Question

Suppose a pure Si crystal has \(5\times10^{28}\) atoms per \(\text{m}^3\). It is doped with \(5\times10^{22}\) atoms per \(\text{m}^3\) of Arsenic. Calculate the majority and minority carrier concentration in the doped silicon. Given: \[ n_i=1.5\times10^{16}\,\text{m}^{-3} \]

Hint:

For an n-type semiconductor: \[ n \approx N_D \] and \[ np=n_i^2. \] Once the majority carrier concentration is known, use the mass action law to calculate the minority carrier concentration.

Answer 1

Concept: Arsenic is a pentavalent impurity. When silicon is doped with arsenic, each arsenic atom contributes approximately one free electron. Therefore, arsenic-doped silicon becomes an n-type semiconductor. In an n-type semiconductor:
• Electrons are the majority carriers.
• Holes are the minority carriers. The carrier concentrations satisfy \[ np=n_i^2, \] where
• \(n\) = electron concentration,
• \(p\) = hole concentration,
• \(n_i\) = intrinsic carrier concentration.

Step 1: Determine the majority carrier concentration. The concentration of donor atoms is \[ N_D=5\times10^{22}\,\text{m}^{-3}. \] Since each donor contributes one electron and donor concentration is much greater than intrinsic concentration, \[ n \approx N_D. \] Hence, \[ \boxed{ n=5\times10^{22}\,\text{m}^{-3} }. \] Therefore, the electron concentration (majority carriers) is \[ \boxed{ 5\times10^{22}\,\text{m}^{-3} }. \]

Step 2: Use the mass action law to calculate minority carrier concentration. Using \[ np=n_i^2, \] we obtain \[ p=\frac{n_i^2}{n}. \] Substituting the values, \[ p= \frac{(1.5\times10^{16})^2} {5\times10^{22}}. \] \[ p= \frac{2.25\times10^{32}} {5\times10^{22}}. \] \[ p= 0.45\times10^{10}. \] \[ p= 4.5\times10^{9}\,\text{m}^{-3}. \]

Step 3: Interpret the result physically. The donor concentration is extremely large compared to the intrinsic carrier concentration. Therefore, the number of free electrons becomes enormously large, while the hole concentration becomes extremely small. This is characteristic of an n-type semiconductor.

Step 4: Write the final answers. Majority carrier concentration (electrons): \[ \boxed{ n=5\times10^{22}\,\text{m}^{-3} } \] Minority carrier concentration (holes): \[ \boxed{ p=4.5\times10^{9}\,\text{m}^{-3} } \]