Solution manual for solid state electronic devices 7th edition by streetman

Solution Manual for Solid State Electronic Devices 7th Edition by Streetman Full file at https://TestbankDirect.eu/ Chapter Solutions Prob 1.1 Which semiconductor in Table 1-1 has the largest Eg? the smallest? What is the corresponding λ? How is the column III component related to Eg? largest Eg : ZnS, 3.6 eV λ= 1.24 = 0.344µm 3.6 smallest Eg : InSb, 0.18 eV 1.24 = 6.89µm 0.18 T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) λ= Al compounds Eg > corresponding Ga compounds Eg > the corresponding In compounds Eg Prob 1.2 For a bcc lattice of identical atoms with a lattice constant of Å, calculate the maximum packing fraction and the radius of the atoms treated as hard spheres with the nearest neighbors touching Each corner atom in a cubic unit cell is shared with seven neighboring cells; thus, each unit cell contains one-eighth of a sphere at each of the eight corners for a total of one atom The bcc cell contains one atom in the center of the cube Thus, we have × 52 + 52 + 52 = 4.330 Å Nearest atoms are at a separation Radius of each atom = × 4.330 Å = 2.165 Å Volume of each atom = Full file at https://TestbankDirect.eu/ π ( 2.165) = 42.5 Å3 Solution Manual for Solid State Electronic Devices 7th Edition by Streetman Full file at https://TestbankDirect.eu/ Number of atoms per cube = + × = Packing fraction = 42.5 × (5)3 = 68% Therefore, if the atoms in a bcc lattice are packed as densely as possible, with no distance between the outer edges of nearest neighbors, 68% of the volume is filled This T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) is a relatively high percentage compared with some other lattice structures Full file at https://TestbankDirect.eu/ Solution Manual for Solid State Electronic Devices 7th Edition by Streetman Full file at https://TestbankDirect.eu/ Prob 1.3 Label planes (6 3) (2 2) x x _x 1/2 y 1/3 z_ 1/4 _x 1/2 y 1/4 z_ 1/2 T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) Prob.1.4 Sketch a body centered cubic unit cell with a monoatomic basis If the atomic density is 1.6x1022 cm-3, calculate the lattice constant What is the atomic density per unit area on the (110) plane? What is the radius of each atom? What are interstitials and vacancies? # of atoms in unit cell a = × + = =1.6×1022cm-3 a ⎛ ⎞ a = ⎜ ⎟ ×10−7 cm = 5≈ ⎝ ⎠ ( ) Area of (110) plane = a a = ( 25 ) ≈ # of atoms = 1+4× = ⇒ Density = = Å −2 ( 25) 0.707×2×1016 #/cm2 =5.6×1014 cm-2 25 Nearest neighbor atoms along body diagonal = Radius = 3a a = 2.17Å Vacancies are missing atoms Interstitials are extra atoms in between atoms (voids) Full file at https://TestbankDirect.eu/ Solution Manual for Solid State Electronic Devices 7th Edition by Streetman Full file at https://TestbankDirect.eu/ Prob 1.5 Calculate densities of Si and GaAs The atomic weights of Si, Ga, and As are 28.1, 69.7, and 74.9, respectively Si: a = 5.43 ⋅10-8 cm, atoms/cell atoms = a 5.43 ⋅10-8cm ( ⋅1022 ) = ⋅1022 cm1 g ⋅ 28.1 mol cm3 6.02 ⋅10 23 mol = 2.33 cmg T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) density = GaAs: a = 5.65 ⋅10-8 cm, each Ga, As atoms/cell 4 = =2.22 ⋅1022 cm1 3 -8 a 5.65 ⋅10 cm ( density = ) 2.22 ⋅1022 cm3 g ⋅ ( 69.7 + 74.9 ) mol 6.02 ⋅10 Prob 1.6 23 mol = 5.33 cmg For InSb, find lattice constant, primitive cell volume, (110) atomic density 3a = 1.44≈ + 1.36≈ = 2.8≈ a = 6.47Å FCC unit cell has lattice points ∴volume of primitive cell = area of (110) plane = 2a density of In atoms = ⋅ 14 +2 ⋅ 12 2a = a same number of Sb atoms = 3.37 ⋅1014 Full file at https://TestbankDirect.eu/ = 3.37 ⋅1014 cm2 cm2 a3 = 67.7≈ Solution Manual for Solid State Electronic Devices 7th Edition by Streetman Full file at https://TestbankDirect.eu/ Prob 1.7 Sketch an FCC lattice unit cell (lattice constant = 5Å) with a monoatomic basis, and calculate the atomic density per unit area on (110) planes What is the atomic density per unit volume? Indicate an interstitial defect in this cell? ( Area of (110) plane = a a ) (( ) ) = 5 Å 1 Number of atoms = × + × = Areal density = atoms/Å 25 (1Å = 10 -8 cm ) T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) = 0.057×1016 atoms/cm = 5.7×1014 atoms/cm Volume of unit cells = a = (125) Å3 = 1.25×10-22 cm3 Number of atoms/unit cell = 8× 1 + 6× = atoms/cm3 1.25×10-22 = 3.2×1022 atoms/cm3 Volume density = An interstitial defect is marked on the FCC sketch above Prob 1.8 Draw direction of diamond lattice This view is tilted slightly from (110) to show the alignment of atoms The open channels are hexagonal along this direction Full file at https://TestbankDirect.eu/ Solution Manual for Solid State Electronic Devices 7th Edition by Streetman Full file at https://TestbankDirect.eu/ Prob 1.9 Show bcc lattice as interpenetrating sc lattices view direction The shaded points are one sc lattice The open points are the interpenetrating sc lattice located a/2 behind the plane of the front shaded points T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) Prob 1.10 (a) Find number of Si atoms/cm2 on (100) surface fcc lattice with a = 5.43Å a=5.43Ǻ number atoms per (100) surface = ⋅ 14 + = atoms atoms per (100) surface area = (b) Find the nearest neighbor distance in InP fcc lattice with a = 5.87Å nearest neighbor distance = a 5.87Å ⋅ 2= ⋅ =4.15Å 2 Full file at https://TestbankDirect.eu/ (5.43Å ) = 6.78 ⋅1014 cm2 Solution Manual for Solid State Electronic Devices 7th Edition by Streetman Full file at https://TestbankDirect.eu/ Prob 1.11 Find NaCl density Na +: atomic weight 23g/mol, radius 1Ǻ Cl -: atomic weight 35.5g/mol, radius 1.8Ǻ unit cell with a = 2.8Å by hard sphere approximation ½ Na and ½ Cl atoms per unit cell = atoms cell g g ⋅ 23 mol + 12 atoms g cell ⋅ 35.5 mol = 4.86 ⋅10-23 cell 6.02 ⋅1023 atoms mol g 4.86 ⋅10-23 cell density = = 2.2 cmg -8 (2.8 ⋅10 cm) cell Prob 1.12 T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) The hard sphere approximation is comparable with the measured 2.17 cmg density Find packing fraction, B atoms per unit volume, and A atoms per unit area A 4Å A 4Å A B A B A A B A B Note: The atoms are the same size and touch each other by the hard sphere approximation radii of A and B atoms are then 1Ǻ number of A atoms per unit cell = ⋅ 18 = A number of B atoms per unit cell = 4Å volume of atoms per unit cell = 1⋅ 4π3 ⋅ (1Å)3 + ⋅ 4π3 ⋅ (1Å)3 = volume of unit cell = (4Å)3 = 64Å3 Å3 π packing fraction = = = 0.13 = 13% 64Å 24 8π B atoms volume density = atom = 1.56 ⋅1022 64Å3 cm3 number of A atoms on (100) plane = ⋅ 14 = A atoms (100) aerial density = atom = 6.25 ⋅1014 (4Å) Full file at https://TestbankDirect.eu/ cm2 8π Å3 Solution Manual for Solid State Electronic Devices 7th Edition by Streetman Full file at https://TestbankDirect.eu/ Prob 1.13 Find atoms/cell and nearest neighbor distance for sc, bcc, and fcc lattices sc: sc lattice atoms/cell = ⋅ 18 = a nearest neighbor distance = a bcc: atoms/cell = ⋅ 18 + = bcc lattice a⋅ nearest neighbor distance = a √3 a fcc: T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) a √2 atoms/cell = ⋅ 18 + ⋅ 12 = nearest neighbor distance = Prob 1.14 fcc lattice a⋅ 2 Draw cubes showing four {111} planes and four {110} planes {111} planes {110} planes Full file at https://TestbankDirect.eu/ a a √2 a Solution Manual for Solid State Electronic Devices 7th Edition by Streetman Full file at https://TestbankDirect.eu/ Prob 1.15 Find fraction occupied for sc, bcc, and diamond lattices sc: atoms/cell = ⋅ 18 = sc lattice a nearest neighbor = a à radius = a 4π ⎛ a ⎞ π ⋅ a3 atom sphere volume = ⋅ ⎜ ⎟ = ⎝ ⎠ unit cell volume = a 1⋅ fraction occupied = bcc lattice atoms/cell = ⋅ 18 + = T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) bcc: π ⋅ a3 = π = 0.52 a nearest neighbor = a a⋅ a⋅ à radius = 4π ⎛ a ⋅ ⎞ π ⋅ ⋅ a3 atom sphere volume = ⋅ ⎜⎜ = ⎟ ⎝ ⎟⎠ 16 unit cell volume = a a π ⋅ ⋅ a3 2⋅ π⋅ 16 fraction occupied = = = 0.68 a diamond: atoms/cell = (fcc) + (offset fcc) = a⋅ a⋅ à atom radius = nearest neighbour = a √3 a √2 diamond lattice 4π ⎛ a ⋅ ⎞ π ⋅ ⋅ a3 atom sphere volume = ⋅ ⎜⎜ = ⎟ ⎝ ⎟⎠ 128 unit cell volume = a 3 fraction occupied = 8⋅ π⋅ ⋅a π⋅ 128 = = 0.34 a 16 Full file at https://TestbankDirect.eu/ a a √3 a √2 a/2 Solution Manual for Solid State Electronic Devices 7th Edition by Streetman Full file at https://TestbankDirect.eu/ Prob 1.16 Calculate densities of Ge and InP The atomic weights of Ge, In, and P are 72.6, 114.8, and 31, respectively Ge: a = 5.66 ⋅10-8 cm, atoms/cell atoms = a 5.66 ⋅10-8cm ( density = 4.41⋅1022 cm3 ) = 4.41⋅1022 cm1 g ⋅ 72.6 mol 6.02 ⋅1023 mol = 5.32 cmg GaAs: a = 5.87 ⋅10-8 cm, each In, P atoms/cell ( density = Prob 1.17 T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) 4 = =1.98 ⋅1022 3 a 5.87 ⋅10-8cm ) 1.98 ⋅1022 cm3 cm3 g ⋅ (114.8+31) mol 6.02 ⋅10 23 mol = 4.79 cmg Sketch diamond lattice showing four atoms of interpenetrating fcc in unit cell Full Interpenetrating Lattice Full file at https://TestbankDirect.eu/ Four Interpenetrating Atoms in Unit Cell Solution Manual for Solid State Electronic Devices 7th Edition by Streetman Full file at https://TestbankDirect.eu/ Prob 1.18 Find AlSbxAs1-x to lattice match InP and give band gap Lattice constants of AlSb, AlAs, and InP are 6.14Ǻ, 5.66Ǻ, and 5.87Ǻ, respectively from Appendix III Using Vegard’s Law, 6.14Å ⋅ x + 5.66Å ⋅ (1 − x) = 5.87Å → x = 0.44 AlSb0.44As0.56 lattice matches InP and has Eg=1.9eV from Figure 1-13 Find InxGa1-x P to lattice match GaAs and give band gap Lattice constant of InP, GaP, and GaAs are 5.87Ǻ, 5.45Ǻ, and 5.65Ǻ, respectively from Appendix III Using Vegard’s Law, 5.87Å ⋅ x + 5.45Å ⋅ (1 − x) = 5.65Å → x = 0.48 Prob 1.19 T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) In0.48Ga0.52P lattice matches GaAs and has Eg=2.0eV from Figure 1-13 A Si crystal is to be grown by the Czochralski method, and it is desired that the ingot contain 1016 phosphorus atoms/cm3 (a) What concentration of phosphorus atoms should the melt contain to give this impurity concentration in the crystal during the initial growth? For P in Si, kd = 0.35 (b) If the initial load of Si in the crucible is kg, how many grams of phosphorus should be added? The atomic weight of phosphorus is 31 SOLUTION (a) Assume that CS = kdCL throughout the growth Thus the initial concentration of P in the melt should be 1016 = 2.86 × 1016 cm −3 0.35 (b) The P concentration is so small that the volume of melt can be calculated from the Full file at https://TestbankDirect.eu/ Solution Manual for Solid State Electronic Devices 7th Edition by Streetman Full file at https://TestbankDirect.eu/ weight of Si From Appendix III the density of Si is 2.33 g/cm3 In this example we will neglect the difference in density between solid and molten Si 5000 g of Si 2.33 g/cm = 2146 cm3 of Si 2.86 × 1016 cm−3 × 2146 cm3 = 6.14 × 1019 P atoms 6.14 × 1019 atoms × 31 g/mole 6.02 × 10 23 atoms/mole = 3.16 × 10−3 g of P Since the P concentration in the growing crystal is only about one-third of that in the T an his th d wo sa eir is p rk w le co ro is ill o u vi pr de f a rse de ot st ny s d s ec ro p an o te y ar d le d th t o a ly by e s in f th se for Un te is ss th ite gr w in e ity o g us d S of rk ( stu e o tat th inc de f i es e lu nt ns co w d le tr p or in a uc y k g rn to rig an on in rs h d th g in t la is e D t w no W iss ea s t p or em ch in er ld m W ina g itt id tio ed e n We or b) melt, Si is used up more rapidly than P in the growth Thus the melt becomes richer in P as the growth proceeds, and the crystal is doped more heavily in the latter stages of growth This assumes that kd is not varied; a more uniformly doped ingot can be grown by varying the pull rate (and therefore kd) appropriately Modern Czochralski growth systems use computer controls to vary the temperature, pull rate, and other parameters to achieve fairly uniformly doped ingots Full file at https://TestbankDirect.eu/ ... cm2 8π Å3 Solution Manual for Solid State Electronic Devices 7th Edition by Streetman Full file at https://TestbankDirect.eu/ Prob 1.13 Find atoms/cell and nearest neighbor distance for sc, bcc,... https://TestbankDirect.eu/ a a √2 a Solution Manual for Solid State Electronic Devices 7th Edition by Streetman Full file at https://TestbankDirect.eu/ Prob 1.15 Find fraction occupied for sc, bcc, and diamond... between atoms (voids) Full file at https://TestbankDirect.eu/ Solution Manual for Solid State Electronic Devices 7th Edition by Streetman Full file at https://TestbankDirect.eu/ Prob 1.5 Calculate