Assume that an ntype silicon has been doped with a dopant co

Assume that an n-type silicon has been doped with a dopant concentration N_0 = 2.3 times 10^17 cm^-3. Find the electron and hole concentrations at equilibrium when T = 275degreeK, T = 300degreeK, and T = 325degreeK.

Solution

at T= 275⁰k

Given N_D=2.3X10¹⁷cm^-3  to find Electron concentration(N_e) and Hole concentration(Np)

1. At T = 275⁰k the value of intrinsic concentration is n_i = 9.65 X10⁹ cm^-3 and Na = 0

Electron concentration is Ne = 1/2{(N_D-Na)+((N_D-Na)² + 4n_i²)^1/2}

= 1/2{(2.3X10¹⁷-0)+((2.3X10¹⁷-0)² + 4(9.65 X10⁹ )²)^1/2}

= 1/2{(2.3X10¹⁷)+(5.29X10³⁴ + 3.7X10²⁰ )^1/2} = 2.3X10¹⁷ cm^-3

and Hole concentration is Np = n_i²/Ne = (9.65 X10⁹ cm^-3)²/ 2.3X10¹⁷ cm^-3 = 404.8cm^-3

2. At T = 300⁰k the value of intrinsic concentration is n_i = 1.01 X10⁹ cm^-3 and Na = 0

Electron concentration is Ne = 1/2{(N_D-Na)+((N_D-Na)² + 4n_i²)^1/2}

= 1/2{(2.3X10¹⁷-0)+((2.3X10¹⁷-0)² + 4(1.01X10⁹ )²)^1/2}

= 1/2{(2.3X10¹⁷)+(5.29X10³⁴ + 4.08X10¹⁸ )^1/2} = 2.3X10¹⁷ cm^-3

and Hole concentration is Np = n_i²/Ne = (1.01 X10⁹ cm^-3)²/ 2.3X10¹⁷ cm^-3 = .43x10² cm^-3

3. At T = 325⁰k the value of intrinsic concentration is n_i = 6.6 X10¹⁰ cm^-3 and Na = 0

Electron concentration is Ne = 1/2{(N_D-Na)+((N_D-Na)² + 4n_i²)^1/2}

= 1/2{(2.3X10¹⁷-0)+((2.3X10¹⁷-0)² + 4(6.6X10¹⁰ )²)^1/2}

= 1/2{(2.3X10¹⁷)+(5.29X10³⁴ + 4.35X10²¹)^1/2} = 2.3X10¹⁷ cm^-3

and Hole concentration is Np = n_i²/Ne = (6.6 X10¹⁰ cm^-3)²/ 2.3X10¹⁷ cm^-3 = 1.8x10² cm^-3