Chapter 14.12, Problem 85AAP

(a)

To determine

The fraction of current carried by electron for $\text{GaSb}$.

The fraction of current carried by hole for $\text{GaSb}$.

Answer to Problem 85AAP

The fraction of current carried by electron for $\text{GaSb}$ is $0.8333$.

The fraction of current carried by hole for $\text{GaSb}$ is $0.167$.

Explanation of Solution

Write the expression for the fraction of current carried by electrons.

    $F_e=\frac{{\mu }_n}{{\mu }_n+{\mu }_p}$                                                                              (I)

Here, the mobility of the electron is ${\mu }_n$ and the mobility of the hole is ${\mu }_p$.

Write the expression for the fraction of current carried by hole.

    $F_h=\frac{{\mu }_p}{{\mu }_n+{\mu }_p}$                                                                                 (II)

Conclusion:

Refer to table 14.6 “Electrical properties of intrinsic semiconducting compounds at room temperature $300\text{K}$” to obtain the motilities for electron and hole for $\text{GaSb}$ as $0.5{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}$ and $0.1{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}$ respectively.

Substitute $0.5{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}$ for ${\mu }_n$ and $0.1{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}$ for ${\mu }_p$ in Equation (I).

    $\begin{array}{c}{F}_e=\frac{0.5{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}}{\left(0.5{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}+0.1{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}\right)}\\ =\frac{0.5{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}}{\left(0.6{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}\right)}\\ =0.8333\end{array}$

Substitute $0.5{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}$ for ${\mu }_n$ and $0.1{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}$ for ${\mu }_p$ in Equation (II).

    $\begin{array}{c}{F}_p=\frac{0.1{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}}{\left(0.5{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}+0.1{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}\right)}\\ =\frac{0.1{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}}{\left(0.6{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}\right)}\\ =0.1666\\ \approx 0.167\end{array}$

Thus, the fraction of current carried by electron for $\text{GaSb}$ is $0.8333$.

Thus, the fraction of current carried by hole for $\text{GaSb}$ is $0.167$.

(b)

To determine

The fraction of current carried by electron for $\text{GaP}$.

The fraction of current carried by hole for $\text{GaP}$.

Answer to Problem 85AAP

The fraction of current carried by electron for $\text{GaP}$ is $0.667$.

The fraction of current carried by hole for $\text{GaP}$ is $0.334$.

Explanation of Solution

Write the expression for the fraction of current carried by electrons.

    $F_e=\frac{{\mu }_n}{{\mu }_n+{\mu }_p}$                                                                                (III)

Here, the mobility of the electron is ${\mu }_n$ and the mobility of the hole is ${\mu }_p$.

Write the expression for the fraction of current carried by hole.

    $F_h=\frac{{\mu }_p}{{\mu }_n+{\mu }_p}$                                                                                (IV)

Conclusion:

Refer to table 14.6 “Electrical properties of intrinsic semiconducting compounds at room temperature $300\text{K}$” to obtain the motilities for electron and hole for $\text{GaP}$ as $0.03{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}$ and $0.015{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}$ respectively.

Substitute $0.03{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}$ for ${\mu }_n$ and $0.015{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}$ for ${\mu }_p$ in Equation (III).

    $\begin{array}{c}{F}_e=\frac{0.03{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}}{\left(0.03{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}+0.015{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}\right)}\\ =\frac{0.03{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}}{\left(0.045{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}\right)}\\ =0.6666\\ \approx 0.667\end{array}$

Substitute $0.03{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}$ for ${\mu }_n$ and $0.015{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}$ for ${\mu }_p$ in Equation (IV).

    $\begin{array}{c}{F}_p=\frac{0.015{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}}{\left(0.03{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}+0.015{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}\right)}\\ =\frac{0.015{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}}{\left(0.045{\text{m}}^{\text{2}}/\text{V}\cdot \text{s}\right)}\\ =0.3333\\ \approx 0.334\end{array}$

Thus, the fraction of current carried by electron for $\text{GaP}$ is $0.667$.

Thus, the fraction of current carried by hole for $\text{GaP}$ is $0.334$.