Question

Please do Q15-Q16

it Teve of doping per cm 3, if thes wafer with p-type doping (NA) has a resistivity of 1 Ohmdm Hint: use Fig. 5. Q15(b) If all p-type impurities are ionized, what is the hole concentration. Q16 (a) If the Si wafer is n-type and has a resistivity of 1 Ohm-cm, what is the concentration of dopants No? Q16(b) If all donors have donated their electrons and become ionized, what is the electron concentration?

Answer 1

Given rho = 1 ohm - cm conductivity, sigma = mu_n n q + mu_p p q for p - type doping sigma_n is negligible sigma = sigma_p implies sigma = 1/rho = sigma_p = mu_p pq (N_A = p) implies sigma = 1/rho = sigma_p = mu_p pq (N_A = P) implies 1 = (450) N_A(1.6)40^-19 (mu_p = 450 cm^2/v - sec) implies N_A = 1/450 times 1.6 times 10^-10 = 0.001388 times 10^19 N_A = 1.388 times 10^16/cm^3 if all p-type imparts are , p = NA infinity concentration, p = 1.388 times 10^16/cm^3 sigma = mu_n n q + mu_p pq for n - type doping, sigma_p is negligible sigma = sigma_n = ln n q = u_n N_sigma q implies 1 = (1500) (N_D) (1.G 10^-19) (mu_n = 1500 cm^2/V SEC) implies N_D = 1/1500 times 1.6 times 10^-19 = 0.0004166 times 10^19 implies N_D = 4.166 times 10^15/cm^3 if all donors have donated tetrahedrons and ionized than n = N_D = 4.166 times 10^15 /cm^3