Solar Panel Calculator
Estimate the solar system size and the number of panels you need to cover your electricity use. Enter your use from your electric bill and the peak sun hours where you live. The calculator shows the system size in kW, the yearly production, and the area the panels cover.
- Choose US or Metric. This changes the panel sizes shown and the units of the panel area.
- Enter your Electricity use from your electric bill, per day, month, or year. Twelve months of bills give the best average. Leave the Offset target at 100% to cover all of it, or change it to cover part of it, or more for a future electric car.
- Enter the Peak sun hours for your address. See how to find your peak sun hours below.
- Keep the System losses at 20%, or raise it for a hot climate or a shaded roof.
- Choose a Panel. Choose Custom panel to enter the wattage on your panel's spec sheet, and its length and width for the panel area.
- Optionally, enter the Price per watt from a quote to estimate the cost.
Number fields accept whole numbers and decimals, with or without commas,
such as 900, 1,250, or 4.25.
How many solar panels you need
A 1 kW system makes about 1 kWh for each peak sun hour, less the system losses. So the system size is your yearly use divided by what 1 kW makes in a year:
Yearly output per kW = peak sun hours × 365 × (1 − system losses)
System size (kW) = yearly use × offset target ÷ yearly output per kW
Panels = system size × 1,000 ÷ panel wattage, rounded up
The calculator turns a monthly use into a yearly one by multiplying by 12, and a daily use by 365. Dividing a daily use by peak sun hours × (1 − losses) gives the same system size. This is the same calculation as the example in NREL's Technology Tips for Solar + Storage REAP Application Reviews: a 3 kW system with 5 kWh/m² of sun a day and 20% losses makes 3 × 5 × 0.80 = 12.0 kWh a day.
Because the panel count is rounded up, the system you install is a little larger than the size you need. The calculator works out the production and the offset from that installed size:
Installed system size = panels × panel wattage ÷ 1,000
Estimated production = installed system size × yearly output per kW
The production is an estimate for the first year. Panels lose some output as they age, about 0.5% a year according to NREL's PVWatts Version 5 Manual, and the calculator doesn't take that into account.
How to find your peak sun hours
Peak sun hours are the sunlight a panel gets in a day, in kWh/m², counted as hours of full sun at 1,000 W/m². One kWh/m² is one hour of full sun, so 5 kWh/m² per day is 5 peak sun hours. Panel ratings are measured at the same 1,000 W/m².
- Open PVWatts, the free solar calculator from the U.S. Department of Energy's national lab (pvwatts.nlr.gov).
- Enter your address, then the direction your roof faces (azimuth) and its tilt.
- On the results page, use the yearly Solar Radiation value, in kWh/m²/day, as your peak sun hours.
PVWatts's solar radiation is the sun on the face of the panels, so it already includes the direction and tilt you entered. That's why the calculator has no fields for them.
As a rough guide, NREL's REAP review tips give 3.0 to 6.0 kWh/m² per day on flat ground over most of the continental U.S. If you only have that flat-ground value, you can use it, but the result will be on the safe side: panels tilted toward the sun get more, and the calculator still takes the losses off, so it may show a panel or two more than you need.
NREL was renamed the National Laboratory of the Rockies (NLR) in December 2025, and PVWatts moved to the nlr.gov address with it.
System losses
System losses cover everything between the panel rating and the AC power the system delivers: dirt, shading, wiring, the inverter, heat, and more. The default of 20% is the overall loss from the array to the inverter output in NREL's REAP review tips.
PVWatts splits these losses in three. Its system losses default to 14.08%, combined from soiling 2%, shading 3%, snow 0%, mismatch 2%, wiring 2%, connections 0.5%, light-induced degradation 1.5%, nameplate rating 1%, age 0%, and availability 3%. The inverter is a separate input at 96% efficiency, and PVWatts works out heat losses from the cell temperature. Together, the first two come to about 17.5%. The calculator uses one number for all of them, so don't enter PVWatts's 14.08% here, or you'll get too few panels.
Panels make less power as they get hotter, so raise the losses in a hot climate. Raise them too for a roof with shade from trees or chimneys.
Panel wattage and size
Most home solar panels today are about 400 to 440 W. The calculator starts with 400 W, the count you'd need for the lower end of that range. The panel sizes are from typical panels:
- 400 W: 74.0 × 41.1 in (1,879 × 1,045 mm), from the Qcells Q.PEAK DUO BLK ML-G10+ data sheet, a 132-cell panel.
- 430 W: 68.1 × 44.0 in (1,730 × 1,118 mm), from the REC Alpha Pure-R datasheet, a shorter, wider, higher-output panel.
Panel sizes vary by model, so check your panel's spec sheet. For a different panel, choose Custom panel and enter its rated power (Pmax) and size.
The panel area is the panels alone. Your roof needs extra space around and between them, which depends on the roof shape and local rules, so the calculator doesn't work out the roof area you need.
Price per watt
Solar quotes are often compared by price per watt. Divide the total of a quote by the system size in watts, and enter that price to compare other sizes. The cost is the installed system size × 1,000 × the price per watt.
Examples
A home using 900 kWh a month
With 5 peak sun hours and 20% losses, 1 kW makes 5 × 365 × 0.8 = 1,460 kWh a year. The home uses 900 × 12 = 10,800 kWh a year, so it needs 10,800 ÷ 1,460 = 7.40 kW. With 400 W panels, that's 7,397 ÷ 400 = 18.5, rounded up to 19 panels, a 7.60 kW system. It makes about 11,096 kWh a year (925 kWh a month), 102.74% of the use, and the panels cover about 401.58 ft² (37.31 m²). At $3.00 per watt, it costs $22,800.
A Metric example
A home using 4,000 kWh a year, with 2.8 peak sun hours and 20% losses, gets 2.8 × 365 × 0.8 = 817.6 kWh a year per kW, so it needs 4.89 kW. With 430 W panels, that's 12 panels, a 5.16 kW system making about 4,219 kWh a year and covering 23.21 m² (249.83 ft²).
FAQ
About 21 panels of 400 W, with 5 peak sun hours and 20% losses. That's 12,000 kWh a year ÷ 1,460 kWh per kW = 8.22 kW, or 20.5 panels, rounded up to 21. With fewer sun hours, you need more.
About 1.6 kWh a day for a 400 W panel, with 5 peak sun hours and 20% losses: 0.4 kW × 5 × 0.8 = 1.6 kWh, or about 584 kWh a year.
The daily sun on your panels in kWh/m², counted as hours of full sun at 1,000 W/m². Most of the continental U.S. gets 3 to 6. PVWatts gives the value for your address.
Your yearly use divided by peak sun hours × 365 × (1 − losses). A home using 10,800 kWh a year with 5 peak sun hours and 20% losses needs about 7.4 kW.
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