# How many solar panels do I need?

Computes the solar system size in kilowatts and the number of panels that cover a share of a home’s electricity use, from the monthly use, the peak sun hours, the panel rating and the system losses, and the energy they make.

- Page: https://www.acalculator.org/construction/solar-panel-calculator
- JSON spec: https://www.acalculator.org/construction/solar-panel-calculator.json
- Version: 4976926093c8

## Default answer

Example with the default inputs (Electricity use per month (kWh) 900, Peak sun hours per day 4.5, Panel rating (W) 400, System losses 14%, Share of use to cover 100%): To make 100% of 900 kWh a month at 4.5 peak sun hours you need 7.65 kW; 400 W panels to buy: 20, making 941.7 kWh a month.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| kwh | Electricity use per month (kWh) | The electricity the home uses in an average month, in kilowatt hours, from the power bills. |
| sun | Peak sun hours per day | The average daily solar energy at the site, in kWh per m² per day, which equals hours of full 1,000 W/m² sun. |
| w | Panel rating (W) | The rated power of one panel in watts, from its label (its output at standard test conditions). |
| loss | System losses | Energy lost between the panels’ rating and the power the home gets: heat, wiring, the inverter, shade and dirt. |
| offset | Share of use to cover | How much of the monthly use the panels should make, in percent: 100 for all of it. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| panels | Solar panels | The system size ÷ the panel rating, rounded up to whole panels. |
| systemKw | System size needed (kW) | The panel power that covers the share of use: daily use × share ÷ (sun hours × (1 − losses)). |
| installedKw | System size with whole panels (kW) | The number of panels × the panel rating. |
| monthlyKwh | Energy made per month (kWh) | What the whole panels make in an average month: kW × sun hours × (1 − losses) × 365 ÷ 12. |
| yearlyKwh | Energy made per year (kWh) | What the whole panels make in a year: kW × sun hours × (1 − losses) × 365. |
| covered | Share of use covered | The energy made per month ÷ the monthly use, in percent. |
| dailyKwh | Electricity use per day (kWh) | The monthly use ÷ (365 ÷ 12) days. |

## Method

System kW = monthly kWh ÷ (365 ÷ 12) × share ÷ (sun hours × (1 − losses)); panels = system W ÷ panel W, rounded up; energy per month = panels × panel kW × sun hours × (1 − losses) × 365 ÷ 12.

## Assumptions

- Peak sun hours are the average daily solar energy on the panels, in kWh per m² per day, typed for your site and tilt; the calculator uses no location data.
- Panels make their rated power at 1,000 W/m² of sun, less the system losses (14% by default, the PVWatts default).
- An average month is 365 ÷ 12 days, and every month gets the average sun; real output is higher in summer and lower in winter.
- No prices, incentives or battery storage: the calculator sizes the panels only.

## Worked examples

1. kwh = 900, sun = 4.5, w = 400, loss = 14%, offset = 100% gives dailyKwh = 29.589041, systemKw = 7.645747, panels = 20, installedKw = 8, monthlyKwh = 941.7, yearlyKwh = 11,300.4, covered = 104.633333%. Source: NREL PVWatts Version 5 Manual (Dobos 2014), 14% default system losses.
2. kwh = 816.14, sun = 4, w = 300, loss = 14%, offset = 100% gives dailyKwh = 26.832, systemKw = 7.8, panels = 26, installedKw = 7.8, monthlyKwh = 816.14, covered = 100%. Source: NREL PVWatts Version 5 Manual (Dobos 2014).
3. kwh = 1,200, sun = 6, w = 500, loss = 20%, offset = 50% gives systemKw = 4.109589, panels = 9, installedKw = 4.5, monthlyKwh = 657. Source: NREL PVWatts Version 5 Manual (Dobos 2014).

## FAQ

### How many solar panels do I need?

Divide your daily electricity use by the energy one kilowatt of panels makes in a day, then by the panel rating. A home using 900 kWh a month uses 900 ÷ 30.42 = 29.6 kWh a day. At 4.5 peak sun hours with 14% losses, 1 kW of panels makes 4.5 × 0.86 = 3.87 kWh a day, so the home needs 29.6 ÷ 3.87 = 7.65 kW, or 20 panels of 400 W.

### What are peak sun hours?

The solar energy a surface gets in a day, in kWh per square metre, written as hours of full sun at 1,000 W/m². A site that gets 4.5 kWh/m² in a day has 4.5 peak sun hours, even if the sun is up for 12 hours. NREL publishes the average for any US location; its PVWatts tool and solar resource maps give the number for your site and panel tilt.

### What are system losses?

A panel makes its rated watts only at 1,000 W/m² of sun and 25 °C. In use, heat, wiring, the inverter, shading, dirt and aging all take some energy away. NREL’s PVWatts uses 14% total losses by default, and the calculator starts at the same number.

### How much energy does one solar panel make?

Panel watts × peak sun hours × (1 − losses). A 400 W panel at 4.5 sun hours and 14% losses makes 0.4 × 4.5 × 0.86 = 1.548 kWh a day, about 47 kWh in an average month.

### Why is the answer rounded up?

You can only buy whole panels, so the calculator rounds the panel count up. The system with whole panels then makes a little more than the share you asked for; the result shows how much, as the share of use covered.

### Does the calculator include cost or savings?

No. Prices, incentives and power rates change often and differ by place, so the calculator sizes the panels only. Use the system size in kW to compare quotes.

## Sources

- Dobos, A. P. (2014). PVWatts Version 5 Manual. National Renewable Energy Laboratory, NREL/TP-6A20-62641 (default system losses of 14%). https://www.nrel.gov/docs/fy14osti/62641.pdf
- NREL, PVWatts Calculator (solar resource in kWh/m²/day for a location, tilt and azimuth). https://pvwatts.nrel.gov
- IEC 60904-3, Photovoltaic devices, Part 3: Measurement principles for terrestrial photovoltaic solar devices with reference spectral irradiance data (standard test conditions: 1,000 W/m², 25 °C cell temperature). https://webstore.iec.ch/publication/61084