Consider a system consisting of a single impurity atom/ion in a semiconductor. Suppose that the impurity atom has one "extra" electron compared to the neighboring atoms, as would a phosphorus atom occupying a lattice site in a silicon crystal. The extra electron is then easily removed, leaving behind a positively charged ion. The ionized electron is called a conduction electron because it is free to move through the material; the impurity atom is called a donor, because it can "donate" a conduction electron. This system is analogous to the hydrogen atom considered in the previous two problems except that the ionization energy is much less, mainly due to the screening of the ionic charge by the dielectric behavior of the medium.

(a) Write down a formula for the probability of a single donor atom being ionized. Do not neglect the fact that the electron, if present, can have two independent spin states. Express your formula in terms of the temperature, the ionization energy I, and the chemical potential of the "gas" of ionized electrons.

(b) Assuming that the conduction electrons behave like an ordinary ideal gas (with two spin states per particle), write their chemical potential in terms of the number of conduction electrons per unit volume,$\raisebox{1ex}{$N_c$}\!\left/ \!\raisebox{-1ex}{$V$}\right.$.

(c) Now assume that every conduction electron comes from an ionized donor atom. In this case the number of conduction electrons is equal to the number of donors that are ionized. Use this condition to derive a quadratic equation for $N_c$in terms of the number of donor atoms $\left(N_d\right)$, eliminatingµ. Solve for $N_c$using the quadratic formula. (Hint: It's helpful to introduce some abbreviations for dimensionless quantities. Try$x=\raisebox{1ex}{$N_c$}\!\left/ \!\raisebox{-1ex}{$N_d$}\right.$,$t=\raisebox{1ex}{$kT$}\!\left/ \!\raisebox{-1ex}{$l$}\right.$and so on.)

(d) For phosphorus in silicon, the ionization energy is localid="1650039340485" $0.044eV$. Suppose that there are $1017p$atoms per cubic centimeter. Using these numbers, calculate and plot the fraction of ionized donors as a function of temperature. Discuss the results.

We need to find a formula for the probability of a single donor atom being ionized.

Gibbs factor is given as:

$P\left(s\right)=e\raisebox{1ex}{$-\left(\epsilon -\mu \right)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.$

Consider a system that consists of a single impurity (atom/ion) which has three possible states, one is an unoccupied state and the other two are occupied states (spin up and spin down):

• Unoccupied state:
  in this case the chemical potential is zero and the energy of the state is zero also, so the Gibbs factor is:

$P\left(s\right)=e^{-\left(0-0\right)/kT}=1$

• Occupied state (when the electron in the ground state):
  in this case substitute with ionization energy $-I$into $\epsilon$(note that we multiply the factor by $2$since we have two electrons):

$P^{\text{'}}\left(s\right)=2e^{\left(I+\mu \right)/kT}$

The Grand partition function is the sum of the two Gibbs factors, that is:

localid="1650885202394" role="math" $Z=1+2e^{\left(I+\mu \right)/kT}$

The probability that the donor atom is ionized equals it's Gibbs factor divided by the Grand partition function:

localid="1650885211600" role="math" $\begin{array}{c}P\left(s\right)=\frac{P\left(s\right)}{Z}=\frac{1}{1+2e^{\left(I+\mu \right)/kT}}\\ P\left(s\right)=\frac{1}{1+2e^{\left(I+\mu \right)/kT}}\end{array}$

We need to find the number of conduction electrons per unit volume,$\raisebox{1ex}{$N_c$}\!\left/ \!\raisebox{-1ex}{$V$}\right.$.

The chemical potential is given by the equation $6.93$ as :

$\mu =-kT\ln \left(\frac{V{Z}_{\mathrm{int}}}{N_c{v}_Q}\right)$

the internal partition function is $2$because we have two spin states, so the chemical potential can be reduced to:

$\mu =-kT\ln \left(\frac{2V}{N_c{v}_Q}\right)$

We need to solve for $N_c$using the quadratic formula.

If every conduction electron comes from an ionized donor, then probability that the donor atom has is:

$P\left(s\right)=\frac{N_c}{N_d}$

use the result of part (a) to get:

$\begin{array}{rr}\frac{N_c}{N_d}& =\frac{1}{1+2e^{\left(I+\mu \right)/kT}}\\ \frac{N_c}{N_d}& =\frac{1}{1+2e^{I/kT}{e}^{\mu /kT}}\end{array}...\left(1\right)$

now we need to use the results of part (b), as follow:

$\begin{array}{c}-\frac{\mu }{kT}=\ln \left(\frac{2V}{N_c{v}_Q}\right)\\ e^{\mu /kT}=\frac{2V}{N_v}\end{array}$

Substitute into (1) get:

$$\begin{gathered} \frac{N_c}{N_d}=\frac{1}{1+2e{\displaystyle \raisebox{1ex}{$I$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.\left(\frac{N_c{v}_Q}{2V}\right)}} \\ \frac{N_c}{N_d}=\frac{1}{1+a{N}_c} \end{gathered}$$

where,

$a=2e^{I/kT}\left(\frac{v_Q}{2V}\right)$

therefore,
$\begin{array}{c}a{N}_c^2+N_c=N_d\\ a{N}_c^2+N_c-N_d=0\end{array}$

this is a quadratic equation which can be solved using the general law, so:

$\begin{array}{c}{N}_c=\frac{-1\pm \sqrt{1+4a{N}_d}}{2a}\\ N_c=\frac{-1\pm \sqrt{1+8e^{I/kT}\left(\frac{v_Q{N}_d}{2V}\right)}}{4e^{I/kT}\left(\frac{v_Q}{2V}\right)}\end{array}$

the number of conduction electrons is always positive, so:
$$\begin{gathered} =\frac{\sqrt{1+8e^{I/kT}\left(\frac{v_Q{N}_d}{2V}\right)}-1}{4e^{I/kT}\left(\frac{v_Q}{2V}\right)} \\ {N}_c=\frac{V{e}^{-I/kT}}{2v_Q}\left(\sqrt{1+8e^{I/kT}\left(\frac{v_Q{N}_d}{2V}\right)}-1\right) \end{gathered}$$

We need to calculate and plot the fraction of ionized donors as a function of temperature.

The quantum value is given as:

$v_Q={\left(\frac{h^2}{2\pi mkT}\right)}^{3/2}$

the mass of the phosphorus atom is $31u$, where $u=1.66\times {10}^{-27}\text{kg}$substitute with the givens to get (that $h=6.6\times {10}^{-34}J.s$and $k=1.38\times {10}^{-23}J.k$)

$$\begin{gathered} v_Q={\left(\frac{{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{2\pi \left(31\times 1.66\times {10}^{-27}\text{kg}\right)\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)T}\right)}^{3/2} \\ =3.086\times {10}^{-29}{\text{T}}^{-3/2}{\text{m}}^3/{\text{K}}^{3/2} \end{gathered}$$

now we need to simplify the result of part (c), the solution of the equation is:

$N_c=\frac{\sqrt{1+4a{N}_d}-1}{2a}$

using the expansion:

$\sqrt{1+x}\approx 1+\frac{x}{2}-\frac{x^2}{8}$

With $x=4a{N}_d$, then:

$\begin{array}{c}{N}_c=\frac{1}{2a}\left(\frac{4a{N}_d}{2}-\frac{{\left(4a{N}_d\right)}^2}{8}\right)\\ N_c=N_d-a{N}_d^2\\ \frac{N_c}{N_d}=1-a{N}_d\\ \frac{N_c}{N_d}=1-2e^{I/kT}\left(\frac{v_Q{N}_d}{2V}\right)\end{array}$

Suppose we have ${10}^{17}\text{P}$atoms per ${\text{cm}}^3$:

$\frac{N_d}{V}=1\times {10}^{23}atom/{\text{m}}^3$

and the ionization energy for phosphorus in silicon is $I=0.044\text{eV}$substitute into$N_c/N_d$get:

$\begin{array}{c}\frac{N_c}{N_d}=1-e^{0.044\text{eV}/\left(8.62\times {10}^{-5}\text{eV}/\text{K}\right)T}\left(\frac{1\times {10}^{23}\times 3.086\times {10}^{-29}}{T^{3/2}}\right)\\ \frac{N_c}{N_d}=1-e^{510.44\text{K}/T}\left(\frac{3.086\times {10}^{-6}}{T^{3/2}}\right)\end{array}$

To plot this function I used python, and the code is shown in the following picture: