GATE 2024 EC – Question 63
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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Correct answer: 0.17 to 0.19
Explanation
With donor degeneracy 2, the fraction of donors that are neutral (occupied) is $\dfrac{1}{1+\frac12e^{(E_D-E_F)/kT}}=0.05$. So $1+\frac12e^{x}=20$, $e^x=38$ and $x=\ln38=3.64$. Then $E_D-E_F=3.64\times0.02=0.0727$ eV. With $E_C-E_F=0.25$ eV, $E_C-E_D=0.25-0.0727=0.177\approx0.18$ eV.