battery pack design

Hi everybody,
I need to design a battary pack to supply 10 kW at 10 V over 10 minutes. The battery cell has nominal voltage 7.4 V, capacity 0.45 Ah and C-rate 30. So, how can I calculate the number of cells in series and in parallel?

Hi Orcuns91, your specification is highly unlikely. 2 batteries by 3.7V are 7.4V, their maximum voltage is 2x4.35V=8.7V. For 10V, you need at least 3. At 10V, to supply 10kW, you need 1000A, which is not likely. Can you please revise your specification?

Carry for cents bazar said:

Hi Orcuns91, your specification is highly unlikely. 2 batteries by 3.7V are 7.4V, their maximum voltage is 2x4.35V=8.7V. For 10V, you need at least 3. At 10V, to supply 10kW, you need 1000A, which is not likely. Can you please revise your specification?

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these are not values from a real application. It's just a theoretical exercise. Values may vary, but I want to learn how to calculate how to connect multiple cells in series and parallel. Can you give me an example to get a battary pack to supply 10 kW at 10 V over 10 minutes?

Normal 18650 batteries have a nominal voltage of 3.6V to 3.8V, we will accept the middle 3.7V. The maximum voltage is between 4.1V to 4.35V, depending on the battery type. 3x3.7V = 11.1V. The maximum voltage will be 3x4.2V(we accept 4.2V for a standard battery) = 12.6V.



If the batteries are connected in parallel, the voltage is still 3.7V, but the current increases. If the batteries are connected in series, the current is 1000A, but the voltage increases.

In order to have 10V and 1000A, 1 battery must be able to supply 1000A, we connect them in series and we get 11.1V nominal voltage. If we want to use 2 batteries theoretically we can connect 2 cells of 5V and get a nominal of 10V, each battery must supply 1000A, the voltage will increase in series and the current will remain 1000A.

If we connect 2 batteries by 10V each and 500A each, in parallel, the voltage will remain 10V, but the current will increase to 1000A. In real life there are no 5V batteries.

A typical good quality pouch type prismatic 50A-hr cell has about 1m-ohm internal resistance, they are about 3.3 - 2.9 V, so 3 needed in series to get 10V approx

at 0.001 ohm say you can lose 0.2V this is 200A, so 5 sets of the above in parallel required to get 1kA,

At 50A-hr, 200A will last for 0.18 hour = 10.8 minutes - so you are just in ... 15 cells in all