Inverter size calculator
Enter the continuous load in watts, the peak (surge) load, a margin and the inverter's efficiency, then the battery voltage and capacity. The calculator gives the inverter rating to buy, the current drawn from the battery and how many hours the usable part of the battery lasts.
Runs in your browser: nothing you enter is sent anywhere.
How to use it
- Enter the continuous load: the watts of everything that runs at the same time.
- Enter the peak load: the highest short demand, for example when a motor or compressor starts.
- Choose a margin and enter the inverter’s efficiency.
- Enter the battery voltage, its capacity in Ah and the usable share to get the runtime.
- Continuous rating = load × (1 + margin).
- Surge rating = peak load.
- Battery current = load ÷ efficiency ÷ battery voltage (the same with the peak load for the peak current).
- Runtime = Ah × V × usable share ÷ (load ÷ efficiency), in hours.
Worked example
A 1,000 W load with a 2,000 W start, 25 % margin, 90 % efficiency, 12 V 200 Ah battery used to 50 %:
- inverter: 1,250 W continuous, 2,000 W surge;
- battery current: 1,000 ÷ 0.9 ÷ 12 = 92.6 A, and 185.2 A during the start;
- usable energy 200 × 12 × 0.5 = 1,200 Wh, input power 1,111 W, runtime 1.08 h.
Tips
- Cables and fuses must carry the battery current: at 12 V and 1 kW that is close to 100 A. Keep the battery cables short.
- Efficiency falls at light load, and the inverter draws some idle power even with nothing connected.
- Battery capacity shrinks at high current; the Ah rating is usually given for a slow discharge, so the runtime is an upper estimate.
Frequently asked questions
What size inverter do I need?
Add the watts of everything that runs at once, add a margin, and check that the inverter's surge rating covers the highest peak, such as a fridge compressor starting. 1,000 W of load with a 25 % margin needs a 1,250 W inverter.
How much current does an inverter draw from the battery?
I = load ÷ (efficiency × battery voltage). 1,000 W from a 12 V battery through a 90 % efficient inverter draws 1,000 ÷ 0.9 ÷ 12 = 92.6 A. At 24 V the current halves, which is why larger systems use higher battery voltages.
How long will my battery run the inverter?
Runtime = Ah × V × usable share ÷ (load ÷ efficiency). A 200 Ah 12 V battery used to 50 % holds 1,200 Wh; at 1,111 W of input power that is about 1.08 hours.
What usable share should I enter?
It depends on the battery chemistry and how long you want it to last. Lead-acid batteries are often used to about half their capacity, lithium iron phosphate batteries deeper. Use the figure from the battery maker.