I'm studying crystalline solar cell and their working principle, but I still don't completely understand the process.

First of all, I understand that we dope a pure crystalline silicon with boron and phosphorus, for example, in order to create a layer where the hole concentration is larger (p-type) and a layer where the electron concentration is larger (n-type). Then, when a net interface between a p-type and a n-type material is created we create a p-n junction, the building block of the basic solar cell. After this interface has been created some holes will move from the p-type to the n-type, while electron will move in the opposite direction, to join the electron and the hole.

So, at the interface the material loses its electronic neutrality and becomes charged. This results in an electric field called depletion layer.

So far it's all pretty much clear, but now:

When a photon of suitable energy (higher than the semiconductor band-gap, which isn't associated at all with the the built-in potential difference, right?) hits the solar cell active layer, an electron is promoted to the conduction band and leaving a hole in the valence band. Now they don't recombines because of the electric filed at p-n auction, so the electron will move to the n-type part. Now, if we connect a wire between the n-type side and the p-type side the electron will flow through it generating current.

My question is: is the electron moving from the n-type to the p-type? Therefore, is the n-type kind of anode where an oxidation is occurring (an electron is leaving to go to the p-type)? What force is making the flow occurs? I don't really get why the electron should return to the initial place where there is an higher voltage potential.

Hope that my doubts are clear. I'm looking forward for your responses!


2 Answers 2


I don't really get why the electron should return to the initial place where there is an higher voltage potential.

A high potential region will generally repel positively charged particles.

An electron has a negative charge.

Therefore it's acted on the opposite way. It is attracted towards regions with high potential.

is the electron moving from the n-type to the p-type?

Yes. In the depletion region in the n material, the free carriers (electrons) have been swept away, leaving the fixed charges behind (positively charged nuclear sites). Similarly, the depletion region in the p material has a negative charge.

So a free electron in the depletion region will be repelled by the p side of the depletion region and attracted to the n side of the depletion region (which has positive charge)


is the electron moving from the n-type to the p-type?

No, the other way around.

The bulk of the cell is p-type. The n-type is a very thin layer on top. As thin as it can be, because you want the photons to go through it into the p-type below.

When the photoexcitation occurs in the p-side, the electron is in a space filled with holes it could (and will happily) recombine with. However, it's being pulled towards the n-type surface because of the field of the pn-junction.

So it keeps moving until it reaches the the interface, where it finds itself suddenly cramming into all the extra electrons up in the n-type, creating more negative charge. Meanwhile, the hole it left behind is sucking up electrons from the back of the cell, creating a positive charge.

So now you have too much negative on top, and too much positive on the bottom, and they really really want to relieve themselves of that pressure, and will happily power your home for you if you'd be so nice to connect a wire between the two sides of the cell.

For all of this to work well you want to have lots of charge carriers and fast moving electrons. If you don't, as you increase the number of photoexcitations, the chance they'll bump into a left-behind hole goes up. That's why silicon caps out production not much above 1 sun of insolation, and why GaAs cells are used for high-performance and concentrated sunlight systems.


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