Chapter 1, Problem 1.2EP

(a) Calculate the majority and minority carrier concentrations in silicon at $T=300\text{K}$ for (i) $N_d=2\times {10}^{16}{\text{cm}}^{-3}$ and (ii) $N_a={10}^{15}{\text{cm}}^{-3}$ . (b) Repeat part (a) for GaAs. (Ans. (a) (i) $n_o=2\times {10}^{16}{\text{cm}}^{-3}$ , $p_o=1.125\times {10}^4{\text{cm}}^{-3}$ ; (ii) $p_o={10}^{15}{\text{cm}}^{-3}$ , $n_o=2.25\times {10}^5{\text{cm}}^{-3}$ ; (b) (i) $n_o=2\times {10}^{16}{\text{cm}}^{-3}$ , $p_o=1.62\times {10}^4{\text{cm}}^{-3}$ ; (ii) $p_o={10}^{15}{\text{cm}}^{-3}$ , $n_o=3.24\times {10}^{-3}{\text{cm}}^{-3}$ ).

To determine

The majority and minority carrier concentration in silicon at given temperature for different conditions.

Answer to Problem 1.2EP

When $N_d=2\times {10}^{16}{\text{cm}}^{-3}$

• The majority carriers, $n_o=2\times {10}^{16}{\text{cm}}^{-3}$ .
• The minority carriers, $p_o=1.125\times {10}^4{\text{cm}}^{-3}$ .

When $N_a={10}^{15}{\text{cm}}^{-3}$

• The majority carriers, $p_o={10}^{15}{\text{cm}}^{-3}$ .
• The minority carriers, $n_o=2.25\times {10}^5{\text{cm}}^{-3}$ .

Explanation of Solution

Given Information:

Given temperature is 300 K.

  $N_d=2\times {10}^{16}{\text{cm}}^{\text{-3}},N_a={10}^{15}{\text{cm}}^{\text{-3}}$

Calculation:

When $N_d=2\times {10}^{16}{\text{cm}}^{\text{-3}}$ ,

The intrinsic carrier concentration of silicon at T=300K, $n_i=1.5\times {10}^{10}{\text{cm}}^{-3}$ .

Since $N_d>>n_i$ then the electron concentration is given by:

  $n_o\approx N_d=2\times {10}^{16}{\text{cm}}^{-3}$

Above value represents, the majority carrier electron concentration in silicon at given temperature.

In thermal equilibrium, the relation between the electrons and holes is given by:

  $p_o{n}_o={\left(n_i\right)}^2$

So, the minority carrier holes concentration is

  $\begin{array}{c}{p}_o=\frac{n_i{}^2}{N_d}\\ =\frac{{\left(1.5\times {10}^{10}\right)}^2}{2\times {10}^{16}}\\ =\frac{2.25\times {10}^{20}}{2\times {10}^{16}}\\ =1.125\times {10}^4{\text{cm}}^{-3}\end{array}$

Hence, the concentration of minority holes is, $p_o=1.125\times {10}^4{\text{cm}}^{-\text{3}}$ .

When $N_a={10}^{15}{\text{cm}}^{-3}$ ,

The intrinsic carrier concentration of silicon at T=300K, $n_i=1.5\times {10}^{10}{\text{cm}}^{-3}$ .

Since, $N_a>>n_i$ then the hole concentration is given by:

  $p_o\approx N_a={10}^{15}{\text{cm}}^{-3}$

Above value represents the majority carrier proton(hole) concentration in silicon at given temperature.

In thermal equilibrium, the relation between the electrons and holes is given by:

  $p_o{n}_o={\left(n_i\right)}^2$

Minority carrier electron concentration is determined as follows:

  $\begin{array}{c}{n}_o=\frac{n_i{}^2}{p_o}\\ =\frac{n_i{}^2}{N_a}\\ =\frac{{\left(1.5\times {10}^{10}\right)}^2}{{10}^{15}}\\ =\frac{2.25\times {10}^{20}}{{10}^{15}}\\ =2.25\times {10}^5{\text{cm}}^{-\text{3}}\end{array}$

Hence, the concentration of minority carrier electron is, $n_o=2.25\times {10}^5{\text{cm}}^{-\text{3}}$ .

To determine

The majority and minority carrier concentration in GaAs at given temperature for different conditions.

Answer to Problem 1.2EP

When $N_d=2\times {10}^{16}{\text{cm}}^{-3}$

• The majority carriers, $n_o=2\times {10}^{16}{\text{cm}}^{-3}$ .
• The minority carriers, $p_o=1.62\times {10}^{-4}{\text{cm}}^{-\text{3}}$ .

When $N_a={10}^{15}{\text{cm}}^{-3}$

• The majority carriers, $p_o={10}^{15}{\text{cm}}^{-3}$ .
• The minority carriers, $n_o=3.24\times {10}^{-3}{\text{cm}}^{-3}$ .

Explanation of Solution

Given Information:

Given temperature is 300 K.

  $N_d=2\times {10}^{16}{\text{cm}}^{-\text{3}},N_a={10}^{15}{\text{cm}}^{-\text{3}}$

Calculation:

When $N_d=2\times {10}^{16}{\text{cm}}^{-3}$

The intrinsic carrier concentration of GaAS at T=300K, $n_i=1.8\times {10}^6{\text{cm}}^{-\text{3}}$ .

Since $N_d>>n_i$ , the electron concentration is given by:

  $n_o\approx N_d=2\times {10}^{16}{\text{cm}}^{-3}$

Above value represents the majority carrier electron concentration in GaAs at given temperature.

In thermal equilibrium, the relation between the electrons and holes is given by:

  $p_o{n}_o={\left(n_i\right)}^2$

So, the minority carrier holes concentration is

  $\begin{array}{c}{p}_o=\frac{n_i{}^2}{N_d}\\ =\frac{{\left(1.8\times {10}^6\right)}^2}{2\times {10}^{16}}\\ =\frac{3.24\times {10}^{12}}{2\times {10}^{16}}\\ =1.62\times {10}^{-4}{\text{cm}}^{\text{-3}}\end{array}$

Hence, the concentration of minority holes is, $p_o=1.62\times {10}^{-4}{\text{cm}}^{\text{-3}}$ .

When $N_a={10}^{15}{\text{cm}}^{-3}$ ,

The intrinsic carrier concentration of GaAs at T=300K, $n_i=1.8\times {10}^6{\text{cm}}^{-3}$ .

Since $N_a>>n_i$ , the hole concentration is given by:

  $p_o\approx N_a={10}^{15}{\text{cm}}^{-3}$

Above value represents the majority carrier proton(hole) concentration in GaAs at given temperature.

In thermal equilibrium, the relation between the electrons and holes is given by:

  $p_o{n}_o={\left(n_i\right)}^2$

Minority carrier electron concentration is determined as follows:

  $\begin{array}{c}{n}_o=\frac{n_i{}^2}{p_o}\\ =\frac{n_i{}^2}{N_a}\\ =\frac{{\left(1.8\times {10}^6\right)}^2}{{10}^{15}}\\ =\frac{3.24\times {10}^{12}}{{10}^{15}}\\ =3.24\times {10}^{-3}{\text{cm}}^{-\text{3}}\end{array}$

Hence, the concentration of minority carrier electron is, $n_o=3.24\times {10}^{-3}{\text{cm}}^{-\text{3}}$ .