Question

Problem 8 A silicon wafer is doped by Phosphorous via Diffusion at 1090°c. TCC) 1200 1000 Gaussian function. The solid solubility of phosphorous is 10 cm and the measured junction depth is 1 um at a substrate concentration of 106 cm 3. Calculate the diffusion time and the total dopant in the diffused layer. Si Cu Au Slope for E,- 1 ev 2 eV 3 eV t ev 5 eV 10-12 Al In 0.6 0.7 1000T (K-)

Answer 1

Given that solid solubility of phosphorous = 10^17 cm^- 3 Junction depth = 1 mu m 10 - 4 cm substrate concentration = 10^6 cm^- 3 Diffusion time = ? total pant in the diffused layer = ? we have, c(x, t) = s/squareroot pi Dt exp (- x_j/4Dt) c(x, t) = C_s exp (- x_j/4Dt) where D = 2.3 times 10^- 13 cm^2/s diffusibility of phosporus 1 times 10^16 = 10^17 exp (- (10^- 4)^2/4 times 2.3 times 10^- 13 times t) 10^16/10^17 = exp (- 10^- 8/9.2 times 10^- 13 t) ln (0.1) = - 10^- 8/9.2 times 10^- 13t - 2.30258 = - 10869.565/t t = 10869.565/2.30258 therefore this is required diffusion time now we can calculate the total dopant then we have C_s = S/squareroot Dt S = c_s squareroot Dt = 10^17 squareroot 2.3 times 10^- 13 times 4720 = 10^17 squareroot 1.0856 times 10^- 9