Chapter 1, Problem 1.1EP

Calculate the intrinsic carrier concentration in gallium arsenide and germanium at $T=300\text{K}$ . (Ans. GaAs, $n_i=1.80\times {10}^6{\text{cm}}^{-3}$ ; Ge, $n_i=2.40\times {10}^{13}{\text{cm}}^{-3}$ )

To determine

Intrinsic carrier concentration in GaAs (Gallium Arsenide) and Germanium (Ge).

Answer to Problem 1.1EP

Intrinsic carrier concentration in GaAs is $n_{i\text{GaAs}}=1.80\times {10}^6{\text{cm}}^{-3}$ and in Germanium is $n_{i_{\text{Ge}}}=2.40\times {10}^{13}{\text{cm}}^{-3}$ .

Explanation of Solution

Given:

Temperature, $T=300K$ ,

For Gallium Arsenide,

Band gap energy is $E_g=1.4\text{eV}$ .

Specific semiconductor coefficient is $B=2.10\times {10}^{14}{\text{cm}}^{-3}{\text{K}}^{-3/2}$ .

Boltzmann’s constant is $k=86\times {10}^{-6}\text{eV/K}$ .

For Germanium,

Band gap energy is $E_g=0.66\text{eV}$ .

Specific semiconductor coefficient is $B=1.66\times {10}^{15}{\text{cm}}^{-3}{\text{K}}^{-3/2}$ .

Boltzmann’s constant is $k=86\times {10}^{-6}\text{eV/K}$ .

Concept Used:

The intrinsic carrier concentration formula is given as

  $n_i=B{T}^{3/2}{e}^{(-E_g/2kT)}$(1)

Calculation:

For Gallium Arsenide, substitute the given values in equation (1).

  $\begin{array}{c}{n}_i=\left(2.10\times {10}^{14}\right){\left(300\right)}^{\frac{3}{2}}{e}^{\left(\frac{-1.4}{2\left(86\times {10}^{-6}\right)\left(300\right)}\right)}\\ =1.79770\times {10}^6{\text{cm}}^{-3}\\ \simeq 1.80\times {10}^6{\text{cm}}^{-3}\end{array}$

For Germanium, substitute the given values in equation (1).

  $\begin{array}{c}{n}_i=\left(1.66\times {10}^{15}\right){\left(300\right)}^{\frac{3}{2}}{e}^{\left(\frac{-0.66}{2\left(86\times {10}^{-6}\right)\left(300\right)}\right)}\\ =2.4036\times {10}^{13}{\text{cm}}^{-3}\\ \simeq 2.40\times {10}^{13}{\text{cm}}^{-3}\end{array}$