11Profile: After p-well Diffusion where:= well dose= well “Dt”There is a pn-junction xj0 where NA(x) = Csub
12Profile: After n+ Source/Drain Diffusion where:= source/drain dose= source/drain “Dt”BUT: now the well profile has changed to…
13New Well Profile where: = overall effective “Dt” There is a pn-junction xj1 where ND(x) = NA(x)There is a new pn-junction xj2 where NA(x) = Csub (where xj2 > xj0)
14ExampleSuppose we want to design a p-well CMOS diffusion process with a well depth of xj2 = 2.5 mm. Assume the n-type substrate doping is 1015 cm-3.
15Example (cont.)If we start with a boron pre-dep with a dose of 5 × 1013 cm-2, followed by a 1-hr drive-in at 1100 oC, what is the initial junction depth (xj0)? Neglect the depth of the pre-dep. The B diffusivity at this temperature is 1.5 × cm2/s.SOLUTION:where: Sw = 5 × 1013 cm-2(Dt)w = (1.5 × cm2/s)(3600s) = 5.4 × cm2NA(xj0) = 1015 cm-3 =>= 1.24mm
16Example (cont.)Find the necessary (Dt)eff for the p-well to reach the desired junction depth of xj2 = 2.5 mm.SOLUTION (This must be solved by iteration!!!):where: x = xj2 = 2.5 mmNA(xj2) = 1015 cm-3Sw = 5 × 1013 cm-2=> (Dt)eff = 2.46 × 10-9 cm-2
17Example (cont.)What is the approximate p-well drive-in time needed if all steps are carried out at 1100 oC?SOLUTION:= 1.64 × 104 s = min
18Example (cont.)If the n+ source/drain junction depth required is xj1 = mm, what is the p-well doping at the source/drain junction?SOLUTION:where: x = xj1 = 2.0 mm(Dt)eff = 2.46 × 10-9 cm2=> NA(xj1 = 2.0 mm) = 9.76 × 1015 cm-3
19Example (cont.)Suppose the source/drain dose (Ssd) is 5 × 1014 cm-2. What is the surface concentration in the source/drain regions and the source/drain diffusion (Dt)sd?SOLUTION:where: x = xj1 = 2.0 mmND(x = xj1) = 9.76 × 1015 cm-3(Solving by iteration): (Dt)sd = 1.52 × 10-9 cm2(ii) Surface Concentration:= 7.24 × 1018 cm-3
20Example (cont.)The phosphorus source/drain regions are deposited and driven in at 1050 oC. At this temperature, the phosphorus diffusivity is 5.8 × cm2/s. Ignoring the contributions of the pre-dep, what is the approximate source/drain diffusion time (tnd)?SOLUTION:= 2.62 × 104 s = min
21Example (cont.)If the boron diffusivity is 6.4 × cm2/s at 1050 oC, correct for the p-well diffusion time to account for the extra diffusion during the source/drain drive-in. (Neglect the contributions of pre-dep steps).SOLUTION:Dp2tnd = × cm2( “Dt” accumulated by boron during source/drain diffusion).=> Initial p-well drive-in may be reduced by this amount, or:= 86.9 min
22Outline Objectives Double Diffusions Concentration-Dependent Diffusion Diffusion in SiliconLateral Diffusion
23VacanciesWhen host atom acquires sufficient energy to leave its lattice site, a vacancy is created.Vacancy density of a given charge state (# vacancies/unit volume, CV) has temperature dependence similar to carrier density:where Ci = intrinsic vacancy density, EF = Fermi level, and Ei = intrinsic Fermi level
24Vacancy-Dependent Diffusion If diffusion is dominated by the vacancy mechanism, D is proportional to vacancy density.At low doping concentrations (n < ni), EF = Ei, and CV = Ci (independent of doping), soD (which is proportional to CV = Ci ), also independent of doping concentration.At high concentrations (n > ni), [exp(EF – Ei)/kT] becomes large, which causes CV and D to increase.
28Junction Depth For g > 0, D decreases with concentration Increasingly steep box-like profiles resultTherefore, highly abrupt junctions are formedJunction depth is virtually independent ofbackground concentration= 1= 2= 3
29Outline Objectives Double Diffusions Concentration-Dependent Diffusion Diffusion in SiliconLateral Diffusion
30Concentration Dependence Boron, arsenic: g ≈ 1Gold, platinum: g ≈ -2Phosphorus: g ≈ 2 (sort of)
32Phosphorus Diffusion (cont.) When surface concentration is low, diffusion profile is an erfc (curve a).As concentration increases, the profile begins to deviate (b and c).At high concentration (d), profile near the surface is similar b, but at ne, kink occurs, followed by rapid diffusion in tail region.Because of high diffusivity, phosphorus is used to form deep junctions, such as the n-tubs in CMOS.
33Outline Objectives Double Diffusions Concentration-Dependent Diffusion Diffusion in SiliconLateral Diffusion
34The Problem1-D diffusion equation is not adequate at the edge of the mask window.There, impurities diffuse downward and sideways (i.e., laterally).In this case, we must consider a 2-D diffusion equation and use numerical techniques to get the diffusion profiles under different initial and boundary conditions.
35Diffusion ContoursContours of constant doping concentration for a constant Cs, assuming D is independent of concentration.
36InterpretationVariation at far right corresponds to erfc distribution.Example: at C/Cs = 10–4, the vertical penetration is about 2.8 µm, whereas the lateral penetration is about 2.3 µm (i.e., the penetration along the diffusion mask-semiconductor interface).
37ImplicationsBecause of lateral diffusion, the junction consists of a central plane (or flat) region with approximately cylindrical edges with a radius of curvature rj.If the mask has sharp corners, the shape of the junction near the corner will be roughly spherical.Since the electric-field intensities are higher for cylindrical and spherical junctions, the avalanche breakdown voltages of such regions can be substantially lower than that of a plane junction.