Battery Capacity from Runtime Calculator
Load watts + desired runtime hours + battery voltage → required battery capacity in Ah and Wh (with safety margin).
Result
General calculation reads
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How to use this calculator
- Enter the constant load wattage.
- Enter target runtime in hours.
- Pick battery voltage (12 V, 24 V, 48 V common).
- Pick usable DoD based on chemistry.
- Set inverter / DC-DC efficiency (0.90 typical).
About this calculator
Battery sizing from a runtime target is straightforward energy bookkeeping: Wh required = load (W) × runtime (h), inflated by usable depth of discharge and converter efficiency. The DoD adjustment matters because lead-acid loses cycle life when discharged below 50%; lithium-iron-phosphate (LFP) tolerates 80%+; only some lithium-ion chemistries handle 100% safely. The η factor covers inverter losses (for AC loads) or DC-DC converter losses (for buck/boost). Peukert's law adds a small correction for very high discharge rates but is usually negligible at consumer scales.
What this calculator does
This calculator answers: "I want to power a W-watt load for H hours from a battery at V volts — how much capacity do I need?" It accounts for the two real-world inflation factors that take the naive Wh = W × h answer and turn it into a usable battery spec: usable depth of discharge (a function of chemistry) and converter / inverter efficiency. The output is in both Wh and Ah so you can compare against any battery datasheet.
How it works — the formula
Wh_load = P · t
Wh_battery = Wh_load / (DoD · η)
Ah_battery = Wh_battery / VPower times time is energy delivered. To deliver that much, the battery must store more — divided by the depth-of-discharge fraction (so it never goes below the safe-discharge floor) and divided again by the converter / inverter efficiency. Divide by nominal voltage to convert Wh to Ah.
Worked examples
- Inputs:
- P=60, t=8, V=12, DoD=0.8, η=0.90
- Output:
- Wh_battery ≈ 667 Wh; Ah_battery ≈ 55.6 Ah
A 100 Ah LFP would have ample margin.
- Inputs:
- P=1500, t=1, V=12, DoD=0.5, η=0.85
- Output:
- Wh ≈ 3529; Ah ≈ 294
Lead-acid users need much larger banks than lithium for the same usable energy.
- Inputs:
- P=200, t=4, V=48, DoD=0.8, η=0.92
- Output:
- Wh ≈ 1087; Ah ≈ 22.6
Higher-voltage banks need less Ah for the same energy — saves on conductor cost.
When to use this vs other tools
Use this for discharge-side sizing. For full off-grid systems you also need panel sizing and days of autonomy.
- Battery Series/Parallel
Use to figure out how many cells in series and parallel to build a pack of the required voltage and capacity.
- kWh Cost per Month
Use to compare battery payback against just buying power from the grid.
- Voltage Drop
Use to size the conductors between battery and load — long DC runs can lose significant power.
Authority note
The Wh/Ah/DoD conventions used here are standard across the battery industry. IEEE 1188 codifies the practice; Battery University is the most-cited practitioner reference.
Limitations
- Assumes constant load. Variable loads (motors with inrush, refrigerators with cycling compressors) need higher peak-current ratings.
- No Peukert correction — accurate within ~5% at discharge rates ≤ 0.5C, conservative above that.
- Temperature: lead-acid loses ~1% capacity per °C below 25 °C; lithium loses much less but throttles at low temperature.
- Aging: lead-acid loses ~20% at end-of-life; lithium ~20% by cycle count rather than years. Oversize for replacement intervals.
Critical loads (medical, life-safety) require redundancy beyond simple sizing. Consult a licensed electrical engineer for code-compliant standby-power design.
Frequently asked
Why not just divide Wh by voltage?+
How big a margin for safety?+
Peukert's law?+
AC vs DC loads?+
Solar charging assumed?+
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