I have a small solar panel of the following,

  • Area = 0.0078 square meter
  • Rated Power = 1 W

Now, I am interested to calculated the total energy this solar panel can harvest over a day.

I am taking the following parameters,

  • Annual solar irradiance = 1200 KWh/square meter

Converting this for my solar panel, I get (1200 KWh/365)*0.0078 = 25.6 Wh

Is this 25.6 Wh the total energy it can provide? But I think is is not right as solar panel is only rated to 1 watts, how can it provide so much? Is this right or I am missing something else in my calculation?

I also tried the other way to calculated the total energy it can harvest in a day. This is what I did next.

I find the average sun shine hour from here, I did the average calculation and got 2.74 hr

Energy = 1 watt * 2.74 = 2.75 Wh, this is the right way to do it?

I didn't use the sun irradiance in my second calculation, is this correct?

Which one of the method is correct to calculate the energy for my solar panel? or there is some other method?


3 Answers 3


First, the annual solar irradiance is calculated by taking the power incident on a square meter when perpendicular to the sun, and multiplying by 24 hours and then by 365 days. Since the irradiance is typically about 1.366 kW/sq m, when you multiply it out you get 11.9 kW-hr/sq m, which is your number.

But you might have noticed a problem. This assumes the sun is shining 24 hours per day, which is hardly proper.

Second, your second calculation also assumes 24 hours per day sunlight, which is not right. Since the average length of day is about 12 hours, your 25.6 Wh should be divided by 2, giving 12.8 Wh per day.

Finally, this number tells you how much power is falling on the solar cell, not how much the cell puts out. It also assumes the cell will be perpendicular to the sun at all times, and that the atmosphere does not attenuate sunlight. The first may be true, the second is not. Consider that the sun is much dimmer at sunset than at noon.

Let's take the case of noon sunlight. Assuming the air is very clear and you are getting the nominal irradiance, the power falling on the cell will be 1366 w/sq m x .0078, or about 10 watts. Since you have a solar cell rated for 1 watt, this says that you can assume the cell has an efficiency of about 10%, which is about right.

For a 1-watt solar cell which is fixed in position, and is perpendicular to the sun at noon, you can figure on a total output of 4-6 Wh on a clear day. On the one hand, as the sun moves away from the noon position, the effective area decreases (reaching zero at sunrise and sunset when the cell is edge-on to the sun). Also, sunlight gets attenuated by the atmosphere at lower angles, and finally the cells themselves will typically become less efficient at lower intensities.

  • \$\begingroup\$ thanks for your answer but I think there is some mistake in what you mention, 1.366 KW/sq m should be 1.366 W/sq m, isn't it? which mean also the calculation is wrong for the total falling power in 0.007 sq m solar panel, it should be approx 10 mW not W. \$\endgroup\$ Commented Jul 11, 2014 at 12:05
  • \$\begingroup\$ Nope, 1.366 kW it is. Note that that's a decimal point, not a comma. And if your number were right, and your panel puts out 10 mW, you could not get 1 Wh out of it. 1 Wh means an average of 1 watt each second for 1 hour. \$\endgroup\$ Commented Jul 11, 2014 at 15:36

Solar panel are usually rated using the STC method which gives a value at an irradiation of 1kW/m2, which already account for efficiency. It does not matter of how big is the panel, as the value given is in irradiation per 1m2, even if the panel is smaller or bigger.

If you know the area and the STC value (1W) you can calculate the efficiency right away which is 1W / (1'000W/m2 * 0.0078m2) = 12%, which is pretty typical for a c-Si panel.

Now the Irradiation value given Annual solar irradiance = 1200 KWh/square meter is also likely to be expressed in total irradiation for a particular area and should account for weather, time of the day and all the parameters, but this may depend to your source.

Since the STC value is given at 1'000W/m2 then it becomes quite simple 1W * (1'200'000Wh/m2 / 1'000W/m2) = 1200Wh maximal theoretical output.

Then you need to account for other factor like tilt angle, angle of light incidence, dust, spectral mismatch, panel glass and cell diffusive and direct response which makes the calculation fairly complex.

You can use software to perform calculation that like PVsyst


They're both incorrect.

The value of 2.74 you have derived is not in units of hours, it's in kWh/m2/day, so it is the average irradiance per day. Multiply it by the area to get the average input energy. You then need to factor in the efficiency of the panel, which is probably below 20%, to get the output energy. The 1W rating is just the limit of the rate you can supply energy from the panel's output to the load.

  • \$\begingroup\$ this means my input energy is 2.74 KWh*0.007 sq m = 19.18Wh, which means the efficiency = (19.18/1*100)= 5.2% \$\endgroup\$ Commented Jul 11, 2014 at 12:23
  • \$\begingroup\$ please forget about me previous comment it is wrong and couldn't edit in time.I meat this, "this means my input energy is 2.74 KWh*0.007 sq m = 19.18Wh, and if I assume the efficiency of the panel to be 15% = 0.15 then the total output energy= 0.15*19.18 Wh =2.87Wh" \$\endgroup\$ Commented Jul 11, 2014 at 12:31
  • \$\begingroup\$ You have to get the efficiency from the datasheet for the panel you are using. Then you can get the output energy. For example, if the efficiency is 0.1 (10%), then the average output energy per day will be 0.1 * 2.74 * 0.0078 kWh. You only use the rating value of 1W for to check that your load requires less than this (otherwise the panel is no good for it). You would use the power requirement of the actual load to find out how long the load will operate for (in average hours per day). These are only averages though. \$\endgroup\$
    – user49118
    Commented Jul 11, 2014 at 12:36
  • \$\begingroup\$ Yes your second comment is correct (some rounding error though from the area value used). \$\endgroup\$
    – user49118
    Commented Jul 11, 2014 at 12:37

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