1. 1.In a P-type Ge ni = 2.1 × 1019 m-3,density of boron = 4.5 × 1023 atoms m-3.The electron and hole mobility are 0.4 and0.2 m2 V-1 s-1 respectively. What is its conductivity before and after the addition of boron atoms?<br />Given data:<br />ni = 2.1 × 1019 m-3<br />µe= 0.4 m2 V-1 s-1<br />µh= 0.2 m2 V-1 s-1<br />nboron = 4.5 × 1023 atoms m-3<br />Solution:<br /> Before adding boron atoms, the semiconductor is an intrinsic SC<br />σi = ni e (µe + µh)<br />= 2.1×1019×1.6×10-19× (0.4+0.2)<br /> Before adding boron atoms, the semiconductor becomes a P-type SC<br />σi = nboron e µh<br />= 4.5 × 1023×1.6×10-19×0.2<br />2. Silicon contains 5 × 1028 atoms m-3.In a n-type Si sample the donor concentration is 1 atom per 2.5 × 107 Si atoms. Find out the position of Fermi level at 300 K? Assume that effective mass of an electron is equal to the free electron mass.<br />Given data:<br />nSilicon = 5 × 1028 atoms m-3<br />The donor concentration is 1 atom per 2.5 × 107 Si atoms<br />T=300 K<br />Eg(Si) = 1.1 eV=1.1×1.6×10-19 J = 1.76×10-19 J<br />me = mh = mo = 9.1×10-31 kg<br />Solution:<br />Donor atoms per m3 (Nd) = 5 × 10282.5 × 107<br /> =<br />Fermi energy for n-type SC<br />Ef = Ed+Ec2 + kbT lnNd22πmekb Th232<br />= Eg2 + kbT lnNd22πmekb Th232<br />=1.76×10-19 2 + 1.38×10-23×300×ln2 × 102122×3.14×9.1×10-31×1.38×10-23×3006.625×10-34232 <br />=<br />3. Determine the density of the donor atoms which have to be added to intrinsic Ge to make it a n-type material of resistivity 0.19×10-2 Ωm. It is given that the mobility of electrons in the n-type SC is 0.325m2V-1s-1.<br />Given data:<br />Resistivity ρ = 0.19×10-2 Ωm<br />Mobility µe= 0.325 m2 V-1 s-1<br />Solution:<br />σe = ne e µe <br />ne= σeeμe<br />= 1ρeμe<br />= 10.19×10-2×1.6×10-19×0.325<br />4. Consider a sample of n-type silicon with Nd=1021 /m3.Determine the electron and hole densities at 300 K. The intrinsic carrier concentration for Si is 9.8 ×1015/m3.<br />Given data:<br />Nd = 1021 /m3<br />ni = 9.8 ×1015/m3<br />Solution:<br />ni2 = ne×nh<br />It is n-type SC, so Nd = ne<br />(density of donor level electrons = density of electron in conduction band)<br />ne= 1021 /m3<br />nh= ni2ne<br />= 9.8×101521021<br /> <br /> 5. For intrinsic Si, the room temperature electrical conductivity is 4×10-4 Ω-1 m-1.The electron and hole mobilities are 0.14 and 0.04 m2V-1s-1 respectively. Compute electron and hole concentrations at room temperature.<br />Given data:<br />σi = 4 × 10-4 Ω-1 m-1<br />µe= 0.14 m2 V-1 s-1<br />µh= 0.04 m2 V-1 s-1<br />Solution:<br /> σi = ni e (µe + µh)<br />ni= σie(μe+μh)<br />= 4×10-41.6×10-19×(0.14+0.04)<br />6. The intrinsic carrier density at room temperature in Ge is 2.37 ×1019 m-3. If the electron and hole mobilities are 0.38 and 0.18 m2V-1s-1 respectively. Calculate its resistivity.<br />Given data:<br />ni = 2.37 × 1019 m-3<br />µe= 0.38 m2 V-1 s-1<br />µh= 0.18 m2 V-1 s-1<br />Solution:<br /> σi = ni e (µe + µh)<br />ρi= 1σi<br />7. The electron and hole mobilities in In-Sb SC are 6 and 0.2 m2V-1s-1 respectively. At room temperature resistivity of In-Sb is 2×10-4 Ωm. Assuming that the material is intrinsic, determine its intrinsic carrier density at room temperature.<br />Given data:<br />ρi = 2 × 10-4 Ωm<br />µe= 6 m2 V-1 s-1<br />µh= 0.2 m2 V-1 s-1<br />Solution:<br /> σi = ni e (µe + µh)<br />ρi= 1σi<br />ni= 1ρie(μe+μh)<br />8. The energy gap of Si is 1.1 eV. Its electron and hole mobilities at room temperature are 0.48 and 0.13 m2V-1s-1 respectively. Evaluate its conductivity.<br />Given data:<br />Eg = 1.1 eV= 1.1×1.6×10-19 J = 1.76×10-19 J<br />µe= 0.48 m2 V-1 s-1<br />µh= 0.13 m2 V-1 s-1<br />Solution:<br />σi = ni e (µe + µh)<br />ni= 22πmekb Th232e-Eg2kbT<br />Assuming me = mh = mo = 9.1×10-31 kg<br />= 22×3.14×9.1×10-31×1.38×10-23×3006.625×10-34232e-1.76×10-192×1.38×10-23×300<br />9. For Si SC with band gap energy is 1.2 eV. Determine the position of Fermi level at 300K.If me=0.12 mo and mh=0.28mo.<br />Given data:<br />T=300 K<br />Eg(Si) = 1.2 eV=1.1×1.6×10-19 J = 1.92×10-19 J<br />me = 0.12mo <br />mh = 0.28mo <br />Solution:<br />Ef= Eg2 + 3kbT4 lnmhme<br />= 1.92×10-192 + (3×1.38×10-23×300)4 ln0.28 mo0.12mo<br />10. In an intrinsic SC, the energy gap is1.2 eV. What is the ratio between its conductivity at 600 K and that at 300 K.<br />Given data:<br />T1=600 K<br />T2=300 K<br />Eg(Si) = 1.2 eV=1.2×1.6×10-19 J = 1.92×10-19 J<br />Solution:<br />σ = Ce-Eg2kbT<br />Let σ1 be the electrical conductivity at T1 K and σ2 the electrical conductivity at T2 K. <br />σ1 = Ce-Eg2kbT1<br />σ2 = Ce-Eg2kbT2<br /> σ1σ2 = e-Eg2kb1T1- 1T2<br /> = e-1.92×10-192×1.38×10-231600 - 1300<br />Note: logex = lnx =2.302 × log10x<br />11. At 20oc the conductivity of Ge is 2 Ω-1m-1. Find the conductivity of Ge at 40oc.Band gap energy of Ge is 0.72 eV.<br />Given data:<br />T1=20oc =293 K<br />σ1 =2 Ω-1m-1<br />T2=40oc = 313 K<br />Eg(Si) = 0.72 eV=0.72×1.6×10-19 J = 1.152×10-19 J<br />Solution:<br />Let σ1 be the electrical conductivity at T1 K and σ2 the electrical conductivity at T2 K. <br />σ1 = Ce-Eg2kbT1<br />σ2 = Ce-Eg2kbT2<br /> σ1σ2 = e-Eg2kb1T1- 1T2<br />σ2= σ1 eEg2kb1T1- 1T2<br />12. The band gap energy of intrinsic SCs A and B are0.36eV and 0.72 eV respectively. Compare the intrinsic career density of A and B at 300 K.<br />Given data:<br />T= 300 K<br />Eg1 = 0.36 eV=0.36×1.6×10-19 J = 0.576×10-19 J<br />Eg2 = 0.72 eV=0.72×1.6×10-19 J = 1.152×10-19 J<br />Solution:<br />σiA = niA e (µe + µh)<br />σiB = niB e (µe + µh)<br /> σiAσiB= niAe(μe+μh)niBe(μe+μh)<br />σiAσiB= niA niB<br />We know,<br />σiA = Ce-Eg12kbT<br />σiB = Ce-Eg22kbT<br />niA niB= σiAσiB<br />= eEg2-Eg12kbT<br />13. Determine the density of the donor atoms which have to be added to intrinsic Ge to make it an n-type material of resistivity 0.19×10-2Ωm. It is given that the mobility of electrons in the n-type SC is 0.325m2V-1s-1<br />Given data:<br />ρ = 0.19 × 10-2 Ωm<br />µe= 0.325 m2 V-1 s-1<br />Solution:<br />µe= σenee <br />ne= σeµee& σe = 1ρ<br />=10.19×10-2×0.325×1.6×10-19<br />14. The hall coefficient of certain Si specimen was found to be -7.35×10-5 C-1m3 from 100 to 400 K. Determine the nature of the SC if the conductivity was found to be 200 Ω-1m-1.Calculate the density and mobility of the charge carrier.<br />Given data:<br />Rh = -7.35×10-5 C-1m3<br />σe = 200 Ω-1m-1<br />Solution:<br />The negative sign of the hall coefficient indicates that the nature of the SC is n-type.<br />ne=1.18Rhe<br />= 1.187.35×10-5×1.6×10-19<br />µe= σenee <br />15. An electric field of 100 Vm-1 is applied to a sample of n-type SC whose hall coefficient is -0.0125 C-1m3.Determine the current density in the sample, assuming electron mobility to be 0.36 m2 V-1 s-1.<br />Given data:<br />Rh = -0.0125 C-1m3<br />Eh = 100 Vm-1<br />µe= 0.36 m2 V-1 s-1<br />Solution:<br />ne= 1.18Rhe<br />σe= ne e µe<br />J= σe E<br />16.A Si plate of thickness 1mm, breadth 10mm and length 100mm is placed in a magnetic field of 0.5 Wbm-2 acting perpendicular to its thickness. If 10-2 A current flows along its length, calculate the hall coefficient, if the hall voltage developed is 1.83 mV.<br />Given data:<br />t = 1 mm = 0.001 m<br />B= 0.5 Wbm-2<br />I = 10-2 A<br />Vh = 1.83 mV = 1.83 × 10-3 V<br />Solution:<br />Rh= VhtIB<br /> = 1.83×10-3×0.00110-2×0.5<br />