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Sun III - Solar Panel Efficiency


This is the third post on the series about sun energy. This one is about “true” solar panel efficiency, i.e. numbers about existing solar panels.

The goal of this post is to estimate the number of solar panel necessary to cover our needs in electricity. We don’t focus on the life cycle, i.e, we discard:

  • the cost for producing the solar panel
    • Extracting the materials
    • Silicium purification
    • Assembly and transportation
  • the installation cost
  • the cost for dismantling the solar panel

If you haven’t read them, you can check the other articles of this series:

Table of Content

What is Needed

We want to know how many solar (photovoltaic) panels are required to cover our needs. We will use over this post the example of France, using the data of 2020 1. The total electricity consumption was 449 TWh, which is divided intro three areas:

  • 38 %: Domestic use (149.6 TWh)
  • 47 %: Professional use (building mainly, 184.4 TWh)
  • 16 %: Industries (62.2 TWh)

In France, we are about 67 Million people. Therefore, the average consumption for domestic use per inhabitant is about 2200 kWh/year. We will work with this figure all along this article.

For other countries, you can find a list of the consumption per country on Wikipedia 2. However, this list does not make the difference between domestic/industrial use. Here, we focus on domestic use, because this is the energy we use to live comfortably in our house.

True Solar Panel Efficiency

When we use a kettle, we know it consumes electricity only when we heat water. When we have a fridge, we know its 24h/24H, 7/7 days. For a solar panel, we would prefer it to produce as much electricity as possible. In its spec, you know at max what it could produce.

However, there is a big difference between true laboratory production and real conditions. By calculating the number of of solar panel, we can see easily the difference between the theoretical minimum, and the real number of solar panel to cover our needs.

A Note about Units

First of all, we need to describe the units, what they represent, and how to move from one to another. On the Internet, different units are used:

  • kW or W, a instantaneous power
  • kWh, an amount of energy
  • kWh/year, an amount of energy, consumed/produced over the year.
  • kWh/m^2/year, an amount of energy, produced by one m² during one year

W and kW

kW = 1000W is an instantaneous power. 1kW is 1000 Joules used/consumed during 1 second. This unit is often used to present the maximal production rate. If a panel get full sunlight, it deploys at max a power of x kW. As electrical goods are often described by their consumption (a kettle is typically 1 kW), you know that to turn on all your devices together, you need to produce at least the same amount as what is consumed. If you have a solar panel of 1 kW, you cannot turn on your kettle plus your oven.


The W/kW is often used to present the “best supply power case” of a solar pannel, while for consumer devices, the maximal amount of energy used per second. This is a power, i.e. an energy produced/consumed per second.

Over an day, you don’t let the kettle on, otherwise there would be no more water in. You turn it on for 2 minutes, and off the rest of the time. But you will consume energy using different devices: computer, lights, washing machine … All these appliances have their own consumption per second. The true consumption is the sum of the consumption of each goods :

\[\text{Consumption} = \sum_{\text{electrical good} g} \text{PowerNeeded}(g) \times \text{TimeUsed}(g)\]

The TimeUsed depends on the scale you consider: you can check how much energy you consumed over 5 minutes, one hour, one day, one week …

If you want to know how much your 1kW kettle consume over your week, and you use it each day 10 minutes but not on Sunday. In total over the week, you used your kettle for one hour, or 3600 seconds. Then, the energy consumed is 1kW x TimeUsedPerDay x MinToSec x DayUsed = 1kW x 10min x 60sec/min x 6 = 1kW x 3600 sec= 3600J.

As a human, we don’t often count in second: you count the amount of seconds you need to swim 50m, but you count in hour the amount of time you are in the office, your sleep time, the time needed for the train cross the country … Hour is a more convenient unit. So instead of having a factor x 3600 everywhere, we have the hour unit that englobes it. As 1h = 3600sec, 3600J = 1kW x 3600sec = 1kWh.

kWh/year and kWh/year/m²

For statistics, and to average over bad and good cases (we consume more energy in winter and in summer, to heat and for cooling) (solar panel produce more in summer), the production/consumption is often looked over a full year. The kWh/year is the amount of energy produced during one year. If we have a 2kW solar panel, then the production over one hour is 2kWh, and the production over a year is 2kWh x 24 x 365 = 17.520 kWh.

Similarly, for solar panel, as energy production depends on the area, the production is described in terms of surface, in kWh/(year.m²) = kWh/year/m². So if 1m² produces x kWh/year, then k m² produces k.x kWh/year. If I have a solar panel which produce 600W of electricity and measure 2m², then it produces 600/2 = 300W/m². Over a year, it represents 2,628 kWh/m²/year.

We can convert back the kWh/year into kW. For instance, if we take the french 2200 kWh/year, it represents 2,200,000 / (365*24)= 251.1 W.

Typical Yield

On average, we receive from the sun \(1400W.m^{-2}\) 3 (for an area perpendicular to sunlight). If we look at photovoltaic solar panel yield, in test conditions (\(1000W.m^{-2}\) ), the best solar panels have a yield of \(20\%\), i.e. if we get \(1400W/m2\), the panel gets \(1400 \times 0.2 = 280W/m^2\). (For a thermal solar panel, the yield is around \(35-40\%\), because heat is the most basic form of energy).

This (low) yield can be explained by three types of factors 4:

  • Pre-photovoltaic losses: absorption by the atmosphere and dust
  • Module losses: cell efficiency, the energy that hit the panel cell but which is not converted), The best cells have a yield in [30, 40] % 5.
  • System losses related to solar panel non-solar related elements, i.e. AC/DC conversion, wires, micro-controller consumption, …

So in theory, we would just need \(1 m^2\) of solar panels to cover our electricity needs.


  • On average, there are 12 hours on sunlight per day. So we need at least twice the initial number of solar panel.
  • This is without paying attention to weather, where clouds limit the intensity (only \(10\) to \(25 \%\) of the total energy is received 6);
  • Solar panels are not cleaned everyday (think to dust, leaves and guano);
  • Solar panels are most of the time fixed with a given angle all year long (this is the topic of the previous article).

We will check commercial numbers which relates the “good case” energy production of solar panel.

Sun Exposure

On average, we have 12 hours of sun exposure per day over the year, independently of if we are located near the equator or near the poles. The maximal number of sun hours is 365 x 12 = 4380 hours.

However, there are large differences of direct sun exposure between regions on earth. In France 7 8, the south get at max 3000 hours of sun, while in the north, only 1500 hours (because its very cloudy). In the good case, we need \(\frac{2200}{3000 \times 280} = 2.6 m^2\) of solar panels and in the worst case \(5.2\). When looking at this figures, even \(5m^2\) of solar panel per person is not bad. Putting solar panel on all possible roof would help to cover the needs of a house (not for building, but that’s already a good point).

The average yield of our solar panel is decrease. From the 280W/m², we moved to [47, 96] W (\(280 \times \frac{3000}{4380 \times 2} = 96\)). Over the year, it represents [420, 840] kWh/year.

Real-World Efficiency.

On the web, we can gather different sources listing the amount of energy produced over a year:

  • In France 9 a 1 kW solar panel produce between 900 to 1,400 kWh/year.
  • In California 10, a 10 kW solar panel will produce 15,611 kWh/year
  • In Yorkshire (UK)11, A 4 kW solar panel produce 2,850 kWh/year

These rate are of the same order of magnitude: from 712 to 1,561 kWh/ year for 1 kW devices. If we want to know the energy produced by , we need to multiply the production rate for 1 kW devices by 0.280 (which is the estimate for the 20% yield solar panel). So the productivity range per is [252, 437] kWh/m²/year.

Compared to our previous estimate ([420, 840] kWh/m²/year), we have half of it. We need to double again the solar panel area to cover our needs. Therefore, the number of of solar panel per person is [5, 8.7] m².

For a house, this area is manageable. Over a building, the area of the roof is too limited. The cost of a solar panel is around 200 - 300€/m² 12 For a family of 4, the investment is in between [4000; 10,000] €.

With the price of a kWh of 0.1740 € in France in 2022 13, auto-sufficiency would save 383€/person or 1531€ for the family of four. You need in the best case 3 years to get your investment back, or 6.5 years if you don’t have the best illumination conditions.

The return over investement takes time also because electicity in France is cheap. In countries such as Germany where the price is around 0.33€/kWh, the time to ROI is much smaller.


When we move from theoretical yields to true yield, the gap is large.

Starting from a theoretical yield of 20 %, giving 280W/m²:

  • We have first to divide by 2 to consider the fact that there is night.
  • There are 30 to 60 % of loss due to clouds/weather, which depends on the area
  • And divide again by 2 to consider operational loss (fixed angle and others).

This represents between 5 to 9m² of solar panel per person. By looking at the cost, it represents a large investment (up to 10,000€). However, if the surface can be exploited, this is worth taking.


  • Many people live in building, apartment, where it there is not enough roof space available
  • Domestic use represent only 40% of the electricity consumption in France
  • This is only for electricity 14 (1 562 TWh in total, 449 TWh = 29% for electricity)


  • producing locally electricity avoid losses during transportation
  • solar panel produce energy during the day. This is preferable to producing electricity during the night, unless we watch movies all night long. This reduce the needs for storage.

In the next post, we will talk about trees and shades.

IV: Shade from the trees


Other readings

Electricity map

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