Metal-to-semiconductor contacts are of great importance since they are present in every semiconductor device. They can behave either as a Schottky barrier or as an ohmic contact dependent on the characteristics of the interface. This chapter contains an analysis of the electrostatics of the M-S junction (i.e. the charge, field and potential distribution within the device) followed by a derivation of the current voltage characterisitics due to diffusion, thermionic emission and tunneling and a discussion of the non-ideal effects in Metal-Semiconductor junctions.

The barrier between the metal and the semiconductor can be identified on an energy band diagram. To construct such diagram we first consider the energy band diagram of the metal and the semiconductor, and align them using the same vacuum level as shown in Figure 2 (a). As the metal and semiconductor are brought together, the Fermi energies of the metal and the semiconductor do not change right away.

The barrier height, fB, is defined as the potential difference between the Fermi energy of the metal and the band edge where the majority carriers reside. From Figure 2 (b) one finds that for an n-type semiconductor the barrier height is obtained from:📷(1.1)Where FM is the work function of the metal and c is the electron affinity. The work function of selected metals as measured in vacuum can be found in Table 1. For p-type material, the barrier height is given by the difference between the valence band edge and the Fermi energy in the metal:📷(1.2)A metal-semiconductor junction will therefore form a barrier for electrons and holes if the Fermi energy of the metal as drawn on the flatband diagram is somewhere between the conduction and valence band edge.In addition, we define the built-in potential, fI, as the difference between the Fermi energy of the metal and that of the semiconductor.

Dear Muhammad Hafiz Bin Abu Bakar ,

Adding to the colleague above,

You can find the width of the depletion region for any arbitrary doping profile N(x)

On e gets first the electric field distribution E(x) by

E(x)=(1/epsilon) Integral N(x) from x=o at the bulk edge of the depletion region to x=xd at the interface between two materials.

By integrating once again one can get the potential distribution V(x) and the barrier height at x=xd assuming the potential at x=0 is equal to zero. That is V(0)=0

Then V(xd)= Integral E(x) dx from x=0 to x=xd

This voltage is the barrier height = phi- X- phi fn

phi is the metal work function.

X is the electron affinity of the semicondcutor

and phi fn= the energy difference between the fermi level Ef and the conduction band edge.

As you see once you know N(x) you can make double integration to get xd as barrier height is known.

You can solve this problem numerically for arbitrary doping profile.

Best wishes

Dear Professor Hadi Jabbar Alagealy and Professor Abdelhalim abdelnaby Zekry , thank you very much for the suggestions.