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

Calculation of Majority and Minority Carrier Concentration in Doped Silicon
We are given:
- Atom density of pure Si: \( N = 5 \times 10^{28} \, \text{atoms/m}^3 \) (not directly needed here)
- Dopant concentration (Arsenic, n-type dopant): \( N_D = 5 \times 10^{22} \, \text{m}^{-3} \)
- Intrinsic carrier concentration: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \)
Arsenic (As) is a donor impurity, so the silicon becomes n-type.
Step 1: Majority Carrier Concentration (Electrons)
For an n-type semiconductor, the majority carriers are electrons. If the semiconductor is lightly doped such that \( N_D \gg n_i \), the majority carrier concentration \( n \) can be approximated as:
\[ n \approx N_D = 5 \times 10^{22} \, \text{m}^{-3} \]
Step 2: Minority Carrier Concentration (Holes)
The product of electron and hole concentrations is constant and equal to the square of the intrinsic carrier concentration:
\[ n \, p = n_i^2 \]
Therefore, the minority carrier concentration (holes) is:
\[ p = \frac{n_i^2}{n} = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{22}} \]
Step-by-step calculation:
\[ (1.5 \times 10^{16})^2 = 2.25 \times 10^{32} \]
\[ p = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} = 0.45 \times 10^{10} = 4.5 \times 10^9 \, \text{m}^{-3} \]
Step 3: Summary
- Majority carrier concentration (electrons): \( n = 5 \times 10^{22} \, \text{m}^{-3} \)
- Minority carrier concentration (holes): \( p = 4.5 \times 10^9 \, \text{m}^{-3} \)
Conclusion:
Doping silicon with donor atoms like arsenic significantly increases the electron concentration (majority carriers) while drastically reducing the hole concentration (minority carriers), making the semiconductor strongly n-type.