I have a 9.8kW Component that runs at 51.8 Volts. That means I need a battery that has about 9.8kWh of energy to run the component for an hour

For the sake of simplicity, say I have some batteries that just happened to match that voltage (51.8V) simply laying around. How many mAh would that battery need to be to run this for an hour? My instinct is to simply say 98 Ah, but this doesn't seem right..


1 Answer 1


Power (Watts, milliWatts, etc.) is given by:

$$ P = IV \\ $$

where $I$ is current in (milli)Amps, $V$ is voltage in Volts, and $P$ is power in (milli)Watts, respectively. Energy (Watt-hours, milliWatt-hours, etc.) is given by:

$$ E = P\Delta t \\ $$

(assuming constant power output), where $E$ is energy in (milli)Watt-hours and $t$ is time in hours. Finally, electric charge is given by:

$$ Q = I \Delta t \\ $$

where $Q$ is current charge in (milli)Amp-hours and $I$ and $t$ are as defined before.

So, if you have an application where you want 9.8 kW to run for an hour, then ideally you would want an available energy reserve of:

$$ E = P\Delta t \\ E = (9.8\mbox{kW})(1\mbox{hr}) \\ \boxed{E = 9.8 \mbox{kWh}} \\ $$

As you've hinted at, batteries aren't rated in Watt-hours; they're generally rated in Amp-hours. There's an easy conversion, though - substitute the equation for power!

So, instead of:

$$ E = P\Delta t \\ $$

You would use:

$$ E = (IV) \Delta t \\ $$

which you can rearrange for the battery's current charge:

$$ E = V(I \Delta t) \\ E = VQ \\ \boxed{Q = \frac{E}{V}} \\ $$

Now you know the energy reserve you'd like to have: 9.8 kWh, and you know the voltage you're going to operate at: 51.8 V, so plug those in and get the answer:

$$ Q = \frac{9.8 \mbox{kWh}}{51.8\mbox{V}} \\ \boxed{Q = 0.189 \mbox{kAh}} \\ $$

Note the units because they're important. If you use energy as kiloWatt-hours, then the output is kiloAmp-hours. Multiply by 1,000 to get to Amp-hours:

$$ \boxed{Q = 189 \mbox{Ah}} \\ $$

I'll make an important note now about an aspect of this problem that you didn't ask. I mentioned available energy reserves because you generally can't get all of the charge from a battery without physically damaging the battery. Allowable depths of discharge vary by battery style and chemistry, so a deep-cycle lead-acid battery might allow you to get down to 10-20% of full charge without damage, but a lithium ion battery might only let you get to 40% of full charge without damage.

Generally speaking, a quality battery manufacturer should provide a plot of expected battery life in charge/discharge cycles versus depth of discharge. It's up to you, the engineer, to choose what tradeoffs are allowable.

For example, you might choose to use a deep discharge (down to 5-10% of full charge) and knowingly damage the battery in exchange for longer operation between charges and/or a lower pack weight - this might be the case for a robotics competition.

Or, you could use a small discharge (only down to 80% of full charge) because it is acceptable to have a large battery bank - this might be the case for a solar power installation.

Whatever the case, you should recognize that your available charge is some percentage of the rated battery charge:

$$ Q_{\mbox{available}} = \eta Q_{\mbox{rated}} \\ \boxed{Q_{\mbox{rated}} = \frac{Q_{\mbox{available}}}{\eta}} \\ $$

So, if you know that you want to discharge down to 40% of full charge, then you are saying that you want to use 60% of the full charge of the battery. If you know how much charge you would like to have, then you can calculate that you need a battery rated to:

$$ Q_{\mbox{rated}} = \frac{ Q_{\mbox{available}} } {0.6} \\ \boxed{Q_{\mbox{rated}} = (1.67) Q_{\mbox{available}}} \\ $$

You need a battery rated to 67% more (or higher) than what you calculated you need to ensure you only ever discharge to 40%.

  • $\begingroup$ Excellent. If you include some equations to calculate Amp Hours (left) at a certain time t while running, which would be useful for 'fuel' gages, or (more importantly) a short explanation of how coulomb counting works with this, I think it deserves a bounty, which I'll happily give away. Wish I could upvote more than once. $\endgroup$
    – tuskiomi
    Jun 30, 2017 at 19:17
  • 1
    $\begingroup$ @tuskiomi - No bounty required; just accepting is enough :) The only equations I can easily give for Ah left is Coulomb counting. Basically just continuously sample the current leaving the pack, usually with a shunt resistor, then do numeric integration like: currentSpent += measuredCurrent*sampleTime. The charge remaining is then your starting charged less the charge you spent. You can also look up Peukert's law, which provides a more precise form of "dumb" Coulomb counting by accounting for the fact that batteries get less efficient as current (heat, typically) goes up. $\endgroup$
    – Chuck
    Jul 1, 2017 at 0:18
  • $\begingroup$ @Chuck Nice job with the equations! A Mechanical Engineer that does math? ;) (I am an MSEE) $\endgroup$ Jul 1, 2017 at 2:57
  • 1
    $\begingroup$ @markshancock - Yup! I found my interests straddling the ME/EE departments and was lucky enough to be in a program that encouraged multidisciplinary studies. BSME, MSME, but my undergrad capstone project was a battery management system and my master's thesis was radar design. I'm into controls, signal processing, robotics, and automation and could argue those were all under the umbrella of an ME degree so I was allowed to continue :) $\endgroup$
    – Chuck
    Jul 1, 2017 at 13:12
  • $\begingroup$ Important: The ´usable capacity´ of a battery depends on the the ´discharge current´. The question is about a fixed battery lifetime of 1 hour. This simplifies the situation, because the ´rated capacity´ of batterys is for a discharge within 1 hour (=1C). In general: When discharging with higher currents (>1C), the efficiency drops. When discharging slower (<1C), the efficiency is better - until some point where self discharge becomes a significant factor. Exact numbers depend on the chemestry of the battery. Keep in mind: Drawing higher current might deplete your battery much faster! $\endgroup$
    – SDwarfs
    Jan 9, 2018 at 13:23

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