3
$\begingroup$

Since, on average, 1/3 of all sunlight that hits a solar panel is reflected, wouldn't it be possible to set up some kind of etalon system where the reflected light is then absorbed by another solar panel, and then the reflected light from that panel is further absorbed by the first panel, and so forth?

So basically, the amount of sunlight present in the etalon would decrease by 2/3 each time it hits one of the two panels.

Is this feasible and practical?

Improve this question
$\endgroup$
2
  • $\begingroup$ Where did that 1/3 come from? I can only find an estimate of 20%, in: Boden, S.A., Bagnall, D.M., 2006. Bio-mimetic subwavelength surface for near-zero reflection sunrise to sunset. In: Proceedings of the FourthWorld Conference on Photovoltaic Energy Conversion, Hawaii. as cited in DOI:10.1016/j.enpol.2008.09.070 $\endgroup$ – 410 gone Aug 15 '13 at 13:37
  • $\begingroup$ can you explain etalon or - better - provide a sketch? $\endgroup$ – mart Aug 20 '13 at 12:18
4
$\begingroup$

Boden & Bagnall (pdf) looked at this question of reflection, in their paper Bio-Mimetic Subwavelength Surfaces for Near-Zero Reflection Sunrise to Sunset.

Their estimate of the proportion of photons reflected, from sunrise to sunset, in a fixed PV system is 20%, rather than your estimate of one-third.

The proposed etalon is not an efficient or effective means of capturing the energy in reflected photons.

However, there are not-dissimilar ideas being researched at the moment, including a lot of work in plasmonics.

Boden & Bagnall propose a moth-eye style coating. Their simulations predict an increase of 12% in energy output, compared to single-layer anti-reflective coatings:

enter image description here

(source: paper as linked above)

However, PV panels are now largely a commodity product that compete on price (except for some specialist applications, e.g. satellites) - so it's generally cheaper to buy more panels, than to pay for fancy coatings on fewer panels.

Improve this answer
$\endgroup$
2
$\begingroup$

The trouble with an etalon is getting the light into it in the first place. Bearing in mind we want normal inidence to maximise the light intensity our solar panel looks something like:

Panel1

But solar panels aren't transparent (obviously since they reflect two thirds of the incident light) so if you wanted to use a second panel you'd need something like:

Panel2

It's true you've nearly doubled the absorption, from 33% to 56%, but you could have just put the two panels side by side and doubled the absorption. Plus you need the light to come in at an angle and that reduces the intensity and therefore the electricity generated.

It would be nice if you could get the light to bounce repeatedly between the same two panels:

Panel3

But you wouldn't be able to get the light into the gap between them.

Improve this answer
$\endgroup$
4
  • $\begingroup$ What if you curve the solar panels? Wouldn't you be able to set up a situation where the light could repeatedly bounce between the same two panels then? $\endgroup$ – John Roberts Jul 16 '13 at 17:57
  • $\begingroup$ You could form your solar panel into a sphere, then make a hole in the sphere and shine the light in that way. The trouble is that the area of the hole is going to be a lot less than the area of the plate you used to make the sphere. So you'd be better off just unrolling the sphere and putting up with the 33% efficiency. Unless you can come up with some geometry that makes the area of your curved plate less than three times the area of the incident light you'll always lose. I'm not sure I'd swear it's impossible, but the fact no-one has done it is significant! $\endgroup$ – John Rennie Jul 16 '13 at 18:22
  • $\begingroup$ "since they reflect two thirds of the incident light" - [citation needed] (the question claims one-third) $\endgroup$ – 410 gone Aug 15 '13 at 13:20
  • $\begingroup$ Oops, that's a mistake - how I managed to turn absorbs two thirds into reflects two thirds I don't know. $\endgroup$ – John Rennie Aug 15 '13 at 15:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.