Tech calculator
Battery Life Calculator
Estimate how long a battery can supply a steady average load.
Ideal estimate before conversion losses, aging, temperature, and reserve capacity.
Estimated runtime
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What the calculator computes
Enter the battery's rated capacity in milliamp-hours and the device's average current draw in milliamps; the calculator divides one by the other and also converts the result to minutes. A milliamp-hour is a unit of electric charge — a battery rated at 1 mAh can supply 1 mA for one hour, or 2 mA for half an hour, or any other combination that multiplies out to the same number. Dividing mAh by mA cancels the 'h' unit's denominator and leaves plain hours, which is why the formula works.
This is an ideal figure: it assumes the current draw stays perfectly constant for the entire discharge and that every milliamp-hour on the label is usable. Neither assumption holds exactly in real devices, so treat the answer as a ceiling rather than a promise.
A worked example
A 3,000 mAh battery powering a device that draws a steady 500 mA: 3000 ÷ 500 = 6.00 hours, or 360 minutes. A smaller example — a 2,600 mAh light on its high setting drawing 300 mA — runs for 2600 ÷ 300 ≈ 8.67 hours, which the calculator would also express as 8 hours 40 minutes.
Why mAh alone doesn't tell the full story
A milliamp-hour figure only means something once you know the voltage it was measured at, because what's actually conserved when voltage changes is energy, not charge. A 10,000 mAh power bank built around a 3.7-volt lithium cell stores 10,000 × 3.7 = 37,000 mWh (37 Wh) of energy. Boosted up to a 5-volt USB output at a perfect 100% conversion efficiency, that same energy only supports 37,000 ÷ 5 = 7,400 mAh at 5 volts — already 26% less than the number printed on the box, purely from the voltage change, with no losses yet involved. Real boost converters run at roughly 85% efficiency, which brings the usable 5-volt capacity down further to about 7,400 × 0.85 ≈ 6,290 mAh.
Because of this, comparing two batteries fairly means comparing watt-hours (mAh × volts ÷ 1000), not raw mAh: a 6,000 mAh pack built from cells wired for 7.4 volts actually stores 6,000 × 7.4 ÷ 1000 = 44.4 Wh, more energy than a 10,000 mAh pack at 3.7 volts (37 Wh), despite the smaller mAh number.
Other reasons real runtime falls short of the ideal number
Devices don't let a battery discharge all the way to 0 volts — firmware enforces a cutoff voltage (commonly around 3.0V for a lithium cell with a 3.7V nominal rating and roughly 4.2V full charge) to protect the chemistry, which quietly reserves some of the rated capacity as unusable headroom.
Higher current draw pulls out disproportionately more capacity than a simple linear model predicts, an effect formalized as Peukert's law — most pronounced in lead-acid batteries, present to a smaller degree in lithium-ion, and caused by internal resistance wasting more energy as heat the harder a battery is pushed.
Capacity also fades with age: a commonly cited industry figure is that lithium-ion cells retain around 80% of their original capacity after roughly 500 full charge cycles, so a two-year-old device rarely reaches the runtime implied by its original spec sheet. Cold weather compounds this — lithium-ion capacity commonly drops by around 20% near freezing compared with room temperature, so a winter runtime estimate built on the rated mAh will run optimistic.
Situations this calculator is built for
- Flashlight or headlamp planning — a 2,600 mAh light drawing 300 mA on high runs about 8 hours 40 minutes by the ideal formula
- Comparing a phone's active-use battery life against standby — a 4,000 mAh phone battery lasts about 4000 ÷ 800 = 5 hours of heavy screen-on use at 800 mA, versus roughly 4000 ÷ 50 = 80 hours (3.3 days) at a 50 mA standby draw
- Sizing a battery pack for a microcontroller or sensor project — a sensor drawing a steady 120 mA off a 5,000 mAh pack gets about 5000 ÷ 120 ≈ 41.7 hours of continuous logging
- Sanity-checking whether a bigger-mAh power bank is worth its extra weight, once the mAh-to-Wh voltage conversion above is factored in rather than comparing the printed numbers directly
- Converting a laptop's watt-hour rating to an equivalent mAh figure at its pack voltage — a 60 Wh battery at 11.1V nominal is about 60,000 ÷ 11.1 ≈ 5,405 mAh
- Estimating drone or RC flight time from a battery's mAh rating and the motors' typical current draw, while keeping in mind peak throttle draws well above the average used in the estimate
Frequently asked questions
Why does my device last less than the number this calculator shows?
The formula assumes one constant current draw for the entire discharge and full access to the rated capacity. Real devices see rising internal resistance as the battery empties, a firmware cutoff voltage that reserves some capacity, temperature swings, and aging — all of which shorten real runtime below this ideal figure.
Does a bigger mAh number always mean more battery life?
Not by itself — mAh only measures charge at the battery's own voltage. Comparing packs fairly means converting to watt-hours (mAh × volts ÷ 1000): a 6,000 mAh pack at 7.4V stores 44.4 Wh, more energy than a 10,000 mAh pack at 3.7V, which stores only 37 Wh.
What is Peukert's law?
It describes how drawing current faster pulls out disproportionately more capacity than a simple linear model predicts, because internal resistance wastes more energy as heat at higher currents. It's most pronounced in lead-acid batteries and present to a lesser degree in lithium-ion.
Why does the result show minutes as well as hours?
Because a decimal like 8.67 hours is harder to plan around than 8 hours 40 minutes. The calculator multiplies the fractional part of the hour figure by 60 to get a whole-number minute value.
How much capacity does a lithium-ion battery lose over time?
A commonly cited figure is around 20% capacity loss after roughly 500 full charge cycles, which is why an older phone or laptop rarely reaches the runtime its original specification promised, even with identical usage habits.
Can I use this for a lead-acid car or UPS battery?
The division still works, but lead-acid chemistry is more sensitive to discharge rate than lithium-ion under Peukert's law, so treat the result as a looser upper bound for lead-acid batteries than for a lithium cell discharged at a comparable rate.