Notation: p-well Pre-dep Boron doping Pre-Dep: t T pp => D pp, C spp t pp = p-well pre-dep time T pp = p-well pre-dep temperature D pp = p-well diffusion constant at pre-dep temperature C spp = surface concentration for p-well pre-dep
Notation: p-well Drive-in t T pd => D pd t pd = p-well drive-in time T pd = p-well drive-in temperature D pd = p-well diffusion constant at drive-in temperature
Notation: n+ Source/Drain Pre-dep Phosphorus doping Pre-Dep: t T np => D np, C snp ; D p1 t np = n + source/drain pre-dep time T np = n + source/drain pre-dep temperature D np = n + source/drain diffusion constant at pre- dep temperature C snp = surface concentration for n + source/drain pre-dep D p1 = boron diffusion constant at source/drain pre-dep temperature
Notation: n+ Source/Drain Drive-in t T nd => D nd ; D p2 t nd = n + source/drain drive-in time T nd = n + source/drain drive-in temperature D nd = n + source/drain diffusion constant at drive-in temperature D p2 = boron diffusion constant at source/drain drive-in temperature
Profile: After p-well Diffusion where:= well dose = well “Dt” There is a pn-junction x j0 where N A (x) = C sub
Profile: After n+ Source/Drain Diffusion where:= source/drain dose = source/drain “Dt” BUT: now the well profile has changed to…
New Well Profile There is a pn-junction x j1 where N D (x) = N A (x) There is a new pn-junction x j2 where N A (x) = C sub (where x j2 > x j0 ) where: = overall effective “Dt”
Example Suppose we want to design a p-well CMOS diffusion process with a well depth of x j2 = 2.5 m. Assume the n-type substrate doping is cm -3.
Example (cont.) If we start with a boron pre-dep with a dose of 5 × cm -2, followed by a 1-hr drive-in at 1100 o C, what is the initial junction depth (x j0 )? Neglect the depth of the pre-dep. The B diffusivity at this temperature is 1.5 × cm 2 /s. SOLUTION: where: S w = 5 × cm -2 (Dt) w = (1.5 × cm 2 /s)(3600s) = 5.4 × cm 2 N A (x j0 ) = cm -3 => = 1.24 m
Example (cont.) Find the necessary (Dt) eff for the p-well to reach the desired junction depth of x j2 = 2.5 m. SOLUTION (This must be solved by iteration!!!): where: x = x j2 = 2.5 m N A (x j2 ) = cm -3 S w = 5 × cm -2 => (Dt) eff = 2.46 × cm -2
Example (cont.) What is the approximate p-well drive-in time needed if all steps are carried out at 1100 o C? SOLUTION: = 1.64 × 10 4 s = min
Example (cont.) If the n+ source/drain junction depth required is x j1 = 2.0 m, what is the p-well doping at the source/drain junction? SOLUTION: where: x = x j1 = 2.0 m (Dt) eff = 2.46 × cm 2 => N A (x j1 = 2.0 m) = 9.76 × cm -3
Example (cont.) Suppose the source/drain dose (S sd ) is 5 × cm -2. What is the surface concentration in the source/drain regions and the source/drain diffusion (Dt) sd ? SOLUTION: where: x = x j1 = 2.0 m N D (x = x j1 ) = 9.76 × cm -3 (i)(Solving by iteration): (Dt) sd = 1.52 × cm 2 (ii) Surface Concentration: = 7.24 × cm -3
Example (cont.) The phosphorus source/drain regions are deposited and driven in at 1050 o C. At this temperature, the phosphorus diffusivity is 5.8 × cm 2 /s. Ignoring the contributions of the pre-dep, what is the approximate source/drain diffusion time (t nd )? SOLUTION: = 2.62 × 10 4 s = min
Example (cont.) If the boron diffusivity is 6.4 × cm 2 /s at 1050 o C, 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: D p2 t nd = 1.68 × cm 2 ( “Dt” accumulated by boron during source/drain diffusion). => Initial p-well drive-in may be reduced by this amount, or: = 86.9 min
Vacancies When host atom acquires sufficient energy to leave its lattice site, a vacancy is created. Vacancy density of a given charge state (# vacancies/unit volume, C V ) has temperature dependence similar to carrier density: where C i = intrinsic vacancy density, E F = Fermi level, and E i = intrinsic Fermi level
Vacancy-Dependent Diffusion If diffusion is dominated by the vacancy mechanism, D is proportional to vacancy density. At low doping concentrations (n < n i ), E F = E i, and C V = C i (independent of doping), so D (which is proportional to C V = C i ), also independent of doping concentration. At high concentrations (n > n i ), [exp(E F – E i )/kT] becomes large, which causes C V and D to increase.
Intrinsic and Extrinsic Diffusion
Effect on Diffusivity C s = surface concentration D s = diffusion coefficient at the surface = parameter to describe concentration dependence
Junction Depth For > 0, D decreases with concentration Increasingly steep box-like profiles result Therefore, highly abrupt junctions are formed Junction depth is virtually independent of background concentration = 1 = 2 = 3
Phosphorus 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 n e, 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.
The Problem 1-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.
Diffusion Contours Contours of constant doping concentration for a constant C s, assuming D is independent of concentration.
Interpretation Variation 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).
Implications Because of lateral diffusion, the junction consists of a central plane (or flat) region with approximately cylindrical edges with a radius of curvature r j. 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.