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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

PV1x_2017_2.5.4_Continuity_and_Ambipolar_Transport-video.mp4