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): \[ p = 4.5 \times 10^9 \, \text{m}^{-3} \]

Answer 2

Concept: Rather than assuming the majority carrier concentration equals the donor concentration outright, this can be derived rigorously from charge neutrality together with the law of mass action, and the usual approximation then checked.

Step 1: Charge neutrality condition.
Assuming all donor atoms are ionized and there are no acceptor atoms, the crystal must remain electrically neutral, so the free electron concentration exceeds the hole concentration by exactly the donor concentration: \[ n - p = N_D \]

Step 2: Law of mass action.\[ n p = n_i^{2} \]

Step 3: Solve the two equations simultaneously.
From the first equation, \( p = n - N_D \). Substituting into the second: \[ n(n - N_D) = n_i^{2} \implies n^{2} - N_D n - n_i^{2} = 0 \] Solving this quadratic in \( n \): \[ n = \frac{N_D + \sqrt{N_D^{2} + 4n_i^{2}}}{2} \]

Step 4: Evaluate using the given numbers.\[ N_D = 5 \times 10^{22}, \quad n_i = 1.5 \times 10^{16} \implies n_i^{2} = 2.25 \times 10^{32} \] \[ 4 n_i^{2} = 9 \times 10^{32}, \qquad N_D^{2} = 2.5 \times 10^{45} \] Since \( 4n_i^2 \) is smaller than \( N_D^2 \) by about thirteen orders of magnitude, the square root simplifies to essentially \( N_D \): \[ \sqrt{N_D^{2} + 4n_i^{2}} \approx N_D \] \[ n \approx \frac{N_D + N_D}{2} = N_D = 5 \times 10^{22} \, \text{m}^{-3} \]

Step 5: Minority carrier concentration.\[ p = \frac{n_i^{2}}{n} = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} = 4.5 \times 10^{9} \, \text{m}^{-3} \]

Final Answers:\[ n = 5 \times 10^{22} \, \text{m}^{-3}, \qquad p = 4.5 \times 10^{9} \, \text{m}^{-3} \]