Enter your battery's capacity and your circuit's current draw to estimate how long it'll run, with a real-world efficiency factor built in.
Estimating battery runtime seems simple on paper, but actual results are affected by battery chemistry, discharge rate, temperature, age, and how "empty" is defined for a given application. This calculator provides a practical estimate that accounts for efficiency losses, while this section explains the underlying factors in more depth.
Battery capacity is typically rated in milliamp-hours (mAh) or amp-hours (Ah), where 1 Ah = 1,000 mAh. This rating represents how much current the battery can theoretically supply for one hour before being depleted — for example, a 2,000 mAh battery could theoretically supply 2,000 mA (2A) for one hour, or 1,000 mA for two hours, under ideal conditions.
Runtime (hours) ≈ Capacity (Ah) / Load Current (A)
This is the simple, ideal calculation assuming constant current draw and 100% usable capacity. Real-world runtime is almost always shorter due to the efficiency factors discussed below.
Several factors reduce actual runtime below the theoretical ideal: internal resistance causes some energy to be lost as heat rather than delivered to the load; battery voltage sags under load and continues dropping as the battery discharges, which can trigger a device's low-voltage cutoff before the battery is technically "empty"; and not all of a battery's rated capacity is usable — many devices stop operating before 100% of capacity is drawn, to protect the battery or maintain stable voltage.
A 2,000 mAh (2 Ah) battery supplying a constant 500 mA (0.5A) load gives an ideal estimate of 2 / 0.5 = 4 hours. Applying a realistic 80% efficiency factor (accounting for voltage sag and incomplete capacity usage) gives an adjusted estimate of approximately 3.2 hours — a meaningfully different number for planning purposes.
Lithium-ion (Li-ion) batteries maintain a relatively flat voltage output through most of their discharge cycle, then drop off more sharply near depletion — this makes runtime estimates more predictable until near the end. They also handle higher discharge currents more efficiently than alkaline cells, with less capacity loss at high current draw. Alkaline batteries have a voltage that declines more gradually and continuously throughout discharge, meaning a device may lose adequate voltage well before the battery's rated capacity is fully used — and alkaline cells lose a larger fraction of their usable capacity at high discharge currents compared to their rating (which is typically measured at a low, standardized discharge rate).
For lead-acid and some other battery chemistries, actual usable capacity decreases disproportionately as discharge current increases — a phenomenon described by Peukert's Law. A battery rated for 100Ah at a slow 5-hour discharge rate might only deliver the equivalent of 70-80Ah if discharged in 1 hour at a much higher current, because higher currents generate more internal losses. This effect is most significant for lead-acid batteries and less pronounced (though not absent) in modern Li-ion cells, which is one reason Li-ion has become preferred for high-drain applications.
Beyond chemistry-specific effects, several other real-world factors affect runtime: temperature (cold significantly reduces usable capacity in most chemistries, especially alkaline and lead-acid), battery age and cycle count (capacity degrades over repeated charge/discharge cycles), varying load current (many real devices don't draw perfectly constant current, e.g., a phone's screen-on vs. idle power draw), and manufacturing tolerances (actual capacity often varies somewhat from the rated nameplate value).
Does double capacity double runtime? Approximately yes, under the same load and conditions — doubling mAh roughly doubles runtime for a constant current draw, though efficiency and chemistry-specific effects mean it's rarely an exact doubling.
Is mAh the same as energy? No — mAh measures charge capacity, not energy. Energy (in watt-hours, Wh) also depends on voltage: Wh = Ah × V. Comparing batteries of different voltages by mAh alone can be misleading; watt-hours is the more accurate comparison.
Why does my device's battery drain faster at high current draw? Higher discharge currents cause greater internal resistance losses and, in chemistries affected by the Peukert effect, reduce the effective usable capacity — this is normal battery behavior, not necessarily a fault.
Should I use 100% efficiency for my calculation? No — using a realistic efficiency estimate (typically 75-90% depending on chemistry and load) gives a more accurate real-world runtime than the ideal theoretical calculation.
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