3.1531/6152 Homework set 3 solution. fall 2003 1. 23 x 10). A later exposure to elevated temperature could not cause the increased concentration at 0.3 microns without shifting the standard deviation and peak value. Also, while the actual distribution may not be exactly Gaussian, for boron, the skew is toward increased concentration near the surface(due to enhanced backscatter of this light species)not what is asked here Problem 6 R=0. um(300 nm)demands an implant of boron at about 100 keV(Fig. 8-3). At this energy, AR,=85 nm, so the dose giving a peak concentration of 10cm is easily calculated to be 2.13 x 10cm". For a background doping of 1015 cm-3, the junction depth before diffusion is given by inverting Eq. 8. 1: x=558 nm or 0.5583.155J/6.152J Homework set 3 solution, fall 2003 4 1.23 ¥ 1014). A later exposure to elevated temperature could not cause the increased concentration at 0.3 microns without shifting the standard deviation and peak value. Also, while the actual distribution may not be exactly Gaussian, for boron, the skew is toward increased concentration near the surface (due to enhanced backscatter of this light species) not what is asked here. Problem 6 Rp = 0.3 µm (300 nm) demands an implant of boron at about 100 keV (Fig. 8-3). At this energy, DRp ≈ 85 nm, so the dose giving a peak concentration of 1017 cm-3 is easily calculated to be 2.13 ¥ 1012 cm-2. For a background doping of 1015 cm-3, the junction depth before diffusion is given by inverting Eq. 8.1: x = 558 nm or 0.558 µm