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Schottky contacts - band bending

by jonas_nilsson
Tags: band, bending, contacts, schottky
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jonas_nilsson
#1
Jun7-05, 01:47 PM
P: 29
Hello!

I am looking a bit at the Schottky contacts, metal-semiconductor, and I am starting to get a hang of it, but some things are missing.

Specifically:
Why must the distance from the Fermi Energy to the vacuum level in the metal and the distance from conduction band edge to vacuum level in the semiconductor stay constant at the contact point?

I only have a very condensed description, and the argument is "because these distances are crystal properties". But it would make more sense to me if the distances had to stay constant further into the crystals, where the dynamic between the two materials should be of less importance.

That's it. Appreciate all answers!
/Jonas
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ZapperZ
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Jun7-05, 02:25 PM
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Quote Quote by jonas_nilsson
Hello!

I am looking a bit at the Schottky contacts, metal-semiconductor, and I am starting to get a hang of it, but some things are missing.

Specifically:
Why must the distance from the Fermi Energy to the vacuum level in the metal and the distance from conduction band edge to vacuum level in the semiconductor stay constant at the contact point?

I only have a very condensed description, and the argument is "because these distances are crystal properties". But it would make more sense to me if the distances had to stay constant further into the crystals, where the dynamic between the two materials should be of less importance.

That's it. Appreciate all answers!
/Jonas
There is something very confusing with your question. The "Fermi Energy" and "Vacuum Level" are all ENERGY levels. These are states that we artificially "draw" within an energy graph. Thus, your phrase "distance from the Fermi Energy to the vacuum level" is confusing. The vaccum level and the Fermi energy for a metal are separated by an ENERGY scale, not a "distance" scale. In fact, there is a name for this interval - the electron affinity.

What I think you may be asking is the reason for bend bending. Specifically, why the Fermi energy level in the metal and the chemical potential in the semiconductor must "line up". This has nothing to do (at least not directly) with the crystal structure, but rather with thermodynamics. At thermodynamical equilibrium, the contacts must be at the same "temperature", and thus the Fermi energy of the metal (which is the highest energy of the occupied state) will coincide with the statistical characteristic energy of the semiconductor. When this occur, the rest of the bands will follow.

Zz.
jonas_nilsson
#3
Jun7-05, 04:16 PM
P: 29
Hi, thanks for replying! And really fast also!

Quote Quote by ZapperZ
There is something very confusing with your question. The "Fermi Energy" and "Vacuum Level" are all ENERGY levels. These are states that we artificially "draw" within an energy graph. Thus, your phrase "distance from the Fermi Energy to the vacuum level" is confusing. The vaccum level and the Fermi energy for a metal are separated by an ENERGY scale, not a "distance" scale. In fact, there is a name for this interval - the electron affinity.
Well I know we're talking about energies, perhaps it's just my english, but I don't see any problem with talking about distances in energy, just as we talk about phase space volumes and so =)

Quote Quote by ZapperZ
What I think you may be asking is the reason for bend bending. Specifically, why the Fermi energy level in the metal and the chemical potential in the semiconductor must "line up". This has nothing to do (at least not directly) with the crystal structure, but rather with thermodynamics. At thermodynamical equilibrium, the contacts must be at the same "temperature", and thus the Fermi energy of the metal (which is the highest energy of the occupied state) will coincide with the statistical characteristic energy of the semiconductor. When this occur, the rest of the bands will follow.

Zz.
I know the lining up has to to with thermal equilibrium (no applied voltage and no light-induced electron-hole generation), but I don't see why the bands follow the way they do. In my lecture notes is a small remark that the energy-distance (don't know what to call it =) ) from the conduction band edge to vacuum level must still be equal the electron affinity after thermal equilibrium has been achieved, and that is exactly at the metal-semiconductor contact surface. Also there is a small note that this follows from the crystal structure. I would have agreed if this was the case far far away from the contact surface, where the materials "don't know" of each others precense. Yet this wouldn't make any sense either, because as you said, the Fermi energy levels must line up throughout the whole body. Apparantly I am lacking something here =)

And, talking about the Fermi energy, isn't that the energy where the likelihood of a state (if it is allowed) to be occupied is 1/2. How can one say that this is the highest occupied state? I mean, the higher states are also possible, but the likelihood of them being occupied sinks quite fast with higher energy. How did you mean?

ZapperZ
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Jun7-05, 04:31 PM
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Schottky contacts - band bending

Quote Quote by jonas_nilsson
I know the lining up has to to with thermal equilibrium (no applied voltage and no light-induced electron-hole generation), but I don't see why the bands follow the way they do. In my lecture notes is a small remark that the energy-distance (don't know what to call it =) ) from the conduction band edge to vacuum level must still be equal the electron affinity after thermal equilibrium has been achieved, and that is exactly at the metal-semiconductor contact surface. Also there is a small note that this follows from the crystal structure. I would have agreed if this was the case far far away from the contact surface, where the materials "don't know" of each others precense. Yet this wouldn't make any sense either, because as you said, the Fermi energy levels must line up throughout the whole body. Apparantly I am lacking something here =)
Maybe someone can elaborate on this. I am clueless on what was said here.

And, talking about the Fermi energy, isn't that the energy where the likelihood of a state (if it is allowed) to be occupied is 1/2. How can one say that this is the highest occupied state? I mean, the higher states are also possible, but the likelihood of them being occupied sinks quite fast with higher energy. How did you mean?
Er... come again?

The Fermi energy is DEFINED as the topmost occupied state of a fermionic system. What you just described is the occupation density, i.e. n(k), or n(E). Why is this even here?

Zz.
Gokul43201
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Jun7-05, 04:46 PM
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That's the chemical potential [itex] \mu[/itex], which is equal to the Fermi Energy at 0 K and pretty darn close at normal temperatures.

[tex]\mu = \epsilon _F \left(1- \frac{\pi^2}{12} \left( \frac{k_BT}{\epsilon_F} \right) ^2 + ... \right) [/tex]
jonas_nilsson
#6
Jun7-05, 05:06 PM
P: 29
Quote Quote by ZapperZ

Er... come again?

The Fermi energy is DEFINED as the topmost occupied state of a fermionic system. What you just described is the occupation density, i.e. n(k), or n(E). Why is this even here?

Zz.
No I didn't mean that. Perhaps I don't have the right words in English right now, the last year I have spent mixing it up with German, but it was however not what I meant. And if the Fermi Energy was the uppermost occupied state, then NO electrons would be able to pass the barrier into the semiconductor without for example absorbing a photon, right? (that means, if E_F < E_conductionband) This would leave the Schottky contact an isolator, which is clearly not the case. It is quite hard to explain without showing you the picture of the bands that I use, I might be able to upload it tomorrow...

Quote Quote by Gokul43201
That's the chemical potential [itex] \mu[/itex], which is equal to the Fermi Energy at 0 K and pretty darn close at normal temperatures.

[tex]\mu = \epsilon _F \left(1- \frac{\pi^2}{12} \left( \frac{k_BT}{\epsilon_F} \right) ^2 + ... \right) [/tex]
Exactly, so we can agree that at room temperature the Fermi Energy and the chemical potential are essentieally the same, and if you have a look at for example this picture from Wikipedia:
http://en.wikipedia.org/wiki/Image:FD_e_mu.jpg
you see that the Fermi energy is where the distribution function is 1/2.

Quote Quote by ZapperZ
Maybe someone can elaborate on this. I am clueless on what was said here.
Well I guess it might be hard to follow what I meant. My main question is still why the bands must "follow" the way they do.
Gokul43201
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Jun7-05, 05:22 PM
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The Fermi Energy is the uppermost "occupied" state. But how do you tell if a state is occupied when it has an accupation probability < 1 ? The problem resolves itself at most working temperatures...by the fact that the energy difference to go from a high probability of occupation to a low probability is small compared to the Fermi Energy.

As to the rest of the question; I'm not sure I understand it at first glance. Does it not suffice to say that the chemical potentials of the two bodies in thermodynamic equilibrium must be equal ? If there were a difference in chemical potentials, there would be a flow of particles (electrons/holes) across the interface till the difference disappeared.
jonas_nilsson
#8
Jun7-05, 05:34 PM
P: 29
Quote Quote by Gokul43201
The Fermi Energy is the uppermost "occupied" state. But how do you tell if a state is occupied when it has an accupation probability < 1 ? The problem resolves itself at most working temperatures...by the fact that the energy difference to go from a high probability of occupation to a low probability is small compared to the Fermi Energy.
Ok, now I'm confused. I guess I don't get what you mean with "the uppermost occupied state" in this case. But we're actually gonna do that on tuesday, I'm right now doing the statistical physics course. But the expected number of electrons in an energy state in (E,E+dE) is N(E)f(E)dE, so I don't see any problems with probabilities < 1, and sure energy states above E_F are also occupied. Perhaps not full, but anyway electrons will be there too, right? But as I said, next week it might get clearer

Quote Quote by Gokul43201
As to the rest of the question; I'm not sure I understand it at first glance. Does it not suffice to say that the chemical potentials of the two bodies in thermodynamic equilibrium must be equal ? If there were a difference in chemical potentials, there would be a flow of particles (electrons/holes) across the interface till the difference disappeared.
Yes I'm allright with equilibrium and all that, but the shape of the bands is the thing. It isn't elaborated at all in the course I do, it's just some theory for a solar cell lab course I just did, but there's this one point where they say that they for certain know where the conduction band edge must be, relative to the vacuum level, at the contact surface between metal and silicon. This is what I would like to know how it works.
ZapperZ
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Jun7-05, 07:38 PM
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Quote Quote by jonas_nilsson
Ok, now I'm confused. I guess I don't get what you mean with "the uppermost occupied state" in this case. But we're actually gonna do that on tuesday, I'm right now doing the statistical physics course. But the expected number of electrons in an energy state in (E,E+dE) is N(E)f(E)dE, so I don't see any problems with probabilities < 1, and sure energy states above E_F are also occupied. Perhaps not full, but anyway electrons will be there too, right? But as I said, next week it might get clearer



Yes I'm allright with equilibrium and all that, but the shape of the bands is the thing. It isn't elaborated at all in the course I do, it's just some theory for a solar cell lab course I just did, but there's this one point where they say that they for certain know where the conduction band edge must be, relative to the vacuum level, at the contact surface between metal and silicon. This is what I would like to know how it works.
You are confusing two different quantities in two different ways: the chemical potential and the Fermi energy; the Fermi energy for METALS and the Fermi energy/chemical potential for SEMICONDUCTORS.

In metals, using the degenerate free electron gas, the Fermi energy is defined as the top of the occupied state at T=0K. This is NOT the same way it is defined for an intrinsic semiconductor, where the middle of the gap is the chemical potential at T=0K.

Zz.
Gokul43201
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Jun8-05, 09:24 AM
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Quote Quote by jonas_nilsson
I'm right now doing the statistical physics course. But the expected number of electrons in an energy state in (E,E+dE) is N(E)f(E)dE, so I don't see any problems with probabilities < 1, and sure energy states above E_F are also occupied. Perhaps not full, but anyway electrons will be there too, right? But as I said, next week it might get clearer
N(E) is the density of states; it conveys no information about the number of electrons in an energy level. That comes from the FD distribution function. Electrons are Fermions - how many can you put in a level ? If for a certain energy level, f(E) = 0.3 say, is it occupied or not ?
jonas_nilsson
#11
Jun8-05, 02:04 PM
P: 29
Quote Quote by Gokul43201
N(E) is the density of states; it conveys no information about the number of electrons in an energy level. That comes from the FD distribution function. Electrons are Fermions - how many can you put in a level ? If for a certain energy level, f(E) = 0.3 say, is it occupied or not ?
Aaah now I see what you asked before. This is great, you're asking very pedagogic questions, thanks alot!

But I don't have any problem saying that f(E) is 0.3. It doesn't give us any information about occupancy of course, I totally agree. But we can say that with probability 0.3 the state is occupied (if the state is at all allowed). How many states with the energy E that are possible -not occupied - per unit volume is described by N(E).

In N(E) the pauli principle has to be taken into account, I guess. How many that can be in one energy state depends on the Hamilton operator, is my second guess. If the eigenenergies are degenerate more states are associated to each energy, and we get higher density of states.

But how does distance come into the picture here? Let's imagine that we have a set of electrons in for example a semiconductor, and all of them are in unique quantum states. Suppose that there are states that are unoccupied, and that one electron jumps into one those free states. How is then the information carried over to the other electrons that this state is now occupied? And is really any information carried over, that would also mean transfer of energy, wouldn't it? And it can't happen instantly, and it can't be so that two electrons at "great" distance from each other couldn't have the exact same set of quantum numbers.
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Jun8-05, 02:46 PM
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Quote Quote by jonas_nilsson
But how does distance come into the picture here? Let's imagine that we have a set of electrons in for example a semiconductor, and all of them are in unique quantum states. Suppose that there are states that are unoccupied, and that one electron jumps into one those free states. How is then the information carried over to the other electrons that this state is now occupied? And is really any information carried over, that would also mean transfer of energy, wouldn't it? And it can't happen instantly, and it can't be so that two electrons at "great" distance from each other couldn't have the exact same set of quantum numbers.
Yes they can! Once BE or FD statistics kicks in, it means that there's significant overlap of ALL the particles involved that they all know how to behave properly AS A WHOLE. In a superconductor, ALL of the condensed electrons have long-range coherence and thus, are essentially described via a single state wavefunction. For an entangled pair, as long as they maintain coherence, they could be on the other side of the universe if they wish and would "know" what the other is doing (EPR-type experiment).

This is diverging away from the main topic of this thread, but once quantum statistics kicks in, one needs to be aware that one is now dealing with the whole glob of things. The "individuality" of each particle is gone because they are no longer distinguishable. The exclusion principle is the result of such indistinguishibility.

Zz.
jonas_nilsson
#13
Jun8-05, 03:04 PM
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Quote Quote by ZapperZ
Yes they can! Once BE or FD statistics kicks in, it means that there's significant overlap of ALL the particles involved that they all know how to behave properly AS A WHOLE. In a superconductor, ALL of the condensed electrons have long-range coherence and thus, are essentially described via a single state wavefunction. For an entangled pair, as long as they maintain coherence, they could be on the other side of the universe if they wish and would "know" what the other is doing (EPR-type experiment).

Wow, this is truly incredible. I'm prepared to join Einstein and other sceptics, this is getting ridiculous =)

Quote Quote by ZapperZ
This is diverging away from the main topic of this thread, but once quantum statistics kicks in, one needs to be aware that one is now dealing with the whole glob of things. The "individuality" of each particle is gone because they are no longer distinguishable. The exclusion principle is the result of such indistinguishibility.

Zz.
You are absolutely true, but as it diverges, it gets interesting aswell =). I'm gonna read some about EPR, and perhaps it's suitable to discuss it in another thread later on. But it's to interesting to stop, I just can't help it!

I really can't accept that just because we're talking quantum mechanics all of a sudden information can travel infinitely fast. But I sense that my problem here is being stuck in the classical view. Talking about one specific electron changing state isn't anymore relevant. Nevertheless, I feel that the electron would have to "tell" the others to keep out of this state!

But there's also another catch isn't it? You couldn't send any information by using this phenomena, could you?
ZapperZ
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Jun8-05, 04:37 PM
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Quote Quote by jonas_nilsson
I really can't accept that just because we're talking quantum mechanics all of a sudden information can travel infinitely fast.
Whoa!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Where did I say there was information moving infinitely fast?

There's nothing "moving" at all between the two entangled entities. That's the WHOLE point. If there is, we should be able to detect it on its way from one particle to another! We detect NO such thing! Furthermore, within QM's formulation, there's NOTHING that is being transfered! Special Relativity (and Einstein) is NOT being violated here!

But I sense that my problem here is being stuck in the classical view. Talking about one specific electron changing state isn't anymore relevant. Nevertheless, I feel that the electron would have to "tell" the others to keep out of this state!
There is a huge amount of risk to associate human behavior to either "god", or inanimate entities like "electrons". This usually comes about because of lack of understanding of how a system is described, evolves, and what exactly is involved in such a description. If you think an electron "tells" the others to keep out of a state, how does it tell them? What mechanism is involved? What exactly is the nature of this signal? Etc.... because frankly, there's nothing in the physics that contains all of this. You might as well make anything up since these things obviously cannot be tested to be true. If we go by that, then all bets are off because *I* can make things up as I go along too. And in terms of imagination, mine can be as wild as anything you can come up especially if I'm not being restrained by any established physics rule.

Now would you want the thread to deteriorate into such a thing?

Zz.
jonas_nilsson
#15
Jun8-05, 05:14 PM
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Quote Quote by ZapperZ
Whoa!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Where did I say there was information moving infinitely fast?

.....If you think an electron "tells" the others to keep out of a state, how does it tell them? What mechanism is involved? What exactly is the nature of this signal? Etc.... because frankly, there's nothing in the physics that contains all of this.
Well that's why I said I can't accept it and tried to adress the problem.
Quote Quote by ZapperZ
You might as well make anything up since these things obviously cannot be tested to be true. If we go by that, then all bets are off because *I* can make things up as I go along too. And in terms of imagination, mine can be as wild as anything you can come up especially if I'm not being restrained by any established physics rule.
It's obvious I haven't grasped the idea of identical particles yet, but this sort of lecturing is just humiliating and boring to read.

Quote Quote by ZapperZ
Now would you want the thread to deteriorate into such a thing?
Zz.
Oh, please.
Gokul43201
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Jun9-05, 01:26 AM
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Let's get back to the original question, what ?

jonas, do you agree that the Fermi energies of the metal and SC want to match each other ? Unless you match chemical potentials, you can not have thermal equilibrium. The way this matching happens is through a transfer of electrons across the interface. This transfer results in a net charge gradient in the SC adjacent to the metal, which gives rise to an electric field in the SC. It is this electric field which causes the bands to bend.

Recall what happens to a band when you apply a bias voltage. This is virtually the same effect here.
jonas_nilsson
#17
Jun9-05, 01:45 AM
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Quote Quote by Gokul43201
Let's get back to the original question, what ?
Yes, I would love to!
Quote Quote by Gokul43201
jonas, do you agree that the Fermi energies of the metal and SC want to match each other ? Unless you match chemical potentials, you can not have thermal equilibrium. The way this matching happens is through a transfer of electrons across the interface. This transfer results in a net charge gradient in the SC adjacent to the metal, which gives rise to an electric field in the SC. It is this electric field which causes the bands to bend.
Yes I agree with you, but why do they bend they do? Right now the shape isn't that interesting to me, but from the pictures I have, it seems like the cunduction band at the contact surface stays untouched. How can this be? Following the band edge into the semiconductor there is a bend, of course, but at the contact surface, E_vacuum - E_c is the same as before.

Quote Quote by Gokul43201
Recall what happens to a band when you apply a bias voltage. This is virtually the same effect here.
Is it really virtually the same? I mean when we apply a bias we will have a gradient also in the Fermi energy, and the barrier at the interface will also become smaller/greater depending on how we apply the bias. This at least holds for the pn-junction, but it shouldn't be much different here I think.

Now I have to get to the University again (don't know where you are, but here it's morning right now), but I'm gonna try to find a picture to show you tonight.

Cheers!
Jonas
marlon
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Jun9-05, 02:34 AM
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Band bending itself occurs because at the interface between a semiconductor and a metal (just as an example), the semiconductor has charges at that surface. These charges are responsible for the actual band bending which can be studied by solving the Poisson equation. Just look at the MOS-example (SiO2/p-type semiconductor).

When you apply a negative volage onto the gate (ie metal layer on the silicon), the SiO2 will have negative charges at the interface and the semi conductor will have postive charges at the interface. Basically what happens is an accumulation of holes (ie the majority carriers) at the inteface. These positive charges are responsible for the band bending at the surface. Starting from the interface, the bands are bended downward.

Now let us rise the applied voltage bias. The holes in the semiconductor will be repelled because the voltage becomes more positive. At a certain V, all holes have been depleted from the interface region and what remains are the negative acceptor ions. Again these negative charges cause the band bending in opposite direction of the previous example : starting from the interface, the bands will go up. The positive charges at the silicon are accounted for by the negative acceptor ions. This is the depletion regime because all majority carriers of the semi conductor have been pushed out of the interface region

If we keep on rising the potential, the depletion zone gets bigger up to a certain point where there is enough energy to bring electrons (ie the minority carriers) at the interface. These electrons for the inversion layer and the bands are bended more strongly as in the depletion case. The reason why electrons arise at the surface has to do with the fact that np must remain constant. Since p (the holes-concentration as a function of the energy) is lowered at the surface due to the positive potential, the n must rise. In the depletion regime, the number of acceptor ions is big enough to provide the necessary n-value, however now, we also need electrons in order to fulfill this np = constant demand.

This is just the general picture but i think it is clear as to why band bending occurs. Flat bands occur when the semiconductor is neutral everywhere, so when there are no regions where acceptor ions are unaccounted for (ie "missing their holes") or no regions of inversion and accumulation.

And at thermodynamic equilibrium both fermi levels need to be aligned because if you take a test-charge and you move it from the silicon to the semiconductor and back, you should not have to do any work. Thus, no energy can be gained by the test charge if you move it through the surface, otherwise current would flow...because don't we all like to gain energy ?

Also keep in mind that in the MOS example there never is any flow of current due to the big band gap of the intermediate SiO2 layer with respect to the metal and the semi conductor.

marlon


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