Understanding the Charge Distribution at the Metal-Semiconductor Interface

[Robotduck]

TL;DR Summary

Metal Semiconductor Contact

Assume n type semiconductor:
1) Can the fermi level of metal change when it makes contact with the n type Semiconductor ? What assumptions do we make in ideal situation ?
2) Is the Schottky Barrier in Metal Semiconductor contact remains constant with an applied forward or reverse voltage ?
3) On what reasons, do we say that the metal semiconductor junction is forward biased when applying positive on the metal side and negative at the semiconductor side, since both have electrons as a majority carriers .

Thank you in advance !

 

[Lord Jestocost]

a) Even if some electrons are transferred from the n-type semiconductor to the metal, the Fermi energy of the metal doesn’t change because of the extremely high electron density in metals.
b) The Schottky barrier - a barrier for electrons to move from the metal to the semiconductor - is the difference between the work function of the metal and the electron affinity of the semiconductor. It doesn't change under an applied forward or reverse voltage.
c) Forward and reverse bias define whether the electrons are injected from the semiconductor to the metal or vice versa. The forward and reverse currents differ from each other because the injected electrons have different barriers to surmount.

Have, for example, a look at
Lecture 9: Metal-semiconductor junctions - nptel

 

[Robotduck]

a) Even if some electrons are transferred from the n-type semiconductor to the metal, the Fermi energy of the metal doesn’t change because of the extremely high electron density in metals.
b) The Schottky barrier - a barrier for electrons to move from the metal to the semiconductor - is the difference between the work function of the metal and the electron affinity of the semiconductor. It doesn't change under an applied forward or reverse voltage.
c) Forward and reverse bias define whether the electrons are injected from the semiconductor to the metal or vice versa. The forward and reverse currents differ from each other because the injected electrons have different barriers to surmount.

Have, for example, a look at
Lecture 9: Metal-semiconductor junctions - nptel

Thank you so much for the reply.

But Can you please elaborate more on this:
"b) The Schottky barrier - a barrier for electrons to move from the metal to the semiconductor - is the difference between the work function of the metal and the electron affinity of the semiconductor. It doesn't change under an applied forward or reverse voltage."

Why does not it change ?

"

c) Forward and reverse bias define whether the electrons are injected from the semiconductor to the metal or vice versa. The forward and reverse currents differ from each other because the injected electrons have different barriers to surmount.

So, you mean : if the electrons are injected from the semiconductor to metal: that is Forward biased case and if the flow is from metal to semiconductor then this is reverse biased case ?
Also, electrons from semiconductor to metal is considered as diffusion but how does electrons from metal to semiconductor flow is drift ? How do we get the Delta function of negative charge at the surface of the metal ?

Thank you so much

 

[Lord Jestocost]

Why does not it change ?

In an ideal case, the work function of the metal and the electron affinity of the semiconductor are material properties.

 

[Robotduck]

Thank you so much.
One last question on this:

How do we get a negative charge sheet on the metal side in a Metal -n type semiconductor contact ?

Thank you

 

[Lord Jestocost]

"Since the density of free carriers is very high in a metal, the thickness of the charge layer in the metal is very thin. Therefore, the potential across the metal is several orders of magnitude smaller than that across the semiconductor, even though the total amount of charge is the same in both regions."

see Fig. 3.3.1 in https://ecee.colorado.edu/~bart/book/book/chapter3/ch3_3.htm

 

[Robotduck]

"Since the density of free carriers is very high in a metal, the thickness of the charge layer in the metal is very thin. Therefore, the potential across the metal is several orders of magnitude smaller than that across the semiconductor, even though the total amount of charge is the same in both regions."

see Fig. 3.3.1 in https://ecee.colorado.edu/~bart/book/book/chapter3/ch3_3.htm

Please correct me if I am wrong:
With positive bias on metal side and negative on semiconductor- it will attract the electrons from the metal leaving immobile positive ions at the metal -semiconductor interface on the metal side ( this positive ion layer ) and on the semiconductor side - due to negative bias we will have negative charge ( not ions) on the semiconductor side. Do I have this right ?