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Apparently, according to EarthScience.SE, the measurement of how "bright" a given day is, is measured in units of kWh/m^2, known simply as "solar radiation". Apparently, 3 kWh/m^2 is the average brightness of an American summer day (useless fact).

This solar panel claims an output of 6V at 330mA. Obviously, solar panels will not perform the same on cloudy, darker days.

What I'm looking to do is to put together an equation/algorithm that will take solar power (in Watts) and solar radiation ("brightness"; in kWh/m^2) as input, and tell me what the adjusted power output is for that panel, based on the current brightness.

As an example, using that particular solar panel and a given point in time when, say, there is only 2 kWh/m^2 of solar radiation in the given vicinity, then the equation might yield something like this:

// Example only
AdjustedPowerOutput(normalOutput, solarRadiation) = normalOutput * (solarRadition / 3) = 1.98 Watts * (2 kWh/m^2 / 3 kWh/m^2) = 1.98 Watts * .667 = 1.32 Watts

So, if this equation was correct, at that point in time in the given day, the panel will only output 1.32 Watts.

How can this "adjusted power output" actually be calculated?

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    \$\begingroup\$ Can you explain what your end goal is? This doesn't really seem to be an electronics design question. In any event, the maximum available output power for a given panel mainly depends on two things. First, how much light is striking the panel (pointing angle effects this as well as brightness of the sun) and panel temperature (lower means more power). If the temperature is held constant, the maximum available power output is fairly linear with incident light level. You can probably find this information in the datasheet for a quality panel. \$\endgroup\$
    – user57037
    Commented Jun 12, 2015 at 19:20
  • \$\begingroup\$ The reason pointing angle matters is because it affects how many photons actually hit the panel. So if the sun is 90 degrees to the panel surface, that is the max. As the panel is rotated toward 0 degrees, the output will drop toward zero. So, basically, you take the solar brightness, multiply by sine of the angle, correct for temperature, and that will be a pretty good approximation. \$\endgroup\$
    – user57037
    Commented Jun 12, 2015 at 19:34
  • \$\begingroup\$ Thanks @mkeith (+1) - In my mind this is an electronics design question because I'm attempting to figure out a way to calculate a solar panel's output as a function of the ambient surroundings. I was only thinking in terms of light/brightness (this "solar radiation" measurement). According to you, panel temperature also plays a role. So at the end of the day I'm looking for a way to say "Given solar radiation S and panel temperature T, a solar panel with max output power P will actually only provide P'." \$\endgroup\$
    – smeeb
    Commented Jun 12, 2015 at 19:34
  • \$\begingroup\$ And pointing angle. ;-) \$\endgroup\$
    – user57037
    Commented Jun 12, 2015 at 19:36

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Normally Solar panels are rated with the power output when irradiated with 1kW/m^2 - this is close to the max solar irradiance at noon on a clear day. You seem to be mixing up energy (kWh) with power (kW).

The current output of a PV cell will be fairly linear with the intensity of solar radiation. The voltage will reduce with temperature but if you keep the cell cool the power will then be proportional to the intensity of radiation.

This assumes that the sunlight falls on the cell at 90 degrees - as the sun moves across the sky that cannot be maintained unless you have a tracking PV array. If not there will be a Cosine correction that will need to be applied.

The total amount of energy you get per day from a cell will depend upon how long the light is present i.e. how long the day is which will vary with latitude and season and how clear the air is (e.g are there clouds).

The link you gave has a map showing the integrated average energy per day for various locations.

This graph shows how the current output of a PV cell varies with light intensity and voltage.

enter image description here

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  • \$\begingroup\$ The other thing is that the max power only occurs at one place on the curve. So if you want to get maximum power, you have to have a DC-DC converter or charge controller that tracks the maximum power point. Google MPPT. \$\endgroup\$
    – user57037
    Commented Jun 12, 2015 at 19:40

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