1. For a material with a surface recombination velocity of 0 cm per sec, a diffusion length of 10-3 cm, an absorption coefficient of 103 cm-1, a depth of surface of n-type material of 10-3 cm, an energy gap of 1.5 eV, and which is irradiated by light (possessing energy in excess of the energy gap) of 5 x 1017 photons per second per cm?, find the short circuit current density. m A/cm Given: Recombination velocity: s= 0 cm per sec Diffusion length: L = 10-3 cm Absorption coefficient: a = 103 cm-1 Depth of surface (n-type): 1 = 10–3 cm Energy gap: Eg = 1.5 eV Irradiated by light: nph = 5 * 1017 photons second * cm2 Base e: el-al) = 2.718281 Charge of e:e= 1.602 * 10-19 Coulombs Using Table 6.2, we can find the collection efficiencies s= 0, L = 10-3, a = 103, the collection efficiency is nco = 0.47 Using Equation 6-33, we can obtain the index of refraction Eg()4 = 173 eV + (1.5 eV) ()4 = 173 eV (7)4 = 115.33 +7 3.27 Using Equation 6-32, we can obtain r, the reflection coefficient r = 4+1)* ? (3.29+1)* = 0.28 Plugging in our numbers, we can find the short circuit density using Equa- tion 6-29 Js = nco(1 - 1)[1 – el-al)]enph Js = (0.47)(1 -0.28) (1 - 2.718281) * (1.6 * 10-19) (5 * 1017ph/sec * cm²) = +Js = (0.47)(1 – 0.28)[1 – 2.718281] * (1.6)(5 * 10-2ph/sec * cm²)(100cm) =