A 10,000 mAh power bank feeding a 500 mA load
- Input
- Capacity = 10,000 mAh, Load = 500 mA, Efficiency = 90%
- Working
- Usable capacity = 10,000 × 0.90 = 9,000 mAh. Runtime = 9,000 ÷ 500 = 18 hours.
- Result
- Runtime = 18 hours (0.75 days)
Battery Life Calculator
Estimate how long a battery lasts under a load — capacity in milliamp-hours and load in milliamps, adjusted for efficiency, gives the runtime in hours and days. Runs entirely in your browser — your numbers never leave this device.
Estimated runtime
18 hours
About 0.75 days of continuous load
- Runtime
- 18 h
- Runtime (days)
- 0.75
- Usable capacity
- 9000 mAh
Runtime assumes a constant load and a steady efficiency. Real batteries deliver less at high discharge rates and as they age, so treat this as an upper-bound estimate.
How it works
Runtime (hours) = capacity (mAh) × efficiency (%) ÷ 100 ÷ load (mA)
A battery's milliamp-hour rating is a budget of current × time: a 10,000 mAh pack can in principle deliver 500 mA for 20 hours or 100 mA for 100. Runtime is therefore capacity divided by load — with one honest adjustment. No real setup spends the whole budget: voltage converters waste a share as heat, battery protection cuts off the last portion, and capacity itself is rated under gentler conditions than most loads apply. The efficiency field captures that haircut as a percentage — 90% is a fair default for a decent power bank, 80–85% for a converter-fed project — and the result is usable capacity divided by load.
Read the figure as an upper-bound estimate for a constant load. Loads that pulse, batteries discharged at a high rate relative to their size, cold temperatures and ageing all shorten real runtime, sometimes substantially; a device that sleeps most of the time will run far longer than its active current alone suggests, and its average current — active current weighted by the duty cycle — is the load figure to enter. For a runtime in days, divide the hours by 24; the calculator shows both.
Worked examples
A 2,000 mAh cell at 100 mA with losses
- Input
- Capacity = 2,000 mAh, Load = 100 mA, Efficiency = 85%
- Working
- Usable capacity = 2,000 × 0.85 = 1,700 mAh. Runtime = 1,700 ÷ 100 = 17 hours.
- Result
- Runtime = 17 hours
A 3,000 mAh battery at 200 mA, 80% efficiency
- Input
- Capacity = 3,000 mAh, Load = 200 mA, Efficiency = 80%
- Working
- Usable capacity = 3,000 × 0.80 = 2,400 mAh. Runtime = 2,400 ÷ 200 = 12 hours.
- Result
- Runtime = 12 hours (0.5 days)
Frequently asked questions
What efficiency should I enter?
100% only for the ideal arithmetic. A USB power bank's boost converter typically delivers 85–90% of the cell's rated capacity at the port; a project drawing through a linear regulator can be far lower, since a linear regulator wastes the voltage difference as heat. When in doubt, 85% is a defensible planning figure and the difference is visible immediately in the result.
My load is in watts, not milliamps. How do I use this?
Convert at the battery's voltage: load in mA = watts ÷ volts × 1000. A 5 W load on a 5 V output is 1,000 mA. But be careful which side of a converter you measure on — a power bank rated 10,000 mAh at its 3.7 V cell holds about 37 Wh, and at a 5 V output that is nearer 7,400 mAh before losses, which is exactly what the efficiency field is for.
Why does my real runtime come out shorter than the estimate?
Usually one of three reasons: the load is not constant and its peaks or average were understated, the discharge rate is high relative to the battery (capacity shrinks at high drain, an effect called Peukert's law in lead-acid cells), or the battery is cold or aged. The calculator models the ideal constant-load case and the efficiency field is where those real-world losses belong.
How do I estimate runtime for a device that sleeps?
Use its average current: multiply the active current by the fraction of time it is awake and add the sleep current for the rest. A sensor drawing 50 mA for 1% of the time and 0.1 mA otherwise averages about 0.6 mA — enter 0.6 mA as the load and a small battery suddenly lasts for months, which is how such devices are designed.
Does a bigger load drain the battery proportionally faster?
In the ideal model, yes — double the milliamps, half the hours. In reality high drains are slightly worse than proportional, because internal resistance wastes more energy as heat at high current. That non-linearity is another reason to leave the efficiency below 100% for heavy loads.
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