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Solar Energy

# Absorptance and cell thickness

Conventional crystalline silicon PV cells are made from wafers that are 100s of micrometers thick. These types of cells are known as “thick-film” cells. Even though 100s of micrometers might seem thin, it is much thicker than some silicon cells that are being developed and cells made from newer PV materials. This quiz will explore why PV cells are designed to have a particular thickness, and look at some advantages we can gain by moving to thinner cells.

When designing a PV cell, how is thickness determined? In general, we want cells to be thin, as it provides cost and performance benefits. A thinner cell takes less PV material, which makes producing the cells cheaper. A thinner cell also means that an excited charge carrier travels less distance to the contacts, which reduces the chance for it to be lost to recombination.

What’s the advantage of making cells thicker?

If an object is thin, light can pass through it without being absorbed. Take paper as an example: a paper lantern covering a light bulb only absorbs a small amount of light, and much of the light is still transmitted through to the room. However if the paper lantern was made with ten layers of paper instead of just one, it would be dark, since those ten layers would absorb most of the light.

The same is true for PV cells. If a PV cell isn’t thick enough, solar photons can pass through the cell without being absorbed. The plot below of silicon absorption depth vs. photon wavelength shows the approximate thickness of silicon required to absorb light of different wavelengths.

The chance that a photon will transmit through a material of thickness $$t$$ with an absorption depth of $$d$$ is given by the Beer-Lambert law: $T = e^{-t/d}$ We can see that the thicker the material, the lower the chance for transmission. This equation takes the form of exponential decay, which makes intuitive sense. Suppose a material absorbs half of the light shining through it over $$\SI{1}{\milli\meter}$$. After each additional millimeter of travel, the light in the material is halved, so while a lot of light is absorbed in the beginning (because there is a lot of light to be absorbed), after a few millimeters there is little light left to be absorbed, so the amount of light absorbed is much less.

Note: the Beer-Lambert law can also be expressed in terms of a material’s absorption coefficient, which is just the inverse of the absorption depth $$a = 1/d$$: $T = e^{-at}$

What is the probability (in percent) that a photon with a wavelength of $$\SI{1050}{\nano\meter}$$ is absorbed before passing through $$\SI{100}{\micro\meter}$$ of silicon? The absorption depth of silicon for $$\SI{1050}{\nano\meter}$$ is $$d = \SI{6e-4}{\meter}$$.

Silicon absorbs shorter wavelength photons ($$< \SI{1000}{\nano\meter}$$) easily and a small thickness is sufficient to absorb all incident sunlight for these wavelengths. However, remembering what we learned in the Shockley-Queisser limit chapter, photons near silicon’s bandgap of $$\SI{1100}{\nano\meter}$$ can be converted to electricity the most efficiently, so we’d like to absorb as many of them as we can. If we want to absorb 95% of photons with a wavelength of $$\SI{1050}{\nano\meter}$$, how thick (in $$\si{\micro\meter}$$) do we need to make the PV cell?

From the last question, we can see why Si PV cells need thicknesses of 100s of micrometers in order to absorb most of the incident sunlight, especially for photons close to the bandgap energy. An important part of what makes c-Si require thicker cells than other materials has to do with the idea of “direct” vs. “indirect” bandgaps. So far we’ve shown energy band diagrams with just energy levels and gaps, but the states that electrons are allowed to occupy contain more information than just the energy of the electron. States that electrons can occupy also have certain allowed electron momenta, so more complex band energy diagrams can show allowable electron states as energy vs. momentum.

Photons have very little momentum, so it is easiest for them to excite an electron to an energy level with the same or similar momentum. Suppose a photon with an energy of $$\SI{2}{\electronvolt}$$ is incident on two materials with the following energy band diagrams:

Which material will absorb the photon more easily?

In materials where the peak of the valence band and the valley of the conduction band are aligned at the same momentum, we call this a “direct-bandgap”. If the peak and valley occur for different momenta, this is an “indirect bandgap”.

In indirect bandgap materials, typically an electron needs a photon (to provide energy) and thermal energy from the lattice (to provide momentum) in order to get promoted to the conduction band, whereas in direct bandgap materials, only a photon is required. In direct bandgap materials, photons are absorbed easily since the photon is all that’s required to excite the electron. Two pieces are required simultaneously for indirect materials, so photons will usually travel further through an indirect bandgap material before being absorbed.

Crystalline silicon (c-Si) is an indirect bandgap material, which is why it needs hundreds of micrometers of thickness in order to absorb photons near the bandgap. The following plot shows the absorption depth as a function of wavelength for c-Si, as well as the absorption depth as a function of wavelength for gallium arsenide (GaAs).

Do you think GaAs is a direct or indirect bandgap material? The bandgap of GaAs corresponds to photons with a wavelength of $$\SI{870}{\nano\meter}$$, so you should focus on comparing absorption of photons with wavelengths $$< \SI{870}{\nano\meter}$$ in GaAs to absorption of photons with wavelengths $$< \SI{1100}{\nano\meter}$$ in c-Si.

Thin-film cells that could be produced with direct bandgap materials like GaAs can use much less material than their thick-film counterparts, which reduces production costs. Another benefit to thin-film cells is that they can be flexible. The flexibility of a device depends on material properties as well as the device geometry.

For a beam in bending, the maximum stress can be given by: $\sigma_{max} = \frac{Et}{2R}$ Where $$E$$ is the Young’s modulus of the material, $$t$$ is the thickness of the beam, and $$R$$ is the radius of curvature of the beam.

[It's not too hard to "derive" this expression. It would probably take one or two extra panes or a short wiki entry. Is that worth doing?]

If crystalline silicon is damaged when subjected to stresses over $$\SI{100}{\mega\pascal}$$, what is the minimum radius of curvature (in meters) that a $$\SI{200}{\micro\meter}$$ thick silicon cell could be bent to without breaking? The Young’s modulus of silicon is $$\SI{150}{\giga\pascal}$$.

Suppose we could reduce the thickness of our silicon PV cell to $$\SI{10}{\micro\meter}$$ while still maintaining reasonable absorption (we’ll look at strategies for achieving this in the next quiz). What is the new minimum radius of curvature (in meters) that the PV cell could be bent to without breaking?

Thinner cells are much more flexible than thick ones, and can be bent further without breaking. If you have an efficient thin-film cell which absorbs effectively either by increasing the absorptance of silicon or using a direct bandgap material, it opens up new possibilities. Thick-film cells needs to be attached to stiff, heavy glass panes so they don’t bend when they are handled and installed. Thin-film cells could be attached to a thin, flexible, clear plastic sheet instead. This might allow for easier and cheaper installation, and could open the potential for new applications.

There are many advantages to thin-film PV cells, but a PV cell must be thick enough to absorb the incident sunlight. Indirect bandgap silicon cells need to be fairly thick, but using direct bandgap materials or improving the absorptance of indirect gap materials can allow for the fabrication of thin-film cells. The following quiz will look at a technique for improving cell absorptance, which could help enable the production of efficient, thin-film cells.

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