Question

helpp

Problem 5b. - 10 Points total A semiconductor material has an energy gap of 0.75 eV, effective masses mn= 0.04 mo and mp= 0.22 mo, where mo is the free electron mass = 9.11 x 103 [kg]. Assume complete ionization. a) Let the temperature be T = 350 °K. The material is un-doped. Find the intrinsic Fermi level EFi and carrier concentration ni- pi (4 points) b) Let the temperature be T = 350 K. The material is doped with donors at concentration No 5 x 1015 [cm3]. Assuming Boltzmann statistics applies, find: 1) the electron concentration n, (2 points) (2 points) (2 points) 2) the hole concentration p and 3) the position of the Fermi level EF.

Answer 1

Energy gap = 0.75 ev E_mid gap = 0.75/2 = 0.375 ev m*n = 0.040 m_0 m_0 = 9.11 times 10^-31 kg m*p = 0.22 m_0 (a) E_fi = E_mid gap + KT/2 ln (m_p/mn)^3/2 E_fi = 0.375 + 350/11600 times 2 ln (0.22/0.04)^3/2 E_fi = 0.375 + 0.015 times ln 12.90 E_fi = 0.413 ev intrinsic fermi level now ni^2 = N_C N_V e^-Eg/KT N_c N_v = 2 (2 pi m* n KT/h^2)^3/2 N_v = 2 (2 pi m* p KT/h^2)^3/2 N_c N_v = 4 (2 pi KT/h^2)^3 (m*_n m* p)^3/2 N_c N_v = 4 (2 pi times 1.38 times 10^-23 times 350/(6.62 times 10^-34)^2)^3 (m* n m* p)^3/2 N_c N_v = 4 times 3.32 times 10^140 times (0.04 m_0 times 0.22 m_0)^3/2 N_c N_v = 4 times 3.32 times 0.624 times 10^140 times 10^-93 N_c N_v = 8.286 times 10^47 = 82.86 times 10^46 now ni^2 = N_c N_v e^- EG/KT