Inverter Sizing Calculator
An inverter has to pass two separate tests, and either one can decide the answer. It has to carry your running load indefinitely, and it has to survive the instant a motor starts, which can draw seven times its running current. This works out both, tells you which one is binding, and gives you the DC input current the battery cable and fuse have to be sized for.
2.0 applies to low-frequency, transformer-based inverters (Victron, Schneider, Outback and similar). High-frequency transformerless units, which is most compact and budget kit, usually manage only 1.3 to 1.5. Check the datasheet before leaving this at 2.
The inverter's LVD. Leave blank for 87.5% of nominal (10.5 V on 12 V, 21 V on 24 V, 42 V on 48 V). DC current peaks here, not at nominal.
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DC cable ampacity with the ambient and bundling corrections, how far each size runs at 12, 24 and 48 V, PV string fuse sizes from Isc, and battery-to-inverter current by inverter size. Three pages, made to pin by the battery bank.
How It Works
Two requirements, one answer
Surge requirement = Surge W ÷ Surge capability
Inverter rating = the larger of the two, rounded up to a size sold
The surge requirement is divided rather than multiplied because the inverter's surge capability is a property of the inverter, not of the load: a 3,000 W unit rated to hold double for five seconds can start a 6,000 W surge. Working backwards, a 7,200 W surge needs a unit rated at 3,600 W or more. On systems with a well pump, a compressor or an air conditioner without soft start, this is almost always the requirement that decides the purchase, and sizing on running watts alone leaves you with an inverter that trips every time the pump kicks in.
Worked example: a cabin with a well pump
A 2,400 W continuous load with a 7,200 W surge, 25% headroom, a 48 V bank, 92% inverter efficiency and a 240 V output:
- Continuous requirement = 2,400 × 1.25 = 3,000 W
- Surge requirement = 7,200 ÷ 2 = 3,600 W
- The surge is larger, so it binds: the inverter must be at least 3,600 W, giving a 4,000 W unit
- DC input at full rating = 4,000 ÷ (0.92 × 48) = 90.6 A
- DC current during the surge = 7,200 ÷ (0.92 × 48) = 163 A for a second or two
- AC output at the running load = 2,400 ÷ 240 = 10 A
Had the continuous figure alone been used, a 3,000 W inverter would have looked sufficient and would have failed to start the pump. That is the commonest inverter complaint on off-grid systems, and it is a sizing error rather than a fault.
Why the DC current matters more than the AC current
The AC side of a 4,000 W inverter at 240 V carries about 17 A at full output, which is ordinary house wiring. The DC side of the same inverter at 48 V carries 90 A, and at 12 V it would carry 362 A. That is the number that decides your battery cable, your busbars and your inverter fuse, and it is why the DC input figure above is quoted at the inverter's full rating rather than at your load: the cable has to survive whatever the inverter is capable of asking for, not what you intend to run.
Size that cable with the solar cable size calculator, choose the fuse with the PV fuse and breaker sizing calculator, and check the drop over the run with the DC voltage drop calculator. The AC output side is a different problem with different rules, and it is ordinary alternating-current wiring: for the circuit and cable sizing on that side, use the AC wire size calculator on Electrical Calculators.
Surge Factors by Load Type
Multiply the running wattage of your largest motor load by the factor below to get its starting draw, then add whatever else is already running to it. Only one motor normally starts at a time, so there is no need to add every surge together; the realistic worst case is the biggest starter plus the steady load.
| Load type | Starting factor | Notes |
|---|---|---|
| Kettle, toaster, heater, hob | 1.0 | Purely resistive, no starting surge at all |
| LED lighting, laptops, chargers | 1.0 | Inrush is measured in milliseconds and does not trouble an inverter |
| Television, router, electronics | 1.2 | Switched-mode supplies charging their input capacitors |
| Microwave | 1.5 | Draws roughly 1.5 times its cooking rating from the supply |
| Compressor fridge or freezer | 3.0 | Starts many times a day, so it must never be marginal |
| Washing machine | 3.0 | Drum motor start; the heating element is resistive on top |
| Power tools, saws, grinders | 3.0 | Universal motors, brief but sharp |
| Air conditioner with soft start | 2.5 | A soft-start module is far cheaper than a bigger inverter |
| Air conditioner, no soft start | 5.0 | The single commonest cause of inverter overload trips |
| Submersible or jet well pump | 5.0 | Starts against a head of water |
| Deep well pump, direct on line | 7.0 | Usually the load that decides the whole system |
Where a surge factor pushes the inverter two sizes up, price a soft starter first. Fitting one to an air conditioner typically drops its factor from 5 to under 2.5, and it costs a fraction of the difference between a 3,000 W and a 6,000 W inverter, as well as reducing the strain on the battery bank at every start.
Frequently Asked Questions
What size inverter do I need?
Take the largest set of loads that could realistically run together, add 20 to 25% headroom, and then check that against your biggest motor start divided by the inverter's surge capability. Buy whichever number is larger, rounded up to a size sold. For most off-grid homes that lands between 2,000 W and 5,000 W; for a campervan it is usually 1,000 W to 2,000 W unless you intend to use an induction hob or a kettle.
Should I add up every appliance I own?
No. Size the inverter on what genuinely runs at the same time, which in a house is far less than the total connected load. The kettle and the microwave are rarely on together, and the washing machine and the power tools are almost never simultaneous. A van is the exception: with only a handful of loads, the chance of them coinciding is high enough to size on the full total. The off-grid load calculator gives both figures so you can judge the difference.
Is a bigger inverter always safer?
No, because inverters draw idle power whether anything is switched on or not, and that draw scales with size. A 5,000 W unit can consume 25 to 50 W doing nothing, which is over 1 kWh a day, and on a small bank that can be a larger load than the appliances. Size for what you need plus sensible headroom, and if you have a large occasional load and a tiny background one, two inverters or a unit with a proper search or standby mode is more efficient than one large one left running.
What does power factor do to inverter sizing?
Power factor affects the current at a given wattage, not the wattage itself, so it changes your AC cable and breaker sizing rather than the inverter's power rating. A 2,400 W load at 0.8 power factor draws the current of a 3,000 VA load, and an inverter rated in VA rather than W has to be judged on that basis. Check whether your inverter is rated in watts or volt-amperes before comparing figures, because the two are only equal at unity power factor.
Why is the DC input current so much larger than the AC output current?
Power is roughly conserved through the inverter, so a low DC voltage means a high DC current. A 3,000 W inverter delivers about 12.5 A at 240 V AC but pulls around 68 A from a 48 V bank and 272 A from a 12 V one. That current sets the battery cable, the busbar and the inverter fuse, and it is the main practical reason larger systems use 48 V rather than 12 V.
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