50 mg of Arsenic As was added in 120 Kg of Si melt beginning

50 mg of Arsenic (As) was added in 120 Kg of Si melt beginning the crystal growth by Czochralski method. Calculate the free electron and hole concentrations in the Si wafer cut from the ingot at a position when 80% of the ingot is grown. Given: Density of Si, = 2.33 gm/cm3, Segregation coefficient of As, k0 = 0.3, Atomic weight of As = 74.92. Consider, T = 300K and intrinsic concentration in Si, ni = 1.5x10^10 cm-3.

Solution

k₀ segregation coefficient

C_s concentration of impurities in the solid

C_o impurity concentration in the solid

f =0 .8

C_s= C_o k_o (1--f)^k_o-1

Finding the value of C₀

Number of Arsenic atoms in the 50 mg arsenic =( ( 50 * 10 ^-3 g)* 6.022 *10²³) / 74.92 amu

=4.01 *10²⁰ atoms

Concentration of impurities in the melt(C₀) =Number of impurities /volume of the melt

Volume of the melt (Si)=Volume of 120 kg of Si= mass /density = (120 *10³) / 2.33 gm/cm³

= 51.502 *10³ cm³

C₀ = 0 .077861 * 10¹⁷ / cm³

From the eqn C_s =0 .072064* 10¹⁷ /cm³

^N_d =.072064* 10¹⁷

Na=0

Arsenic is a n type dopant (pentavalent impurity) which is called a donor as it gives free electrons to the semiconductor.

For an n type semiconductor

Majority carrier electron concentration is given by

n₀= 1/2 { N_d-N_a) +(N_d--N_a)² +4 n_i² }^1/2

   =3.603* 10¹⁵ cm^--3

minority carrier hole concentration is given by

p₀= n_i²/n_o=6.244 *10⁴ cm ^--3