CH274 - Electrons in Solids and Materials Exercise 5.1
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CH274 - Electrons in Solids and Materials
Exercise 5.1 – contact potential in p-n junction
Consider a p-n junction made of Ge doped with the same concentration, nd = 5×1014 cm3, of Al and Sb atoms in the two halves of the junction, respectively. Assume that NC = NV = 7×1018 cm3, where NC is the effective density of states at the bottom of the conduction band and NV is the effective density of states at the top of the
valence band. Further assume that the energy gap of germanium is Eg = 0.66 eV. What is the value of the contact potential of the p-n junction at 300 K?
Model answer
The contact potential C is due to the electric field generated by the unscreened atomic ions in the depletion region.
The contact potential is equal to the difference of the two Fermi levels of the p and the n doped parts of the junction before putting them in contact.
The position of the Fermi level in the n part of the junction before contact (let’s call it EF,n) can be obtained from the equation
nd = NC e一(EC 一EF ,n )/ kT ,
where NC is the effective density of states at the conduction band edge, EC is the energetic position of bottom of the conduction band and nd is the donor density. By solving this equation with respect to EF,n:
EF ,n = EC 一 kT ln(|| NC )| .
Similarly, the position of the Fermi level in the p part of the junction before contact (let’s call it EF,p) will be given by the following equation:
EF ,p = EV + kT ln(|| NV )| ,
where NV is the effective density of states at the valence band edge, EV is the energetic position of top of the valence band and pa is the acceptor density.
As a consequence,
C = EF ,n 一 EF ,p = EC 一 kT ln))|| 一 EV 一 kT ln))|| = (EC 一 EV )一 kT ln))|| .
Under the conditions given by the problem data, i.e. NC = NV and nd = pa,
C = Eg 一 2kTln))|| = 0.66 一 2 . 1.38 〉10一 23 . 300 . ln ))||/1.6〉10一19 = = 0.166 eV.
2022-09-19