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I am a little confused on how to determine the duration that a battery can power a device if you know the load's current requirements and the Amp-Hr rating of the battery...

Also, how do you determine the current you can draw from a battery if you know the time that a device needs to operate and the Amp-Hr rating of the battery?

Thanks

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  • \$\begingroup\$ There is a very useful (and accurate) battery life calculator here. Check it out. :-) \$\endgroup\$ Apr 30, 2015 at 19:50
  • \$\begingroup\$ OP - All the calculator TimH linked does is reduce the entered capacity by 15%, then divide Amp-hour by Amp to get hours, and convert that to days. So basically the same as Kyran's answer, it takes a calculation that doesn't take into account inefficiencies, temperature, etc, and subtracts a percentage to get a realistic estimate. In a professional environment this might not be close enough, but it works for a close approximation. \$\endgroup\$
    – I. Wolfe
    Apr 30, 2015 at 20:50

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If you look at the units, Amp-hour (capacity of battery) and amps (current draw of device) you can divide the capacity by the load current to get the hours it will run for. This is super naive though, and does not take into account the effect of how large the load current might be (certain battery chemistry do not like high load currents) and temperature effects.

A 2500mAh battery with a 100mAh load current SHOULD last, according to the simple specs given, 25 hours. In reality, depending how the load changes as the battery voltage discharges, the operating temperature, battery manufacturing tolerances etc.. The actual life of the battery may be much less.

If you have the required life of device and a battery capacity (maybe you've already chosen a battery you want to use?) you can work out a reasonable target (average, maximum) load current estimate by simply dividing the battery capacity by the required working life in hours you want.

For example: If you have a 5 Amp-hour battery, and you want at least a 24 hour life out of it, 5/24 = 208 milliamps. To be safe I would aim to consume 25% less than the estimated number.

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