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I'm trying to design a large battery pack, that should provide a given amount of power, with a reasonable amount of heat dissipations. To do so, I have three degrees of freedom:

  • indidividual cell capacity
  • number of cells in parallel
  • number of cells in series

Given an internal resistance, I was thinking of having a lot of parallel branches with cells of low capacity, in order to lower the current (and thus the heat dissipations in RI²).

However, I heard that the product of internal resistance times capacity is constant, meaning that internal resistance increases when capacity decreases.

Is that true? I could not find any confirmations on the net. Is there a more accurate relationship between these two?

Thank you for your help!

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    \$\begingroup\$ Of a cell. But you aren't assembling a cell. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 21 '16 at 13:40
  • \$\begingroup\$ I'm assembling a large number of cells, and I want to minimize the global heat dissipation (which is the sum of each cell's dissipation) \$\endgroup\$ – tintindu34 Jul 21 '16 at 13:45
  • \$\begingroup\$ Putting them in parallel reduces the amount of current each supplies proportionally. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 21 '16 at 13:47
  • \$\begingroup\$ Yes, this is what I thought. But when you reduce each cell's capacity (because you need smaller amount of current in each branch), internal resistance rises and dissipations as well. That's why I'm trying to figure out by how much R increases when C decreases \$\endgroup\$ – tintindu34 Jul 21 '16 at 13:51
  • \$\begingroup\$ @IgnacioVazquez-Abrams Though in total there will be roughly the same amount of energy dissipated.. \$\endgroup\$ – Eugene Sh. Jul 21 '16 at 13:52
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internal resistance increases when capacity decreases

these parameters are not directly connected.

Is more correct to say that internal resistance is related to battery discharge current. Indeed, a battery with higher discharge current will have a smaller internal resistance.

For example, a LiPo prismatic cell of 3000mAh used to have a bigger discharge current than a cylindrical LiIon with the same capacity.

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I think you should go with higher voltage and low current if you want to achieve low heat dissipation.

You should put many cells in series to create high voltage per branch. And then you can connect these hugh voltage branches in parallel.

So your power will be same but it will be in high volt and low current. Afterwards at the time of actual application you can use dc-dc to convert it to actual requirement.

This configuration will give you low I2R losses wich will reduce the heat dissipation and lower load current per cell means you can utilize relatively higher capacity from the available battery capacity.

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  • \$\begingroup\$ Not sure... Following that reasonning, I should get even more parrallel branches to reduce the current in each branch. That lead me to a very large number of branches. But I'm pretty sure that there is another phenomenon that limitates the number of parallel branches. \$\endgroup\$ – tintindu34 Nov 20 '17 at 16:30
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internal resistance increases when capacity decreases

I think this is correct because if you take two 2000 mAh capacity cells in parallel with 100mΩ each, the effective resistance is 50mΩ. So a single 4000 mAh cell of the same chemistry should have the same 50mΩ internal resistance as two 2000 mAh cells in parallel. One can argue that a single 4000 mAh cell is in effect the same as two 2000 mAh cells in parallel.

This is why the 4000 mAh cell can output 4A and still be have a discharge rate of 1C whereas a single 2000 mAh cell can only do 2A for 1C discharge rate.

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