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Solar Ray depiction

Consider I have a solar panel setup as in the above picture. The top sketch is sketching sun rays in the early morning around 6 AM in a clear bright sky. The bottom sketch is sketching the sun ray hitting the solar panel at around 8 AM.

I installed a solar panel on a mountain, where we usually see the sun more than 12 hours a day. I found that even if I turn the solar panel to face the early morning rays perpendicularly, it doesn't produce much electricity. But when it reaches around 8 AM with the solar panel to perpendicularly facing the sun, the electricity it produces will increase to almost maximum performance, not very different to the productivity at noon. Quantitatively, this is typically an increase in voltage from around 5V to 16V, of 18V maximum. In this case, it is the same sun, the same place, the same solar panel, and the same condition (it faces perpendicular to the sunray, as in the picture). So what is different in the early mornings here so that the output is so different?

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When the sun is near the horizon, the sun rays have to travel through more air to get to you than when it's directly overhead. This phenomenon is known as atmospheric extinction, and this page has a nice cartoon diagram to illustrate it:

enter image description here

This effect can be quite large. This page has a graph of the approximate effect as a function of zenith angle, which is the angle from directly overhead; a zenith angle of 90° is therefore an object on the horizon.

enter image description here

This graph is logarithmic; every unit on the vertical axis corresponds to a factor of about 2.5 times less energy getting to the ground. We can see that anything closer than about 15° to the horizon will have its light diminished substantially, certainly enough that you would notice on a power meter.

In addition, sunlight near the horizon is reddened substantially; blue light is more likely to be scattered by molecules in the air, while red light is more likely to travel straight through the atmosphere to your solar panel. (This is why sunsets are red & orange compared to the light you see during the day.) Roughly speaking, your solar panels take particles of light (photons) and turn them into electrical energy. But because of the properties of solar cells, there is a minimum energy that a photon must have (called the band gap) to excite any electricity at all. Redder photons have less energy than bluer photons, so the red photons that are prevalent in sunlight from near the horizon may not be able to generate any electricity.* In other words, there's a double whammy: less photons are getting to your solar panel, and the ones that do get to your solar panel have less energy on average and can't generate electricity.

*I say "may not" because solar panel engineers are clever folks, and there are some more modern types of solar panel that are better at converting sunlight into electricity over a wide range of photon energies. Look up "multi-junction cells" for more information.

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    $\begingroup$ Quote: "Sunlight near the horizon is reddened substantially; blue light is more likely to be scattered by molecules in the air, while red light is more likely to travel straight through the atmosphere to your solar panel." I think this is the answer, as I believe this is about photoelectric as this is solar panel. $\endgroup$
    – AirCraft Lover
    Aug 19, 2021 at 16:35
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    $\begingroup$ @AirCraftLover I wonder how you can be so confident. It's definitely not obvious to me that the greater scattering of blue light has more of an effect than overall atmospheric extinction. In fact, my gut says that the latter effect is more important (bu of course I don't really know either way). $\endgroup$
    – d_b
    Aug 20, 2021 at 0:54
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    $\begingroup$ @AirCraftLover: If you provide more details about the manufacturer & model of your solar panel, I might be able to ascertain what the band gap is and how much of an effect the reddening might have. (No guarantees, though.) $\endgroup$
    – Michael Seifert
    Aug 20, 2021 at 2:23
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    $\begingroup$ @AirCraftLover The function for atmospheric extinction includes 3 factors, one of which is the "reddening" of light via Raleigh scattering(weight 0.14). The other 2 factors are aerosol scattering(weight 0.12), and molecular absorption(weight 0.02). This last one is effectively the non-transparency factor of clean air. $\endgroup$
    – PcMan
    Aug 20, 2021 at 7:18
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    $\begingroup$ @OscarBravo, yes, the blue and the above frequency like indigo and violet. The higher frequency, the more energy it can created. So, by the absent of the higher frequency rays, will significantly decrease the energy in the solar panel. Here is a good explanation that the blue light is no more significant in the afternoon. $\endgroup$
    – AirCraft Lover
    Aug 20, 2021 at 13:09
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If I understand your drawing, at all measurement times the sun is fully abvoe the horizon and the solar panel is always perpendicular to the line-of-sight to the sun.
In this case, what you are observing is the length of the path thru the atmosphere between "space" and the solar panel. The atmosphere absorbs a decent amount of light per km of travel (especially at lower altitudes), leading to the lower power/m^2 received at the panel. When the sun is close to overhead, this absorption path is much shorter.

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  • $\begingroup$ I edited the picture so I will not be misinterpreted, the solar panel is not remain on top just as like in the first sketch. As I mentioned, it is the same place. Everything are same except the time. Of course both situation are line of sight, it is clear, as I said, in a clear bright sky. As this is photoelectric effect, I believe this is must be analyzed from quantum mechanic. $\endgroup$
    – AirCraft Lover
    Aug 19, 2021 at 16:23
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It is the earth only you should be rotating, not the ray. The reason the amount of power produced in the morning is less than that of mid day, is due to the amount of light per unit area being less in the morning. This you should see by rotating the earth only, the number of ray lines hitting your detector will decrease.

If you're tracking the ray with your solar panel, then the amount of light hitting the panel within the range that it is sensitive to diminishes due to it having to travel through a longer path in the atmosphere.

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    $\begingroup$ I don't think so - he says his panel is tracking, so the rays are always perpendicular to the panel, and assuming the sun is fully above his horizon, the exoatmospheric irradiance "pointing towards" the array is constant. $\endgroup$
    – Carl Witthoft
    Aug 19, 2021 at 15:19
  • $\begingroup$ @CarlWitthoft Ah yes, so it could be that the amount of light in the spectrum that the solar panel is sensitive to is diminished due to the extra amount of atmosphere it needs to travel through? $\endgroup$
    – jamie1989
    Aug 19, 2021 at 15:22
  • $\begingroup$ Quote: "The reason the amount of power produced in the morning is less than that of mid day, is due to the amount of light per unit area being less in the morning." What is the physic explanation of this the amount of light per unit area being less in the morning? $\endgroup$
    – AirCraft Lover
    Aug 19, 2021 at 16:26
  • $\begingroup$ Btw, I corrected the second sketch to avoid wrong interpretation. As I mentioned in my question body, it is the same place. So, the place are same, only different in time. $\endgroup$
    – AirCraft Lover
    Aug 19, 2021 at 16:28

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