I am part of a team and we build a robot. I know how many watts all the subsystems of the robot will need and I also know how much time we want to operate it. Our system will need 323.344 Watt and we want to work for 30 min.

I asked a professor to help me and he gave me this formula :

$$C_{bat} = P_{bat}* 1h / ΔSoc$$

$ΔSoc$ told me that is the maximum acceptable Voltage dipping and that I can find this information in the Datasheets.

We decided to use Li-Po batteries but I cannot find datasheet for any Li-Po.

How do you calculate the needed capacity for your project? How can I calculate it ? Do you know this formula?


1 Answer 1


30 minutes is half an hour. 323.344 Watts is approximately 324 Watts.

If you want 324 Watts for half an hour, then you want a (324 * 0.5) Watt-hour battery. This assumes, though, that your battery will go from 100% charge to 0% charge after exactly 30 minutes.

Discharging a battery down to a state of charge of 0% can physically damage batteries, preventing them from ever charging again, and the battery voltage will usually fall quite dramatically as the state of charge approaches 0%, so you might wind up browning out your electronics before you actually hit 0%, leaving the last few percent of state of charge uncaptured and thus falling short of your desired 30 minutes of run time.

What might be a better approach would be to assume that you want your batteries sized such that you can supply the full (324*0.5) Watt-hours of capacity when the batteries are discharged from, for example, 80% of full charge down to 20% of full charge. That is, you want:

$$ (0.80 - 0.20) * \mbox{Capacity} = 324*0.5 \mbox{Watt-hours}\\ $$

The difference between your starting and ending state of charge is called, not surprisingly, the change or delta ($\Delta$) state of charge, and you can solve for capacity by:

$$ \Delta \mbox{SOC} * \mbox{Capacity} = (324 * 0.5) \mbox{Watt-hours}\\ $$

$$ \mbox{Capacity} = \frac{324*0.5}{\Delta\mbox{SOC}} \mbox{Watt-hours} \\ $$

In the example I gave, where you are designing your system to supply your desired capacity between 80 and 20 percent of full state of charge, your $\Delta\mbox{SOC} = 0.8-0.2 \boxed{= 0.6} $, so you wind up with a target battery capacity of:

$$ \mbox{Capacity} = \frac{324*0.5}{0.6} \mbox{Watt-hours} \\ \boxed{\mbox{Capacity} = 270 \mbox{Watt-hours}} \\ $$

Remember that $P = IV$, so 270 Watt-hours could be rewritten 270 volt-amp-hours. You can divide your target capacity by the nominal battery voltage to get the actual battery size in amp-hours.

  • $\begingroup$ Thank you very much @Chuck. So , if I use a Li-Po 11.1V 2700mAh and we have 24.32Ah - > 24324 mAh I need like 24342/2700 about 9 of these batteries. Right ? $\endgroup$ Commented Oct 28, 2019 at 18:58
  • 1
    $\begingroup$ @MichaelDuth - 270 Watt-hours is 270 volt-amp-hours. If your pack voltage is nominally 11.1V, then you need (270/11.1) Amp-hours of capacity, which is 24.32 Ah. Since your battery is 2.7 Ah, you need 24.32/2.7 = 9.009 batteries. Technically you should be using 10, since the calculation is >9.000, but you've got 40% state of charge held back as reserve in the calculations, so it's up to you. Sorry to re-work your math, but I just wanted to make sure everyone else reading here could follow along easily. $\endgroup$
    – Chuck
    Commented Oct 28, 2019 at 19:03
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    $\begingroup$ Ok, thank you ! $\endgroup$ Commented Oct 28, 2019 at 19:14

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