Solar Battery Calculator
Battery kWh = daily kWh × days ÷ (DoD × inverter efficiency): 5 kWh for 1 day with LiFePO4 at 80% and a 90% inverter needs 6.94 kWh (144.7 Ah at 48 V). Panels = daily Wh ÷ (peak sun hours × 0.75): at 4 sun hours that is 1,667 W, or 5 × 400 W panels.
How to use
- Enter your daily use in kWh and the largest load running at once in watts, or switch to the appliance list and add each device with its watts and hours per day.
- Choose the days of autonomy (how long the battery must last without sun), the battery voltage and chemistry. LiFePO4 is usually cycled to 80 to 90% depth of discharge, lead-acid to 50%.
- Enter the peak sun hours for your location and season, the panel rating and the system losses. For off-grid use, take the peak sun hours of your worst month.
How it works
Battery capacity (kWh) = daily kWh × days of autonomy ÷ (depth of discharge × inverter efficiency). Battery Ah = kWh × 1000 ÷ system voltage.
Solar array (W) = daily Wh ÷ (peak sun hours × (1 − system losses)). Panel count = array W ÷ panel W, rounded up.
Inverter continuous rating = largest simultaneous load × 1.25. Motor loads (fridge, pump, fan, power tools) can draw about 3 times their running watts for a moment at start-up, so the surge rating must cover that.
Charge controller current ≈ installed panel W ÷ battery voltage × 1.25. This is a rough figure for an MPPT controller; check the panel open-circuit voltage against the controller limit as well.
Worked example
- Load 5 kWh/day, 1 day of autonomy, 48 V LiFePO4 at 80% depth of discharge, inverter 90%: 5 ÷ (0.8 × 0.9) = 6.94 kWh, or 144.7 Ah at 48 V.
- Peak sun hours 4 and 25% losses: 5000 ÷ (4 × 0.75) = 1,667 W of panels, so 5 panels of 400 W (2,000 W) that produce about 6 kWh on an average day.
- Largest load 1000 W: inverter at least 1,250 W. Charge controller ≈ 2,000 ÷ 48 × 1.25 = 52.1 A.
Typical peak sun hours
Rough ranges only. Use the value for your exact site and worst month from NREL PVWatts / NSRDB or the World Bank Global Solar Atlas.
| Climate | Peak sun hours per day |
|---|---|
| High latitudes in winter | 1-2 |
| Cloudy temperate climate, yearly average | 3-4 |
| Sunny mid-latitudes, yearly average | 4-5 |
| Deserts and sunny tropics | 5-7 |
Assumptions
- The load is AC power through an inverter, so the inverter efficiency is applied to the battery. For DC loads, set it to 100%.
- The battery must cover the whole daily load on sunless days; it is not reduced by any solar input on those days.
- Panel sizing assumes the panels fully recharge the battery from one day's sun. To recover from several cloudy days faster, add panels.
- This is a design estimate for planning. Have a qualified professional confirm the final system.
FAQ
How big a battery do I need for 5 kWh a day?
With one day of autonomy, a LiFePO4 battery used to 80% and a 90% efficient inverter: 5 ÷ (0.8 × 0.9) = 6.94 kWh, which is 144.7 Ah at 48 V. For two days on lead-acid at 50% it grows to 22.22 kWh.
What are peak sun hours?
The number of hours per day the sun would need to shine at 1000 W/m² to deliver the day's solar energy. It is not the same as daylight hours. Look up your location in NREL PVWatts or the Global Solar Atlas and use the lowest month if you need power all year.
Why 25% system losses?
Panels rarely produce their label rating. Heat, dust, wiring, mismatch and the charge controller all take a share. NREL PVWatts uses 14% before temperature effects; 20 to 30% in total is a common, slightly cautious assumption for small off-grid systems.
Should I choose 12 V, 24 V or 48 V?
The energy in kWh is the same, but a higher voltage means lower current, thinner cables and a smaller charge controller. 12 V suits small systems below about 1 kW, 24 V the middle range and 48 V larger home systems.
How large should the inverter be?
At least 1.25 times the largest load you run at the same time, with a surge rating that covers motor start-up. A 150 W fridge can briefly draw around 450 W.
Can I rely on this result to build a system?
Use it for planning and budgeting. Cable sizes, fuses, panel string voltage, temperature and local electrical rules need a qualified installer or electrician.