Imagine that you are in the lab and you can decide the thickness of the Si layer of your solar cell. You want to optimize the solar cell performance for a wavelength of λ=1000nm, for which the absorption coefficient is α(1000nm)=102cm−1.
Which of the following thicknesses dSi would give a better performance? Take into account that you already know two things:
(1) Beer-Lambert's law.
(2) For silicon, the minority carrier diffusivity is around D=27cm2/s and the minority carrier lifetime is around τ=15μs.
a) 100μm
b) 180μm
c) 300μm
180
Using Beer-Lambert's law we can calculate the intensity of the light after passing through the silicon layer:
II0=e−α(1000nm)dSi
For dSi=100μm, we have II0=0.37, so 63% of the light is absorbed in the silicon layer.
For dSi=180μm, we have II0=0.17, so 83% of the light is absorbed in the silicon layer.
For dSi=300μm, we have II0=0.05, so 95% of the light is absorbed in the silicon layer.
It would seem that dSi=300μm is the best option. However, we can calculate the diffusion length as
Ld=Dτ−−−√=27∗15∗10−6−−−−−−−−−−−√=0.02cm=200μm
If we choose dSi=300μm, we would have a silicon layer much thicker than the diffusion length, hence an inefficient collection of the charge carriers. Therefore the best option would be dSi=180μm, which achieves a slighlty lower absorption but better collection, since it is smaller than the diffusion length.
Well, if I were in the lab, I would definitely go for the thickest option because everybody knows that bigger is better! So, my choice would be c) 300μm. I mean, who needs a thin layer of silicon when you can have a chunky one, right? Plus, with a thickness like that, you'll be able to see your reflection in the solar cell - it's like a mirror and a power source all in one!
To determine which thickness of the Si layer would give a better performance for a wavelength of λ=1000nm, we need to consider the absorption of light in the material.
According to Beer-Lambert's law, the absorption of light in a material is related to the thickness of the material and the absorption coefficient. The law is given by the equation:
I = I0 * e^(-α * d)
where:
I0 is the initial intensity of the light,
I is the intensity of the light after passing through the material,
α is the absorption coefficient of the material, and
d is the thickness of the material.
In this case, we know that the absorption coefficient α at a wavelength of 1000nm is 102cm−1.
To optimize the solar cell performance, we want to minimize the amount of light that gets absorbed by the material. This means that we want the value of e^(-α * d) to be as close to 1 as possible.
Let's calculate the values of e^(-α * d) for the given thicknesses and see which one is closer to 1:
a) For dSi = 100μm = 0.1cm:
e^(-α * d) = e^(-102 * 0.1) = e^(-10.2) ≈ 1.539 x 10^(-5)
b) For dSi = 180μm = 0.18cm:
e^(-α * d) = e^(-102 * 0.18) = e^(-18.36) ≈ 1.034 x 10^(-8)
c) For dSi = 300μm = 0.3cm:
e^(-α * d) = e^(-102 * 0.3) = e^(-30.6) ≈ 1.973 x 10^(-14)
Comparing the values, we can see that the value of e^(-α * d) is closest to 1 for option (a) with dSi = 100μm. Therefore, a thickness of 100μm would give better performance in terms of minimizing the absorption of light for a wavelength of λ=1000nm.