Question

In an extrinsic semiconductor, the hole concentration is given to be $1.5 n_i$ where $n_i$ is the intrinsic carrier concentration of $1 \times 10^{10} \text{ cm}^{-3}$. The ratio of electron to hole mobility for equal hole and electron drift current is given as ___________ (rounded off to two decimal places).

Hint:

Remember the mass-action law, $np=n_i^2$, which is fundamental for finding carrier concentrations in extrinsic semiconductors. Also, recall the drift current density formulas: $J_{drift} = \text{charge density} \times \text{mobility} \times \text{field} = (\text{q} \times \text{carrier conc.}) \times \mu \times E$.

Answer 1

Correct Answer: 2.2

Given data: Hole concentration, $p = 1.5 n_i = 1.5 \times (1 \times 10^{10}) = 1.5 \times 10^{10} \text{ cm}^{-3}$. Intrinsic carrier concentration, $n_i = 1 \times 10^{10} \text{ cm}^{-3}$.
Step 1: Find the electron concentration, n. Using the mass-action law for semiconductors in thermal equilibrium: $np = n_i^2$. $n = \frac{n_i^2}{p} = \frac{(1 \times 10^{10})^2}{1.5 \times 10^{10}} = \frac{1 \times 10^{20}}{1.5 \times 10^{10}} = \frac{1}{1.5} \times 10^{10} = \frac{2}{3} \times 10^{10} \text{ cm}^{-3}$.
Step 2: Use the condition of equal drift currents. The electron drift current density is $J_n = qn\mu_n E$. The hole drift current density is $J_p = qp\mu_p E$. We are given that the drift currents are equal, which implies their densities are equal: $J_n = J_p$. $qn\mu_n E = qp\mu_p E$.
The terms $q$ (elementary charge) and $E$ (electric field) cancel out. $n\mu_n = p\mu_p$.
Step 3: Calculate the required ratio of mobilities. We need to find the ratio of electron mobility to hole mobility, which is $\frac{\mu_n}{\mu_p}$. Rearranging the equation from Step 2: $\frac{\mu_n}{\mu_p} = \frac{p}{n}$.
Substitute the values for p and n: $\frac{\mu_n}{\mu_p} = \frac{1.5 \times 10^{10}}{\frac{2}{3} \times 10^{10}} = \frac{1.5}{2/3} = \frac{3/2}{2/3} = \frac{3}{2} \times \frac{3}{2} = \frac{9}{4} = 2.25$.
The ratio is 2.25. Rounded to two decimal places, this is 2.25.