltage.
If you have 3 cells in parallel, and one is having a lower capacity, the other one will compensate and make the voltage difference less significant.
So, if you have the cells in serie first, you have triple the number of individual cells to measure their voltage and see if they are equal.
Last edited by obcd on Apr 13 2020 3:21pm, edited 1 time in total.
Re: Basic question??
If I read the question correctly, either way you’ll have an 18V 7Ah pack. You have to have a third string in parallel to make a 10.5Ah pack. And you should definitely join the cells into parallel groups. That connection doesn’t have to carry very much power, so it can have a small conductor.
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Re: Basic question??
none.ridgerunner wrote: ↑Apr 13 2020 12:57pmIf I wire five 3.6V 3500mAh cells in series I get an 18V 3500mAh battery (pretty easy) now if I parallel this with another 18V battery (five 3.6V in series):
1. what is the difference if I ONLY tie the top (18V) AND bottom (ground) together vs tying each of the 4 taps together also? For that matter can I do each of these cases?
2. In question 1 are both considered 5s2p?
3. Which instance of question 1 will give me 3500mA*3 or 10500mA?
to get something times three, you must have three of them.
you have only two.
additionally, you cannot get ma (current capability) from any of the information you provided. you can only get mah (capacity).
with only two, you only get 7000mah.
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Re: Basic question??
That is one "String" of 5 cells, 5S (1P implied), 18V 3.5Ahridgerunner wrote:If I wire five 3.6V 3500mAh cells in series I get an 18V 3500mAh battery
> if I parallel this with another 18V battery (five 3.6V in series)
Now you have two strings, 5S2P, 18V 7A
Add a third you have 5S3P, 18V 10.5Ah
Now, to simplify things, unless you **need** modular sub-packs like that, best to create a pack composed of a single string, first connecting 3P into each cell "Group" at the lowest level.
Still 18V 10.5Ah, but 3P5S rather than 5S3P.
Hope that helps.