Question

Calculate the intrinsic carrier concentration for GaAs at 300 K, given that,
* 0.56 ; * 0.068 and 1.42
mh = me me = me Eg = eV.

Answer 1

Answer on Question #51241, Physics, Solid State Physics

Calculate the intrinsic carrier concentration for GaAs at $300\mathrm{K}$, given that,

* 0.56; * 0.068 and 1.42

$\mathrm{mh} = \mathrm{me} \quad \mathrm{me} = \mathrm{me} \quad \mathrm{Eg} = \mathrm{eV}$

$$T = 300\mathrm{K}$$ $$m_c = 0.56 m_e$$ $$m_c = 0.068 m_e$$ $$\Delta W = 1.42 \mathrm{eV}$$

Solution

The own concentration of charge carriers is given by Eq.(1)

$$n_i = \sqrt{N_c N_v} \exp\left(-\frac{\Delta W}{2kT}\right)$$

where $k = 1.38 \cdot 10^{-23} J / K$ is the Boltzmann constant; $T = 300K$ is the temperature.

Effective density of states for electrons in the conduction band is given by Eq.(2).

$$N_c = \frac{2(2 \cdot \pi \cdot m_e \cdot k \cdot T)^{3/2}}{h^3} = \frac{2(2 \cdot \pi \cdot 0.56 \cdot 9,1 \cdot 10^{-31} kg \cdot 1,38 \cdot 10^{-23} J / K \cdot 300K)^{3/2}}{(6,62 \cdot 10^{-34})^3} = 1.05 \cdot 10^{25} m^{-3}$$

where $h = 6,62 \cdot 10^{-34} J \cdot s$ is the Planck constant.

Effective density of states for holes in the valence band is given by Eq.(3).

$$N_v = \frac{2(2 \cdot \pi \cdot m_e \cdot k \cdot T)^{3/2}}{h^3} = \frac{2(2 \cdot \pi \cdot 0.068 \cdot 9,1 \cdot 10^{-31} \cdot 1,38 \cdot 10^{23} T)^{3/2}}{(6,62 \cdot 10^{34})^{-3}} = 4.45 \cdot 10^{23} m^{-3}$$

The intrinsic concentration

$$n_i = 10^{23} \sqrt{1.05 \cdot 100 \cdot 4.45} \exp\left(-\frac{1,42}{2 \cdot 8,625 \cdot 10^{-5} \cdot 300}\right) \approx 2.6 \cdot 10^{12} m^{-3}.$$

**Answer**: $n_i \approx 2.6 \cdot 10^{12} m^{-3}$

http://www.AssignmentExpert.com/