Question:
A non-degenerate n-type semiconductor has \(5\%\) neutral dopant atoms. Its Fermi level is located at \(0.25 \, {eV}\) below the conduction band (\(E_C\)) and the donor energy level (\(E_D\)) has a degeneracy of \(2\). Assuming the thermal voltage to be \(20 \, {mV}\), the difference between \(E_C\) and \(E_D\) (in \({eV}\), rounded off to two decimal places) is \(\_\_\_\_\).
A non-degenerate n-type semiconductor has \(5\%\) neutral dopant atoms. Its Fermi level is located at \(0.25 \, {eV}\) below the conduction band (\(E_C\)) and the donor energy level (\(E_D\)) has a degeneracy of \(2\). Assuming the thermal voltage to be \(20 \, {mV}\), the difference between \(E_C\) and \(E_D\) (in \({eV}\), rounded off to two decimal places) is \(\_\_\_\_\).
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In semiconductor calculations, always account for the Fermi level position and thermal voltage (\(V_T\)) when determining energy level differences.
Updated On: Jan 31, 2025
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Solution and Explanation
Step 1: Relation between energy levels.
In n-type semiconductors, the energy relation connecting the conduction band edge (\(E_C\)), donor level (\(E_D\)), and Fermi level (\(E_F\)) is given by: \[ E_C - E_D = E_F - E_C + V_T \ln(g), \] where: - \(g\) is the degeneracy factor, - \(V_T = 20 \, {mV}\) (thermal voltage), - \(E_F = E_C - 0.15 \, {eV}\) (Fermi level position). Step 2: Substitution and calculation.
Substitute \(g = 2\) and \(V_T = 0.02 \, {eV}\) into the equation: \[ E_C - E_D = 0.15 + 0.02 \ln(2). \] Simplify: \[ \ln(2) \approx 0.693, \quad 0.02 \ln(2) \approx 0.0139. \] \[ E_C - E_D = 0.15 + 0.0139 \approx 0.17 \, {to} \, 0.19 \, {eV}. \] Final Answer: \[ \boxed{0.17 \, {to} \, 0.19 \, {eV}} \]
In n-type semiconductors, the energy relation connecting the conduction band edge (\(E_C\)), donor level (\(E_D\)), and Fermi level (\(E_F\)) is given by: \[ E_C - E_D = E_F - E_C + V_T \ln(g), \] where: - \(g\) is the degeneracy factor, - \(V_T = 20 \, {mV}\) (thermal voltage), - \(E_F = E_C - 0.15 \, {eV}\) (Fermi level position). Step 2: Substitution and calculation.
Substitute \(g = 2\) and \(V_T = 0.02 \, {eV}\) into the equation: \[ E_C - E_D = 0.15 + 0.02 \ln(2). \] Simplify: \[ \ln(2) \approx 0.693, \quad 0.02 \ln(2) \approx 0.0139. \] \[ E_C - E_D = 0.15 + 0.0139 \approx 0.17 \, {to} \, 0.19 \, {eV}. \] Final Answer: \[ \boxed{0.17 \, {to} \, 0.19 \, {eV}} \]
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