How Many Solar Panels Do I Need? The Formula, a Worked Example and a Quick Sizing Table

How many solar panels do I need? Divide your yearly electricity use by what one kilowatt of panels produces at your location, then divide the result by the wattage of the panel you plan to buy. With a 400 W panel, an average US home that buys about 10,360 kWh a year needs roughly 18 panels (about 7.2 kW) at 5 peak sun hours and about 23 panels at 4 peak sun hours. That is very close to the typical home solar system size of 7.15 kW that NREL uses in its analyses, according to the DOE Homeowner’s Guide to Solar.

How many solar panels do I need? The 3-step formula

This is how to calculate solar system size by hand, the same logic any solar system size calculator uses.

1. Find your annual usage in kWh. Add up 12 months of bills.
2. Find the system size in kW. Annual kWh ÷ (peak sun hours × 365 × derate factor).
3. Find the panel count. System kW × 1,000 ÷ panel wattage (Pmax on the datasheet), then round up. Rounding down leaves you short.

The derate factor covers real-world losses: dirt (soiling), shading, wiring, panel mismatch, inverter conversion and aging, plus heat on hot days. This guide uses 0.80, meaning a system delivers about 80% of what its nameplate rating times the sun hours would suggest. That matches a July 2024 NREL training presentation for USDA (NREL/PR-7A40-90582), which lists overall losses from array to inverter output of about 20%. The PVWatts V8 documentation treats system losses as a separate input and its default inverter efficiency is 96%, so a detailed model may land a little above or below 0.80.

Step 1: Find your annual kWh

Your utility bill shows kilowatt-hours used each month. Add the last 12 months, because one summer or winter bill can be far off the annual average.

If you do not have the bills yet, use a national benchmark. In the EIA Electric Sales, Revenue, and Average Price tables with data for 2024, the average US residential customer bought 863 kWh per month. Multiplied by 12, that is about 10,360 kWh per year. The state averages range widely, from about 495 kWh a month in Hawaii and 503 in California to about 1,202 in Louisiana in the same table, so your own bills matter more than the national figure.

One caveat from EIA’s own FAQ on home electricity use: these numbers are electricity purchased from the utility, so homes that already have net-metered solar look smaller than they really are.

Step 2: Find your peak sun hours

A peak sun hour is one hour of sunlight at 1,000 watts per square meter. The same NREL presentation notes that a kWh per square meter of sunlight is sometimes called a “sun-hour”, gives about 4.9 kWh/m² per day on a horizontal surface as typical for Colorado, and says horizontal solar resource varies from 3.0 to 6.0 kWh/m² per day over most of the continental US.

Panels that are tilted toward the sun usually receive more than the horizontal value, which is why this guide shows 4, 5 and 5.5 hours as planning scenarios. For your real number, enter your address in PVWatts from NLR (the lab formerly known as NREL). It uses National Solar Radiation Database weather data and reports solar radiation for your tilt and azimuth, which makes it a reliable free “how many solar panels do I need calculator” for a first pass.

Worked example: an average home with 400 W panels

Take the 2024 EIA average and a site with 5 peak sun hours:

• One kW of panels produces about 5 × 365 × 0.80 = 1,460 kWh per year.
• System size: 10,360 ÷ 1,460 = 7.1 kW.
• Panels: 7,100 W ÷ 400 W = 17.7, rounded up to 18 panels.
• Check: 18 × 400 W = 7.2 kW, and 7.2 × 1,460 = 10,512 kWh a year, which covers the 10,360 kWh target.

At 4 peak sun hours, one kW produces 4 × 365 × 0.80 = 1,168 kWh a year. The same home then needs 10,360 ÷ 1,168 = 8.9 kW, or 8,870 ÷ 400 = 22.2, rounded up to 23 panels. At 5.5 hours it needs 6.45 kW, or 17 panels.

So for an average home, the answer to “how many solar panels do I need to power my house” is 17 to 23 panels of 400 W, and the spread comes from the climate.

Quick sizing table: panels needed by usage and sun hours

All figures assume 400 W panels and a 0.80 derate factor. The kW shown is the calculated size before rounding up to whole panels.

Middle cell: 9,000 ÷ (5 × 365 × 0.80) = 6.16 kW; 6,164 ÷ 400 = 15.4, so 16 panels. At 5.5 hours, 14 panels would make 5.6 × 1,606 = 8,994 kWh, just short of 9,000, so the table shows 15.

How many square feet of solar panels do I need?

Roof area is where many plans change. The CertainTeed 400 W module CTM10400HC11-06 measures 1,722 × 1,134 mm on its manufacturer datasheet, about 67.8 × 44.6 inches. That is 1.95 m², or about 21 sq ft per panel.

• Per kW: 21 sq ft ÷ 0.4 kW = about 52.5 sq ft of panel area per kW.
• For the worked example: 18 panels × 21 sq ft = about 378 sq ft of panels.

Plan for more roof than that: fire code setbacks, vents, chimneys, skylights and gaps between rows take space, and rules vary locally. The DOE homeowner guide notes that panels typically perform best on south-facing roofs with a slope between 15 and 40 degrees, though other roofs may work. If the good roof planes cannot fit the panel count, you either accept a smaller system, use higher-wattage panels, or move part of the array to the ground.

Factors that change how to size a solar system

• Shading. Trees, chimneys and nearby buildings cut output. Raise the loss percentage in PVWatts, and trust a site survey over any rule of thumb.
• Tilt and orientation. East- or west-facing planes produce less per kW than south-facing ones in the US, so they need more panels for the same kWh.
• Panel wattage and efficiency. Higher efficiency means more watts from the same footprint (see below).
• New loads. An electric vehicle or a heat pump can add thousands of kWh a year. For an EV, multiply your expected annual miles by the vehicle’s kWh per 100 miles from fueleconomy.gov, then divide by 100. Add that load before you size the system.
• Net metering and export rules. How your utility credits exported power affects whether it pays to cover 100% of your usage or a smaller share. Rules differ by state and utility and change over time, so ask your utility or installer for the current tariff in writing.
• Incentives. Incentives change, so check IRS.gov and the DSIRE database (dsireusa.org) for current programs. According to IRS.gov, the Residential Clean Energy Credit (25D) is not available for property placed in service after December 31, 2025.

Before you sign a contract, make sure the quote shows these assumptions. Our guide on what a home solar proposal should explain before you sign lists the usage, production and shading assumptions to ask for.

How many watt solar panels do I need?

Panel wattage does not change how much energy you need, only how many modules and how much roof it takes.

The JA Solar JAM54D41 440/MB datasheet lists a 440 W module at 1,722 × 1,134 mm, practically the same footprint as the 400 W example above. For the worked example:

• 7,100 W ÷ 400 W = 17.7 → 18 panels, 7.2 kW, about 378 sq ft.
• 7,100 W ÷ 440 W = 16.1 → 17 panels, 7.48 kW, about 357 sq ft.

The 440 W panel saves one module here and fits about 10% more kW on the same roof (21 sq ft ÷ 0.44 kW = about 48 sq ft per kW). Because of rounding, the saving is not always proportional. Higher-wattage residential modules are often bifacial; on a flush roof mount, the rear side adds little, as explained in our article on whether bifacial solar panels are worth it.

Grid-tied vs off-grid: what size solar system do I need?

The formula above sizes a grid-tied system on annual energy, because the grid absorbs summer surplus and covers winter shortfall.

An off-grid or battery-backed system is sized differently:

• Size the panels on the worst month, usually December, not on the annual average. Use the monthly solar radiation values from PVWatts for that month in the same formula.
• Size the battery on daily use and days of autonomy. An average home’s 10,360 kWh a year is about 28.4 kWh a day (10,360 ÷ 365), before any battery and inverter losses.
• For backup only, size the battery for the circuits you need during an outage.

Our deep dive on batteries, warranties and backup power explains how storage and backup choices affect the design.

Bigger properties, barns and small farms

The math is the same on larger properties. A shop, well pump or small farm that uses 30,000 kWh a year needs 30,000 ÷ 1,460 = 20.5 kW at 5 peak sun hours, or 20,548 ÷ 400 = 51.4, rounded up to 52 panels of 400 W.

At that scale, buying full pallets can lower the price per watt. Our guide to solar panels by the pallet covers pallet quantities, costs and delivery checks. You can compare modules by wattage and size in the A1 SolarStore solar panels catalog.

How to determine solar system size: a final checklist

• Add 12 months of kWh from your bills, plus any planned EV or heat pump load.
• Get your peak sun hours for your roof’s tilt and direction from PVWatts.
• Divide annual kWh by (sun hours × 365 × 0.80) to get kW.
• Divide watts by panel wattage and round up.
• Check the panel count against your usable roof area at about 21 sq ft per 400 W panel.
• Compare the result with installer quotes and ask about big differences.

Frequently Asked Questions