Arsenic is ion implanted into a uniformly doped, bare 𝑝-type Si substrate doped to NA =1016 cm-3. The implant energy is 𝐸 = 100 keV and the implant dose is 𝜙 = 8x1014 cm-2. For aprojected range of 𝑅𝑝 = 0.06 m and a projected straggle of Δ𝑅𝑝= 0.02 m, calculate:(a) the peak As density (in cm-3) [10 Pts](b) the surface As density (in cm-3) [10 Pts](c) the junction depth 𝑥𝑗 (in m). [10 Pts]𝑁(𝑥) = 𝜙√2𝜋∆𝑅𝑝𝑒𝑥𝑝 (− (𝑥 − 𝑅𝑝)22∆𝑅𝑝2

For a 𝑝-type substrate doped to NA = 4.5x1015 cm-3 and diffusion of arsenic from a constantsurface density of 1.5x1021 cm-3, calculate (use k = 8.62x10-5 eV/K):(a) the junction depth 𝑥𝑗 (in m) for diffusion at T = 1100 °C for t = 90 min. [10 Pts](b) the time (in min) required for a junction depth of 𝑥𝑗= 1 m at 1100 °C. [10 Pts]Use the diffusion equation below:𝑁(𝑥, 𝑡) = 𝑁0𝑒𝑟𝑓𝑐 ( 𝑥2√𝐷𝑡)where𝑁0 is the surface density (#/cm3).𝐷 is the diffusion coefficient and given by 𝐷 = 24exp(-Ea/kT) cm2/s for arsenicin silicon. Ea = 4.08 eV.𝑡 is the length of time the diffusion takes place (s)The complementary error function erfc(y) can be found in many math apps includingExcel

Find the resistance of 1 cm 3 pure silicon crystal. What is the resistance when the crystal is doped with arsenic if the doping is 1 in 10 9 , that is 1 part per billion(ppb). Given data: Atomic concentration in silicon is 5×10 22  cm −3 ,n i​ =1.0×10 10  cm −3 ,μ c​ =1350 cm 2  V −1  s −1 &μ h​ =480 cm 2  V −1  s −1

Substrate doping ND = 1015 cm-3, polysilicon gate doping density ND = 1020 cm-3, gateoxide thickness tx = 650 A, and oxide-interface charge density Nox = 2 x 1010 cm-2 . UseEsi = 11.7EO and Eox = 397EO for the dielectric coefficients of silicon and silicon-dioxide,respectively.(a) Calculate the threshold voltage V, for VSB = 0.(b) Determine the type and the amount of channel ion implantation which are necessaryto achieve a threshold voltage of V = - 2 V

6. A careless person has disposed of arsenic in a well used for drinking water. The arsenic sits at the bottomof the well and diffuses into the water. This diffusion obeys the following equation:Yt = DYxx (1)where Y is the concentration of arsenic. The surface of the well is at x = 0 and the bottom at x = L =20m. At x = 0 the concentration is fixed at Y = 0 and at x = L the concentration is fixed at Y = 1.(a) [2]What is the steady state concentration of arsenic in the well?Solution: The steady-state condition is:Yt = 0Therefore the governing equation becomes:0 = DYxxIntegrating twice:0 = DY + c1x + c2Where c1 and c2 are constants of integration. Substituting the boundary conditions Y (0, t) = 0and Y (L, t) = 1:Y = xL(b) The arsenic is removed from the bottom after the concentration has reached a steady state. Unpol-luted ground water now keeps the concentration at x = L fixed at Y = 0 while the concentration atx = 0 remains fixed at Y = 0. You need to find how the concentration of arsenic changes in time.i. [2]Sketch the initial and boundary conditions.Solution: The initial and boundary conditions look like this

A careless person has disposed of arsenic in a well used for drinking water. The arsenic sits at the bottomof the well and diffuses into the water. This diffusion obeys the following equation:Yt = DYxx (1)where Y is the concentration of arsenic. The surface of the well is at x = 0 and the bottom at x = L =20m. At x = 0 the concentration is fixed at Y = 0 and at x = L the concentration is fixed at Y = 1.(a) [2]What is the steady state concentration of arsenic in the well