Chapter 14.12, Problem 74AAP

A silicon wafer is doped with 2.50 × 10¹⁵ phosphorus atoms/cm³, 3.00 × 10¹⁷ boron atoms/cm³, and 3.00 × 10¹⁷ arsenic atoms/cm³. Calculate (a) the electron and hole concentrations (carriers per cubic centimeter), (b) the electron and hole mobilities (use Fig. 14.26), and (c) the electrical resistivity of the material.

To determine

The concentration of electrons and holes.

Answer to Problem 74AAP

The concentration of electrons is $2.50\times {10}^4{\text{cm}}^{\text{-3}}$.

The concentration of holes is $9\times {10}^{15}{\text{cm}}^{-3}$.

Explanation of Solution

Write the expression to calculate the concentration of holes.

    $p_p=N_a-N_d$                                                                      ...... (I)

Here, concentration of boron atoms is $N_a$, the sum of the concentration of phosphorus and arsenic is $N_d$ and the concentration of holes is $p_p$.

Write the expression to calculate $N_d$.

    $N_d=\text{concentration}\text{of}\text{phosphorus}+\text{concentration}\text{arsenic}$                          ...... (II)

Write the expression to calculate the concentration of electrons.

    $n_p=\frac{n_i{}^2}{p_p}$                                                                          ...... (II)

Here, the intrinsic carrier concentration is $n_i$ and the concentration of electrons is $n_p$.

Conclusion:

Substitute $2.5\times {10}^{15}{\text{cm}}^{-3}$ for concentration of phosphorus and $3\times {10}^{17}{\text{cm}}^{\text{-3}}$ for concentration of arsenic in equation (II).

    $\begin{array}{c}{N}_d=2.5\times {10}^{15}{\text{cm}}^{-3}+3\times {10}^{17}{\text{cm}}^{\text{-3}}\\ =\left(2.5+300\right)\times {10}^{15}{\text{cm}}^{\text{-3}}\\ =3.025\times {10}^{17}{\text{cm}}^{\text{-3}}\end{array}$

Substitute $3.025\times {10}^{17}{\text{cm}}^{-3}$ for $N_d$ and $3\times {\text{10}}^{17}{\text{cm}}^{-3}$ for $N_a$ in equation (I).

    $\begin{array}{c}{p}_p=3.025\times {10}^{17}{\text{cm}}^{-3}-3\times {\text{10}}^{17}{\text{cm}}^{-3}\\ =\left(3.025-3\right)\times {10}^{17}{\text{cm}}^{\text{-3}}\\ =2.5\times {10}^{15}{\text{cm}}^{\text{-3}}\end{array}$

Substitute $2.5\times {10}^{16}{\text{cm}}^{-3}$ for $p_p$ and $1.5\times {10}^{10}{\text{cm}}^{-3}$ for $n_i$ in equation (II).

    $\begin{array}{c}{n}_p=\frac{{\left(1.5\times {10}^{10}{\text{cm}}^{-3}\right)}^2}{2.5\times {10}^{15}{\text{cm}}^{-3}}\\ =\frac{2.25\times {10}^{20}{\text{cm}}^{\text{-6}}}{2.5\times {10}^{15}{\text{cm}}^{-3}}\\ =9\times {10}^4{\text{cm}}^{\text{-3}}\end{array}$

Thus, the concentration of electrons is $2.50\times {10}^{15}{\text{cm}}^{\text{-3}}$.

Thus, the concentration of holes is $9\times {10}^4{\text{cm}}^{-3}$.

To determine

The electron and hole motilities.

Answer to Problem 74AAP

The mobility of electrons is $400{\text{cm}}^{\text{2}}/\text{V}\cdot \text{s}$.

The mobility of holes is $125{\text{cm}}^{\text{2}}/\text{V}\cdot \text{s}$.

Explanation of Solution

Conclusion:

Refer to the Figure-14.26, “The effect of total ionized impurity concentration on the mobility of charge carriers in silicon at room temperature.” to obtain the value of total impurity concentration as $C_T=6.025\times {10}^{17}\text{ions}/{\text{cm}}^{\text{3}}$.

Refer to the Figure-14.26, “The effect of total ionized impurity concentration on the mobility of charge carriers in silicon at room temperature.” to obtain the value of electron mobility at $C_T=4.1\times {10}^{16}\text{ions}/{\text{cm}}^{\text{3}}$ as ${\mu }_n=400{\text{cm}}^{\text{2}}/\text{V}\cdot \text{s}$ and holes mobility as ${\mu }_p=125{\text{cm}}^{\text{2}}/\text{V}\cdot \text{s}$.

Thus, the mobility of electrons is $400{\text{cm}}^{\text{2}}/\text{V}\cdot \text{s}$.

Thus, the mobility of holes is $125{\text{cm}}^{\text{2}}/\text{V}\cdot \text{s}$.

To determine

The electrical resistivity of the material.

Answer to Problem 74AAP

The electrical resistivity of the material is $6.25\Omega \cdot \text{cm}$.

Explanation of Solution

Conclusion:

Write the expression to calculate the resistivity for n-type semiconductor.

    $\rho =\frac{1}{q{\mu }_p{p}_p}$                                                                                ...... (III)

Substitute $1.6\times {10}^{-19}\text{C}$ for $q$, $400{\text{cm}}^{\text{2}}/\text{V}\cdot \text{s}$ for ${\mu }_n$ and $2.50\times {10}^{15}{\text{cm}}^{\text{-3}}$ for $n_n$ in equation (III).

    $\begin{array}{c}\rho =\frac{1}{\left(1.6\times {10}^{-19}\text{C}\right)\times \left(400{\text{cm}}^{\text{2}}/\text{V}\cdot \text{s}\right)\times \left(2.50\times {10}^{15}{\text{cm}}^{\text{-3}}\right)}\\ =\frac{1}{\left(640\times {10}^{-19}\text{C}\cdot {\text{cm}}^{\text{2}}/\text{V}\cdot \text{s}\right)\left(2.50\times {10}^{15}{\text{cm}}^{\text{-3}}\right)}\\ =\frac{1}{1600\times {10}^{-4}{\left(\Omega \cdot \text{cm}\right)}^{-1}}\\ =6.25\Omega \cdot \text{cm}\end{array}$

Thus, the electrical resistivity of the material is $6.25\Omega \cdot \text{cm}$.