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Electronics · Power

Battery Life Calculator

Estimate how long a battery lasts for a given load, with a realistic derating factor.

mAh
Charge stored in the battery.
mA
Average current the device draws.
Efficiency: 0 < x ≤ 1 (default 0.8).
Try a scenario
Estimated runtime
8hours

About 8h 0m of real use — ideal (no derating) would be 10 h (10h 0m). This is a rough estimate.

Runtime vs load current (lower load lasts longer)
Estimated runtime falling as load current rises16 h4 h100 mA400 mA

Estimated battery life ≈ capacity ÷ load current × derating. For a 2000 mAh battery driving a 200 mA load, the ideal runtime is 2000 ÷ 200 = 10 h; applying a 0.8 derating factor gives a realistic 8 h (8h 0m). It is only an estimate — real runtime is usually a bit lower.

How the estimate works

Divide the battery’s charge capacity (in mAh) by the average current your device draws (in mA) and you get the ideal runtime in hours. That figure assumes a perfect battery that delivers every last milliamp-hour at full voltage — which never happens. A derating factor between 0 and 1 shaves the ideal number down to something closer to reality, accounting for converter losses, self-discharge, and the fact that a device cuts off before the cell is truly empty. A common starting point is 0.8, meaning you expect to use about 80% of the nameplate capacity. Because the derating is a judgement call, treat the output as a ballpark, not a guarantee.

runtime (h) = capacity (mAh) ÷ load current (mA) × derating

capacity = battery charge (mAh), load current = average draw (mA), derating = efficiency factor between 0 and 1

Worked example

Estimate how long a 2000 mAh battery lasts driving a steady 200 mA load, using a 0.8 derating factor:

  1. 1
    Match your units. Put capacity in mAh and load current in mA so they cancel to hours. A 2 Ah battery is 2000 mAh.
  2. 2
    Divide capacity by current. 2000 mAh ÷ 200 mA = 10 h. This is the ideal, best-case runtime.
  3. 3
    Apply a derating factor. Multiply by your efficiency estimate: 10 h × 0.8 = 8 h of realistic runtime.
  4. 4
    Convert to a friendly duration. 8 h reads as 8h 0m; a 30 h estimate would show as 1d 6h 0m.

Ideal runtime for a 2000 mAh battery

Capacity ÷ current before any derating. Multiply by your derating factor (e.g. × 0.8) for a realistic figure.

Load currentIdeal runtime (2000 ÷ mA)At 0.8 derating
20 mA100 h80 h
50 mA40 h32 h
100 mA20 h16 h
200 mA10 h8 h
500 mA4 h3.2 h

Why real battery life is shorter

The nameplate capacity is a lab number. Manufacturers rate cells at a gentle discharge current and a friendly temperature. Draw more current and the usable capacity falls — the Peukert effect means a high, sustained load empties a battery faster than the simple ratio predicts. Cold temperatures raise internal resistance and cut capacity further, while heat accelerates self-discharge.

Devices also quit early. Most electronics stop working at a cutoff voltage well above the battery’s absolute floor, leaving usable charge stranded. Add DC-DC converter losses and a load that spikes rather than staying constant, and you can see why a derating factor is essential. For anything critical, measure the real current draw and treat this calculator as a rough first estimate.

What is the battery life formula?
Runtime in hours ≈ battery capacity (mAh) ÷ load current (mA) × derating factor. Keep capacity and current in matching milli-units so they cancel cleanly to hours.
Why is real battery life shorter than the calculation?
The Peukert effect reduces usable capacity at higher currents, cold cuts capacity, self-discharge and converter losses waste charge, and devices stop at a cutoff voltage before the cell is empty. That is exactly what the derating factor accounts for.
What derating factor should I use?
Start around 0.8 for a typical device at a moderate load. Use 0.85–0.9 for a light, steady low-power load, and drop toward 0.5–0.7 for high-current, spiky, or cold-weather use where the Peukert effect bites harder.
How do I convert amp-hours to milliamp-hours?
Multiply amp-hours by 1000. A 2.5 Ah cell is 2500 mAh, and an 18650 rated 3.4 Ah is 3400 mAh. Then divide by the load current in mA to get hours.
Does this work for milliwatt-hours or watt-hours?
Not directly — this tool uses charge (mAh), not energy. To use energy ratings, divide watt-hours by the device’s power in watts to get hours, since mAh already assumes a fixed voltage.
Does a higher load always mean proportionally shorter life?
Roughly, but not exactly. Doubling the current more than halves runtime once the load is high, because the Peukert effect shrinks the usable capacity. That is why a lower derating factor suits heavy loads.