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.

Gibb's Factor $=exp\left[-\frac{1}{kT}\left(E(s\right)-\mu N\left(s\right))\right]......\left(1\right)$$=exp\left[-\frac{1}{kT}\left(E\right(s)-\mu N(s\left)\right)\right].....\left(1\right)$

Here, $k$is the Boltmann's constant, $T$is the temperature, $E\left(s\right)$is the energy of the state $s$, $N\left(s\right)$is the number of atoms for the state $s$, and $\mu$is the chemical potential.

Let us consider the system to be a single donor atom, there are three possibilities:

(1)

Unoccupied state which means one ionized state with no electron: In this case, the energy of the state is equal to zero

$E=0$

The occupation number of atoms is equal to zero.

$N=0$

Substitute $0$for $E$and $0$for $N$in the equation (1) and simplify.

Gibbs factor $=exp\left[-\frac{\left(0-\mu (0\right))}{kT}\right]$

$$\begin{gathered} =exp\left(0\right) \\ =1 \end{gathered}$$

In this case, the energy of the state is equal to $-I$

$E=-I$

Here, $I$is the ionization energy.

The occupation number of atoms is equal to $1$.

$N=1$

Substitute $1$for $E$and $1$for $N$in the equation (1) and simplify.

$$\begin{gathered} Gibbsfactor=exp\left(\frac{-(-I-\mu )}{kT}\right) \\ =exp\left(\frac{I+\mu }{kT}\right) \end{gathered}$$

Since the electron has two independent state. So the degeneracy is $2$, then

Gibbs factor $=2exp\left(\frac{\left(I+\mu \right)}{kT}\right)$

The grand partition function is sum of the Gibbs factors.

$Z=1+2exp\left(\frac{\left(I+\mu \right)}{kT}\right)$

$P_{ionized}=\frac{1}{Z}$

Substitute $1+2exp\left(\frac{\left(I+\mu \right)}{kT}\right)$for $Z$.

$P_{ionized}=\frac{1}{1+2exp\left({\displaystyle \frac{I+\mu }{kT}}\right)}$

Hence, the probability that the donor atom is ionized is$\frac{1}{1+2exp\left({\displaystyle \frac{\left(I+\mu \right)}{kT}}\right)}$

(b)

$Z_{int}=2$

Use equation 7.10 and then the expression for chemical potential is as follows:

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

Here, $V$is the total volume, $Z_{int}$is the sum over all relevant internal states, $N_c$is the number of the conduction electrons, and $v_Q$is the quantum volume,

Substitute $2$for $Z_{int}$.

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

Hence, the chemical potential is$kT\ln \left(\frac{N_c{v}_Q}{2V}\right)$

(c)

$P_{ionized}=\frac{N_c}{N_d}$

Here $N_c$is the number of conduction electrons, $N_d$is the number of donor atoms.

Rearrange the equation $\mu =kT\ln \left(\frac{N_c{v}_Q}{2V}\right)$for $\frac{N_c{v}_Q}{2V}$.

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

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

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

Substitute $\frac{1}{1+2exp\left({\displaystyle \frac{I+\mu }{kT}}\right)}$for $P_{ionized}$in the equation role="math" $P_{ionized}=\frac{N_c}{N_d}$and solve for $\frac{N_c}{N_d}$.

$$\begin{gathered} \frac{N_c}{N_d}=\frac{1}{\left(1+2exp\left({\displaystyle \frac{I+\mu }{kT}}\right)\right)} \\ =\frac{1}{1+2exp\left({\displaystyle \frac{\mu }{kT}}\right)exp\left({\displaystyle \frac{I}{kT}}\right)} \end{gathered}$$

Substitute $\frac{N_c{v}_Q}{2V}$for $exp\left(\frac{\mu }{kT}\right)$

$$\begin{gathered} \frac{N_c}{N_d}=\frac{1}{1+2\left({\displaystyle \frac{N_c{v}_Q}{2V}}\right)exp\left({\displaystyle \frac{I}{kT}}\right)} \\ =\frac{1}{1+\left({\displaystyle \frac{N_c{v}_Q}{2V}}\right)exp\left({\displaystyle \frac{I}{kT}}\right)} \end{gathered}$$

$x=\frac{N_c}{N_d},y=\left(\frac{N_d{v}_Q}{V}\right)exp\left(\frac{I}{kT}\right)$and then the equation

$\frac{N_c}{N_d}=\frac{1}{1+\left({\displaystyle \frac{N_c{v}_Q}{V}}\right)exp\left({\displaystyle \frac{I}{kT}}\right)}$becomes as follows:

$$\begin{gathered} x=\frac{1}{1+xy} \\ {x}^{2}y+x-1=0 \end{gathered}$$

Solve the above quadratic equation.

$x=\frac{-1\pm \sqrt{1+4y}}{2·y}$

Substitute $\frac{N_c}{N_d}$for $x,\left(\frac{N_d{v}_Q}{V}\right)exp\left(\frac{I}{kT}\right)$for $y$.

$$\begin{gathered} \frac{N_c}{N_d}=\pm \frac{-1\sqrt{1+4{\displaystyle \frac{N_d{v}_Q}{V}}exp}\left({\displaystyle \frac{I}{kT}}\right)}{2·\left({\displaystyle \frac{N_d{v}_Q}{V}}\right)exp\left({\displaystyle \frac{I}{kT}}\right)} \\ {N}_c=\pm \frac{V}{2v_{Q}exp\left({\displaystyle \frac{I}{kT}}\right)}\left[\sqrt{1+4\frac{N_d{v}_Q}{V}exp\left(\frac{I}{kT}\right)}-1\right] \end{gathered}$$

The number of conduction electrons always a positive number. Hence, the number of donor atoms is as follows:

$N_c=\frac{V}{2v_{Q}exp\left({\displaystyle \frac{I}{kT}}\right)}\left[\sqrt{1+4\frac{N_d{v}_Q}{V}exp\left(\frac{I}{kT}\right)}-1\right]$

(d) Quantum volume $v_Q={\left(\frac{h}{\sqrt{2\mathrm{\pi mkT}}}\right)}^3$

Substitute $6.625\times {10}^{-34}J.s$for $h,1.66\times {10}^{-27}Kg$for $m$, and $1.381\times {10}^{-23}J{K}^{-1}$for $k$.

$$\begin{gathered} v_Q={\left(\frac{6.625\times {10}^{-34}J.s}{\sqrt{\left(2\mathrm{\pi }\right)\left(32\right)(1.66\times {10}^{-27}\mathrm{Kg})(\mathrm{i}.381\times {10}^{-23}{\mathrm{JK}}^{-1})\mathrm{T}}}\right)}^3 \\ =\frac{2.938\times {10}^{-29}}{{\left(T\right)}^{{\displaystyle \frac{3}{2}}}} \end{gathered}$$

For the phosphorous in silicon, the ionization energy is as follows:

$$\begin{gathered} I=0.044eV\left(\frac{1.6\times {10}^{-19}J}{1.0eV}\right) \\ =0.0704\times {10}^{-19}J \end{gathered}$$

Substitute $0.704\times {10}^{-19}J$for $I,1.381\times {10}^{-23}J{K}^{-1}$for $k$in the above equation we get

$$\begin{gathered} \frac{I}{kT}=\frac{0.0704\times {10}^{-19}J}{(1.381\times {10}^{-23}J.K^{-1})T} \\ =\frac{509.775}{T} \end{gathered}$$

Simplify the equation$$\begin{gathered} x=\frac{-1\pm \sqrt{1+4y}}{2.y} \\ x=\pm \frac{\left[-1+\sqrt{1+4y}\right]}{2y} \\ =\frac{1}{2y}\left[-1+\left[1+\frac{4y}{2}+\frac{{\left(4y\right)}^2\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}\right]\right] \\ =\frac{1}{2y}\left(2y-\frac{4^2{y}^2}{4.2!}\right)(neglecthigherterms) \\ x=(1-2y) \end{gathered}$$

$$\begin{gathered} N_d={10}^{17}Patoms/cc \\ ={10}^{17}\times {10}^{6}Patoms.Cubicmeter \\ ={10}^{23}Patoms.Cubicmeter \end{gathered}$$

Substitute $\frac{N_c}{N_d}$for $x,\left(\frac{N_d{v}_Q}{V}\right)exp\left(\frac{I}{kT}\right)$for $y,\frac{3.084\times {10}^{-10}}{\sqrt{T}}$for $v_Q$in the equation $x=(1-2y)$and simplify.

$$\begin{gathered} \frac{N_c}{N_d}=\left[1-2\times \frac{N_d}{V}exp\left(\frac{I}{kT}\right)\right] \\ =\pm \left[1-\left(2\times \frac{{10}^{23}\times 2.938\times {10}^{-29}}{{\left(T\right)}^{{\displaystyle \frac{3}{2}}}}\right)\times exp\left(\frac{2\times 509.775}{T}\right)\right] \\ =\pm \left[1-\frac{5.876\times {10}^{-6}}{{\left(T\right)}^{{\displaystyle \frac{3}{2}}}}exp\left(\frac{\left(1019.55\right)}{T}\right)\right] \\ =\left[\frac{6.168\times {10}^{-6}}{{\left(T\right)}^{{\displaystyle \frac{3}{2}}}}exp\left(\frac{\left(1019.55\right)}{T}\right)-1\right] \end{gathered}$$

On the axis, one unit is equal to $100K$