Question

Doping changes the Fermi energy of a semiconductor. Consider silicon, with a gap of 1.11 eV between the top of the valence band and the bottom of the conduction band. At 300 K the Fermi level of the pure material is nearly at the midpoint of the gap. Suppose that silicon is doped with donor atoms, each of which has a state 0.13 eV below the bottom of the silicon conduction band, and suppose further that doping raises the Fermi level to 0.11 eV below the bottom of that band (see the figure below). For (a) pure and (b) doped silicon, calculate the probability that a state at the bottom of the silicon conduction band is occupied. (c) Calculate the probability that a donor state in the doped material is occupied.

Answer 1

Given

     Temperature T = 300 K

    Boltzmann constant k = 8.62*10^-5 eV / K

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Part a)

the probability that astate at the bottom of the pure silicon conduction band isoccupied is given by

         P(t) =

Here, Energy ofenergy gap E = 1.11 eV

energy of the Fermi state, E_F = E/2 = 0.555 eV

substituting the values, we get

P(t) = 4.78*10^-10

Part b)

the probability that a state of doped silicon is occupied is given by

         P '(t) =

      Here E - E_F = 0.11 eV

substituting the values, we get

P'(t) = 0.014

Part c)

the probability thata donor state in the doped material is occupied is given by

         P " (t)=

      Here,

E = 1.11 - 0.13 = 0.98 eV

E_F = 1.11- 0.11 = 1.0 eV

substituting the values, we get

P''(t) = 0.68