Fermi energy level is defined highest energy level below which all energy levels are filled at ok. But in the case of a semiconductor there is no allowed energy level between the valence band and the Fermi energy level. How is this possible?
In general ,we say that Ef corresponds to that level which has probability 1/2 being occupied. So for semiconductors Ef must located between Ev &Ec. It is assumed that width of Vb and Cb are very small as compared to Eg. Let each band consist of Z number of possible states per unit volume. At T=0 ,all the states in Ev are filled while all the states in the Cb are empty. At T>0 ,density of electron in Cb is nc = z / {exp{[Ec-Ef] /KT} +1} and density of electrons in the Vb is given as
nv = z / {exp{[Ev-Ef] /KT} +1}. Practically all the electrons in Cb are from Vb. So nc +nv = z. So adding the equations we have ( z / {exp{[Ec-Ef] /KT} +1} ) + ( z / {exp{[Ev-Ef] /KT} +1}) = z . Solving this equations we get Ef =( Ec+Ev )/2 so it is midway between the Ev and Ec. Actually this is valid approximation. If we consider detailed calculation of nc & nv then effective values of elecrton mass (Me) , hole mass (Mh) and T affect on calculation giving result
Ef = { ( Ec+Ev)/2} + (3/4) KT log (Mh / Me) under the condition ne = nh for intrinsic semiconductor.
If we assume Me = Mh the second term vanishes giving same result Ef =( Ec+Ev )/2 .
Ref- Chapter 12 , The Electron Distribution in insulators and semiconductors. from Solid State Physics by A. J. Dekker. MACMILLAN INDIA LIMITED.
From - Sanjay Gadekar , PESJMJ , University of pune.
answer of this question lies in the mathematics so you requires a lot brain exercise and to convince yourself about this...what i did...but it is true due to "kT"...Good Luck
The fermi level does not only lie in the center of the bandgap, it can be shifted up and down depending on the dopant concentration.
In effect, the fermi level is the level to which electrons can fill up to. So if the fermi level (Fl) lies between the conduction band (Ec) and the valence band (Ev), what it means is that if there were states in that region, between Ev and Fl they would be filled. And there are states that can exist between Ec and Ev
These states are normally things like trap states and defects that have an energy level that lies in this forbidden region and the possibility of filling these states is determined by this Fl.
The Fl is also very important when drawing band diagrams as Fl matching is what gives the eventual band diagram (with band bending).
I have a question here. Why does Fermi level decrease when temperature is increased or when a material is doped with a dopant?
When the material is doped, the density of states in the conduction/valence band changes.
I'm not sure as to the actual physical mechanism involved in the lowering of the Fermi level, but when I think about it, I imagine there being more states, so it pulls the Fermi level towards that direction. If you dope Silicon with Boron, then you create holes, pulls the fermi level down and makes the material p type.
But the important thing is that the Fermi level is an imaginary 'filling' level that is very useful in understanding materials, but as far as i know, there is no compatible physical analogue, not like the conduction band and valence band which have actual physical meaning. (Correct me if I am wrong here)
Finally, I don't know what you mean by Fermi level changing with temperature. The band gap changes with temperature according to the Varshni equation, and the accurate position of the Fermi level in an intrinsic semiconductor material is slightly affected by the temperature, but not by much and generally negligible when solving questions.
If you are referring to doped materials, then I would argue that any temperature variations of the Fermi level will be dominated by the dopant effect.
Hope this helps.
thanks a lot. Could you recommend to me a book on this subject? I'm searching on p type (perovskite structure) and I would like to know the mechanism of conductance in such material by increasing temperature or by dopant...
Yeah you are correct... Actually I doped a material and then analyze its conductance by increasing temperature. For example in comparing dopant (impure) one with pure one the conductance is very different, because the pure one is p type and the material with dopant is a p type too, so the density of holes is increased. So why a material can be show conduct in lower temperature, could I say the holes may be participating?
This affirmation is true only for the intrinsic semiconductor. The valence energy band of this material is full at T=0 while the conducting band is empty. At T>0 some electrons jump from the valence band to the conducting one, and the number of these electrons is exactly equal to the number о remaining holes. Precisely in order to satisfy this condition, the Fermi level must lie in the center of the gap.
Dear Ajitesh Porwal. It seems to me that you're a first degree student in solid state physics or physics of semiconductors. Thus, my best bet is that you must, first of all, go to the classroom and not miss your courses. I'm sure that this question will be answered in your solid state or semiconductors course. This is not the place to answer the students homeworks. Good luck!
And... before I forget it, the same advice goes to Zahra Derakhshi.
The fermi level only lie in the center of the bandgap which is the case of undoped semiconductor (ideal case). As a usual, the feremi level can be shifted up and down depending on the dopant concentration and temperature.
I think researchgate is a place where you can ask anything related with scientific fields, of course, including the fundamentals. Considering the diversity of teaching levels around the world, it should be encouraged a lot for the students who want to learn something here.
The Fermi level represents the highest energy level that is filled at 0 Kelvin, or the energy level that has a 50% probability of being filled. Note though that in a semiconductor without defect levels in the band gap, there exist no states to fill near the Fermi level. For an intrinsic semiconductor, the Fermi level is at the midgap. However, for an n-doped semiconductor the Fermi level is above the intrinsic level, and for a p-doped semiconductor it is below the intrinsic level. This means statistically that the distribution of energy levels filled will shift upwards for an n-type and down for a p-type. If you think of this in terms of the teal curve in the link I posted (approximately the 0 Kelvin case), you will shift the step to the right in the n-type case, reaching higher energy levels.
The reason the FL exists even though there are no states in the gap (for an intrinsic, very pure case) is that the Fermi level is simply a probability of IF STATES EXIST, they are filled or not. The F-D distribution does not imply that states exist between the VB maximum and the FL, but only gives information about whether states (if they exist) would filled or not. It is just a statistical measure.
To answer Zahra's question: The Fermi level, as far as I know, generally does not shift due to changing temperature. However, the Fermi-Dirac distribution changes as a function of temperature and thus the Fermi level takes on a slightly different meaning. At low temperatures it is reasonable to say that all levels below the FL are filled and all above are empty, however as temp increases this becomes less and less true because the F-D distribution "smears." At high temperature, many states above the FL WILL be filled an many below WILL be empty. The specific reason why the Fermi level decreases as you dope it (specifically p dope!) is that p-dopants are acceptors, meaning that they are missing an electron. Shallow p dopants introduce EMPTY states near the valence band, which means that the FL will decrease because it is a statistical measure of the probability of filled states.
http://en.wikipedia.org/wiki/File:FD_e_mu.svg
The Fermi energy is a concept in quantum mechanics usually referring to the energy of the highest occupied quantum state in a system of fermions at absolute zero temperature. Confusingly, the term "Fermi energy" is often used to describe a different but closely related concept, the Fermi level (also called chemical potential). The Fermi energy and Fermi level are the same at absolute zero, but differ at other temperatures.
In quantum mechanics, a group of particles known as fermions (for example, electrons, protons and neutrons) obey the Pauli exclusion principle. This states that two fermions can not occupy the same (one-particle) quantum state. The states are labeled by a set of quantum numbers. In a system containing many fermions (like electrons in a metal), each fermion will have a different set of quantum numbers. To determine the lowest energy a system of fermions can have, we first group the states into sets with equal energy, and order these sets by increasing energy. Starting with an empty system, we then add particles one at a time, consecutively filling up the unoccupied quantum states with the lowest energy. When all the particles have been put in, the Fermi energy is the energy of the highest occupied state. What this means is that even if we have extracted all possible energy from a metal by cooling it to near absolute zero temperature (0 kelvin), the electrons in the metal are still moving around. The fastest ones are moving at a velocity corresponding to a kinetic energy equal to the Fermi energy. This is the Fermi velocity.
The Fermi energy (EF) of a system of non-interacting fermions is the increase in the ground state energy when exactly one particle is added to the system. It can also be interpreted as the maximum energy of an individual fermion in this ground state. The chemical potential at zero temperature is equal to the Fermi energy.
Dear Goodenough Cui,
You must learn to identify when somebody is asking for resolve its homework. Ask and discuss about scientific and technological issues will always be welcomed in a website like this. To resolve the student's homework, is not. The best way to help students to learn, is NOT to resolve their homework. A chinese proverb says: teach the people to fish, do not give them fish. The fundamentals of solid state physics can be found EVEN at the wikipedia (see the answer by Gazi Aliev above), explained at a level good enough to help resolve your questions, and translated to several languages. We expect that every ResearchGate's member is able to do, at least, a basic research on the web. Is common place that, at present, students all around the world prefer to come here (or another places) to ask ALL, instead to try to LEARN by themselves.
I was incorrect. The Fermi level is weakly dependent on temperature, not independent. As temperature increases greatly, the Fermi level tends towards the intrinsic level.
Fermi level exist due to kT i.e. thermal energy. the picture of th conduction and valance band with a precise band gap is only for vacuum. so in actual practice (at room temperature) some of the charge carriers will have sufficient energy to escape from the valance band according to fermi dirac distribution. due to this reason fermi level exists.!!!!!
In general ,we say that Ef corresponds to that level which has probability 1/2 being occupied. So for semiconductors Ef must located between Ev &Ec. It is assumed that width of Vb and Cb are very small as compared to Eg. Let each band consist of Z number of possible states per unit volume. At T=0 ,all the states in Ev are filled while all the states in the Cb are empty. At T>0 ,density of electron in Cb is nc = z / {exp{[Ec-Ef] /KT} +1} and density of electrons in the Vb is given as
nv = z / {exp{[Ev-Ef] /KT} +1}. Practically all the electrons in Cb are from Vb. So nc +nv = z. So adding the equations we have ( z / {exp{[Ec-Ef] /KT} +1} ) + ( z / {exp{[Ev-Ef] /KT} +1}) = z . Solving this equations we get Ef =( Ec+Ev )/2 so it is midway between the Ev and Ec. Actually this is valid approximation. If we consider detailed calculation of nc & nv then effective values of elecrton mass (Me) , hole mass (Mh) and T affect on calculation giving result
Ef = { ( Ec+Ev)/2} + (3/4) KT log (Mh / Me) under the condition ne = nh for intrinsic semiconductor.
If we assume Me = Mh the second term vanishes giving same result Ef =( Ec+Ev )/2 .
Ref- Chapter 12 , The Electron Distribution in insulators and semiconductors. from Solid State Physics by A. J. Dekker. MACMILLAN INDIA LIMITED.
From - Sanjay Gadekar , PESJMJ , University of pune.
As a approximation Ef is mid way and temperature independent . But if we consider effective masses Me and Mh then Ef moves up with temperature.
I express my thanks to Sanjay Gadekar for the clear explanation on why Ef=1/2(Ec+Ev) theoretically and simply.
Very good answer Atefeh Shaabani. A lot of people must follow your approach (may be a little different): I don't know, but at first I will try to find it; if I fail, then I will ask you to help me. WELL DONE!
The answers by Gazi Aliev and Sanjay Gadekar above show us that a simply research is enough in must cases to resolve our doubts. There are some people out there that simply wants that others make the work and provide them the answers.
Fermi energy which divides the energy gap between the filled states and unfilled states
You can see this point better in books of Statistical Mechanics. The Fermi level if defined as the last occupied level is not well defined for a semiconductor.
We can define the Fermi level as the limit for T going to 0 of the chemical potential.
For an intrinsic (non-doped) semiconductor this limit is at the center of the gap.
For (not strongly) doped semiconductors the Fermi level lies inside the impurity band that is close to the conduction band for n-type semiconductors and to the valence band for p-type semiconductors, by definition.
I recommend the old book "Statistical Mechanics” by Reif.
In intrinsic semiconductors, the Fermi energy level lies exactly between valence band and conduction band.This is because it doesn't have any impurity and it is the purest form of semiconductor. An example of intrinsic semiconductor is Germanium whose valency is four and hence covalent bond is being formed between these atoms. Where as in the case of n-type semiconductor, the fermi energy level lies nearer to the conduction band because of n-type impurity (pentavalent) e.g. Arsenic, which is nothing but donor energy level. Similarly in the case of p-type semiconductor, the fermi energy level lies nearer to the valence band because of p-type impurity (trivalent)e.g. Boron, which is nothing but acceptor energy level. AT 0K an intrinsic semiconductor behaves like an insulator and above 0K equal number of hole -electron pairs are produced and hence both hole and electron contribute for electrical conductivity.
Sanjay's analysis is correct. An alternative point of view is from the definition of electronic chemical potential, u=dG/dn for small values of dn, where G is the Gibbs free energy and n is the number of electrons. At T=0 K, the electron chemical potential is the Fermi energy. For a solid where the electron density is over 10E24 cm^-3, u is linear around dn = 0 +/- 1. At dn=-1 and dn=+1, u is equal to the ionization potential (I) and electron affinity (A) of the solid, ie., the top of the valence band and bottom of conduction band, respectively. Hence, at dn=0, u=(I+A)/2=(Ec+Ev)/2.
Well, may I add to already comprehensive discussion... The issue is originated from some misunderstanding betweem semiconductor and metal communities. The Fermi level is rigorously defined at T = 0, Then, it is the topmost occupied energy level at T = 0 and, in semiconductors, corresponds to the top of the valence band. In contrast, the electrochemical potential, which is also usually called Fermi level, lies in the gap, as explained in detail in above comments.
Its a quite older post but I was just searching something and came across this post. Actually it is very interesting and you may think it was just taken. But this is not the case. Moreover intrinsic Fermi level is not always at the mid of the band gap, actually it deviates from the midgap depending on effective density of states.
In intrinsic semiconductor under equilibrium condition you know n=p=n_i.
Say E_i is the fermi level in intrinsic semiconductor.
Then n=N_c*exp(-(E_c-E_i)/kT)
p=N_v*exp(-(E_i-E_v)/kT)
using n=p; you will get
E_i = (E_c-E_v)/2 - kT ln(N_c/N_v)/2
This equation clearly says only if effective density of state of electron and holes in the CB and VB respectively are same only then we have intrinsic fermi level at the midgap otherwise not. Usually 2nd term in above equation is negligible and fermi level is approximately at the midgap. This was the mathematical argument. If we look from physical prospective then we can say that if density of states of CB and VB are same and number of electrons leaving the VB has to get space in the CB, then probability of vacancy of at VB edge should be equal to the probability of occupancy at CB edge. Which is possible only if fermi level is at the midgap of the semiconductor. (Note the assumption made in the argument). From this argument also we can say if N_c is higher than N_v then less probability of occupancy at CB edge as compared to probability of vacancy at VB edge will create number of electron in CB equal to vacancy in VB and vice versa.
Hope these argument help you to understand intrinsic fermi level in better way.
First of all the Fermi level can be considered as an hyphothethical energy level of an electron, in which it is 50% of occupancy.This Fermi level thus, doesnot necesserly corresponds to actual energy level. If you look at the equation for the probability of finding electron at a given energy level E, F(E) = 1/[1+exp(E-Ef)/KT; consider at T different from zero and E=Ef, then you find that F(E) =1/2.
Hope I answer your question!
See any elementary textbook from physics of condensed matter or stat. physics, for example.
The discussion up to date concerns the mathematics and misses the essential point that the Fermi level (aka, the chemical potential) is fixed by the reservoir of particles with which the system (here, the semiconductor) exchanges particles (here, electrons). The reservoir are the metallic electrodes which are attached to the semiconductor. The Fermi level of the infinite reservoir of charge represented by the electrodes (relative to the vacuum level) does not change - only the position of the bands relative to this Fermi level move up or down (due to temperature change, doping, electric field). Viewed from inside the semiconductor, it looks like the Fermi level is changing, whereas, in fact, it is the bands that move. So, there are particles at the Fermi level in the electrodes that are in electrical contact with the semiconductor.
i agree with Joshua. At T=0 K and when there is no gap between CB and VB then Fermi level that is on the top of VB band represent the probability of an electron to be 50% in conduction band and 50% in Valence band. Generally it represent the energy reference. Now the same case with T not equal to 0 K and with a gap between CB and VB, here also Fermi level represent the same case of probability. Now Consider n type semiconductor where donors available to donate electrons and the donor level lie just below the CB, at that situation Fermi level goes up because of some greater probability of electron of this side and this process just reverse for the p type semiconductor. i hope you got the point why in intrinsic semiconductor Fermi level lies in middle.
Best regards, A. Ahad
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