Tiêu đề 1.Numericals_Conduction in semiconductors.pdf ✅

Unit 1: Conduction in Semiconductors 2020-2021 Department of Physics, KLE Technological University, Hubballi Page 1 Review Question Problems Solution Problem no. 14: Find the resistivity of intrinsic germanium at 300K. Given that the intrinsic density of carriers is 2.5x1019 /m3 , electron mobility is 0.38 m2 /V-s, hole mobility is 0.18 m2 /Vs. Solution: Given- ni = 2.5 x 10 19/ m3 , μe = 0.38 m2 /Vs, μh = 0.18 m2 /Vs ௘ߤ)௜݁݊ = ௜ߪ (௛ߤ + ρ୧ = 1 en୧൫μୣ + μ୦ ൯ = 1 1.6 x 10ିଵଽx 2.5 x 10ଵଽx (0.38 + 0.18) ρ୧ = 0.446 Ωm. Problem No. 15: In a p-type germanium, ni = 2.1 x 1019/m3 , density of boron = 4.5 x 1023 atoms m-3. The electron and hole mobility are 0.4 and 0.2 m2 /Vs respectively. What is its conductivity before and after the addition of boron atoms? Solution: ni = 2.1 x 1019/m3 , n boron = 4.5 x 1023 atoms/m3 , μe = 0.4 m2 /Vs, μh = 0.2 m2 /Vs Before adding boron atoms, the semiconductor is intrinsic in nature ௘ߤ)௜݁݊ = ௜ߪ (௛ߤ + = 2.1 x 10ଵଽx 1.6 x 10ିଵଽx (0.4 + 0.2) σ୧ = 2.016 Ωିଵ݉ିଵ After adding boron atoms, the semiconductor becomes a p type σୠ୭୰୭୬ = nୠ୭୰୭୬ e μ୦ = 4.5 x 10ଶଷx 1.6 x 10ିଵଽx (0.2) = 1.44 x 10ସ Ωିଵ݉ିଵ Problem No. 16: Consider a sample of n-type silicon with Nd = 1021 /m3 . Determine the electron and hole densities at 300K. The intrinsic carrier concentration for Si is 9.8 x 10 15 /m3 . Solution: ni = 9.8 x 1015/m3 , Nd = 1021 /m3 ௜݊ = ݌ ݊. ଶ Therefore, n = Nd = 1021 /m3 ௜݊ = ݌ ୢ.N ଶ = ݌ ௡೔ మ ୒ౚ = (ଽ.଼ ௫ ଵ଴భఱ) మ ଵ଴మభ = 9.604 x 1010/ m3 Electron density at 300K = n = 1021/m3 Hole density at 300K= p = 9.604 x 1010/m3 .
Unit 1: Conduction in Semiconductors 2020-2021 Department of Physics, KLE Technological University, Hubballi Page 2 Problem No. 17: A sample of silicon is doped with 4x1016/cm3 of Gallium (a group III atom). What are the concentrations of electrons and holes? (Given ni = 1.45 x 1010 /cm3 ) Solution: As a group III atom, gallium functions as a acceptor, so the doped material is p-type. The density of holes is essentially equal to the density of acceptors, so p = NA = 4 x 1016 /m3 The density of electrons is given by the mass action law as, ݊ = ௡೔ మ ୮ = ௡೔ మ ୒ఽ = (ଵ.ସହ ௫ ଵ଴భబ) మ ସ ௫ ଵ଴భల n = 5.3 x 103 /cm3 Problem No. 18: The electron and hole mobilities of Si sample are 0.135 and 0.048 m2 /Vs respectively. Density of silicon atoms ni = 1.5 x 1016 m-3. Determine the conductivity of Intrinsic Si at 300 K. The sample is then doped with 1023 phosphorous atom/m3 . Determine the equilibrium hole concentration and conductivity. Solution: μe = 0.135 m2 /Vs, μh = 0.048 m2 /Vs ni = 1.5 x 1016 m-3 σSi = ? at 300K σ୧ = en୧൫μୣ + μ୦ ൯ = 1.5 x 10ଵ଺x 1.6 x 10ିଵଽx (0.135 + 0.048) σ୧ = 0.44 x 10-3 Ω ିଵ݉ିଵ The sample doped with 1023 phosphurus atoms /m3 n = Nd = 1023 /m3 σୢ୭୮ୣୢ = Nୢμୣ e = 10ଶଷx 1.6 x 10ିଵଽ (0.135 ) σୢ୭୮ୣୢ = 2.16 x 10ଷ Ωିଵ݉ିଵ Equilibrium hole concentration, ݌ = ௡೔ మ ୒ౚ = (ଵ.ହ ௫ ଵ଴భల) మ ଵ଴మయ p = 2.25 x109 /m3 Problem No. 19: Calculate the drift velocity for electrons and holes in a 1mm length of silicon at 27 oC when the terminal voltage is 10V. (Given μe = 1500 cm2 /Vs and μh = 500 cm2 /Vs) Solution: l = 1mm, V=10 V, μe = 1500 cm2 /Vs = 0.15 m2 / Vs μh = 500 cm2 /Vs = 0.05 m2 / Vs Drift current velocity for electrons, ݒ= ௘ ିఓ೙௏ ௟ = ି଴.ଵହ଴଴ ௫ ଵ଴ ଵ ௑ ଵ଴షమయ ݏ݉/ −1500 = ௘ݒ
Unit 1: Conduction in Semiconductors 2020-2021 Department of Physics, KLE Technological University, Hubballi Page 3 Drift current velocity for electrons, ݒ= ௛ ܸ݌ߤ− ݈ = ି଴.ଵହ଴଴ ௫ ଵ଴ ଵ ௑ ଵ଴షమయ ݏ݉/ 500 = ௛ݒ Problem No. 20: An n-type semiconductor sample has a donar density of 10 21/m3 . It is arranged in a Hall experiment having magnetic field of 0.5 T and the current density is 500 A/m2 . Find the Hall voltage if the sample is 3mm wide. Solution: Nd = 1021 /m3 , B = 0.5 T, J = 500 A/m2 , w = 3mm, VH =? Hall Voltage, ܸு = ܤܫுܴ ݐ = ௃.௪.௧.஻ ௧.௡.௘ J = I/A = ହ଴଴ ௫ ଴.ହ ௫ ଷ ௫ଵ଴షయ 1.6 x 10−19 x ଵ଴మభ I = J.A ܸு = 4.6875 ݔ 10ିଷ ܸ I = J.w.t and ܴு = − ଵ ௡௘ and n = ND Problem No. 21: Hall coefficient of a specimen of doped silicon found to be 3.66 × 10–4m 3C –1. The resistivity of the specimen is 8.93 x 10-3 m. Find the mobility and density of the charge carriers Solution: RH = 3.66 × 10–4m 3C –1, ρ = 8.93 x 10-3 m Find μ and p Since RH is positive, the semiconductor is p type ܴு = ଵ ௣௘ therefore, = ݌ ଵ ோಹ௘ = ଵ ଵ.଺ ୶ ଵ଴షభవ୶ ଷ.଺଺ ୶ ଵ଴షర = 1.7 x 10 22/ m3 σ = 1/ρ μh= σ/pe = ଵ ஡.୮.ୣ = ଵ 1.6 x 10−19x 8.93 x 10−3 x 1.7 x 1022 = 0.0412 m2 / Vs