Continuous watts, surge watts and VA are three different numbers
An inverter carries at least two ratings and often three, and they answer different questions.
Continuous watts is what the unit can deliver indefinitely without its heatsink reaching shutdown temperature. This is the rating your steady load must fit inside, with headroom. Manufacturers sometimes qualify it by ambient temperature, and a unit rated 3,000 W at 25 °C may be rated 2,400 W at 40 °C.
Surge watts is what it can deliver for a second or two while a motor gets moving. Induction motors draw far more current at zero speed than at running speed, because the rotor has not yet developed back-EMF. The nameplate locked-rotor amps figure — LRA — is that current, and dividing it by the running amps gives the starting factor this calculator uses. Typical surge ratings are twice the continuous rating on a high-frequency inverter and three times or more on a low-frequency transformer-based one.
Volt-amperes is what the unit can deliver when the load's current and voltage are out of phase. Watts are the real power a load consumes; VA is the product of RMS volts and RMS amps regardless of phase. An inverter's output stage is limited by current, so a 3,000 VA inverter running a load at 0.8 power factor delivers only 2,400 W. If your equipment list is dominated by motors and switch-mode supplies, size against the VA figure and treat the watt figure as advertising.
The fourth number nobody quotes is the DC input current, and it is the one that determines what the installation costs. Everything on the battery side — cable, fuse, disconnect, busbar, lug — is sized from it, and at 12 V it is four times what the same load draws at 48 V.
How the four numbers are derived
Continuous rating. Add the running wattage of everything that can be on simultaneously, then apply a design margin. Twenty to twenty-five percent is the normal figure, and it exists for two reasons: inverter efficiency falls off near full output, and continuous operation at 100% of rating leaves nothing for the ambient temperature rise inside a battery box in August.
Apparent power. Divide the continuous watts by the load's power factor. A resistive load — a kettle, a filament lamp, a heating element — has a power factor of 1 and needs no adjustment. Induction motors run near 0.8, and cheap switch-mode supplies without power factor correction can present 0.6. Where you are unsure, 0.9 is a defensible whole-house average and 0.8 is a conservative one.
Surge rating. The largest motor starts while everything else is already running, so the peak demand is the other loads plus the motor's starting power. Because the continuous figure already includes the motor's running power, the surge requirement is the larger of the two numbers rather than their sum. If your motor is small relative to the rest of the load, or your starting factor is low because the motor has a soft starter, the continuous rating can end up being the binding constraint — and the calculator says so when it happens.
DC input current. Power in equals power out divided by efficiency, so Idc = Pac ÷ (Vdc × η). Note that this uses the nominal bank voltage. A real lead-acid bank sags toward 11.5 V per 12 V nominal under heavy load, and a battery protection circuit may cut off there, so the actual peak current is higher than the nominal calculation shows. That is one reason the cable ampacity check applies a 1.25 factor.
Cable. Two constraints, and you take the larger conductor. Ampacity must cover 1.25 times the continuous DC current, using the 75 °C copper column of NEC Table 310.16. Cross-sectional area must be big enough that 2·ρ·L·I ÷ cmil stays inside your voltage-drop budget, with ρ = 12.9 for copper and the factor of two accounting for both the positive and negative conductors. On short battery-to-inverter runs the ampacity rule usually governs; past about ten feet the drop rule takes over. The voltage drop calculator handles the same arithmetic for AC circuits.
Worked example: an off-grid cabin with a well pump
The cabin runs 1,400 W of lights, laptop, fridge controls and a small fan continuously. The well pump draws 1,000 W running and has a nameplate LRA three times its running amps. The bank is 48 V, the inverter is 90% efficient, the cable run is five feet, and you allow 2% drop on it. Power factor across the mix is 0.9 and you want a 25% design margin.
- Total running load. 1,400 + 1,000 = 2,400 W.
- Continuous rating. 2,400 × 1.25 = 3,000 W. A 3,000 W inverter is the minimum; a 3,500 W unit gives room for a future addition.
- Apparent power. 3,000 ÷ 0.9 = 3,333 VA. If the unit you are considering is quoted only in VA, it must be at least this.
- Load while the pump starts. 1,400 + (1,000 × 3) = 4,400 W, which is larger than the 3,000 W continuous figure, so 4,400 W is the surge requirement. A 3,000 W inverter with a 2× surge rating covers 6,000 W and passes comfortably.
- DC input current. 3,000 ÷ (48 × 0.90) = 3,000 ÷ 43.2 = 69.4 A. During the pump start it briefly reaches 4,400 ÷ 43.2 = 101.9 A.
- Cable by ampacity. 1.25 × 69.4 = 86.8 A. In the 75 °C copper column, 4 AWG carries 85 A — just short — and 3 AWG carries 100 A. So 3 AWG.
- Cable by voltage drop. The budget is 2% of 48 V = 0.96 V. Required area = 2 × 12.9 × 5 × 69.4 ÷ 0.96 = 8,958 ÷ 0.96 = 9,332 circular mils, which is only 10 AWG. Ampacity governs, so 3 AWG stands, and its actual drop is 2 × 12.9 × 5 × 69.4 ÷ 52,620 = 0.170 V, or 0.35%.
Now move the same system to a 12 V bank. The DC current becomes 3,000 ÷ (12 × 0.90) = 277.8 A, the ampacity requirement becomes 347 A, and you are into 400 kcmil or paralleled 4/0 cable with a 400 A class-T fuse. That single change is why almost every modern off-grid system above 2 kW is built at 48 V.
How to read the result before buying
Compare the surge figure to the inverter's surge specification, not to its continuous rating. Manufacturers state surge as a wattage and a duration — for example 6,000 W for 20 seconds, or 9,000 W for 100 milliseconds. A well pump takes one to three seconds to reach speed, so a millisecond-scale peak rating tells you nothing useful. Ask for the figure at one second.
Oversizing has a cost. Every inverter draws idle current whether or not anything is running, typically 10 to 30 W for a 3 kW unit, and that no-load draw scales roughly with the unit's size. On an off-grid system with a small bank, an inverter twice the size you need can consume a meaningful share of your daily production doing nothing. Size for the load you have plus one planned addition, and use the inverter's search or standby mode when the load is intermittent.
Low-frequency and high-frequency inverters are not interchangeable at the same rating. A low-frequency unit uses a heavy iron transformer and typically sustains three times its continuous rating through a motor start; a high-frequency unit is lighter and cheaper but may manage only twice, briefly. If your surge requirement is more than twice your continuous requirement, the transformer-based design is usually the honest choice.
The DC current dictates the whole battery side. Fuse, disconnect switch, busbar and battery terminal hardware are all rated from it, and NEC 690.9 and Article 706 govern the overcurrent protection for battery circuits in a permitted installation. The battery bank sizing calculator checks that the bank can actually deliver that current without excessive voltage sag, which is a different question from whether it stores enough energy.
DC input current by inverter output and bank voltage
| AC output | 12 V bank | 24 V bank | 48 V bank |
|---|---|---|---|
| 1,000 W | 92.6 A | 46.3 A | 23.1 A |
| 1,500 W | 138.9 A | 69.4 A | 34.7 A |
| 2,000 W | 185.2 A | 92.6 A | 46.3 A |
| 3,000 W | 277.8 A | 138.9 A | 69.4 A |
| 4,000 W | 370.4 A | 185.2 A | 92.6 A |
| 5,000 W | 463.0 A | 231.5 A | 115.7 A |
| 6,000 W | 555.6 A | 277.8 A | 138.9 A |
Cable losses scale with the square of current, so the same load on a 48 V bank dissipates one sixteenth of the cable heat it would on a 12 V bank through the same conductor. That, not the inverter price, is what makes higher bank voltages cheaper above about 2 kW.
Where inverter sizing goes wrong
- Adding surge watts to running watts for every appliance at once. Two motors almost never start in the same instant, and sizing as though they do produces an inverter twice the size you need. Take the largest single starting load and add it to the others' running load.
- Reading the peak rating off the box. The largest number on an inverter's packaging is usually a fraction-of-a-second peak. Size against the continuous rating and check the surge specification at a duration of one second or more.
- Ignoring power factor. A 3,000 W inverter feeding a 0.8 power factor load runs out of current at 2,400 W. Compute VA and compare it to the VA rating where one is published.
- Using nominal battery voltage for the fuse. A loaded 48 V bank can sag to 44 V or lower, which raises the current for the same output power. Sizing the DC overcurrent device at 1.25 times the nominal-voltage current absorbs part of that, but a bank with high internal resistance needs the sag measured, not assumed.
- Forgetting the inverter's own idle consumption. Ten to thirty watts around the clock is 240 to 720 Wh a day, which on a small off-grid system can exceed the load you actually cared about.
- Sizing cable on ampacity alone on a long run. Past about ten feet at typical off-grid currents, voltage drop rather than heating becomes the binding constraint, and a cable that is legal can still cause nuisance low-voltage shutdowns.
- Assuming a generator can be sized the same way. Generators have their own motor-starting behaviour driven by alternator design and engine governor response; see the generator sizing calculator rather than reusing this result.
Where the starting factor comes from
The only authoritative source for a motor's starting current is its nameplate — either an LRA figure in amps or an NEMA code letter that maps to a kVA-per-horsepower range. Divide LRA by running amps to get the factor this calculator wants. Where the nameplate is unavailable, practitioners commonly assume around three for a scroll or soft-started compressor and five to six for a capacitor-start induction motor such as a shallow-well pump; treat those as rules of thumb to be replaced by real data, not as design values. A variable-frequency drive or a listed soft starter changes the picture entirely, often bringing the starting factor close to one and removing the surge constraint from the design.
What this calculation does not decide
Sizing the inverter answers how much power you can draw at once. It says nothing about how long — that is the battery's job, and it depends on capacity, depth of discharge and the bank's ability to hold voltage at the current you are pulling. Run the battery runtime calculator next; a 5 kW inverter on a 100 Ah 48 V bank is a perfectly valid pairing that will run a 4 kW load for well under an hour.
It also says nothing about replacement energy. On an off-grid system the array has to put back everything the loads take out plus the round-trip losses, and that sizing runs through the solar panel array sizing calculator and the solar charge controller sizing calculator. Inverter capacity and daily energy are independent specifications; systems fail on either.
Finally, a grid-interactive or hybrid inverter is sized differently again, because its AC output can be supplemented from the grid during a surge and its continuous rating may be split between pass-through and inverting capacity. The numbers here apply cleanly to a standalone inverter running from a battery, which is the configuration in a cabin, a van, a boat, or a whole-house backup system in island mode.
Key terms
- Locked rotor amps (LRA)
- The current a motor draws at the instant of start, before the rotor turns and develops back-EMF. Found on the motor nameplate; dividing it by running amps gives the starting factor used here.
- Power factor
- The ratio of real power in watts to apparent power in volt-amperes. It is 1 for resistive loads and lower where current and voltage are out of phase or the current waveform is distorted.
- Surge rating
- The output an inverter can sustain for a stated short period. Always paired with a duration; a peak quoted without one is not a usable specification.
- No-load draw
- The DC power an inverter consumes when it is on but supplying nothing. Typically 10–30 W on a 3 kW unit, and the reason standby or search mode exists.
- Bank voltage
- The nominal DC voltage of the battery string — 12, 24 or 48 V. For a given AC output, DC current is inversely proportional to it, and cable heat is inversely proportional to its square.