Question

Draw the band diagram (the relative positions of conduction band edge EC, valence band edge Ev, Fermi level EF) for the four following cases. Clearly note EC −EF, EF −EV, Ei −EF, EG = EC −EV. Ei is the intrinsic Fermi level. Take NC=NV =1025 m−3, EG=1.1 eV, ni=1.5×1016 m−3, kT=0.026 eV.

(Q1.1) p-type, NA=5×1023 m−3.

(Q1.2) p-type, NA=5×1021 m−3.

(Q1.3) n-type, ND=5×1023 m−3.

(Q1.4) n-type, ND=5×1021 m−3.

Q2 Draw the band diagram (the relative positions of conduction band edge Ec, valence band edge EU, Fermi level EF) for the four following cases. Clearly note Ec - EF, EF - Ev, Ei - EF, EG- Ec - Ev. E is the intrinsic Fermi level. Take Nc Ny-1025 m-3, Ec1.1 eV, n 1.5x1016 m-3, kT-0.026 eV. [80 pts] (Q1.1) p-type, ŅA=5x1023 m-3 (Q1.2) p-type, Л-5x 1021 m-3 (Q1.3) n-type, ND-5x1023 m3. (Q1.4) n-type, ND-5x1021 m-3.

Answer 1

P type = N_A = 5 times 10^negative 23 m^n 3 N_c = N_v = 10^25 m m^negative 3 E_g = E_c - E_v = 1.1 eV n_i = 1.5 times 10^n16 m^negative 3 k_B T = 0.026 eV n = n_i e^CE_p - E_i)/k_B T P = n_i eTE_i - E_g)/k_B T So E_f = NA = 5 times 10^23 m^negative 3 as p = N_A - N_D/2 = squareroot (N_A - N_D/2)^2 + n^2_u = 5 times 10^23/2 + squareroot (5 times 1-0^23/2)^2 + (1.5 times 10^16)^2 As N_D ~ - P = 2.5 times 10^23 + 2.5 times 10^23 = 5 times 10^23 m^negative 3 n = n^2_i/N_A =
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