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Battery Life Calculator

Battery runtime from mAh, mA and both voltages: 2,500 mAh at 3.7 V into a 500 mA, 5 V load through an 85% converter runs 3 h 9 min, not 5 h. How it works

Use the Battery Life Calculator

Rated capacity printed on the battery. Choose Wh when the pack states energy: it is used as is, not multiplied by the voltage.

Steady load is one average current. Sleep and active mode averages a device that wakes briefly and sleeps the rest of the time.

The current the device draws at the device voltage below, not at the battery voltage.

Nominal voltage: 3.7 V for one Li-ion or LiPo cell.

The voltage the device runs on, for example 5 V. Equal to the battery voltage when there is no converter.

Energy loss in a boost or buck converter. 85 % is a typical figure; use 100 % when there is no converter.

Share of the rated capacity you can actually draw (cut-off voltage, ageing, cold). 100 % means all of it.

Estimated runtime
3h 9m
Hours total 3.15

Capacity divided by current alone would give 5.00 hours. The voltage conversion, converter efficiency and usable capacity you entered turn that into the figure above: the battery holds 9.25 Wh and the device draws 2.50 W.

Runtime = capacity × battery voltage × usable % × converter efficiency ÷ (current × device voltage). It assumes a steady load and that the rated capacity applies at your discharge rate. Batteries give less at a high rate (the Peukert effect) and in the cold, and lose capacity with age and self-discharge, so treat the result as an upper estimate.

About the Battery Life Calculator

A 2,500 mAh, 3.7 V battery running a 500 mA, 5 V device through an 85% efficient boost converter lasts about 3 h 9 min, not the 5 hours that 2,500 ÷ 500 suggests. The battery life calculator compares energy: capacity × battery voltage × usable share × converter efficiency, divided by the device current × device voltage.

Runtime of a 3.7 V battery into a 5 V device at 85% efficiency

Capacity100 mA250 mA500 mA1,000 mA2,000 mA
1,000 mAh6 h 17 min2 h 31 min1 h 15 min38 min19 min
2,500 mAh15 h 44 min6 h 17 min3 h 9 min1 h 34 min47 min
5,000 mAh1 d 7 h 27 min12 h 35 min6 h 17 min3 h 9 min1 h 34 min
10,000 mAh2 d 14 h 54 min1 d 1 h 10 min12 h 35 min6 h 17 min3 h 9 min
20,000 mAh5 d 5 h 48 min2 d 2 h 19 min1 d 1 h 10 min12 h 35 min6 h 17 min

Load current at 5 V, 100% usable capacity. Three cells are above 1 C, where the calculator warns that the real runtime will be shorter: the 1,000 mAh battery at 1,000 and 2,000 mA, and the 2,500 mAh battery at 2,000 mA.

Frequently asked questions

How long will a 2500 mAh battery last?
At 500 mA, 2,500 ÷ 500 = 5 hours when the device runs straight from the battery with no converter: set the device voltage to the battery voltage and the efficiency to 100%, because the 85% default would take 15% off. Through a 3.7 V to 5 V converter at 85% it is 3 h 9 min, and at 100 mA 15 h 44 min.
Why is the runtime shorter than mAh divided by mA?
The device draws its current at 5 V, not 3.7 V, and the converter loses 15%. 500 mA at 5 V is 2.50 W; the battery supplies 2.5 ÷ 3.7 ÷ 0.85 = 0.795 A, and 2.5 Ah ÷ 0.795 A = 3.15 hours.
How long will a 12V 100Ah battery last?
Set both voltages to 12 V, efficiency to 100% and usable capacity to the share you will draw. At 5 A with 50% usable, 100 Ah lasts 10 h 0 min (600 Wh available). A 60 W load at 120 V through a 90% inverter (0.5 A, device voltage 120 V) gets 9 h 0 min from the same battery.
What does the 1 C warning mean?
C-rate is the battery current divided by its capacity in amp-hours, and 1 C empties a battery in about an hour. A 1,000 mAh cell feeding a 1,000 mA, 5 V load draws 1.59 A, 1.59 C, and many batteries give less than their rated capacity at that rate.
What should I enter as usable capacity?
The share of the rated capacity the device will actually use: less than 100% when it cuts off above empty, or the battery is old or cold. At 80% the default case drops from 3 h 9 min to 2 h 31 min.
How long will a battery last on a device that sleeps most of the time?
Choose Sleep and active cycle and enter both currents and times. The average is (awake current × awake time + sleep current × sleep time) ÷ (awake time + sleep time): 80 mA for 1 s and 20 µA for 59 s average 1.353 mA. The default 2,500 mAh, 3.7 V battery into a 5 V device at 85% then runs 1,162.23 h, about 48 days 10 hours.

Why mAh ÷ mA overstates the runtime

A power bank's mAh rating is for its 3.7 V cells, but the USB port delivers 5 V, so the same milliamps carry more energy on the way out, and the boost converter loses some on top. A 10,000 mAh bank feeding a 1,000 mA load at 5 V through an 85% converter runs about 6 h 17 min, not the 10 hours the label suggests.

When the device runs at the battery's own voltage, such as a 12 V pump on a 12 V battery, enter the same voltage twice and set the efficiency to 100%. Leaving the 85% default in place would cut the answer by 15% for a loss that is not there.

How the runtime is calculated

runtime (h) = capacity (Ah) × battery V × usable share × efficiency ÷ (current (A) × device V)

  • usable share: the part of the rated capacity you let the battery give up, 100% by default

  • efficiency: the loss in a boost or buck converter, 85% by default, 100% with no converter

Example: 2,500 mAh at 3.7 V is 9.25 Wh, and a 500 mA load at 5 V is 2.50 W. 9.25 × 0.85 ÷ 2.5 = 3.15 h, about 3 h 9 min. Capacity ÷ current alone would say 5 h.

For a device that sleeps and wakes, the sleep and active cycle uses average current = (Ia × ta + Is × ts) ÷ (ta + ts), awake (a) and asleep (s). 80 mA for 1 s and 20 µA for 59 s average 1.353 mA: 1,162.23 h, about 48 days, with the other defaults. A capacity in Wh is used as the energy directly.

Above 1 C (battery current ÷ capacity in Ah) the page warns that many batteries cannot give their rated capacity. High discharge rates, cold, self-discharge and ageing are not modelled, so the result is an upper estimate.

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