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I don't understand how cells behave when connected in parallel. So what if I have two Li-Ion cells: one ist "rated" at 15 amps (which means that the cell is able to discharge at 15 amps without problems , right?) The other cell ist rated to 5 amps. Now assume both have appropriate same Voltage to connect them both in parallel. What happens if you connect a load which would draw 10 amps now? Would both cells try to discharge with the same 10A (and thus the 5A-cell would overheat and fail...), or would the current be spread up eavenly? Is it actually forseeable at all what would happen, since it depends on the internal resistance of both cells, doesn't it? Does someone know what would happen?

Then another question: what if you put two cells of different capacity in parallel? Isn't the cell with the higher capacity supposed to give more current? Because the discharge curve of the high-capacity-cell is above the discharge curve of the low-capacity cell. So the high-capacity-cell would have to discharge more of its current until it decreases its voltage to the same level as the low-capacity-cell? So would that mean that the internal resistance gets lower with a higher capacity, is it proportional?

Hopefully my explanations aren't to confusing and hopefully there are people who are bothered by the same question :) Thank you!

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  • \$\begingroup\$ well, you don't put Li-ion cells with different ratings in parallel. \$\endgroup\$
    – Wesley Lee
    Feb 17, 2017 at 19:05
  • \$\begingroup\$ In your example each one will discharge at 5A.. But yes, if you connect them to a higher load (say 15A) the one with 5A rating will be overloaded. \$\endgroup\$
    – Eugene Sh.
    Feb 17, 2017 at 19:08
  • \$\begingroup\$ It all depends on the internal resistance of the cells (and that's usually not specified). \$\endgroup\$
    – BobU
    Feb 17, 2017 at 19:15

2 Answers 2

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If we use two cells of the same type build with narrow tolerances connected in parallel the current will distribute equally to both cells. The capacity, voltage and internal resistance are the same for both cells. If the charge state and the cell temperature are the same we may assume the total current is shared in a 50 % ratio.

But if we use different cells with different capacity and rated current and unknown internal resistance we don't know how the current is shared.

Can you tell us more about those two cells, what about capacity and internal resistance? A datasheet would be helpful.

Cells may be optimized in different ways, for high current and a little more weight or large capacity with small weight and low current. This would result in a low internal resistance for the first cell and higher resistance for the second.

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  • \$\begingroup\$ Sure, it's always better to spread out the capacity and current, but I justed wanted to know whether it's possible because theoretically it's also an interesting question. So that's why I don't have a datasheet. \$\endgroup\$
    – GeoKwi
    Feb 17, 2017 at 21:14
  • \$\begingroup\$ I plan to build an e-bike battery and was wondering whether you can put different cells in parallel. Also you said that high-current cells have lower internal resistance. So the current rate is basicly just the current at which the heat can be delivered away without causing much damage? (Because in this case the current would be spread a little bit between the cells) I still don't get the connection between internal resistance capacity and current-rating. Is there any? (I'm sorry, I don't want to be annoying, just asking...) \$\endgroup\$
    – GeoKwi
    Feb 17, 2017 at 21:22
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Pretty sure batteries are idealised voltage sources with a resistor in series. So by adding two in parallel, you add the reciprocal of their internal resistances, add their current capacities, but not their voltages. This means that they share the current drawn equally if they have the same voltage, but as one decreases in voltage it's current decreases and the other higher capacity cell will compensate. This doesn't mean that it will charge the other one though, it will only increase the time it takes for the lower capacity cell to fully discharge. For example, if you connected one 2305 mAh cell with one 1155 mAh cell you would have a total current capacity of 3460 and resistance can be obtained with the following mathematical abstraction; 1/R = 1/r_1 + 1/r_2. It is because the voltage distributes itself across both paths and is kept constant between the two cells, and because the total resistance actually decreases as seen in the formula.

If you connected a 5A cell with a 15A cell and tried to draw 10A, the would share the current drawn depending on the ratio of their current capacities. In the above example it would be 67% from the higher capacity cell and 33% from the lower cell. If the higher capacity was your 5A cell that would be a problem. At least I think that's how it works.

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