How Many Solar Panels Do I Need?

How the calculator counts panels

Start with the electricity you use. Your bill shows it in kilowatt-hours (kWh) for the month, and many bills also print the last 12 months. US homes averaged 863 kWh a month in 2024 (EIA), but homes differ a lot, so use your own figure. Type it per month or per year, then choose the share of it you want the panels to cover.

The calculator runs the same steps you see on the rating plate:

1. Target: your use for a year times the share to cover. 900 kWh a month is 10,800 kWh a year.
2. System size: the target ÷ 365 ÷ (peak sun hours × performance factor). That is the kW of panels that makes the target on an average day.
3. Panels: the system size ÷ one panel’s kW, rounded up. Rounding down would leave you short, so the count never rounds down, and an exact fit stays exact.
4. Check: whole panels make a little more than the target. The plate shows that yearly figure rounded down to 10 kWh, and the share of your use it covers, also rounded down.

For the example on this page, Texas at 5.22 sun hours and a factor of 0.75, the target of 10,800 kWh needs 7.56 kW. That is 18.89 panels of 400 W, so 19 panels and a 7.60 kW system.

Peak sun hours and the performance factor

A peak sun hour is one hour of full sun at 1,000 W per square meter. A place with 5.22 peak sun hours gets as much sunlight on the panels in a day, on average over the year, as 5.22 hours of that full sun. The number changes with the season: winter months make less than the yearly average, and summer months make more.

The performance factor covers what is lost between that sunlight and the power you can use. Hot panels make less, dust and shade block some light, the wires waste a little, and the inverter that turns the panels’ DC into household AC has its own loss. PVWatts counts 14.08% of system losses by default, then adds the effect of heat and the inverter on top.

Each state on this page carries its own factor from the same model run as its sun hours. Hot states sit lower: Arizona 0.73, Alabama 0.72. Cold states sit higher: Wisconsin, Illinois and North Dakota 0.80. Pick your state and both numbers fill in. If you know your own sun hours, for example from a PVWatts run for your address or an installer’s report, type them in. The state box switches to Custom and the factor stays where it was.

Where the state numbers come from

Each state row is one NREL PVWatts v8 run for the state’s largest city: a 1 kW system, standard panels, a fixed roof mount at 20° facing due south (azimuth 180°), the default 14.08% losses, and NSRDB typical-year weather. Sun hours are the model’s yearly average sunlight on the panels. The factor is the yearly AC output divided by sun hours × 365. As a check, the EU’s PVGIS model, run on separate weather data, agrees within about 5% for most states. A single city stands in for a whole state, so your roof can differ by 10% or more inside a state.

How much roof the panels take

The plate shows two areas. The first is the panels themselves: the count times the area of one panel, rounded up to a whole square foot, with square meters beside it (1 ft² = 0.09290304 m²). Check the size on your panel’s datasheet; two common 400 W class panels measure 21.1 ft² and 20.8 ft².

The second is a planning figure of 100 ft² of roof per kW of panels, from the University of Arizona Extension and the Center for Sustainable Energy. It leaves room for the edges, vents and paths that installers keep clear. In the example, 19 panels cover 401 ft², but the plan calls for about 760 ft² of roof.

What this estimate leaves out

It sizes the panels and nothing else. Batteries, incentives, shade from trees or chimneys, a roof that faces east or west, panel aging and your utility’s rules all move the real number. That is also why two quotes for the same house can show different system sizes: they start from different sun hours, losses and roof layouts. Use this page to check that a quote is in a sensible range, and let the installer’s design settle the final system.

Worked example

1. Inputs: 900 kWh a month, Texas (Houston, 5.22 peak sun hours, factor 0.75), 100% of use, 400 W panels of 21.1 ft².
2. Target = 900 kWh × 12 × 100% = 10,800 kWh a year
3. Per day = 10,800 kWh ÷ 365 = 29.589 kWh
4. kW needed = 29.589 kWh ÷ (5.22 h × 0.75) = 7.5579 kW
5. Panels = 7.5579 kW ÷ 0.4 kW = 18.89, rounded up = 19
6. System = 19 × 400 W = 7.60 kW
7. Makes = 7.60 kW × 5.22 h × 0.75 × 365 = 10,860.2 kWh a year
8. Panel area = 19 × 21.1 ft² = 400.9 ft² (37.2 m²)
9. Roof to plan = 7.60 kW × 100 ft² = 760 ft²

Questions people ask

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