Suppose a ‘n’-type wafer is created by doping Si crystal having 5 × 1028 atoms/m3 with 1ppm concentration of As. On the surface 200 ppm Boron is added to create the ‘P’ region in this wafer. Considering ni = 1.5 × 1016 m–3, (i) Calculate the densities of the charge carriers in the n & p regions. (ii) Comment which charge carriers would contribute largely for the reverse saturation current when the diode is reverse biased.
When As (pentavalent) is added to Si the n-type water is created So the number of majority carrier in n-type water; Ne=ND and DSi=1061×5×1028=5×1022/m3 For number of minority carriers nh ne.nh=ni2 nh=neni2=5×10221.5×1016×1.5×1016(ni=1.5×1016/m3) (Given) =0.3×1.5×1032−22=0.45×10+10 per m3 When boron (trivalent) is implanted in Si crystal, p-type water is formed with number of holes nh=(ND×n of Si) =106200×5×1028=1000×1028−6 nh=1×1025 per m3 Minority carrier in p type water ne.nh=ni2 ne=nhni2=10251.5×1016×1.5×1016=2.25×1032−25 =2.25×107 electrons per m3 When reversed bias is appled on p-n junction then the minority charge carrier moves toward depletion layer ie holes nh=(0.45×1010 per m3) from n side and ne=2.25×107/m3 from p side moves towards junction and make the depletion layer thicker .
