Ion implantation, diffusion 4. Phosphorus is implanted at 100 keV to a dose of Q=2.5 x 104cm2 in a p-type Si wafer, NA=10cm". (Use data below for implanting into Si) 2000 Range 5000 4000 安1500 3000 Arsenic 2000 Gallium Energy( kev) Energy(keV) a. At what depth is the implanted P concentration a maximum? [41 b. What is the peak phosphorus concentration? [41 c. What is the depth of the junction before diffusion? 141 d. What happens to the value of C, and its location if the implant is subjected to a diffusion process at an elevated temperature? [4 e. With what mathematical form does the junction depth evolve with time upon diffusion at an elevated temperature? [41 a. From figure. R=1250 A(1.25 x 10 cm) b.AR2=350A,C1=Q/v(2T)ARl=285×10°cm3 c. The implant concentration profile is equal to the background acceptor concentration: C(=)=C, expl (x-RpL=10 8 Inverting to solve for x 2△R gives:x=R,+v2AR, In(28.5)=0.22 um d. The value of Cn decreases but its position remains the same( to first order) upon diffusion. If you took"its location "to mean that of the junction answer is"junction moves deeper into Si e. The implant profile advances deeper into the wafer like V(Dr)4 Ion implantation, diffusion 4. Phosphorus is implanted at 100 keV to a dose of Q = 2.5 ¥ 1014 cm-2 in a p-type Si wafer, NA = 1018 cm-3. (Use data below for implanting into Si). a. At what depth is the implanted P concentration a maximum? . [4] b. What is the peak phosphorus concentration? [4] c. What is the depth of the junction before diffusion? [4] d. What happens to the value of Cp and its location if the implant is subjected to a diffusion process at an elevated temperature? [4] e. With what mathematical form does the junction depth evolve with time upon diffusion at an elevated temperature? [4] A a. From figure, Rp = 1250 Å (1.25 ¥ 10-5 cm). b. DRp = 350 Å, Cp = Q/[÷(2π) DRp] = 2.85 ¥ 1019 cm-3. c. The implant concentration profile is equal to the background acceptor concentration: † C(x) = Cp exp - (x - Rp ) 2 2DRp 2 Ê Ë Á Á ˆ ¯ ˜ ˜ =1018. Inverting to solve for x gives: † x = Rp + 2DRp 2 ln(28.5) = 0.22 mm d. The value of Cp decreases but its position remains the same (to first order) upon diffusion. If you took “its location” to mean that of the junction, the answer is “junction moves deeper into Si”. e. The implant profile advances deeper into the wafer like ÷(Dt)