Doping changes the Fermi energy of a semiconductor. Consider silicon, with a gap of 1.11eV between the top of the valence band and the bottom of the conduction band. At 300K the Fermi level of the pure material is nearly at the mid-point of the gap. Suppose that silicon is doped with donor atoms, each of which has a state 0.15eV below the bottom of the silicon conduction band, and suppose further that doping raises the Fermi level to 0.11eV below the bottom of that band (Fig. 41-22). 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 state in the doped material (at the donor level) is occupied.

a) Energy gap of silicon,$E_g=1.11eV$

b) Temperature value, $T=300K$

c) Fermi energy of silicon, $E_f=0.555eV$

d) At the given temperature T = 300K , the Fermi level is at the midpoint of the gap.

e) Silicon is doped with donor atoms; each has energy state $∆E=0.15eV$below the bottom of the silicon conduction band.

f) Doping raises Fermi level to $E=0.11eV$below the bottom of that silicon band.

When we willingly introduce certain impurities into the pure semiconductor to modify its electrical or optical properties is known as doping. The pure semiconductor is also known as an intrinsic semiconductor and the doped one is known as an extrinsic semiconductor.

Formula:

The probability of the condition that a particle will have energy E according to Fermi-Dirac statistics,

$P\left(E\right)=\frac{1}{e^{\left(E-E_F\right)kT}+1}..............................\left(1\right)$

where$∆E=E-E_F$

In the case of pure silicon, the requirement is given by:$∆E=E_F$

Thus, the value of the occupancy probability is given as-

$$\begin{gathered} P\left(E\right)=\frac{1}{e^{\left(0.555eV\right)/\left(8.62\times {10}^{-5}eV/K\right)\left(300K\right)}+1} \\ =4.79\times {10}^{-10} \end{gathered}$$

Hence, the value of the probability is $4.79\times {10}^{-10}$.

In case of doped silicon, the required is given by:$∆E=0.11eV$ .

Thus, the value of the occupancy probability is given using this data in equation (i) as follows:

$$\begin{gathered} P\left(E\right)=\frac{1}{e^{\left(0.11eV\right)/\left(8.62\times {10}^{-5}eV/K\right)\left(300K\right)}+1} \\ =1.40\times {10}^{-2} \end{gathered}$$

Hence, the value of the probability is$1.40\times {10}^{-2}$ .

The energy of the donor state relative to the top of the valence band is given by,

$$\begin{gathered} E=1.11eV-0.15eV \\ =0.96eV \end{gathered}$$

Now, the Fermi energy of the material is given by:

$$\begin{gathered} E_f=1.11eV-0.11eV \\ =1.00eV \end{gathered}$$

Thus, the occupancy probability is given using equation (i) as follows:

$$\begin{gathered} P\left(E\right)=\frac{1}{e^{\left(0.96eV-1eV\right)/\left(8.62\times {10}^{-5}eV\right)\left(300K\right)}+1} \\ =0.824 \end{gathered}$$

Hence, the probability is 0.824 .