[SOLVED] Is Fermi level in Metals- constant ? and how do Quasi Fermi levels in PN junctions ?

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Hello,

a) Does the Fermi level of the metal changes when applying a voltage across Metal Oxide Semiconductor Capacitor ?

b) Why does the Quasi Fermi level moves up with decrease in potential and vice versa ? For Example : In case of F.B., p-side potential is higher so E(FP)<E(FN). Why ?

Thank you !!
 

Too many questions!!

1. Fermi level in metal is a constant at absolute zero. At any finite (non-zero) temperatures, the top energy levels are distorted (they are partially filled) and only these electrons are responsible for conduction. All the electrons in filled orbitals are not available for conduction.

2. As you go up higher in energy level, the degeneracy increases. That means there are many electrons with the same energy.

3. What happens if you put the metal in an external electric field? Instead of writing equations, I shall explain very qualitatively. When a hydrogen atom is placed in an external electric field or magnetic field, the degeneracy is lifted. That means electrons that had same energy (before applying the electric field) do not have the same energy any more (after the application of electric field). That is Stark effect (and Zeeman effect). Because the system is adiabatic, half of the energy levels will be increased and and another half will have their levels reduced (by the same amount).

4. For a metal, valence band and the conduction bands shift in presence of an external electric field (when you apply a potential across a pn- junction). How they shift? That is complicated (It means I have to look up).

5. But what happens when you create a junction? Say a pn junction of the same material (crystal structure) but different carriers on two different sides of the junction? Here you have a p type material that have conduction band, valence band and an impurity band. It is the impurity band that we are concerned. Current is carried by p-type (or n-type) carriers on one side. Valence band and the conduction band will match nicely (because the underlying base is same and is responsible for the valence and conduction bands). But the impurity bands will not. In fact they will now be mis-aligned because they are different impurities. One will be at a higher level and the other will be (obvious now) at a lower level. Electrons (or holes) will move from the higher level to the lower one (and recombine- just disappear) but this causes a potential to be produced.The side that loses n-type carrier will be positive and the side that loses p-type carrier will be negative. A steady state is produced when the junction potential prevents further diffusion.
 

1st point :
"In the metal, C.B and V.B. overlaps. Fermi Level is in the C.B., so you are talking about the states above the Fermi level which are empty at absolute zero. At non zero temperature, some of the states above Fermi level will be filled and will be responsible for the conduction along with the holes below the Fermi Level. Am I right ?"

4th point:

" I thought- since metal has abundance of electrons, so pulling some electrons out or putting some into the metal will not change the Fermi Level. Isn't it so ?"
" I follow this too but my question was- how this Quasi Fermi Level change with the external bias ? Positive Bias means the Quasi level will go down in energy whether its on p side or n side, why ?"

5th point:
I understand the formation of the Quasi Fermi levels but how do they change with an external bias ? Its an inverse relation. Increase in potential will cause the Fermi level to go down in energy. I think you did explain it a bit but I could not follow it. I am sorry
 

1st query: right. Levels deep down will have no effect on the electrical conduction.

4th query: All solids have abundance of electrons but for most, they are not available for conduction. For metals, electrons at the top of the Fermi level are available for conduction (at non zero temperature). Because the conduction band overlaps with the valence band, even a small potential is sufficient for conduction. External potential (electric field) does distort both conduction and valence bands. For metals, both electrons and holes conduct (approx) equally. Your second part (Quasi level will go down in energy whether its on p side or n side, why?) applies for pn junction and not for a homogeneous metal. The exact nature of the change depends on the crystal structure, electron density (search Fermi surface of Copper, Gold or any other metal) etc. In general, the band will be split and the splitting depends on the electric field strength.

5th query: a pn junction is made in the same crystal but different impurity bands in different parts. The impurity bands are therefore localized (say p type on the left and n type on the right of a semiconductor rod). Assume the junction to be sharp (for discussion purposes). Because the impurity bands are localized, there will be majority carriers and minority carriers (of opposite types on either side). At the junction, there will be some holes and electrons recombination and that will a potential difference. On the p-type side, there will be negative charge and on the n-type side there will be a positive charge. This makes an electric field that causes the bands to distort and merge in a continuous curve (the distortion will be only near the junction). The potential will prevent further diffusion and recombination of electrons and holes. This is a steady state in absence of external potentials.

When you add an external potential what happens? It is simple to consider the junction as a capacitor (this is a real effect because a very thin layer at the junction is devoid of majority carriers- they have recombined). You add +v to the p-side, the junction is having -ve charge on that side. The electrons are taken away and the junction potential decreases and that helps more diffusion of electrons and holes. Same thing happens to the -ve added to the n-side. Basically you are trying to reduce the junction potential but this causes flow of majority carriers on both sides.
 
 

Did I get this right ?
Click to expand...

yes, that is what exactly happens (it is somewhat more complex than that but this is the simplest).

When you reverse bias the junction, you add electrons to the p-side, they reach the junction and the negative charge increases. And at the same time you are removing electrons from the n side, the positive charge on the junction increases on that side. The net result is that the junction potential increases and this blocks the flow of current (till the junction breaks down because of excessive electric field and the thin layer).

The junction is not sharp, it formed by diffusion of electrons and holes (the majority carriers from the respective sides) but with a reverse bias the layer becomes thinner. Opposite thing happens when you forward bias the junction.

For LEDs, the peculiar shape of the band at the junction is interesting and causes recombination of electrons and holes with emission of light.

Forget about the LEDs. Just focus on a point contact diode. It has a metal semiconductor interface that is very small.

We did not discuss about the minority carriers. But they are there all the time. Basically they come from the intrinsic semiconductor. But remember that a diode that has a lower forward drop will have higher reverse current. Interesting, but true.

Conventional diode drop is essentially the junction potential - at higher current the double layer at the junction becomes more diffuse but the overall electric field is reduced.

Any reverse biased diode can be used as a vericap. Only manufacturing details change but not the physics.

At the junction, there will be some holes and electrons recombination and that will a potential difference
Click to expand...

In absence of external potential, some holes and electrons recombine. Hence a thin layer becomes devoid of charge. But the p-side that has lost some holes due to recombination becomes negatively charged. And the n-side that has lost some electrons becomes positively charged. That is because before the recombination, each side was electrically neutral. Once sufficient potential is developed at the junction, diffusion of holes and electrons stops (the negative charge on the p-side holds back further movement of holes and the same principle works on the other side).
 
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