Question

Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

Hint:

For doped semiconductors:

n-type → \( n \approx N_D \)
p-type → \( p \approx N_A \)
Always use \( np = n_i^2 \) for minority carriers.

Answer 1

Concept: Arsenic is a pentavalent impurity → produces n-type semiconductor. Key relations:

Majority carriers (electrons): \( n \approx N_D \)
Mass action law: \[ np = n_i^2 \]

Step 1: Identify semiconductor type. Arsenic (Group V) donates electrons → n-type. So:

Majority carriers → electrons
Minority carriers → holes

Step 2: Majority carrier concentration. Donor concentration: \[ N_D = 5 \times 10^{22} \, \text{m}^{-3} \] Since \( N_D \gg n_i \): \[ n \approx N_D = 5 \times 10^{22} \, \text{m}^{-3} \]
Step 3: Minority carrier concentration. Using mass action law: \[ np = n_i^2 \] \[ p = \frac{n_i^2}{n} \] Substitute values: \[ n_i = 1.5 \times 10^{16} \] \[ n_i^2 = (1.5)^2 \times 10^{32} = 2.25 \times 10^{32} \] \[ p = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} = 0.45 \times 10^{10} = 4.5 \times 10^9 \, \text{m}^{-3} \] Final Answers:

Majority carrier concentration (electrons): \[ n = 5 \times 10^{22} \, \text{m}^{-3} \]
Minority carrier concentration (holes): \[ p = 4.5 \times 10^9 \, \text{m}^{-3} \]