Solar Energy
# PV: Engineering and Advanced Concepts

As we learned over the last few quizzes, we can use anti-reflective coatings and design front contacts to maximize the chance that an incident photon makes it into the PV material, but we also want to maximize the chance that photons in the material are absorbed.

This is especially true of indirect bandgap materials like silicon, which tend to be worse at absorbing photons than direct gap materials.

Increasing cell thickness will increase absorptance, but as we learned in the previous quiz, we’d prefer to keep the cell as thin as possible. Fortunately, there are strategies for increasing the photon path length without increasing cell thickness.

If we make the back contact of the cell reflective, what is the probability (in percent) that a photon traveling through our original \(\SI{200}{\micro\meter}\) thick layer of silicon will be absorbed?

In the previous quiz where the Fresnel equations were introduced, we only considered reflection for normal incidence, but in practice photons will intercept interfaces at many different incident angles. When light transmits through an interface, the incident angle and the transmitted angle can be different.

The relationship between the incident and transmitted angles depend on the indices of refraction of the two materials, and is given by Snell's law: \[n_i \sin{\theta_i} = n_t \sin{\theta_t}\] However, something strange happens for light traveling from a material with a high index of refraction to one with a low index of refraction for large incident angles. For example, consider the case where \(n_i = 2\), \(n_t = 1\), and \(\theta_i = \pi/4\). If we try to solve for the transmitted angle, we get \(\sin{\theta_t} = \sqrt{2} \approx 1.414\), which is problematic because the sine function has a maximum value of 1. Since a transmitted angle that would satisfy Snell's law is impossible, no light is transmitted and there is a 100% chance for reflection - this is called "total internal reflection".

For light going from a high to low index of refraction, there is a critical angle \(\theta_c\) above which all light will be totally internally reflected. If silicon has an index of refraction \(n_{Si} = 3.6\), what is the critical angle (in radians) between silicon and air?

Photons reflected back towards the top of the PV cell at angles greater than the critical angle must be reflected back into the cell - they are trapped. This gives us a much better chance of absorption. Therefore, we don’t want our back contact to have specular (mirror-like) reflection - any light that makes it into the silicon from air will have a transmitted angle less than the critical angle, so if it specularly reflects off the back contact it will reflect back to the silicon/air interface at an incidence angle less than the critical angle.

We'd rather have photons reaching the back contact reflect into large incidence angles. Unfortunately, there's no optical design that allows us to force light from all incidence angles to reflect back at large incidence angles. What we can have is a diffuse back reflector. Diffuse reflection (as opposed to mirror-like specular reflection) means that photons are reflected in random directions.

One particular case of diffuse reflection is called Lambertian reflection. In Lambertian reflection, the reflected intensity from a surface is proportional to the cosine of the reflected angle \(\theta\). This means that if you shine light on a particular point of a Lambertian reflector and divided the hemisphere surrounding that point into strips based on the reflected angle, the number of photons intercepted by each strip divided by that strip's area would be proportional to \(\cos\theta\): \[I = \frac{N_{photons}}{A} \propto \cos\theta\]

Note: the next few panes use ideas from probability and continuous random variables that are not covered elsewhere in the solar energy course. If you are unfamiliar with these concepts, it may be worth going through this quiz from the probability course.

It would be useful to know how much a Lambertian back reflector will increase the number of times a photon will travel through the silicon layer before leaving. To figure out the probability for light reflected from a Lambertian back reflector to be totally internally reflected when it reaches the silicon/air interface, we first need to know the probability density function (PDF) of reflected angles from a Lambertian surface.

The reflected angle can vary from \(0\) (normal to the surface) to \(\pi/2\) (tangent to the surface), so the PDF \(f_X\) is a function that gives the probability of a reflected angle lying between two values: \[P (\theta_1 \leq X \leq \theta_2) = \int_{\theta_1}^{\theta_2} f_X(\theta) d\theta\] \[P (0 \leq X \leq \pi/2) = 1 = \int_0^{\pi/2} f_X(\theta) d\theta\]

We know that for a Lambertian surface, the reflected intensity is proportional to the cosine of the reflection angle: \[\frac{N_{photons}}{A} = \cos\theta\] and that the number of photons reflected at the reflection angle \(\theta\) is proportional to the PDF: \(N_{photons}(\theta) \propto f_X(\theta)\).

Which expression gives the probability density function of reflected angles from a Lambertian surface?

What we'd really like to know is the probability that the reflected angle from a Lambertian reflector will be less than the critical angle. We can figure out the probability of the reflected angle being less than a particular value from the cumulative distribution function (CDF) \(F_X\), which is defined by: \[F_X(x) = P(X \leq x)\] For a distribution given by the probability density function \(f_X\), the probability that a value lies between \(a\) and \(b\) can be found by integrating \(f_X\) from \(a\) to \(b\): \[P(a \leq X \leq b) = \int_a^b f_X(x) dx\] So the corresponding probability using the CDF can be found from: \[P(a \leq X \leq b) = F_X(b) - F_X(a)\] and the CDF can be found by integrating the probability density function: \[F_X(a) = \int_{-\infty}^a f_X(x) dx\]

What is the expression for the probability that light reflected from a Lambertian reflector will have a reflected angle \(\leq \theta\)? In other words, what is the CDF of reflected angles from a Lambertian surface?

For a given material, we now have the probability that a photon will remain trapped in the cell after reflecting from a Lambertian surface. However each time the photon is trapped, it reflects back to the Lambertian surface, reflects at a different angle, and has a new opportunity to escape the cell. If the probability that a particular reflection will keep the the photon trapped is given by \(p\), then the probability that the photon will be trapped at least twice is given by \(p^2\), the probability that the photon will be trapped at least three times is given by \(p^3\) and so on.

We can calculate the expected number of passes the photon will have through the cell \(N_{passes}\) by adding the probabilities of the photon being trapped numerous times. We know the photon gets at least one pass, and then there is a \(p\) chance it will get trapped for a second pass, a \(p^2\) chance it will get trapped for a third pass, and so on: \[N_{passes} = 1 + p + p^2 + p^3 + ... \]

This gives us a geometric sum, or a polylogarithm. The sum of this polylogarithm is given by: \[N_{passes} = \sum\limits_{i=0}^{\infty} p^i = \frac{1}{1 - p} \]

From the previous problem, we found that the probability for a photon to be trapped after reflecting from a Lambertian surface is given by \(p = 1 - (1/n)^2\). Plugging this into the equation above yields: \[N_{passes} = \frac{1}{1 - (1 - (1/n)^2)} = \frac{1}{(1/n)^2} = n^2\]

Thus, by using a Lambertian back reflector and light trapping, we can cause photons to take \(n^2\) passes through the cell on average, greatly increasing the effective path length! For silicon, this corresponds to approximately 13 passes.

The geometric sum tells us that on average, photons will take \(n^2\) passes through the cell, which is already a significant increase in the effective path length that a photon takes through the cell. However, the effective path length is improved by over \(n^2\) times compared to the thickness of the cell. For one, each pass travels through the thickness of the cell twice, which doubles the effective path length. Additionally, many reflections will be at large angles, which increases the effective path length further.

What is the improvement factor beyond \(n^2\) that light trapping applies to the effective path length? In other words, if a cell of thickness \(t\) that makes use of light trapping achieves an average effective path length of \(cn^2t\), find \(c\).

Light trapping with a perfect Lambertian reflector leads to an overall enhancement factor of \(4n^2\), which means that the effective path length photons take through the material on average are \(4n^2\) times greater than the actual thickness of the cell. This \(4n^2\) enhancement factor is called the light trapping limit, or Yablonovitch limit. The physicist Eli Yablonovitch proved that this is the maximum enhancement possible with light trapping through using the detailed balance principle.

With light trapping, the chance to absorb photons even very close to silicon's band gap energy can approach 100% for a \(\SI{200}{\micro\meter}\) thick cell, and many silicon PV cells make use of light trapping to improve their absorptance. With light trapping, even cells that are only 10s of micrometers thick can absorb effectively, as can be seen in the plot below:

However, it is still difficult to make silicon cells this thin due to other fabrication challenges. With further advances, however, light trapping should allow us to make thin-film cells, even out of indirect band gap materials like silicon.

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