Chapter 3, Problem 3.1P

To determine

The intrinsic carrier $n_i$ for silicon and the fraction value of atom ionized at $T=-{55}^{\circ }\text{C}$ , $T=0^{\circ }\text{C}$ , $T={20}^{\circ }\text{C}$ , $T={75}^{\circ }\text{C}$ and $T={125}^{\circ }\text{C}$ .

Answer to Problem 3.1P

The value of intrinsic carrier $n_i$ at $T=-{55}^{\circ }\text{C}$ is $26.84\times {10}^5\text{carriers}/{\text{cm}}^3$ , $T=0^{\circ }\text{C}$ is $1.523\times {10}^9\text{carriers}/{\text{cm}}^3$ , $T={20}^{\circ }\text{C}$ is $8.595\times {10}^9\text{carriers}/{\text{cm}}^3$ , $T={75}^{\circ }\text{C}$ is $3.7001\times {10}^{11}\text{carriers}/{\text{cm}}^3$ and $T={125}^{\circ }\text{C}$ is $4.7227\times {10}^{12}\text{carriers}/{\text{cm}}^3$ and the fraction of atom ionized at $T=-{55}^{\circ }\text{C}$ is $5.368\times {10}^{-17}$ , $T=0^{\circ }\text{C}$ is $3.04\times {10}^{-14}$ , $T={20}^{\circ }\text{C}$ is $1.719\times {10}^{-13}$ , $T={75}^{\circ }\text{C}$ is $7.4002\times {10}^{-12}$ and $T={125}^{\circ }\text{C}$ is $9.4455\times {10}^{-10}$ .

Explanation of Solution

Calculation:

The intrinsic carrier density is given by

  $n_i=B{T}^{\frac{3}{2}}{e}^{\frac{-E_g}{2kT}}$ …… (1)

Here, $n_i$ is the intrinsic carrier density, $T$ is the given temperature, $B$ is the material dependent parameter, $B=7.3\times {10}^{15}{\text{cm}}^{-3}{\text{K}}^{-\frac{3}{2}}$ , $E_g$ is the bandgap energy, $E_g=1.12\text{eV}$ and $k$ is the Boltzmann’s constant, $k=8.62\times {10}^{-5}\text{eV}/\text{K}$ .

The fraction of atom ionized is calculated as,

  $F=\frac{n_i}{5\times {10}^{22}}$ …… (2)

Here, $n_i$ is the concentration, $F$ is the fraction of ionized atom and $5\times {10}^{22}$ is the atoms in a silicon crystal.

For $-{55}^{\circ }\text{C}$ into $\text{K}$ is given by,

  $\begin{array}{c}-{55}^{\circ }\text{C}=-{55}^{\circ }\text{C}+273\text{K}\\ =\text{218}\text{K}\end{array}$

Substitute $7.3\times {10}^{15}{\text{cm}}^{-3}{\text{K}}^{-\frac{3}{2}}$ for $B$ , $1.12\text{eV}$ for $E_g$ , $8.62\times {10}^{-5}\text{eV}/\text{K}$ for $k$ and $218\text{K}$ for $T$ in equation (1).

  $\begin{array}{c}{n}_i=\left(7.3\times {10}^{15}\right){\left(218\right)}^{\frac{3}{2}}{e}^{\left(\frac{-1.12}{2\left(8.62\times {10}^{-5}\right)\left(218\right)}\right)}\\ =\left(7.3\times {10}^{15}\right)\left(3218.73\right)e^{\left(\frac{-1.12}{3758.32\times {10}^{-5}}\right)}\\ =\left(7.3\times {10}^{15}\right)\left(3218.73\right)e^{-29.8}\\ =\left(23496.74\times {10}^{15}\right)\left(1.1423\times {10}^{-13}\right)\end{array}$

Solve it further as

  $n_i=26.84\times {10}^5\text{carriers}/{\text{cm}}^3$

Substitute $26.84\times {10}^5\text{carriers}/{\text{cm}}^3$ in equation (2).

  $\begin{array}{c}F=\frac{26.84\times {10}^5}{5\times {10}^{22}}\\ =5.368\times {10}^{-17}\end{array}$

For $0^{\circ }\text{C}$ into $\text{K}$ is given by

  $\begin{array}{c}{0}^{\circ }\text{C}={\text{0}}^{\circ }\text{C}+273\text{K}\\ =\text{273}\text{K}\end{array}$

Substitute $7.3\times {10}^{15}{\text{cm}}^{-3}{\text{K}}^{-\frac{3}{2}}$ for $B$ , $1.12\text{eV}$ for $E_g$ , $8.62\times {10}^{-5}\text{eV}/\text{K}$ for $k$ and $273\text{K}$ for $T$ in equation (1).

  $\begin{array}{c}{n}_i=\left(7.3\times {10}^{15}\right){\left(273\right)}^{\frac{3}{2}}{e}^{\left(\frac{-1.12}{2\left(8.62\times {10}^{-5}\right)\left(273\right)}\right)}\\ =\left(7.3\times {10}^{15}\right)\left(4510.7\right)e^{\left(\frac{-1.12}{4706.2\times {10}^{-5}}\right)}\\ =\left(32928.112\right)e^{-23.79677}\\ =\left(23496.74\times {10}^{15}\right)\left(4.625\times {10}^{-11}\right)\end{array}$

Solve it further as

  $n_i=1.523\times {10}^9\text{carriers}/{\text{cm}}^3$

Substitute $26.84\times {10}^5\text{carriers}/{\text{cm}}^3$ in equation (2).

  $\begin{array}{c}F=\frac{1.523\times {10}^9}{5\times {10}^{22}}\\ =3.04\times {10}^{-14}\end{array}$

For ${20}^{\circ }\text{C}$ into $\text{K}$ is given by

  $\begin{array}{c}{20}^{\circ }\text{C}={\text{20}}^{\circ }\text{C}+273\text{K}\\ =\text{293}\text{K}\end{array}$

Substitute $7.3\times {10}^{15}{\text{cm}}^{-3}{\text{K}}^{-\frac{3}{2}}$ for $B$ , $1.12\text{eV}$ for $E_g$ , $8.62\times {10}^{-5}\text{eV}/\text{K}$ for $k$ and $293\text{K}$ for $T$ in equation (1).

  $\begin{array}{c}{n}_i=\left(7.3\times {10}^{15}\right){\left(293\right)}^{\frac{3}{2}}{e}^{\left(\frac{-1.12}{2\left(8.62\times {10}^{-5}\right)\left(293\right)}\right)}\\ =\left(7.3\times {10}^{15}\right)\left(5015.35\right)e^{\left(\frac{-1.12}{5051.32\times {10}^{-5}}\right)}\\ =\left(7.3\times {10}^{15}\right)\left(5015.35\right)e^{-22.1724}\\ =\left(36612.071\times {10}^{15}\right)\left(2.347\times {10}^{-10}\right)\end{array}$

Solve it further as

  $n_i=8.595\times {10}^9\text{carriers}/{\text{cm}}^3$

Substitute $26.84\times {10}^5\text{carriers}/{\text{cm}}^3$ in equation (2).

  $\begin{array}{c}F=\frac{8.595\times {10}^9}{5\times {10}^{22}}\\ =1.719\times {10}^{-13}\end{array}$

For ${75}^{\circ }\text{C}$ into $\text{K}$ is given by,

  $\begin{array}{c}{75}^{\circ }\text{C}={\text{75}}^{\circ }\text{C}+273\text{K}\\ =\text{348}\text{K}\end{array}$

Substitute $7.3\times {10}^{15}{\text{cm}}^{-3}{\text{K}}^{-\frac{3}{2}}$ for $B$ , $1.12\text{eV}$ for $E_g$ , $8.62\times {10}^{-5}\text{eV}/\text{K}$ for $k$ and $348\text{K}$ for $T$ in equation (1).

  $\begin{array}{c}{n}_i=\left(7.3\times {10}^{15}\right){\left(348\right)}^{\frac{3}{2}}{e}^{\left(\frac{-1.12}{2\left(8.62\times {10}^{-5}\right)\left(348\right)}\right)}\\ =\left(7.3\times {10}^{15}\right)\left(6491.85\right)e^{\left(\frac{-1.12}{5999.52\times {10}^{-5}}\right)}\\ =\left(47390.547\times {10}^{15}\right)e^{-18.66}\\ =\left(47390.547\times {10}^{15}\right)\left(7.807\times {10}^{-9}\right)\end{array}$

Solve it further as

  $n_i=3.7001\times {10}^{11}\text{carriers}/{\text{cm}}^3$

Substitute $26.84\times {10}^5\text{carriers}/{\text{cm}}^3$ in equation (2).

  $\begin{array}{c}F=\frac{3.7001\times {10}^{11}}{5\times {10}^{22}}\\ =7.4002\times {10}^{-12}\end{array}$

For ${125}^{\circ }\text{C}$ into $\text{K}$ is given by,

  $\begin{array}{c}{125}^{\circ }\text{C}={\text{125}}^{\circ }\text{C}+273\text{K}\\ =\text{398}\text{K}\end{array}$

Substitute $7.3\times {10}^{15}{\text{cm}}^{-3}{\text{K}}^{-\frac{3}{2}}$ for $B$ , $1.12\text{eV}$ for $E_g$ , $8.62\times {10}^{-5}\text{eV}/\text{K}$ for $k$ and $398\text{K}$ for $T$ in equation (1).

  $\begin{array}{c}{n}_i=\left(7.3\times {10}^{15}\right){\left(398\right)}^{\frac{3}{2}}{e}^{\left(\frac{-1.12}{2\left(8.62\times {10}^{-5}\right)\left(398\right)}\right)}\\ =\left(7.3\times {10}^{15}\right)\left(7940.07\right)e^{\left(\frac{-1.12}{6861.52\times {10}^{-5}}\right)}\\ =\left(57962.54796\times {10}^{15}\right)e^{-16.32291}\\ =\left(57962.54796\times {10}^{15}\right)\left(8.1479\times {10}^{-8}\right)\end{array}$

Solve it further as

  $n_i=4.7227\times {10}^{12}\text{carriers}/{\text{cm}}^3$

Substitute $26.84\times {10}^5\text{carriers}/{\text{cm}}^3$ in equation (2).

  $\begin{array}{c}F=\frac{4.72276\times {10}^{12}}{5\times {10}^{22}}\\ =9.4455\times {10}^{-10}\end{array}$