91Ó°ÊÓ

Gibb's Factor $=exp\left[-\frac{1}{kT}\left(E(s\right)-\mathrm{μ}N\left(s\right))\right]......\left(1\right)$$=exp\left[-\frac{1}{kT}\left(E\right(s)-\mathrm{μ}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 $\mathrm{μ}$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 $=exp\left[-\frac{\left(0-\mathrm{μ}(0\right))}{kT}\right]$

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

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.

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

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

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

The grand partition function is sum of the Gibbs factors.

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

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

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

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

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

(b)

$Z_{int}=2$

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

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

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 $Z_{int}$.

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

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

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

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

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

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

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

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

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

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

$x=\frac{-1Â\pm \sqrt{1+4y}}{2·y}$

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

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

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 $v_Q={\left(\frac{h}{\sqrt{2\mathrm{Ï€³¾°ì°Õ}}}\right)}^3$

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

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

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

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

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

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

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

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

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

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

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