Solar System Size Calculator
Estimate the photovoltaic system size, panel count and roof area needed to cover your monthly electricity use
Energy & Location Data
Check your latest utility bill for your average monthly kWh
Estimates assume unshaded, south-facing roof sections. Your local solar installer will conduct a detailed site audit.
How to use this tool
- Choose m² (Metric) or sq ft (US) for the roof area result.
- Enter your average monthly electricity use in kWh from a utility bill.
- Enter the peak sun hours per day for your location, then pick the panel power and the system efficiency.
- Read the system size in kW, the number of panels, the estimated yearly output and the roof area needed.
Worked example
A home using 500 kWh a month consumes about 16.45 kWh a day. At 80% system efficiency the array must produce 16.45 ÷ 0.80 = 20.56 kWh, and with 4.5 peak sun hours that means 4.57 kW. With 400 W panels that is 4,569 ÷ 400 = 11.4, rounded up to 12 panels, producing about 6,312 kWh a year on paper and needing roughly 252 sq ft of roof. The output is a little above the 6,000 kWh the household uses because panels come in whole units. Real results depend on shading, roof angle and local conditions, so treat this as a planning figure.
Why this tool exists and where its numbers come from
Households want to know roughly how many panels would offset their electricity use. The tool sizes the array from consumption, sun hours and system losses; its 75–85% efficiency range brackets the roughly 82.5% overall derate implied by the PVWatts defaults from NREL (14% system losses and a 0.96 inverter), and shading, orientation and local rules still need a professional survey.
Sources
- NREL: Availability and Performance Loss Factors for U.S. PV Fleet Systems (TP-5K00-88769)
- NREL: PVWatts Version 5 Manual (default 14% system losses, 0.96 inverter efficiency)
For the full method, see the Methodology section just below.
Methodology
The calculator sizes a photovoltaic (PV) array from your consumption using a simple insolation model: monthly kWh is converted to a daily average (monthly / 30.4), divided by system efficiency, then divided by peak sun hours. Worked example: 500 kWh per month is 16.45 kWh per day; at 80% efficiency the array must produce 20.56 kWh of DC-rated energy, and with 4.5 peak sun hours that is 4.57 kW.
Solar sizing at a glance, plus Priya’s cloudy-roof case
These rows use the tool’s own formula with 4.5 peak sun hours, 400 W panels and 80% efficiency. Panels are always rounded up, so the output sits slightly above what the household uses.
| Monthly use (kWh) | System (kW) | Panels | Roof (m²) | Output (kWh/yr) |
|---|---|---|---|---|
| 200 | 1.83 | 5 | 9.8 | 2,630 |
| 350 | 3.20 | 8 | 15.6 | 4,208 |
| 800 | 7.31 | 19 | 37.1 | 9,993 |
| 1,200 | 10.96 | 28 | 54.6 | 14,727 |
| 2,000 | 18.27 | 46 | 89.8 | 24,194 |
Priya’s case. She also uses 500 kWh a month, but her roof gets only 3.0 peak sun hours. The array grows to 6.85 kW, which means 18 panels of 400 W, about 35.1 m² of roof and 6,312 kWh a year. With 5.5 sun hours the same household needs 3.74 kW, 10 panels and 19.5 m². Switching the efficiency from 80% to 75% at 3.0 hours adds a panel (19 instead of 18); 85% saves one (17). Sun hours matter more than any other input, so check yours.
Planning figure only, not a quote or a performance guarantee. The sun-hours value is whatever you type in, so take it from a local solar map or your installer rather than guessing, and remember that shading or a different roof angle can move the real output either way.
Smart Home & Energy Monitoring
A power strip, a plug-in watt meter and an indoor camera. The watt meter helps you measure real consumption before sizing panels.
Anker 12-Outlet Power Strip with Surge Protector
A 12-outlet power strip with 2100J surge protection and USB ports, according to the listing.
Digital Electricity Usage Monitor & Watt Meter
A plug-in watt meter that shows power, voltage and current, according to the listing. Measured watts give the calculator better inputs than nameplate figures.
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Frequently asked questions
What are “Peak Sun Hours”? ▼
Peak sun hours are not the total hours of daylight. They are the equivalent number of hours per day at a solar irradiance of 1,000 watts per square meter. Even if the sun is up for 12 hours, you may only get 4 to 5 peak sun hours.
Why is system efficiency only 80%? ▼
Solar panels produce direct current (DC), but your home uses alternating current (AC). Inverter conversion, voltage drop in the wiring, dust and the efficiency loss on hot summer days reduce the usable AC output to roughly 80% of the panels’ DC rating.
Do I need a battery for my solar system? ▼
If your utility offers 1:1 net metering, you can send excess energy to the grid during the day and use those credits at night without a battery. If your grid has frequent outages or no net metering, a battery storage system is worth considering.
Why does the yearly output differ from my yearly consumption, and what does the estimate ignore? ▼
Panels come in whole units, so the array is rounded up and produces slightly more than the target on paper. The calculation also uses one average peak sun figure for the whole year, while real output swings with season, weather, roof tilt and compass direction, and panels lose a little output as they age. It does not include battery storage, time-of-day tariffs, export rules or incentives, all of which decide whether a given system pays off. Treat the number of panels and roof area as a starting point for quotes, and ask installers for a site-specific production estimate.