Topic: Electronic Devices and Circuits
In an N-type semiconductor, the Fermi level $E_F$ equals:
- A. $E_C - kT \ln(N_C/N_D)$
- B. $E_C - kT \ln(N_D/N_C)$
- C. $E_V + kT \ln(N_C/N_D)$
- D. $E_V - kT \ln(N_C/N_D)$
Answer: A. $E_C - kT \ln(N_C/N_D)$
Explanation: In an n-type semiconductor at thermal equilibrium, assuming complete ionization of donor atoms, the electron concentration in the conduction band is approximately equal to the donor concentration, $n \approx N_D$.
The carrier concentration is also given by $n = N_C e^{-(E_C - E_F)/(kT)}$.
Equating the two: $N_D = N_C e^{-(E_C - E_F)/(kT)}$.
Taking the natural logarithm of both sides: $\ln(N_D/N_C) = -(E_C - E_F)/(kT) \implies E_C - E_F = -kT \ln(N_D/N_C) = kT \ln(N_C/N_D)$.
Thus, $E_F = E_C - kT \ln(N_C/N_D)$.