Sizing a battery for a UPS, a camper van, a boat, a solar system or an off-grid cabin starts with two numbers: how much power the load draws and how long it must run. Dividing the power by the battery voltage gives the current, and current multiplied by time gives the amp-hours the load consumes.
A battery should not be emptied completely. Lead-acid batteries last far longer if at least half their charge stays in them, while lithium batteries are comfortable down to about 20%. The calculator scales the capacity up so that the chosen share of charge remains at the end.
Lead-acid batteries have a second handicap: they deliver less than their rated capacity when discharged quickly. Their capacity is usually rated over 20 hours, so a battery drained in 10 minutes gives up only around 60% of its label value. The lead-acid formula uses the factor 0.6 + 0.02 × t (hours), capped at 1 for 20 hours or more, to allow for that. Lithium batteries hold their capacity at high rates, so no derating is applied. If the load runs on AC through an inverter, enter its efficiency (85–95% is typical) so the battery current includes the conversion losses.
How to use the Battery Size Calculator
- Load power
- Battery type
- Battery voltage
- Required run time
- Charge to leave in the battery — About 50% for lead-acid and 20% for lithium is recommended.
- Inverter or converter efficiency — Use 100% for DC loads; 85–95% when running AC devices through an inverter.
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Enter the load power, battery type and voltage, the required run time, the share of charge to keep and the inverter efficiency. The calculator converts power to battery current, multiplies by time and scales the result for the reserve and, for lead-acid, for the discharge rate.
Worked example
A 200 W desktop computer must keep running for 10 minutes on a 12 V lead-acid UPS battery, leaving 50% of the charge. The load current is 200 ÷ 12 = 16.67 A and 10 minutes is 0.1667 h, so the load uses 2.78 Ah. The rate factor is 0.6 + 0.02 × 0.1667 = 0.6033, so the battery needs 100 × 16.67 × 0.1667 ÷ (50 × 0.6033) = 9.21 Ah. A lithium battery keeping 20% would need only 2.78 ÷ 0.8 = 3.47 Ah.
How to use the result
Round up to the next standard battery size, add margin for ageing (capacity falls to about 80% over the battery’s life) and check that the battery can supply the current – its maximum discharge rate in C must exceed the “discharge rate” shown.
Inputs explained
| Field | What to enter |
|---|---|
| Load power | Units: W, kW, hp |
| Battery type | Choose one: Lead-acid (flooded, AGM, gel), Lithium-ion / LiFePO4 |
| Battery voltage | In V |
| Required run time | Units: min, h, days |
| Charge to leave in the battery | In %. About 50% for lead-acid and 20% for lithium is recommended. |
| Inverter or converter efficiency | In %. Use 100% for DC loads; 85–95% when running AC devices through an inverter. |
What each result means
- Battery current (load current) = [Load power]÷([Battery voltage]×[Inverter or converter efficiency]÷100)
- Charge used by the load = [Battery current (load current)]×[Required run time]
- Required battery capacity = 100×[Battery current (load current)]×[Required run time]÷((100 – [Charge to leave in the battery])×[Rate derating factor])
- Battery energy rating = [Required battery capacity]×[Battery voltage]
- Discharge rate = [Battery current (load current)]÷[Required battery capacity]
- Number of 100 Ah batteries in parallel = ceil([Required battery capacity]÷100)
Example values
How the main result changes when Load power changes and everything else stays at the example values above:
| Load power (W) | Required battery capacity (Ah) |
|---|---|
| 100 | 4.6041 |
| 150 | 6.9061 |
| 200 | 9.2081 |
| 250 | 11.5101 |
| 300 | 13.8122 |
| 400 | 18.4162 |
Formula
I = P ÷ (V × η) · Lithium: B = 100 × I × t ÷ (100 − Q) · Lead-acid: B = 100 × I × t ÷ [(100 − Q) × min(1, 0.02t + 0.6)] (t in hours, Q = % left in the battery)
Worked example
Inputs — Load power: 200 W; Battery type: Lead-acid (flooded, AGM, gel); Battery voltage: 12 V; Required run time: 10 min; Charge to leave in the battery: 50 %; Inverter or converter efficiency: 100 %.
Results — Battery current (load current) = 16.667 A; Charge used by the load = 2.778 Ah; Rate derating factor = 0.603; Required battery capacity = 9.2081 Ah; Battery energy rating = 110.5 Wh; Discharge rate = 1.81 C; Number of 100 Ah batteries in parallel = 1.
Key terms
- Load current
- Current drawn from the battery, I = P ÷ (V × efficiency).
- Depth of discharge
- Share of the capacity actually used; the rest is the reserve.
- 20-hour rate
- The discharge time at which lead-acid capacity is usually rated.
- Inverter efficiency
- Share of battery power that reaches the AC load.
Frequently asked questions
What size battery do I need for a 1,000 W load?
At 12 V the load draws about 83 A. For 1 hour with lithium (20% reserve) you need about 104 Ah; with lead-acid (50% reserve) about 270 Ah.
How long will a 100 Ah battery run a 500 W inverter?
About 100 × 12 × 0.8 × 0.9 ÷ 500 ≈ 1.7 hours for lithium; roughly half that for lead-acid.
Why does lead-acid need a bigger battery?
You should use only about half its capacity, and it delivers less than its rated capacity at high currents.
Should I use 12 V, 24 V or 48 V?
Higher voltages halve the current for the same power, allowing thinner cables and lower losses – 24 V or 48 V is better above about 1–2 kW.
What is the 0.02t + 0.6 factor?
A simple approximation of how much rated capacity a lead-acid battery delivers when discharged over t hours instead of the 20-hour rating.
Do I need to include inverter losses?
Yes for AC loads – an 88% efficient inverter takes about 14% more from the battery than the load uses.