Why Does the Metal Fermi Level Remain Constant in Schottky Barriers?

[paul_harris77]

Dear All

I am having problems with how to draw band diagrams with band bending for Schottky Barriers and PN junctions. My issue is that I don't know how you determine which Fermi level remains fixed at its equilibrium level and which Fermi level moves to align with it when the two materials are brought into contact.

My textbook gives the following diagram for an N-type semiconductor in contact with a metal with a higher work function than the semiconductor:

[URL]http://homepage.ntlworld.com/beehive77/images/ntypeschottky.jpg[/URL]

It looks to me here that the metal fermi level remains constant at its equilibrium value, and the n-type semiconductor moves downwards to meet it. I understand the direction in which the bending must take place and the theory behind it, but I don't understand why it is the metal fermi level that remains constant, and not the n-type fermi level.

Could it not just as easily be the n-type fermi level remaining constant and the metal fermi level moving upwards to meet it?

My textbook then gives me the following diagram for a p-type semiconductor in contact with a metal with a lower work function than the semiconductor:

[URL]http://homepage.ntlworld.com/beehive77/images/ptypeschottky.jpg[/URL]

This time, the semi conductor fermi level remains constant and the metal fermi level moves up to align with it. Why is it the opposite way round?

I originally assumed that both fermi levels would move to meet each other and meet at some "middle ground" energy level, but the textbook seems to suggest otherwise!

Please could someone shed some light onto what happens and why?

Any answers would be greatly appreciated!

Many thanks

Paul

 

[jsgruszynski]

The thing is that the Work Function (~ the "metal Fermi level") and semiconductor Fermi level must equilibrate because of the boundary condition of contact. For an N type semiconductor, the Fermi level is nearer the top of the band (because of doping) while it is near the bottom of the band for P type. Thus the above or below depends on this. This is what bends the interface gap down or up.

The reason why it bends is that you have formed a junction vaguely like a PN junction. Except the depletion is only on the semiconductor side - there is effectively an infinite amount of carriers on the metal side (metals have an "electron gas") so its "depletion layer" is infinitesimally small. This is why the metal never "bends" its work function - tons of electrons as the carrier concentration in a metal is like 10²⁵ to 10²⁶ (closer Avogadro's number or the number of atoms) rather than the 10¹⁸ for even a highly doped semiconductor. That's a lot of carriers to recombine any possibly minority carrier injection into the metal. Essentially naught depletion or other effect on the metal.

Remember that the Fermi level is a statistical parameter related to carrier concentration. And then remember that the minority carrier concentration of a PN junction drops exponentially from the interface - kind of like how the bending of the band on the semiconductor side occurs.

Iff (if and only if) the Fermi level was exactly at the Work Function energy to begin with would there be no bending. This is basically what you are trying to achieve with an ohmic contact but you don't exactly get that since you are at the whims of the periodic table. But if the difference is small it's not much of a barrier (but is the source of "contact resistance").