English Subtitles for PV1x_2017_3.2.1_Continuity_and_Ambipolar_Transport-video

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It’s hard to keep track of charge in a solar cell.

Generation, recombination, and spatial movement are all occurring at the same time. How can we keep track of all these processes? Well, it all comes down to one equation: the ambipolar transport equation, which we will discuss in detail in this video. The ambipolar transport equation in fact the continuity equation for charge carriers in semiconductors.

Positive as well as negative charge carriers carry the current in a semiconductor, hence the name ambipolar transport. In this video we will first visualize the one-dimensional ambipolar transport equation, then we will apply it to an example to improve our understanding, and finally we will expand the equation to three dimensions. In this slide a volume element is shown of a semiconductor, with sides of dx, dy, and

dz. The transport equation is in essence a way of bookkeeping of charge carriers. In our daily lives many of us are familiar with bookkeeping, so let’s first see how that works by looking at a bank account. The bank account itself can be seen as the volume element. Money coming in such as salary, and going out as expenses.

This influx and outflux of money is the bank account equivalent of charge carrier flowing in and out of the volume element. If you have managed to have a good bank account, you may get some interest on your balance, in other words, money is generated. This interest rate can be compared to the generation in a semiconductor. And finally, the bank charges costs for using the account.

The fact that you have a bank account costs money and this resembles recombination in the volume element. Now let’s turn to the ambipolar transport equation. In this lecture we will not derive this equation mathematically, but we will conceptualize using the volume element shown earlier. The ambipolar transport equation is given by the following relation.

This equation relates all possible charge transport processes in the volume element to the time rate change of the carrier concentration. Charge is flowing in and out of this volume element by drift and diffusion. Inside this volume element charge is generated. And charge is also recombining. This rather complex formula is derived in the solar energy book, and for more information

I refer you to chapter 6.5 of the book. The same derivation can be applied for holes, leading to more or less the same equation, except that the drift term is now negative. This is because of the opposite effect the electric field has on positively charged holes, compared to negatively charged electrons. We will now look at an example.

Imagine a slab of n-type material of length ‘L’. We illuminate the material with a uniform illumination, but shade half of it. This results in a uniform generation in the left-hand half of the material, and no generation in the right half. For the sake of simplicity we assume an infinite carrier lifetime, which leads to no recombination. Of course it is not realistic to assume the lifetime to be infinite, but it will allow

us to understand the application of the ambipolar transport equation without delving in detailed math. For this example we also assume that the electric field strength is zero. And finally, we take the excess carrier concentration to be equal to zero at the boundaries of the slab, which is often the situation at the interface between a metal and semiconductor. We start with the illuminated part.

Since there is no electric field, no recombination, and we consider a static or steady-state situation, we can remove several terms from the ambipolar transport equation, reducing this to the following differential equation. When we solve this differential equation, we obtain the following equation, where ‘C_1’ and ‘C_2’ are integration constants. We do the same for the shaded part.

First removing several terms from the ambipolar transport equation, then solving the differential equation, leading to the following equation. Here ‘C_3’ and ‘C_4’ are integration constants also. We now have obtained two equations with four unknown constants. To solve this we need four boundary conditions. It it given that the excess carrier concentration is equal to zero at the edges of the material,

and thus results in two boundary conditions. But an important step is to consider the boundary between the illuminated and the shaded part. At this boundary the carrier concentration needs to be continuous. This means that the excess carrier concentration left of the boundary needs to be equal to the carrier concentration right of the boundary. And because of continuity, the space derivative of the carrier concentration on both sides

of the boundary need to be the same as well. Now we have four boundary conditions, and we can solve the two equations. This is left as an exercise for the viewer, pause the video now to try to work it out yourself as I will show you the final solution after this short pause . Now that we have the final solution, we can analyse it further by plotting it, resulting in the following graph.

We see that at the edges the excess carrier concentration is indeed zero. And at the interface between illumination and shading, we see that both the value and slope of the carrier concentration are continuous. The carrier concentration peaks in the illuminated part, which is as expected since this is the part where charge is being generated. It also shows a parabolic behaviour.

In the shaded part, only diffusion is present, resulting in a linear behaviour. Now that we have gone through an example in one dimension, we can expand the ambipolar transport equations to three dimensions. Let’s start with the equation for electrons. The only terms which are related to the dimensions of the control volume are the diffusion and drift terms, which show a dependence on ‘x’.

When we expand these terms to three dimensions, we obtain the following relation. It is the cross product of the Nabla operator and the current density vector ‘J_n’, divided by the charge ‘q’. For holes a similar equation is obtained. The only difference is the charge, which results again in a minus sign. Now to summarize, we have visualized the one-dimensional ambipolar transport equation, applied it to

an example of an n-type material, and finally expanded the equation to three dimensions. With this equation at our disposal, we can tackle many semiconductor physics problems, and design better solar cells.

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