By Tim King · Updated
Battery Life Calculator
Run Time from Capacity and Load
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How the Battery Life Calculator works
This battery life calculator estimates how long a battery will run a device. Enter the battery's capacity and voltage, the average load in milliamps, amps or watts, and an efficiency and reserve figure, and the calculator estimates the run time in hours, days or months. It also shows the usable capacity in mAh and Wh, the average draw and the discharge rate as a C-rate, and lists run times at a quarter, half, double and four times the load so you can see how much a change in consumption buys.
The formula. Run time in hours = capacity in mAh × efficiency × (1 − reserve) ÷ load in mA. For example, a 2,000 mAh cell at 50 mA with 85% efficiency and a 10% reserve gives 2,000 × 0.85 × 0.9 ÷ 50 = 30.6 hours. A second example: a 100 Ah leisure battery feeding a 5 A load at 90% efficiency with a 50% reserve lasts 100 × 0.9 × 0.5 ÷ 5 = 9 hours. The load is the figure that decides how many hours or days a device will last.
Battery run time is capacity divided by current: a 2,000 mAh cell supplying 150 mA lasts about 13 hours in theory. The theory rarely survives contact with a real device, which is why the calculator has efficiency and reserve fields. Use both honestly and the estimate will be close; leave them at 100 and 0 and it will be optimistic by a third or more.
Capacity units. Milliamp-hours and amp-hours describe charge, and only translate to energy when multiplied by voltage. That is why the voltage field matters: 2,000 mAh at 3.7 V is 7.4 Wh, while 2,000 mAh at 12 V is 24 Wh. Power banks quote the mAh of their 3.7 V cells, which is why a "10,000 mAh" bank charges a 3,000 mAh phone only two or three times once the 5 V conversion and losses are accounted for. If the label gives watt-hours, choose Wh and enter the voltage anyway so loads in milliamps can be converted.
Loads in watts. Many devices are rated in watts rather than amps. The calculator divides watts by the battery voltage to get the current the battery actually supplies, so a 5 W device on a 12 V battery draws about 420 mA from it. For a device that uses a different voltage through a converter, enter the device's watts and the battery's voltage, and let the efficiency field cover the converter loss.
Efficiency. Voltage regulators, inverters and boost converters waste some of the energy as heat. Small switching converters manage 85 to 92 percent, linear regulators far less, and inverters from 12 V to mains around 85 to 90 percent at a fair load and much worse at a light one. Peukert's effect also reduces the effective capacity of lead-acid batteries at high discharge rates, which you can fold into this figure.
Reserve. Most batteries should not be run flat. Lithium cells cut out around 3.0 V with some capacity notionally left; lead-acid batteries are damaged by going below 50 percent; and a device will usually shut down before the battery is truly empty. Set the reserve to the share you do not expect to use: 10 percent for a lithium gadget, 50 percent for a lead-acid leisure battery.
Average, not peak. A device that sleeps at 20 µA and wakes for a second at 100 mA every minute averages about 1.7 mA, and it is the average that determines run time. For duty-cycled equipment work out the time-weighted average first and enter that. The table of run times at other loads shows how strongly run time depends on this figure, which is why sleep modes matter far more than battery size for sensors and remotes.
C-rate. The discharge rate line expresses the load as a fraction of capacity per hour: 0.075C means the battery would take just over 13 hours to empty. Most cells are rated for 0.2C to 1C continuous; drawing several C shortens life and reduces the capacity you get. For sizing a battery to a known load rather than the reverse, use the battery sizing calculator.
Battery life calculator by load current
A 2,000 mAh, 3.7 V cell at 85% efficiency with a 10% reserve.
Run time and discharge rate for a 2,000 mAh battery by load
| Load | Run time | C-rate |
|---|---|---|
| 25 mA | 2.6 days | 0.01C |
| 50 mA | 30.6 hours | 0.03C |
| 75 mA | 20.4 hours | 0.04C |
| 100 mA | 15.3 hours | 0.05C |
| 125 mA | 12.2 hours | 0.06C |
| 150 mA | 10.2 hours | 0.08C |
| 175 mA | 8.7 hours | 0.09C |
| 200 mA | 7.7 hours | 0.1C |
| 225 mA | 6.8 hours | 0.11C |
| 250 mA | 6.1 hours | 0.13C |
| 275 mA | 5.6 hours | 0.14C |
| 300 mA | 5.1 hours | 0.15C |
| 325 mA | 4.7 hours | 0.16C |
| 350 mA | 4.4 hours | 0.18C |
| 375 mA | 4.1 hours | 0.19C |
| 400 mA | 3.8 hours | 0.2C |
| 425 mA | 3.6 hours | 0.21C |
| 450 mA | 3.4 hours | 0.23C |
| 475 mA | 3.2 hours | 0.24C |
| 500 mA | 3.1 hours | 0.25C |
| 525 mA | 2.9 hours | 0.26C |
| 550 mA | 2.8 hours | 0.28C |
| 575 mA | 2.7 hours | 0.29C |
| 600 mA | 2.6 hours | 0.3C |
Sources: Ampere hour.
Checked against: 2,000 mAh at 100 mA = 20 hours before losses
FAQ
How long will a 2,000 mAh battery last?
Divide capacity by the load current and apply a realistic efficiency and reserve. At 150 mA, 2,000 mAh gives 13.3 hours in theory and about 10 hours with 85 percent efficiency and a 10 percent reserve, which is what the calculator's defaults show.
How do I convert mAh to Wh?
Multiply by the voltage and divide by 1,000. A 3,000 mAh cell at 3.7 V is 11.1 Wh. The calculator shows the usable capacity in both units, and if your battery label is in Wh you can enter it directly by choosing Wh as the unit.
Why does my power bank charge my phone fewer times than the maths suggests?
Because its mAh figure is at 3.7 V, while the phone charges at 5 V or more, and 10 to 20 percent is lost in conversion each way. A 10,000 mAh bank holds 37 Wh; a 3,000 mAh phone battery needs about 13 Wh delivered, and closer to 16 Wh drawn, so two full charges is typical.
What efficiency should I enter?
Around 85 percent for a device with a switching regulator or a small inverter, 90 percent or better for direct DC loads with only a protection circuit, and 60 to 75 percent for old linear regulators or inverters running well below their rated power.
Should I include a reserve?
Yes. Lithium cells lose life if run fully flat and lead-acid batteries are damaged below about half charge. Ten percent suits lithium gadgets and 50 percent is the right figure for a lead-acid leisure or solar battery.
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