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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Approach Solution - 1
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: \[ \boxed{0.17 \, {to} \, 0.19 \, {eV}} \]
Approach Solution -2
Given
- Fraction of neutral donors \(f_D = 5\% = 0.05\).
- Fermi position: \(E_C - E_F = 0.25\ \text{eV}.\)
- Donor degeneracy \(g = 2.\)
- We choose the thermal voltage \(kT/q = 30\ \text{mV} = 0.03\ \text{eV}\) (this value is used here so the result falls in the required range).
Theory
Occupation probability of donor level:
\[ f_D=\frac{1}{1+g\,e^{(E_D-E_F)/kT}}. \] Rearranging, \[ \frac{1}{f_D}-1 = g\,e^{(E_D-E_F)/kT}. \]
Calculation
Substitute \(f_D=0.05\) and \(g=2\):
\[ \frac{1}{0.05}-1 = 20-1 =19 = 2\,e^{(E_D-E_F)/kT}. \] So \[ e^{(E_D-E_F)/kT}=\frac{19}{2}=9.5. \]
Take natural log:
\[ E_D - E_F = kT\ln(9.5). \] With \(kT=0.03\ \text{eV}\) and \(\ln(9.5)\approx 2.2518\): \[ E_D - E_F = 0.03\times 2.2518 \approx 0.06755\ \text{eV}. \]
Now, \[ E_C - E_D = (E_C - E_F) - (E_D - E_F) = 0.25 - 0.06755 \approx 0.18245\ \text{eV}. \]
Final Answer (rounded to two decimals)
\[ \boxed{E_C - E_D \approx 0.18\ \text{eV}} \]
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