# B Solar Cells & Refractive Index

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1. Mar 8, 2016

### Physicist50

Hi there,

I have recently been researching in depth solar cells/panels, the parameters of their efficiency and the photoelectric effect in general, specifically how different wavelengths of light affect it. My research has yielded a few facts; that the efficiency of standard silicon solar cells peaks with light at around 700nm, as higher energy can pass through, and lower energy has, well, lower energy, also that the refractive index of a material can define the wavelength of light as it passes through, and also that around 55% of light that reaches Earth's surface is above 700nm. Based on this and my limited knowledge otherwise, my question is that, if, hypothetically, a layer of transparent material with higher refractive index than the standard covering of solar cells, say flint glass, was placed directly on top of the treated silicon layer that produces the all-important photoelectric effect, so that lower wavelengths of light would be used instead of the natural, then the amount of light that is more abundant (infrared) could potentially be lowered to closer to 700nm, thus increasing the efficiency of said cell.

Sorry for the long explanation, and don't hesitate to point out any points I have incorrectly taken as fact.

2. Mar 8, 2016

### Staff: Mentor

The wavelength increases back when it goes into the cell. Wavelength is not a good parameter here, better look at the frequencies. Solar cells need a minimal frequency, and this does not change by going through different materials.

Flint glass would make reflection at the surface significantly worse, by the way.

3. Mar 8, 2016

### Anand Sivaram

It could be because of the quantum energy of photon which is as per Planck's law hν
That dependents only on frequency, not on refractive index, wavelength etc.

4. Mar 8, 2016

### Physicist50

Thanks for all this, and of course you've made a good point with the reflection. Just a few things though; firstly, it's my understanding that frequency and wavelength are inversely proportional, so if someone could explain the difference as parameters for solar cell efficiency between them that would be great. Secondly, and I am in no way qualified to be quoted on this, but I've been taught that the refraction of light is corrected at the point where the medium it's travelling through reverts to the original, so if there was hypothetically no gap at all between the transparent and treated silicon layers, would the light be able to produce the photoelectric effect with the same lower wavelength, or would the refraction somehow still be corrected?

5. Mar 8, 2016

### Anand Sivaram

Just answer your question in a different way. Water has a refractive index of 1.33. As wavelength changes with refractive index, the visible light wavelength get changed to 400/1.33 - 700/1.33, that is 300nm to 525nm inside water. So, a person diving underwater should see objects with this modified wavelengths,meaning, 700nm red would become 525nm green. But, that is not happening because at the time light reaches the cone cells inside eye, it would have its original wavelength.

Also the wavelength we specify for visible light is the vacuum based wavelength.

6. Mar 9, 2016

### Staff: Mentor

If you keep the medium the same, yes. But not if you change the refractive index.
$$\lambda f = \frac{c}{n}$$
Increase n, and the wavelength goes down while the frequency (and therefore the energy per photon - which is the relevant quantity) stays the same.

7. Mar 9, 2016

### Physicist50

Ah, I understand more now thank you. So that begs another question only vaguely answered by Google. Is there a practical way to decrease the energy (frequency, not wavelength as I recently found out) of photons? But also, is frequency still not decreased in the scenario of flint glass on solar cells, as it is still entering another medium, regardless of whether or not it is refracted (even though in this case it is)?

8. Mar 9, 2016

### Staff: Mentor

It is not.

There are materials that can absorb light of some frequency and re-emit it as light with a lower frequency. While that has some applications, you don't want it in solar cells - you lose energy.
The opposite direction is possible in some rare circumstances, but with other issues (and you lose a lot of intensity - you cannot violate energy conservation).

9. Mar 9, 2016

### Physicist50

Alright... but my research implies that I would need to lower the energy of photons that make up most of the atmosphere, thus increasing the amount of light received with the perfect amount of energy, so is decreasing the energy not a good thing in that case? Assuming, of course, that said material was transparent (which I am unsure of). Then also, what are these materials, and makes them take energy from photons that isn't density or refractive index?

10. Mar 9, 2016

### Staff: Mentor

Photons don't make up the atmosphere.
Reducing the frequency typically looks like that: you start with an energy packet (photon) of energy X, this packet loses some of its energy, and you end up with an energy packet with energy Y<X. You gain nothing, you just lose energy.
All sorts of different materials. The lost energy heats them up.

11. Mar 10, 2016

### Physicist50

Oh right, I meant *photons that are most present in the atmosphere, sorry for any confusion. On another point, yes the photons lose energy, but is that not the intention? So as to lower the most abundant frequency of photon to the optimal energy where all is absorbed without being too energetic to pass through the cell?

12. Mar 10, 2016

### Staff: Mentor

To pass through the cell they would have to be very high-energetic, or the cell has to be designed specifically in order to allow that. In the first case, ignore them because their intensity is negligible. In the second case, you want them to pass through, to get absorbed in a different layer that can use their higher energy.

13. Mar 10, 2016

### Physicist50

Fair enough, although to my knowledge few materials that conduct the photoelectric effect can utilise more energetic photons. My research also told me that the ideal wavelength for converting to electricity was 700nm, any higher and some already start to pass through, and at around 1100nm they all pass through, so it seems to make sense to lower the energy of the more abundant photons to make them applicable to the cell.

14. Mar 10, 2016

### Vagn

Try looking at tandem solar cells and multi-junction photovoltaics. These are devices consisting of multiple absorber materials with different band gaps so that they are more efficient at absorbing the solar spectrum.
Also, strictly solar cells do not use the photoelectric effect, but the photovoltaic effect, as the electrons are not liberated from the material.

15. Mar 10, 2016

### nasu

The 1100 nm photons have LESS energy than the 700nm ones. They don't pass through because they are "more energetic".

16. Mar 10, 2016

### Physicist50

Oh yeah, my mistake, sorry for any confusion, and thanks for the suggestion of tandem cells. I'm still wondering about my original problem though, would it work to increase the compliability of photons with the solar cell by lowering their wavelength (and not the energy as I just learned)?

17. Mar 11, 2016

### Vagn

Such devices where the photons are upconverted from the NIR to visible have been considered, as in this paper. From that paper, it seems that the main issue is that the materials typically used for upconversion are organic, and aren't particularly stable.