How to use it
- Type your Yearly consumption from the bills added up for a year.
- Check the Peak sun hours for your area, and the Power of one panel with its length and width from the data sheet.
- Keep the System losses at 20 %, or type a figure for your system, and type the Roof area for the panels.
- Read Panels to install, then Roof check to see whether the panels fit.
How the panels are counted
The page works out how much energy one kilowatt-peak (kWp) of panels gives in a year at your sun hours: the sun hours × 365 × (1 − losses). It divides your yearly consumption by that to get the kWp you need, converts it to watts and divides by the power of one panel, rounding up to whole panels.
The roof area needed is the number of panels times the area of one panel, so you can check it against the roof.
yield per kWp (kWh a year) = sun hours × 365 × (1 − losses)
kWp needed = yearly consumption ÷ yield per kWp
panels = kWp needed × 1 000 ÷ watts per panel (rounded up)
roof area needed = panels × panel length × panel widthSun hours and losses
The default of 2.9 kWh/m² a day is the average for London in the PVGIS database of the European Commission's Joint Research Centre, which gives the monthly horizontal radiation from 2005 to 2023. Other cities have other figures, and the figure for your place is better: type it in place of the default.
The default losses of 20 % follow the Wikipedia article on photovoltaic power stations, which says that modern solar parks should have a performance ratio in excess of 80 %. The losses of your own system depend on its equipment and on its shade.
What this calculator does not include
- Shade from trees, chimneys and neighbouring buildings, which lowers the energy in a way the page does not model.
- The tilt and direction of the roof: the radiation figure is for a flat surface, so a roof can give more or less than the page shows.
- Batteries, the cost of the system and the rules for selling any surplus to the grid.
Worked example: 4,000 kWh a year, 2.9 kWh/m² a day, 20 % losses
A yearly use of 4,000 kWh needs 4,000 ÷ (2.9 × 365 × 0.80) = 4,000 ÷ 846.8 = 4.72 kWp. Nine panels of 550 W make 4.95 kWp, which gives 4.95 × 846.8 = 4,192 kWh a year: about 105 % of the use.
Nine panels of 2.28 × 1.13 m take 23.2 m² of roof, so on a 60 m² roof they fit with room to spare.
| Value | |
|---|---|
| Panels | 9 of 550 W |
| System size | 4.95 kWp |
| Energy per year | 4,192 kWh |
| Roof area the panels need | 23.2 m² |
Frequently asked questions
How many solar panels do I need for my home?
Divide your yearly consumption by the yearly energy that one kWp gives in your area, then round up to whole panels. The calculator does this for the consumption, sun hours and losses you type.
Where do I find my yearly consumption?
Add up the kWh of the last twelve bills, or read the yearly total in your supplier's account. The 4,000 kWh in the example is a placeholder, not an average.
What are peak sun hours?
They are the average daily solar radiation on a flat surface, in kWh per square metre a day. A value of 4.2 means that on an average day the sun delivers the same energy as 4.2 hours of full sunshine at 1 kW per square metre.
Why is the number of panels rounded up?
A panel cannot be split. The page takes the next whole panel, so the system covers a little more than your use, and the share shown in the result tells you by how much.
Does the tool store what I enter?
No. The calculation runs in your browser and nothing you type is sent to us or stored.
Sources
- PVGIS, European Commission Joint Research Centre: monthly horizontal radiation for London, 2005 to 2023
- Wikipedia: Photovoltaic power station (performance ratio above 80 % in modern solar parks)
Figures and text last checked on Friday, November 13, 2026.