How this instrument works
A battery's amp-hour (Ah) rating tells you how much current it can supply for how long, but sizing one for a real application takes more than dividing power by voltage. You also have to decide how deep you're willing to discharge it — running a battery all the way to 0% shortens its life and, for lead-acid especially, is rarely achievable in practice — so a target 'remaining charge' percentage sets a safety margin the battery must be big enough to leave in reserve.
Lithium-ion and lead-acid batteries are sized differently because they discharge differently. Li-ion delivers close to its full rated capacity regardless of how fast you draw it down, so its required-capacity formula is a simple scale-up: B = 100 x IL x t / (100 - Q), where IL is the load current (power divided by voltage), t is the runtime in hours, and Q is the percentage charge you want left over. Lead-acid batteries lose usable capacity faster the harder and longer you draw them (a real, well-documented effect related to Peukert's law), so their formula adds a correction term in the denominator: B = 100 x IL x t / [(100 - Q)(0.02t + 0.6)], which inflates the required capacity for longer discharge times to compensate.
That correction term is why the same load and runtime need noticeably more lead-acid capacity than lithium-ion capacity to hit the same depth-of-discharge target — it isn't a rounding difference, it's the chemistry itself delivering less usable energy the slower and longer you draw it down.
- Enter Load / application power draw in watts and Battery voltage in volts — together these set the load current.
- Enter Required runtime in hours, the length of time the battery must supply that load.
- Set Charge remaining allowed at end of runtime (%) — a depth-of-discharge safety margin, not the amount you plan to use.
- Toggle Battery is lithium-ion on for Li-ion chemistry, or leave it off for the lead-acid formula's extra correction term.
- Read Required battery capacity in amp-hours (Ah) — the minimum-rated capacity that meets your runtime and reserve target.
Worked example — a 100W, 12V lead-acid setup for 5 hours
A 100W load running at 12V draws IL = 100/12 = 8.33A. You need 5 hours of runtime, want to leave 20% charge remaining (an 80% depth-of-discharge limit, typical lead-acid practice), and you're using a lead-acid battery. Applying B = 100 x IL x t / [(100-Q)(0.02t+0.6)] = 100 x 8.33 x 5 / [(80)(0.02x5+0.6)] = 4166.67 / (80 x 0.7) = 4166.67 / 56 = 74.40 Ah.
Switch that same 5-hour, 20%-reserve scenario to lithium-ion instead and the simpler formula B = 100 x IL x t / (100 - Q) applies, with no time-dependent correction term — Li-ion's required capacity scales linearly with runtime rather than growing faster than linearly the way lead-acid's does.
Questions
Why does lead-acid need a bigger battery than lithium-ion for the same load and runtime?
Lead-acid chemistry delivers progressively less of its rated capacity the faster or longer you draw current from it, an effect the industry commonly ties to Peukert's law. The (0.02t + 0.6) correction term in the lead-acid formula inflates the required amp-hour rating to compensate for that loss, while lithium-ion's simpler formula assumes close to its full rated capacity is usable regardless of discharge rate — so a lead-acid battery sized for the same job will always come out larger than a lithium-ion one under this model.
What percentage should I use for 'charge remaining allowed'?
It's your depth-of-discharge safety margin, not how much you plan to use — a common lead-acid practice is to stop discharging at around 50% remaining (a 50% depth of discharge) to protect cycle life, while many lithium-ion packs can safely go down to 10-20% remaining or lower without meaningful damage. Check your specific battery's datasheet for its recommended minimum state of charge, and use that as your 'remaining' figure here rather than assuming a one-size-fits-all number.
How do I find my load's power draw in watts?
Check the device's nameplate or datasheet for a wattage rating, or calculate it from voltage x current if you know the operating current draw instead. If several devices share one battery, add up each device's watts to get the total load power to enter here — this instrument sizes for the combined draw, not any single device in isolation.
Does this account for battery aging or temperature effects?
No — this instrument sizes for a fresh battery at its rated capacity and doesn't adjust for the capacity loss that comes with age, cold temperatures, or repeated deep-discharge cycles, all of which reduce a real battery's usable capacity over time. For a safety margin against those real-world effects, consider sizing to a somewhat larger capacity than the bare result here, especially for lead-acid batteries used in cold environments.
Can I use this for a battery bank made of multiple cells or batteries wired together?
Yes — enter the combined load and target runtime as usual, and the required capacity result is the total Ah your bank needs to provide. If you're wiring batteries in parallel to reach that total, divide the result by each battery's individual rated capacity to find how many you need; if wiring in series for higher voltage, each battery in that series string needs the full calculated capacity, since series wiring doesn't change amp-hour capacity.