Solid State Electronic Devices SOLUTIONS MANUAL - Streetman

Solid State Electronic Devices (6th Edition)

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Chapter 1 Prob.I.1 Which semiconductor in Table 1-1 has the largest E 9 '? the smallestr I-Yhat is the corresponding).? HOIl) is the column III component related to E 9 '? largest E 9 ; ZnS, 3.6 eV:: 1.24 ). = 3:6 = O.344Jl11'l smallest E y : lnSb, 0.18 eV. ). = 6.89Jlm Note Al compounds have larger E 9 than the corresponding: Ga compounds, which are larger than In compounds. Prob.l.2 Here we need to calculate the l7lOXimum packingfraction, treating the atoms as hard spheres. Nearest atoms are at aseparation lX~(5X.J2)Z +5]. =4..330A 2 Radius of each atom = lx 4.330A = 2.165A 2 V91umeof each atom = .1 1t (2.16S)3 = 42.SA. 3 3 Number of atoms per cube = I + 8 x.!. = 2 8 P kin fracti 42.5 X2 = 68% ac g on - (S)3 1
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................................... __ ._~._ .. ~ ............ _ ...... <=1""""...- .. (233) z 2 <1 0 0> all edges I 1 ; ~ (1 1 1) z Prob.l.3 (a) Label planes: x y x y x y z x y z 3 3 3 3 2 2 1/3 1/3 1/3 1/3 112 112 2 3 3 (b) Draw equivalent directions in a cube (Need not show atoms) 2
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E> cti;G~ 5(1;) T 'c .... ""'\. I 1 3 to calculate the volume density olSi, its density on the (110) plane and the b~en two (Jdjacent (111) pkmes. <11 1> all body diagonals . with basis of 2 c atomlt-' ( 1 1 of atoms per cube = 8x-+-x6 x2=8 . 8 2 " . 8. ... nA102% ...,3 ./ ty = . 8 3 = OJ.V" X CIIl V" (S.43xIO-) , the (llO) plane we have 4 atoms on comers, 2 on the top and bottom planes, and 2 'or (see Fig. I-a). V . /' f·w.> /' I.~,..~ 3
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4 a. J Prob.l.S Using the hard-sphere model, find the latticeconslant of InSb. the volume of theprim' cell and the atomic density on the (110) plane. 4X..!.+..!.X2+2 (llO)plane:' 4 2_ 9 .59xl0 14 cm- 1 ./ (5.43 >< 10- 8 X..J2 ><5.43XlO- 8 ) Basis of Si crystal at Oand!, !,! which is along [1 1 1). 4 4 4 ~ hk ~) Distance = ~(~.)= 2.39A. :I .J3a . - = 1.44+ 1.36= 2.8A 4 a=6.47A. 3 In FCC,unit cell has 4lattice points. Therefore, volume of primitive ceIl = ~ = 67. 7}...3; Area of (110) plane = ..J2a 2 1 . 1 4><-+2><- ..J2 Density of In atoms = ~ 2 2 - 2 = 3.37 X 10 14 em- 1 2a a Same numberot Sb atoms = 3.37>< 10 14 em- 1 4
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Ptob.1.6 Draw NaC/ lattice (1 0 0) and unit cell. Two PQssiblcUDlt cells arc shown, with either Na'" or cr at the comers. 5 5
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Prob. 1.7 frob. 1.8 _0._ ~ -y-- --9 I I I I I I 0-- --0- --0 6 Thh vi..., h tUee<! aUlhely froc (110) to ahow the alignmene of at088. Th. open ebannela are he"egonal alOIt8 thi. d1rec: t10n. Th. shaded po1nea ara one ae lattice; ths open po1nU ar .. the 1nter.- Penetrating sc:. loc.ated e/2 hhlnd tha pl ..... of the front a.h.ded points. ehh direction 6
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Prpb.1.9 (a) Find the number ofSi atoms/em] on the surface of a (1 0 0) oriented Si wafer. a=5.43A Each a 2 has 1 +!( 4) = 2 atoms on thesurface. " 2atoniS/cell.. =6.78xlO 14 cm- 1 _" (5.43 x IO-S)2 cm 2 /ce ll (b) What is the distance (in 1) between nearest In neighbors in InP? a 2 In atoms are in an fcc sublattice with a .. 5.87 A, nearest neighbors are a r;; 5.87 r;; A -,,2 =-,,2 =4.15 v' 2 2 7 7
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The unit cell contains V2 Na and V2 C1 atoms. Using the hard sphere approximation, a=2.8..\.. Prob.l.10 Find NaCI density. Na+: atom;cwt. 23. radius 1 A. cr: atomitlweight 35.5, radius 1.8A. Prob. 1.12 Find atoms/cell and nearest neighbor dist<mce for sc, bee. and fcc lattices. (see solution to Prob. 1.5) HS):;; 1 a atoms/cell = nearest neighbor = atoms/cell = nearest neighbor = atoms/cell = nearest neighbor = for fcc (see Example I-I) -2 1 (23+ 35.5)/(6.02 x 10 23 ) denstty - 2.2 ft/em3 - (2.8 x 1O- 8 l I!I (1/4) (3/4) (1/4) !(li; ....
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