Chapter 14.12, Problem 76AAP

(a)

To determine

The concentration of majority carriers.

Answer to Problem 76AAP

The concentration of majority carriers is $2.60\times {10}^{19}\text{holes}/{\text{cm}}^{\text{3}}$.

Explanation of Solution

Write the expression to calculate the concentration of majority charge carriers in a p-type semiconductor.

    $p_p=\frac{1}{\rho q{\mu }_p}$                                                                                       ...... (I)

Here, the concentration of majority carriers is $p_p$, charge  of electron is $q$ and the mobility of hole is ${\mu }_p$.

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 hole mobility as $480{\text{cm}}^2/\text{V}\cdot \text{s}$ at $27°\text{C}$.

Substitute $5\times {10}^{-4}\Omega \cdot \text{cm}$ for $\rho$, $1.60\times {10}^{-19}\text{C}$ for $q$ and $480{\text{cm}}^2/\text{V}\cdot \text{s}$ in Equation (I).

    $\begin{array}{c}{n}_n=\frac{1}{\left(5\times {10}^{-4}\Omega \cdot \text{cm}\right)\times \left(1.60\times {10}^{-19}\text{C}\right)\times \left(480{\text{cm}}^2/\text{V}\cdot \text{s}\right)}\\ =\frac{1}{\left(7.5\times {10}^{-4}\Omega \cdot \text{cm}\right)\left(768\times {10}^{-19}\text{C}\cdot {\text{cm}}^{\text{2}}/\text{V}\cdot \text{s}\right)}\\ =\frac{1}{3840\times {10}^{-23}{\text{cm}}^3}\\ =2.60\times {10}^{19}\text{holes}/{\text{cm}}^{\text{-3}}\end{array}$

Thus, the concentration of majority carriers is $2.60\times {10}^{19}\text{holes}/{\text{cm}}^{\text{3}}$.

(b)

To determine

The ratio of boron to silicon atoms.

Answer to Problem 76AAP

Thus the ratio of boron to silicon atoms is $5.21\times {10}^{-4}$.

Explanation of Solution

Write the expression to calculate silicon atoms.

    $N_{\text{Si}}=\frac{{\rho }_{\text{Si}}{N}_o}{\text{atomic}\text{weight}\text{of}\text{silicon}}$                                                 ...... (I)

Here, the number of silicon atoms is $N_{\text{Si}}$, the density of silicon is ${\rho }_{\text{Si}}$ and the Avogadro number is $N_o$.

Write the expression for ratio boron to the silicon atoms.

    $R=\frac{N_{\text{B}}}{N_{\text{Si}}}$                                                                                  ...... (II)

Here, the number of arsenic atoms is $N_{\text{B}}$.

Conclusion:

Substitute $2.33\text{g}/{\text{cm}}^{\text{3}}$ for ${\rho }_{\text{Si}}$, $6.023\times {10}^{23}\text{atoms}/\text{mol}$ for $N_o$ and $28.09\text{g}/\text{mol}$ for atomic weight of silicon in Equation (I).

    $\begin{array}{c}{N}_{\text{Si}}=\frac{\left(2.33\text{g}/{\text{cm}}^{\text{3}}\right)\times \left(6.023\times {10}^{23}\text{atoms}/\text{mol}\right)}{28.09\text{g}/\text{mol}}\\ =\frac{14.03359\times {10}^{23}\text{g}\cdot \text{atoms}/{\text{cm}}^{\text{3}}\cdot \text{mol}}{28.09\text{g}/\text{mol}}\\ =4.99\times {10}^{22}\text{atoms}/{\text{cm}}^{\text{3}}\end{array}$

Substitute $4.99\times {10}^{22}\text{atoms}/{\text{cm}}^{\text{3}}$ for $N_{\text{Si}}$ and $2.60\times {10}^{19}\text{atoms}/{\text{cm}}^{\text{3}}$ for $N_{\text{B}}$ in Equation (II).

    $\begin{array}{c}R=\frac{2.60\times {10}^{19}\text{atoms}/{\text{cm}}^{\text{3}}}{4.99\times {10}^{22}\text{atoms}/{\text{cm}}^{\text{3}}}\\ =5.21\times {10}^{-4}\end{array}$

Thus the ratio of boron to silicon atoms is $5.21\times {10}^{-4}$.