I PV cell: understanding electron movement in the conduction band

[matthias31415]

TL;DR Summary

Usually, explanations of the PV effect focus on what happens at the pn-junction and that photons get electrons from the valence to the conduction band before they are attracted towards the positive charge of the n-layer. But what about the rest? What is the driving force that moves the electrons AWAY from that positive charge? Where does the energy come from that creates a current with the negative pole being located next to the positive charge that is supposed to be the driving force?

The summary says it all: the common explanations of the PV effect could easily describe a perpetuum mobile. Just as if a merry-go-round with negatively charged horses could be set in motion by having a stationary positive charge. Somehow, the energy of the photon must be used in order to create the current, but how? If the energy of the photon is merely used to get the electron from the valence to the conduction band, this will not work, as, in the metaphor above, that would correspond to just releasing the break of the merry-go-round.

For the sake of simplicity, I would like to describe the path of a single electron from the place where it was released from the silicium atom through an external consumer back to the atom where it originally came from. I know that reality is more complicated with multiple electrons performing multiple steps, leading to the impression that holes are "moving" in the opposite direction. However, my gut feeling is, that, if there is no working single-electron-model, the more complicated models will fail as well.

 

[Lord Jestocost]

Notions like “perpetuum mobile” make absolutely no sense in the context of the PV effect!

To principally understand how one can transform light into electricity on base of the photovoltaic effect, one can refer to the lecture “How to Transform Light into Electricity“ by Arno Smets (https://ocw.tudelft.nl/course-lectures/transform-light-electricity/ (OpenCourseWare provided by the TU Delft)).

 

[Yuras]

What is the driving force that moves the electrons AWAY from that positive charge?

By "positive charge" you mean the charge of the depletion region on the n-side of the junction? Or the hole created by the photon?
The electric field across the junction pushes the electron and the hole out of the depletion region, creating a potential difference between the n- and p-layers. That potential difference drives electron though a circuit. Does it answer your question?

The summary says it all: the common explanations of the PV effect could easily describe a perpetuum mobile.

It'd help if you provide a link to such an explanation, and describe how perpetuum mobile follows from it in your opinion. That way we'll have a specific subject to discuss instead of guessing what exactly you are referring to.

 

[matthias31415]

By "positive charge" you mean the charge of the depletion region on the n-side of the junction? Or the hole created by the photon?
The electric field across the junction pushes the electron and the hole out of the depletion region, creating a potential difference between the n- and p-layers. That potential difference drives electron though a circuit. Does it answer your question?

Yes, by "positive charge" I meant the charge of the depletion region on the n-side of the junction.

Let me summarize your answer: The electric field across the junction [with the positive pole at the n-side and the negative pole at the p-side] pushes the electron towards the positive pole (n-side) and the hole towards the negative pole (p-side) (...) creating a potential difference between the n- and p-layer.

But: the newly created potential difference between the n- and p-layer, where does it have its positive pole and where does it have its negative pole? You are saying that this potential difference drives the electron through a circuit. So you are saying that its negative pole is now, all of a sudden, at the n-layer and its positive pole at the p-layer?

 

[matthias31415]

Notions like “perpetuum mobile” make absolutely no sense in the context of the PV effect!
...

Well, of course not. "perpetuum mobiles" do not exist while PV cells do work perfectly.

However, my impression is that the common explanations of a PV cell must miss something crucial, because they do sound like the description of a perpetuum mobile. Is this not clear from my original post?

Anyway, thanks for the lecture. I will look at it!

 

[matthias31415]

...It'd help if you provide a link to such an explanation, and describe how perpetuum mobile follows from it in your opinion. That way we'll have a specific subject to discuss instead of guessing what exactly you are referring to.

I thought that I had included such an explanation in my original post. The analogy was a merry-go-round with negatively charged horses (corresponding to the electrons in the conduction band) and a stationary positive charge next to it (corresponding to the positive charge of the n-layer). This is an analogy that I have invented myself to illustrate why, to my eyes, common explanations of the pv-effect are incomplete.

 

[Yuras]

Let me summarize your answer: The electric field across the junction [with the positive pole at the n-side and the negative pole at the p-side] pushes the electron towards the positive pole (n-side) and the hole towards the negative pole (p-side) (...) creating a potential difference between the n- and p-layer.

Well, probably it's more correct to say that it's chemical potential what separates the electron and the hole, not the electric field.

But: the newly created potential difference between the n- and p-layer, where does it have its positive pole and where does it have its negative pole? You are saying that this potential difference drives the electron through a circuit. So you are saying that its negative pole is now, all of a sudden, at the n-layer and its positive pole at the p-layer?

Then I guess the confusion is in electric potential vs electrochemical one.
Let's ignore photons and just connect a voltmeter across the p-n junction. Will you see 0.7V? I bet not. The electric potential across the junction can't push electrons through the circuit.
On the other hand, the electron and the hole created by a photon and pushed out of the depletion region, create electrochemical potential, which works as emf here.
The key to understanding the difference is to note that 0.7V electric potential (aka built-in potential of the p-n junction) is a thermodynamic equilibrium.

 

[Yuras]

I thought that I had included such an explanation in my original post. The analogy was a merry-go-round with negatively charged horses (corresponding to the electrons in the conduction band) and a stationary positive charge next to it (corresponding to the positive charge of the n-layer). This is an analogy that I have invented myself to illustrate why, to my eyes, common explanations of the pv-effect are incomplete.

The problem is that we don't know what is the "common explanation" you have in mind. If it produces perpetuum mobile, then it's safe to assume that it's wrong. There are infinitely many wrong explanations, we can't tell what is wrong specifically with the one you have in mind without knowing it.

 

[Lord Jestocost]

When light falls on the junction region of a p-n diode and the energy of the photons is greater than the band gap, electrons are freed from the covalent bonds and electron-hole pairs are created in the region. Due to the electric field across the space charge region in the junction, the recombination of these electron-hole pairs can be avoided to some extent when the generated mobile electrons/holes have sufficient time to move in reaction to the electric field across the junction to the n-side and p-side of the diode (electrons in the junction region move to the n-side and holes move to the p-side). Under irradiation with light, the number of electrons in the n-region thus increases and the number of holes in the p-region increases. This leads to a non-equilibrium situation in the p-n diode: A gadient of the "electrochemical potential" is generated across the diode.

The photovoltaic effect is defined as the generation of a potential difference between the two connections of a p-n diode leading to an electric current flow through an external circuit upon irradiation of light.

 

[matthias31415]

(...) Let's ignore photons and just connect a voltmeter across the p-n junction. Will you see 0.7V? I bet not. The electric potential across the junction can't push electrons through the circuit. (...)

I guess it depends on the type of voltmeter (i.e. if that voltmeter depends on the flow of a small current). What if I connect an oscilloscope across the p-n junction? I'd guess it does show 0,7V, and it will show that the n-layer is the positive pole as compared to the p-layer. Right?

Next, we let the sun shine and connect a consumer to the cell, will the oscilloscope still show 0,7V? If yes, my next question is: what will the oscilloscope show if I connect it to the n-layer and to the electrode next to the n-layer?

 

[matthias31415]

The problem is that we don't know what is the "common explanation" you have in mind. If it produces perpetuum mobile, then it's safe to assume that it's wrong. There are infinitely many wrong explanations, we can't tell what is wrong specifically with the one you have in mind without knowing it.

The "common explanation" I am refering to goes like this:
1) the energy of the photon is used to free electrons from the covalent bonds
2) the electric field across the p-n-junction causes free electrons to go in circles (inside the cell: from the p- to the n-layer; outside the cell: from the n- to the p-layer)

Point 2) is what I compare to a perpetuum mobile. In order for this whole thing to work, the energy of the photon must be the driving force of the current. In my (most likely wrong) understanding of the "common explanations", the energy of the photon is simply used to free the electron from the covalent bonds, and in the end, when an electron returns to these covalent bonds, the exact same amount of energy will be released. But if that were the case, where does the energy come from that causes the current?

 

[Cthugha]

For the sake of simplicity, I would like to describe the path of a single electron from the place where it was released from the silicium atom through an external consumer back to the atom where it originally came from. I know that reality is more complicated with multiple electrons performing multiple steps, leading to the impression that holes are "moving" in the opposite direction. However, my gut feeling is, that, if there is no working single-electron-model, the more complicated models will fail as well.

Are you sure that this should be an advanced level thread and you are on grad student level?
In a semiconductor, the electron is not removed from an atom by optical excitation. It goes from the valence to the conduction band. In neither of those bands the electron can be assigned to a single atom. Any standard lecture on solid state stuff directly tells you that your gut feeling is completely off. The electrons are not released from an initial atom.

The "common explanation" I am refering to goes like this:
1) the energy of the photon is used to free electrons from the covalent bonds
2) the electric field across the p-n-junction causes free electrons to go in circles (inside the cell: from the p- to the n-layer; outside the cell: from the n- to the p-layer)

Point 2) is what I compare to a perpetuum mobile. In order for this whole thing to work, the energy of the photon must be the driving force of the current. In my (most likely wrong) understanding of the "common explanations", the energy of the photon is simply used to free the electron from the covalent bonds, and in the end, when an electron returns to these covalent bonds, the exact same amount of energy will be released. But if that were the case, where does the energy come from that causes the current?

There is no common legitimate explanation which goes like that. Both electrons are delocalized. They are just in delocalized states of different energy (and of course mobility). You do not break bonds here and I never heard any reliable source stating so (however, there may of course be pop-sci sources claiming that - especially laymen explanations on the pages of solar cell companies can be horrible).

Any standard textbook on band structure formation and solid state physics might be helpful here to get a real understanding of the basics.

 

[matthias31415]

Are you sure that this should be an advanced level thread and you are on grad student level?
(...)

Well, solid state physics (obviously) is not my primary field of expertise, in that sense "advanced level" would be inappropriate.

However, I tried to resolve my questions with collegues who have a deeper understanding of solid state physics and they were not able to specifically address my questions. I therefore assumed my question to be somewhat tricky.

If anyone feels that "advanced level" is inappropriate, please feel free to change it (I don't know if this is possible once the thread has started).

 

[Lord Jestocost]

The "common explanation" I am refering to goes like this:
1) the energy of the photon is used to free electrons from the covalent bonds
2) the electric field across the p-n-junction causes free electrons to go in circles (inside the cell: from the p- to the n-layer; outside the cell: from the n- to the p-layer)

Point 2) is what I compare to a perpetuum mobile. In order for this whole thing to work, the energy of the photon must be the driving force of the current. In my (most likely wrong) understanding of the "common explanations", the energy of the photon is simply used to free the electron from the covalent bonds, and in the end, when an electron returns to these covalent bonds, the exact same amount of energy will be released. But if that were the case, where does the energy come from that causes the current?

“We start with a discussion of energy harvesting from a single-junction photovoltaic cell consisting of a semiconductor photodiode. When facing a hot object such as the sun, i.e., under positive illumination, such a photodiode can absorb a net in-flux of photons with an energy above its bandgap to produce electrical power. The theoretical efficiency limit of a single-junction cell for harvesting incoming radiation from the sun (Fig. 1a) is well established as the Shockley–Queisser limit1. Under full concentration or angle restricted absorption and emission, a maximum efficiency of 40.7% can be obtained at an optimal bandgap energy of 1.08 eV when the cell is maintained at a temperature of 300 K.”

From “Thermodynamic limits for simultaneous energy harvesting from the hot sun and cold outer space” by Wei Li, Siddharth Buddhiraju and Shanhui Fan (Light: Science & Applications volume 9, Article number: 68 (2020))

 

[matthias31415]

Are you sure that this should be an advanced level thread and you are on grad student level?
In a semiconductor, the electron is not removed from an atom by optical excitation. It goes from the valence to the conduction band. In neither of those bands the electron can be assigned to a single atom. Any standard lecture on solid state stuff directly tells you that your gut feeling is completely off. The electrons are not released from an initial atom.
(...)

I am aware that these "explanations" are oversimplifications. My idea was that they are still of some use when modified. With respect to electrons being removed from an atom: probably wrong, but not very far off from the notion that an excited electron leaves behind a positively charged "hole". See, e.g. Wikipedia: "illuminating the material creates an electric current because excited electrons and the remaining holes are swept in different directions by the built-in electric field of the depletion region." (photovoltaic effects / solar cells).

So let me rephrase my (wrong) explanation:
1) the energy of the photon is used to exicte an electron which leads to the creation of an electron-hole pair.
2) the electric field across the p-n-junction causes electrons and holes to move in opposite directions, so that in the end, a current is flowing, which can be used to power a consumer.

Still this sounds weird. So, should one completely let go of the notion the the photon energy is used to create an electron-hole-pair?

 

[Lord Jestocost]

.... such a photodiode can absorb a net in-flux of photons with an energy above its bandgap to produce electrical power ....

 

[matthias31415]

.... such a photodiode can absorb a net in-flux of photons with an energy above its bandgap to produce electrical power ....

I don't know if I completely get the message...

First, I thought: well, obviously, the energy of the photon needs to be above the bandgap, because if it were below the gap, it wouldn't lift any electrons from the valence band to the conduction band.

Or is the message: the energy needs to be above the bandgap, because if it were exactly equal to the bandgap, it would only lift electrons from the valence band to the conduction band but not produce any additional power?

 

[Yuras]

I guess it depends on the type of voltmeter (i.e. if that voltmeter depends on the flow of a small current). What if I connect an oscilloscope across the p-n junction? I'd guess it does show 0,7V, and it will show that the n-layer is the positive pole as compared to the p-layer. Right?

Hmm, I don't see why it would be 0.7V. Doesn't oscilloscope "depend on the flow of a small current" as well?

In my (most likely wrong) understanding of the "common explanations", the energy of the photon is simply used to free the electron from the covalent bonds, and in the end, when an electron returns to these covalent bonds, the exact same amount of energy will be released. But if that were the case, where does the energy come from that causes the current?

Ah, now the question actually makes sense to me. Basically you are asking the following: electron is exited, which requires $E_{ex}=h\nu$ energy. Then electron goes through the circuit and performs work $\Delta W$. Finally, it's recombined with the hole releasing $E_{rec}=h\nu$. As a net result we have extra $\Delta W$ of energy. Do I understand it correctly?
Unfortunately I don't understand the subject in enough details to give a satisfactory answer, only few comments. When considering just a single electron, you can (I think) use the equilibrium model of the p-n junction, but then the current is pretty match zero, so $\Delta W$ is zero as well. You can think about already established photovoltaic current and consider a single electron in the current, but then you can't use the equilibrium model - e.g. you have to consider Fermi level splitting, which I think will make $E_{rec}$ smaller.

Or is the message: the energy needs to be above the bandgap, because if it were exactly equal to the bandgap, it would only lift electrons from the valence band to the conduction band but not produce any additional power?

IIRC all the energy above the gap gets lost to phonons due to thermalization...

 

[Lord Jestocost]

Ah, now the question actually makes sense to me. Basically you are asking the following: electron is exited, which requires $E_{ex}=h\nu$ energy. Then electron goes through the circuit and performs work $\Delta W$. Finally, it's recombined with the hole releasing $E_{rec}=h\nu$. As a net result we have extra $\Delta W$ of energy. Do I understand it correctly?

In cace of a circuit with resistance $R\equiv0$.

 

[matthias31415]

(...)

First, I thought: well, obviously, the energy of the photon needs to be above the bandgap, because if it were below the gap, it wouldn't lift any electrons from the valence band to the conduction band.

Or is the message: the energy needs to be above the bandgap, because if it were exactly equal to the bandgap, it would only lift electrons from the valence band to the conduction band but not produce any additional power?

@Lord Jestocost: thanks for your thumbs up! I was wondering if it refers to the first or the second paragraph?

 

[matthias31415]

(...)
Ah, now the question actually makes sense to me. Basically you are asking the following: electron is exited, which requires $E_{ex}=h\nu$ energy. Then electron goes through the circuit and performs work $\Delta W$. Finally, it's recombined with the hole releasing $E_{rec}=h\nu$. As a net result we have extra $\Delta W$ of energy. Do I understand it correctly?
Unfortunately I don't understand the subject in enough details to give a satisfactory answer (...)

Yes, this looks like what I tried to say, except that I feel a bit unsure about some details (e.g. with respect to $E_{rec}=h\nu$, does that imply that the energy from recombination is released in the form of a photon?).

While being relieved that someone finally understands me, at the same time, I am a bit puzzled. My feeling is that I am missing the very core of the photovoltaic effect, while you seem to think that a satisfactory answer is a matter of details...

 

[matthias31415]

(...)

IIRC all the energy above the gap gets lost to phonons due to thermalization...

I have read this also, but either:
(1) the energy of the photon is above the gap and the excess energy is (at least in part) converted into electrical energy or
(2) The energy to move the electron from the valence band to the conduction band is larger than the energy that is released when, in the end, the electron moves back from the conduction band to the valence band or
(3) The solar cell does not produce electrical energy

 

[Cthugha]

I am aware that these "explanations" are oversimplifications. My idea was that they are still of some use when modified. With respect to electrons being removed from an atom: probably wrong, but not very far off from the notion that an excited electron leaves behind a positively charged "hole". See, e.g. Wikipedia: "illuminating the material creates an electric current because excited electrons and the remaining holes are swept in different directions by the built-in electric field of the depletion region." (photovoltaic effects / solar cells).

So let me rephrase my (wrong) explanation:
1) the energy of the photon is used to exicte an electron which leads to the creation of an electron-hole pair.
2) the electric field across the p-n-junction causes electrons and holes to move in opposite directions, so that in the end, a current is flowing, which can be used to power a consumer.

Still this sounds weird. So, should one completely let go of the notion the the photon energy is used to create an electron-hole-pair?

I do not exactly get what you consider weird about that. Maybe you can clarify.

Indeed in solar cells you create an exciton (bound electron-hole pair) or free electrons and holes. If you leave them unattented and just wait, the electron will reduce its energy by scattering and move to the bottom of the conduction band and the hole will reduce its energy and go to the top of the valence band. Then (assuming a direct band gap), they may recombine (so the electron returns to the valence band) and emit a photon with an energy equal to the energy of the band gap. This is how you "get back" the energy initially put into the system and this is how semiconductor diodes work.

Now you instead have an electric field (usually built-in due to the pn-junction and not caused by external bias) and rip the particles apart. In very simplified language, it becomes easier for the electron to return to the valence band by going through the circuit to recombine with the hole as compared to go across the pn-junction directly. In that way, you get back (at least a part of) the initial photon energy by extracting work from the electron instead of getting the energy out as a photon.

I have read this also, but either:
(1) the energy of the photon is above the gap and the excess energy is (at least in part) converted into electrical energy or
(2) The energy to move the electron from the valence band to the conduction band is larger than the energy that is released when, in the end, the electron moves back from the conduction band to the valence band or
(3) The solar cell does not produce electrical energy

I still do not get your point. It is not the excess energy that is converted into electrical energy. It is the total photon energy (minus some losses due to scattering or similar events of course).

I am not sure what you mean by stating the energy "that is released when, in the end, the electron moves back from the conduction to the valence band". This energy is not released in the sense that it is dissipated. It is this very energy that is converted into electrical energy. If it was not, you would have to get this energy out as a photon.

 

[matthias31415]

(...) Indeed in solar cells you create an exciton (bound electron-hole pair) or free electrons and holes. If you leave them unattented and just wait, the electron will reduce its energy by scattering and move to the bottom of the conduction band and the hole will reduce its energy and go to the top of the valence band. Then (assuming a direct band gap), they may recombine (so the electron returns to the valence band) and emit a photon with an energy equal to the energy of the band gap. This is how you "get back" the energy initially put into the system and this is how semiconductor diodes work.

Now you instead have an electric field (usually built-in due to the pn-junction and not caused by external bias) and rip the particles apart. In very simplified language, it becomes easier for the electron to return to the valence band by going through the circuit to recombine with the hole as compared to go across the pn-junction directly. In that way, you get back (at least a part of) the initial photon energy by extracting work from the electron instead of getting the energy out as a photon. (...)

Thanks, Cthugha, this clarifies a lot! From your answer, I understand that ("in very simplified language"), going through the circuit is a way for the electron to get from the conduction band back to the valence band. Is it then correct to say that at the end of the circuit, i.e. at the electrode of the pv-cell, all electrons are already back in the valence band (at least those electrons that performed the work)?

Bingo, now everything makes sense to me! Of course, at the other electrode, where holes are moving, the electrons are back in the valence band. Thats the difference between moving electrons and moving holes!

The only thing that I am still missing: why does this electric field across the p-n-junction favour one way of the electrons (through the circuit) over the other (directly back to the hole)? If the electric field is strong enough to rip electrons and holes apart, why isn't this electric field strong enough to hold electrons back from going through the circuit?

 

[sophiecentaur]

Doesn't oscilloscope "depend on the flow of a small current" as well?

Amplifiers all have a finite input impedance so, even in what we would regard as a Voltage - Driven amplifier there is current. The input amplifier on a regular 'scope will be about 1MΩ and a 'times ten' probe increases it to 10MΩ. You can. of course, go higher .