Question

In a non-degenerate bulk semiconductor with electron density $n = 10^{16}$ cm$^{-3}$, the value of $(E_C - E_{Fn}) = 200$ meV, where $E_C$ and $E_{Fn}$ denote the conduction band edge and electron Fermi level, respectively. Assume thermal voltage as 26 meV and intrinsic carrier concentration as $10^{10}$ cm$^{-3}$. For $n = 0.5 \times 10^{16}$ cm$^{-3}$, the closest approximation of $(E_C - E_{Fn})$, among the given options, is ________________.

• A. 226 meV
• B. 174 meV
• C. 218 meV
• D. 182 meV

Hint:

For non-degenerate semiconductors, Fermi level shifts follow a logarithmic dependence on carrier concentration: $\Delta(E_C - E_F) = V_T \ln(n_1/n_2)$.

Answer 1

The Correct Option is C

For a non-degenerate semiconductor, electron density is related to the Fermi level by \[ n = n_i \exp\left( \frac{E_F - E_C}{V_T} \right) \quad\Longrightarrow\quad E_C - E_F = V_T \ln\left( \frac{n}{n_i} \right), \] where $V_T = 26$ meV. Given for $n_1 = 10^{16}$ cm$^{-3}$: \[ E_{C1} - E_{F1} = 200\ \text{meV}. \] For the new electron density $n_2 = 0.5 \times 10^{16}$ cm$^{-3}$, we compute: \[ E_{C2} - E_{F2} = E_{C1} - E_{F1} + V_T \ln\left( \frac{n_1}{n_2} \right). \] Since \[ \frac{n_1}{n_2} = \frac{10^{16}}{0.5 \times 10^{16}} = 2, \] we have: \[ E_{C2} - E_{F2} = 200\ \text{meV} + 26\ \text{meV} \times \ln(2). \] Using $\ln(2) \approx 0.693$: \[ 26 \times 0.693 \approx 18.0\ \text{meV}. \] Thus, \[ E_{C2} - E_{F2} \approx 200 + 18 = 218\ \text{meV}. \] Final Answer: 218 meV