91Ó°ÊÓ

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(\mathrm{Î\mu }-\mathrm{μ}\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 $\mathrm{Î\mu }$(note that we multiply the factor by $2$since we have two electrons):

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

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

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

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

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

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 :

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

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

$\mathrm{μ}=−kT\ln \left(\frac{2V}{N_c{v}_Q}\right)$

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:

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

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

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

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:

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

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:

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

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

$$\begin{gathered} v_Q={\left(\frac{{\left(6.626×{10}^{−34}\text{J}â‹\dots \text{s}\right)}^2}{2\mathrm{Ï€}\left(31×1.66×{10}^{−27}\text{kg}\right)\left(1.38×{10}^{−23}\text{J}/\text{K}\right)T}\right)}^{3/2} \\ =3.086×{10}^{−29}{\text{T}}^{−3/2}{\text{m}}^3/{\text{K}}^{3/2} \end{gathered}$$

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:

$\sqrt{1+x}≈1+\frac{x}{2}−\frac{x^2}{8}$

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 ${\text{cm}}^3$:

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

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

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

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