Question

2. Make a "wire" out of intrinsic InP that has a length of 0.20 m and a radius of 0.50 mm. a) Determine the number of atoms per m3 in this wire. b) How far above or below the middle of the bandgap is the Fermi energy in InP at 300K? Express your answer in eV e) Determine the number of intrinsic charge carriers in this wire at 300K d) Determine the conductivity of intrinsic InP and the resistance of this wire (at 300K)

Answer 1

(a) The number of atoms per m³ in this wire is

The density of InP is 4.81 g/cm³

No.of atoms per unit cell of InP are 8

Lattice constant a = 5.868 $\small A^{\circ}$ = 5.868x10^-10 m

Atomic density : N_u/V_u = 8/a³ = 8 / (0.5868 x 10 ^-6)³ = 3.959 x10¹⁹ atoms/m³

(b)

For intrinsic semiconductor Fermi energy level from conduction band can be found by following equation

   ------------------(1)

Effective density of states in the conduction band of InP, N_c = 1.1x10¹⁴ T^3/2 (cm^-3)

  Effective density of states in the valence band of InP, N_v = 2.2 x10¹⁵T^3/2 (cm^-3)

Energy gap E_G = 1.344eV

Boltzman constant, k = 1.38064852 × 10^-23 m² kg s^-2 K^-1

T = 300 K

Substituting the above values in eq (1)

c) The intrinsic concentration is

  

d) conductivity is given by below equation

$\small \sigma =ne\mu _e$

no.electrons per cubic meter is n = Nve/Vu = 4(3+5)/a³ = 32/ / (0.5868 x 10 ^-6)³ = 1.58x10²⁰ electrons/m³

e = 1.6 x 10-¹⁹C

mobility of InP is $\small \mu _e$ = 0.46 m²/Vs

$\small \sigma =ne\mu _e$ = 1.58x10²⁰ x 1.6 x 10^-19C x 0.46 m²/Vs = 11.628 ( $\small \Omega$ m)^-1

resistivity is $\small \rho =$ 1/ $\small \sigma$ = 0.086 $\small \Omega$ m

There the resistance of the wire is

R = $\small \rho$ (L/A) =

L is the length of the wire 0.20 m and A = 7.854 x10^-7 m²

R = 0.086 x (0.20/7.854 x10^-7) = 22 k $\small \Omega$