I am working on designing a lithium-ion battery pack that will power a solar car that runs on a brushless DC motor. There are several pack configurations I am trying to choose between, and to do so, I want to find the trade-off between the configuration I choose and how long that battery pack will last at nominal conditions before it drains completely (or to cutoff voltage).

Here are the motor specs: Nominal Power: 2000W Max Power: 5000W Nominal input voltage: 96V

Here are the battery specs: Std. Charge/Discharge (Nominal Capacity): 3500 mAh Nominal voltage: 3.635V Max voltage: 4.2V Standard Discharge: 0.2C (680mA) Cutoff Voltage: 2.5V Max discharge current: 10A

The pack configurations I am choosing between are 27s15p, 29s14p, and 34s12p. So far I have been unsuccessful in deriving an equation that will determine the battery life (run-time) and would greatly appreciate the advice!

  • \$\begingroup\$ Compute battery Watt hours (Amp hour capacity * nominal pack voltage). Then divide the result by the average power consumption of the car, in Watts. The result is the battery life, in hours. So it is: watt-hours / watts = hours. The motor max power does not matter. You need to know the actual average power consumption of everything powered by the battery pack. \$\endgroup\$
    – user57037
    Commented Jan 24, 2022 at 2:16
  • \$\begingroup\$ Are there any other components between the motor and the battery pack? You will need to measure how much power the motor actually uses because it will depend on the actual load - or we can just guess 2000W. The more accurate of an estimation you need, the more information needs to be taken into account. Least accurate: calculate the energy content of a battery, multiply by the number of batteries, divide by the power of the motor, done. \$\endgroup\$ Commented Jan 24, 2022 at 2:25

1 Answer 1


Here is the very roughest of calculations:

One battery contains 3.635V x 3500mAh = 12.7Wh

If the motor uses 2000W constantly it can be powered for 0.006 hours = 23 seconds. Of course you would not power the motor with just one battery - this is just for the calculation.

Your packs have 405, 406 and 408 batteries respectively, basically the same number. You get about 2.6 hours of power from any of these.

This is extremely approximate. A more accurate calculation would include the actual load on the motor (which affects its power usage) and the actual discharge curve of the battery.

Fundamentally, the same number of the same batteries will store the same amount of energy no matter how you arrange them.

The main factor - maybe the only factor to consider in choosing the arrangement is the voltage. My first thought is to choose 27s15p because it matches the nominal voltages quite closely, but in reality you get to use your design creativity here. Unlike a computer chip, you have a wide latitude to run a motor above or below its nominal voltage, which will increase or decrease the power respectively. That's good since your battery will also go above and below its nominal voltage depending on the charge level. Without seeing the motor's datasheet I won't make any absolute statements about how much voltage is okay.


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