Question

The hole concentration in an intrinsic semiconductor is $5 \times 10^8\text{ m}^{-3}$. When it is doped with certain impurity, the electron concentration becomes $4 \times 10^{12}\text{ m}^{-3}$. Find the new value of the hole concentration. Also identify the type of new semiconductor formed after doping.

Hint:

Mass Action Law ($n_e \cdot n_h = n_i^2$) shows that increasing one type of carrier concentration via doping drastically suppresses the minority carrier concentration due to increased recombination rate.

Answer 1

Concept:
• In an intrinsic semiconductor, electron concentration equals hole concentration: $n_e = n_h = n_i$.

• According to Mass Action Law for semiconductors in thermal equilibrium, $n_e \cdot n_h = n_i^2$, regardless of doping level.

Step 1: Given parameters
Intrinsic carrier concentration $n_i = 5 \times 10^8\text{ m}^{-3}$.
New electron concentration after doping $n_e = 4 \times 10^{12}\text{ m}^{-3}$.

Step 2: Calculate new hole concentration
Using mass action law:
\[ n_e \cdot n_h = n_i^2 \]
\[ n_h = \frac{n_i^2}{n_e} \]
Substitute given values:
\[ n_h = \frac{(5 \times 10^8)^2}{4 \times 10^{12}} \]
\[ n_h = \frac{25 \times 10^{16}}{4 \times 10^{12}} \]
\[ n_h = 6.25 \times 10^4\text{ m}^{-3} \]

Step 3: Identify the type of semiconductor
Comparing electron concentration $n_e$ and hole concentration $n_h$:
\[ n_e = 4 \times 10^{12}\text{ m}^{-3} \quad \text{and} \quad n_h = 6.25 \times 10^4\text{ m}^{-3} \]
Since $n_e \gg n_h$, electrons are majority charge carriers and holes are minority carriers.
Therefore, the doped semiconductor is an n-type semiconductor.

Step 4: Conclusion
The new hole concentration is $6.25 \times 10^4\text{ m}^{-3}$ and the resulting doped material is an n-type semiconductor.