半导体物理与器件课后习题
半导体物理与器件第四版课后习题答案4复习进程
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4.11
只供学习与交流
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2.8 1019
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300
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0.0128 eV
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m n 0.55mo
E E Fi
midgap
半导体物理与器件第四版课后习题标准答案
半导体物理与器件第四版课后习题答案————————————————————————————————作者：————————————————————————————————日期：2______________________________________________________________________________________3Chapter 33.1If o a were to increase, the bandgap energy would decrease and the material would beginto behave less like a semiconductor and morelike a metal. If o a were to decrease, the bandgap energy would increase and thematerial would begin to behave more like an insulator._______________________________________ 3.2Schrodinger's wave equation is:()()()t x x V x t x m ,,2222ψ⋅+∂ψ∂-η()tt x j ∂ψ∂=,ηAssume the solution is of the form:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x ηexp , Region I: ()0=x V . Substituting theassumed solution into the wave equation, we obtain:()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧∂∂-t E kx j x jku x m ηηexp 22 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u ηexp()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⋅⎪⎭⎫ ⎝⎛-=t E kx j x u jE j ηηηexp which becomes()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m ηηexp 222()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jk ηexp 2()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u ηexp 22()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+=t E kx j x Eu ηexp This equation may be written as()()()()0222222=+∂∂+∂∂+-x u mE x x u x x u jk x u k ηSetting ()()x u x u 1= for region I, the equation becomes: ()()()()021221212=--+x u k dx x du jk dxx u d α where222ηmE=αQ.E.D.In Region II, ()O V x V =. Assume the same form of the solution:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x ηexp , Substituting into Schrodinger's wave equation, we find:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m ηηexp 222()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jk ηexp 2()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u ηexp 22()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+t E kx j x u V O ηexp()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=t E kx j x Eu ηexp This equation can be written as:______________________________________________________________________________________4()()()2222xx u x x u jk x u k ∂∂+∂∂+- ()()02222=+-x u mEx u mV O ηη Setting ()()x u x u 2= for region II, this equation becomes()()dx x du jkdx x u d 22222+()022222=⎪⎪⎭⎫ ⎝⎛+--x u mV k O ηα where again222ηmE=αQ.E.D._______________________________________ 3.3We have ()()()()021221212=--+x u k dx x du jk dxx u d α Assume the solution is of the form: ()()[]x k j A x u -=αexp 1 ()[]x k j B +-+αexpThe first derivative is()()()[]x k j A k j dxx du --=ααexp 1 ()()[]x k j B k j +-+-ααexpand the second derivative becomes()()[]()[]x k j A k j dxx u d --=ααexp 2212()[]()[]x k j B k j +-++ααexp 2Substituting these equations into the differential equation, we find()()[]x k j A k ---ααexp 2()()[]x k j B k +-+-ααexp 2(){()[]x k j A k j jk --+ααexp 2()()[]}x k j B k j +-+-ααexp ()()[]{x k j A k ---ααexp 22 ()[]}0exp =+-+x k j B α Combining terms, we obtain()()()[]222222αααα----+--k k k k k ()[]x k j A -⨯αexp()()()[]222222αααα--++++-+k k k k k()[]0exp =+-⨯x k j B α We find that00= Q.E.D.For the differential equation in ()x u 2 and theproposed solution, the procedure is exactly the same as above._______________________________________ 3.4We have the solutions()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp for a x <<0 and()()[]x k j C x u -=βexp 2()[]x k j D +-+βexp for 0<<-x b .The first boundary condition is ()()0021u u =which yields0=--+D C B AThe second boundary condition is201===x x dx dudx du which yields()()()C k B k A k --+--βαα ()0=++D k βThe third boundary condition is ()()b u a u -=21 which yields()[]()[]a k j B a k j A +-+-ααexp exp ()()[]b k j C --=βexp ()()[]b k j D -+-+βexp______________________________________________________________________________________5and can be written as()[]()[]a k j B a k j A +-+-ααexp exp ()[]b k j C ---βexp ()[]0exp =+-b k j D βThe fourth boundary condition isbx a x dx dudx du -===21 which yields()()[]a k j A k j --ααexp()()[]a k j B k j +-+-ααexp()()()[]b k j C k j ---=ββexp()()()[]b k j D k j -+-+-ββexp and can be written as()()[]a k j A k --ααexp()()[]a k j B k +-+-ααexp ()()[]b k j C k ----ββexp()()[]0exp =+++b k j D k ββ_______________________________________ 3.5(b) (i) First point: πα=aSecond point: By trial and error, πα729.1=a (ii) First point: πα2=aSecond point: By trial and error, πα617.2=a _______________________________________ 3.6(b) (i) First point: πα=aSecond point: By trial and error, πα515.1=a (ii) First point: πα2=aSecond point: By trial and error, πα375.2=a _______________________________________ 3.7ka a aaP cos cos sin =+'ααα Let y ka =, x a =α Theny x x xP cos cos sin =+'Consider dydof this function.()[]{}y x x x P dyd sin cos sin 1-=+⋅'- We find()()()⎭⎬⎫⎩⎨⎧⋅+⋅-'--dy dx x x dy dx x x P cos sin 112y dydx x sin sin -=-Theny x x x x x P dy dx sin sin cos sin 12-=⎭⎬⎫⎩⎨⎧-⎥⎦⎤⎢⎣⎡+-' For πn ka y ==,...,2,1,0=n 0sin =⇒y So that, in general,()()dkd ka d a d dy dxαα===0 And22ηmE=α SodkdEm mE dk d ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-22/122221ηηα This implies thatdk dE dk d ==0α for an k π= _______________________________________ 3.8(a) πα=a 1π=⋅a E m o 212η______________________________________________________________________________________6()()()()2103123422221102.41011.9210054.12---⨯⨯⨯==ππa m E o η19104114.3-⨯=J From Problem 3.5 πα729.12=aπ729.1222=⋅a E m o η()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J12E E E -=∆1918104114.3100198.1--⨯-⨯= 19107868.6-⨯=Jor 24.4106.1107868.61919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J From Problem 3.5, πα617.24=aπ617.2224=⋅a E m o η()()()()2103123424102.41011.9210054.1617.2---⨯⨯⨯=πE18103364.2-⨯=J34E E E -=∆1818103646.1103364.2--⨯-⨯= 1910718.9-⨯=Jor 07.6106.110718.91919=⨯⨯=∆--E eV_______________________________________ 3.9(a) At π=ka , πα=a 1π=⋅a E m o 212η()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα859.0=a o()()()()210312342102.41011.9210054.1859.0---⨯⨯⨯=πo E19105172.2-⨯=J o E E E -=∆11919105172.2104114.3--⨯-⨯= 2010942.8-⨯=Jor 559.0106.110942.81920=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J At π=ka . From Problem 3.5,πα729.12=aπ729.1222=⋅a E m o η()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J 23E E E -=∆1818100198.1103646.1--⨯-⨯= 19104474.3-⨯=Jor 15.2106.1104474.31919=⨯⨯=∆--E eV_____________________________________________________________________________________________________________________________73.10(a) πα=a 1π=⋅a E m o 212η()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JFrom Problem 3.6, πα515.12=aπ515.1222=⋅a E m o η()()()()2103123422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J 12E E E -=∆1919104114.310830.7--⨯-⨯= 19104186.4-⨯=Jor 76.2106.1104186.41919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JFrom Problem 3.6, πα375.24=aπ375.2224=⋅a E m o η()()()()2103123424102.41011.9210054.1375.2---⨯⨯⨯=πE18109242.1-⨯=J 34E E E -=∆1818103646.1109242.1--⨯-⨯=1910597.5-⨯=Jor 50.3106.110597.51919=⨯⨯=∆--E eV _____________________________________3.11(a) At π=ka , πα=a 1π=⋅a E m o 212η()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=J At 0=ka , By trial and error, πα727.0=a o π727.022=⋅a E m o o η()()()()210312342102.41011.9210054.1727.0---⨯⨯⨯=πo E19108030.1-⨯=J o E E E -=∆11919108030.1104114.3--⨯-⨯= 19106084.1-⨯=Jor 005.1106.1106084.11919=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka , From Problem 3.6, πα515.12=aπ515.1222=⋅a E m o η()()()()2103423422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J 23E E E -=∆191810830.7103646.1--⨯-⨯=______________________________________________________________________________________81910816.5-⨯=Jor 635.3106.110816.51919=⨯⨯=∆--E eV_______________________________________ 3.12For 100=T K,()()⇒+⨯-=-1006361001073.4170.124g E164.1=g E eV200=T K, 147.1=g E eV 300=T K, 125.1=g E eV 400=T K, 097.1=g E eV 500=T K, 066.1=g E eV 600=T K, 032.1=g E eV _______________________________________ 3.13The effective mass is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m ηWe have()()B curve dkEd A curve dk E d 2222>so that ()()B curve m A curve m **<_______________________________________ 3.14The effective mass for a hole is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m p η We have that()()B curve dkEd A curve dk E d 2222> so that ()()B curve m A curve m p p **<_______________________________________ 3.15Points A,B: ⇒<0dkdEvelocity in -x direction Points C,D: ⇒>0dkdEvelocity in +x directionPoints A,D: ⇒<022dk Ednegative effective massPoints B,C: ⇒>022dkEd positive effective mass_______________________________________ 3.16For A: 2k C E i =At 101008.0+⨯=k m 1-, 05.0=E eV Or()()2119108106.105.0--⨯=⨯=E JSo ()2101211008.0108⨯=⨯-C3811025.1-⨯=⇒CNow ()()38234121025.1210054.12--*⨯⨯==C m η 311044.4-⨯=kgor o m m ⋅⨯⨯=--*31311011.9104437.4 o m m 488.0=*For B: 2k C E i =At 101008.0+⨯=k m 1-, 5.0=E eV Or ()()2019108106.15.0--⨯=⨯=E JSo ()2101201008.0108⨯=⨯-C 3711025.1-⨯=⇒CNow ()()37234121025.1210054.12--*⨯⨯==C m η321044.4-⨯=kgor o m m ⋅⨯⨯=--*31321011.9104437.4o m m 0488.0=*_____________________________________________________________________________________________________________________________93.17For A: 22k C E E -=-υ()()()2102191008.0106.1025.0⨯-=⨯--C3921025.6-⨯=⇒C ()()39234221025.6210054.12--*⨯⨯-=-=C m η31108873.8-⨯-=kgor o m m ⋅⨯⨯-=--*31311011.9108873.8o m m 976.0--=* For B: 22k C E E -=-υ()()()2102191008.0106.13.0⨯-=⨯--C382105.7-⨯=⇒C()()3823422105.7210054.12--*⨯⨯-=-=C m η3210406.7-⨯-=kgor o m m ⋅⨯⨯-=--*31321011.910406.7o m m 0813.0-=*_______________________________________ 3.18(a) (i) νh E =or ()()341910625.6106.142.1--⨯⨯==h E ν1410429.3⨯=Hz(ii) 141010429.3103⨯⨯===νλc E hc 51075.8-⨯=cm 875=nm(b) (i) ()()341910625.6106.112.1--⨯⨯==h E ν 1410705.2⨯=Hz(ii) 141010705.2103⨯⨯==νλc410109.1-⨯=cm 1109=nm _______________________________________ 3.19(c) Curve A: Effective mass is a constantCurve B: Effective mass is positive around 0=k , and is negativearound 2π±=k ._______________________________________ 3.20()[]O O k k E E E --=αcos 1 Then()()()[]O k k E dkdE---=ααsin 1()[]O k k E -+=ααsin 1 and()[]O k k E dkEd -=ααcos 2122Then221222*11ηηαE dk Ed m o k k =⋅== or212*αE m η=_______________________________________ 3.21(a) ()[]3/123/24l t dnm m m =*()()[]3/123/264.1082.04o o m m =o dnm m 56.0=*(b)oo l t cn m m m m m 64.11082.02123+=+=*oo m m 6098.039.24+=o cnm m 12.0=*_______________________________________ 3.22(a) ()()[]3/22/32/3lh hh dp m m m +=*______________________________________________________________________________________10()()[]3/22/32/3082.045.0o o m m +=[]o m ⋅+=3/202348.030187.0o dpm m 473.0=*(b) ()()()()2/12/12/32/3lh hh lh hh cpm m m m m ++=* ()()()()om ⋅++=2/12/12/32/3082.045.0082.045.0 o cpm m 34.0=*_______________________________________3.23For the 3-dimensional infinite potential well, ()0=x V when a x <<0, a y <<0, and a z <<0. In this region, the wave equation is:()()()222222,,,,,,z z y x y z y x x z y x ∂∂+∂∂+∂∂ψψψ()0,,22=+z y x mEψηUse separation of variables technique, so let ()()()()z Z y Y x X z y x =,,ψSubstituting into the wave equation, we have222222zZXY y Y XZ x X YZ ∂∂+∂∂+∂∂ 022=⋅+XYZ mEηDividing by XYZ , we obtain021*********=+∂∂⋅+∂∂⋅+∂∂⋅ηmE z Z Z y Y Y x X X Let01222222=+∂∂⇒-=∂∂⋅X k x X k x X X xx The solution is of the form: ()x k B x k A x X x x cos sin += Since ()0,,=z y x ψ at 0=x , then ()00=Xso that 0=B .Also, ()0,,=z y x ψ at a x =, so that()0=a X . Then πx x n a k = where...,3,2,1=x n Similarly, we have2221y k y Y Y -=∂∂⋅ and 2221z k zZ Z -=∂∂⋅From the boundary conditions, we find πy y n a k = and πz z n a k = where...,3,2,1=y n and ...,3,2,1=z n From the wave equation, we can write022222=+---ηmE k k k z y xThe energy can be written as()222222⎪⎭⎫ ⎝⎛++==a n n n m E E z y x n n n z y x πη _______________________________________ 3.24The total number of quantum states in the 3-dimensional potential well is given (in k-space) by()332a dk k dk k g T ⋅=ππ where222ηmEk =We can then writeηmEk 2=Taking the differential, we obtaindE Em dE E m dk ⋅⋅=⋅⋅⋅⋅=2112121ηηSubstituting these expressions into the densityof states function, we have()dE EmmE a dE E g T ⋅⋅⋅⎪⎭⎫ ⎝⎛=212233ηηππ Noting thatπ2h=ηthis density of states function can be simplified and written as______________________________________________________________________________________()()dE E m h a dE E g T ⋅⋅=2/33324πDividing by 3a will yield the density of states so that()()E hm E g ⋅=32/324π _______________________________________ 3.25For a one-dimensional infinite potential well,222222k an E m n ==*πη Distance between quantum states()()aa n a n k k n n πππ=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=-+11Now()⎪⎭⎫ ⎝⎛⋅=a dkdk k g T π2NowE m k n*⋅=21ηdE Em dk n⋅⋅⋅=*2211η Then()dE Em a dE E g n T ⋅⋅⋅=*2212ηπDivide by the "volume" a , so()Em E g n *⋅=21πηSo()()()()()EE g 31341011.9067.0210054.11--⨯⋅⨯=π ()EE g 1810055.1⨯=m 3-J 1-_______________________________________ 3.26(a) Silicon, o nm m 08.1=*()()c nc E E h m E g -=*32/324π()dE E E h m g kTE E c nc c c⋅-=⎰+*232/324π()()kT E E c nc cE E h m 22/332/33224+*-⋅⋅=π()()2/332/323224kT h m n⋅⋅=*π ()()[]()()2/33342/33123210625.61011.908.124kT ⋅⋅⨯⨯=--π ()()2/355210953.7kT ⨯=(i) At 300=T K, 0259.0=kT eV ()()19106.10259.0-⨯=2110144.4-⨯=J Then()()[]2/3215510144.4210953.7-⨯⨯=c g25100.6⨯=m 3- or 19100.6⨯=c g cm 3-(ii) At 400=T K, ()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV()()19106.1034533.0-⨯=21105253.5-⨯=J Then()()[]2/32155105253.5210953.7-⨯⨯=c g2510239.9⨯=m 3-or 191024.9⨯=c g cm 3-(b) GaAs, o nm m 067.0=*()()[]()()2/33342/33123210625.61011.9067.024kT g c ⋅⋅⨯⨯=--π ()()2/3542102288.1kT ⨯=______________________________________________________________________________________(i) At 300=T K, 2110144.4-⨯=kT J()()[]2/3215410144.42102288.1-⨯⨯=c g2310272.9⨯=m 3- or 171027.9⨯=c g cm 3-(ii) At 400=T K, 21105253.5-⨯=kT J()()[]2/32154105253.52102288.1-⨯⨯=c g2410427.1⨯=m 3- 181043.1⨯=c g cm 3-_______________________________________ 3.27(a) Silicon, o p m m 56.0=* ()()E E h mE g p-=*υυπ32/324()dE E E h m g E kTE p⋅-=⎰-*υυυυπ332/324()()υυυπE kTE pE E h m 32/332/33224-*-⎪⎭⎫ ⎝⎛-=()()[]2/332/333224kT h mp-⎪⎭⎫ ⎝⎛-=*π ()()[]()()2/33342/33133210625.61011.956.024kT ⎪⎭⎫ ⎝⎛⨯⨯=--π ()()2/355310969.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215510144.4310969.2-⨯⨯=υg2510116.4⨯=m 3- or 191012.4⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J ()()[]2/32155105253.5310969.2-⨯⨯=υg2510337.6⨯=m 3-or 191034.6⨯=υg cm 3- (b) GaAs, o p m m 48.0=*()()[]()()2/33342/33133210625.61011.948.024kT g ⎪⎭⎫ ⎝⎛⨯⨯=--πυ ()()2/3553103564.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215510144.43103564.2-⨯⨯=υg2510266.3⨯=m 3- or 191027.3⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.53103564.2-⨯⨯=υg2510029.5⨯=m 3-or 191003.5⨯=υg cm 3-_______________________________________ 3.28(a) ()()c nc E E h m E g -=*32/324π()()[]()c E E -⨯⨯=--3342/33110625.61011.908.124πc E E -⨯=56101929.1 For c E E =; 0=c g1.0+=c E E eV; 4610509.1⨯=c g m 3-J 1-2.0+=c E E eV;4610134.2⨯=m 3-J 1- 3.0+=c E E eV; 4610614.2⨯=m 3-J 1- 4.0+=c E E eV; 4610018.3⨯=m 3-J 1-(b) ()E E hm g p-=*υυπ32/324()()[]()E E -⨯⨯=--υπ3342/33110625.61011.956.024E E -⨯=υ55104541.4______________________________________________________________________________________For υE E =; 0=υg 1.0-=υE E eV; 4510634.5⨯=υg m 3-J 1-2.0-=υE E eV;4510968.7⨯=m 3-J 1-3.0-=υE E eV; 4510758.9⨯=m 3-J 1-4.0-=υE E eV;4610127.1⨯=m 3-J 1-_______________________________________ 3.29(a) ()()68.256.008.12/32/32/3=⎪⎭⎫ ⎝⎛==**pnc m m g g υ(b) ()()0521.048.0067.02/32/32/3=⎪⎭⎫ ⎝⎛==**pncmm g g υ_______________________________________3.30 Plot_______________________________________ 3.31(a) ()()()!710!7!10!!!-=-=i i i i i N g N g W()()()()()()()()()()()()1201238910!3!7!78910===(b) (i) ()()()()()()()()12!10!101112!1012!10!12=-=i W 66= (ii)()()()()()()()()()()()()1234!8!89101112!812!8!12=-=i W 495=_______________________________________ 3.32()⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F exp 11(a) kT E E F =-, ()()⇒+=1exp 11E f ()269.0=E f(b) kT E E F 5=-, ()()⇒+=5exp 11E f()31069.6-⨯=E f (c) kT E E F 10=-, ()()⇒+=10exp 11E f()51054.4-⨯=E f_______________________________________ 3.33()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F exp 1111or()⎪⎪⎭⎫⎝⎛-+=-kT E E E f F exp 111(a) kT E E F =-, ()269.01=-E f (b) kT E E F 5=-, ()31069.61-⨯=-E f (c) kT E E F 10=-, ()51054.41-⨯=-E f_______________________________________ 3.34(a) ()⎥⎦⎤⎢⎣⎡--≅kT E E f F F exp c E E =; 61032.90259.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f 2kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.020259.030.0exp F f 61066.5-⨯=kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.00259.030.0exp F f 61043.3-⨯=23kT E c +; ()()⎥⎦⎤⎢⎣⎡+-=0259.020259.0330.0exp F f 61008.2-⨯=kT E c 2+; ()()⎥⎦⎤⎢⎣⎡+-=0259.00259.0230.0exp F f 61026.1-⨯=______________________________________________________________________________________(b) ⎥⎦⎤⎢⎣⎡-+-=-kT E E f F F exp 1111()⎥⎦⎤⎢⎣⎡--≅kT E E F exp υE E =; ⎥⎦⎤⎢⎣⎡-=-0259.025.0exp 1F f 51043.6-⨯= 2kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.020259.025.0exp 1F f 51090.3-⨯=kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.00259.025.0exp 1F f 51036.2-⨯=23kT E -υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.020259.0325.0exp 1F f 51043.1-⨯= kT E 2-υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.00259.0225.0exp 1F f 61070.8-⨯=_______________________________________ 3.35()()⎥⎦⎤⎢⎣⎡-+-=⎥⎦⎤⎢⎣⎡--=kT E kT E kT E E f F c F F exp exp and()⎥⎦⎤⎢⎣⎡--=-kT E E f F F exp 1()()⎥⎦⎤⎢⎣⎡---=kT kT E E F υexp So ()⎥⎦⎤⎢⎣⎡-+-kT E kT E F c exp()⎥⎦⎤⎢⎣⎡+--=kT kT E E F υexp Then kT E E E kT E F F c +-=-+υOr midgap c F E E E E =+=2υ_______________________________________ 3.3622222man E n πη= For 6=n , Filled state()()()()()2103122234610121011.92610054.1---⨯⨯⨯=πE18105044.1-⨯=Jor 40.9106.1105044.119186=⨯⨯=--E eVFor 7=n , Empty state()()()()()2103122234710121011.92710054.1---⨯⨯⨯=πE1810048.2-⨯=Jor 8.12106.110048.219187=⨯⨯=--E eVTherefore 8.1240.9<<F E eV_______________________________________ 3.37(a) For a 3-D infinite potential well ()222222⎪⎭⎫⎝⎛++=a n n n mE z y x πη For 5 electrons, the 5thelectron occupies the quantum state 1,2,2===z y x n n n ; so()2222252⎪⎭⎫ ⎝⎛++=a n n n m E z y x πη()()()()()21031222223410121011.9212210054.1---⨯⨯++⨯=π 1910761.3-⨯=Jor 35.2106.110761.319195=⨯⨯=--E eV For the next quantum state, which is empty, the quantum state is 2,2,1===z y x n n n . This quantum state is at the same energy, so 35.2=F E eV(b) For 13 electrons, the 13th electronoccupies the quantum state______________________________________________________________________________________3,2,3===z y x n n n ; so ()()()()()2103122222341310121011.9232310054.1---⨯⨯++⨯=πE 1910194.9-⨯=Jor 746.5106.110194.9191913=⨯⨯=--E eVThe 14th electron would occupy the quantum state 3,3,2===z y x n n n . This state is at the same energy, so 746.5=F E eV_______________________________________ 3.38The probability of a state at E E E F ∆+=1 being occupied is()⎪⎭⎫ ⎝⎛∆+=⎪⎪⎭⎫ ⎝⎛-+=kT E kT E E E f F exp 11exp 11111 The probability of a state at E E E F ∆-=2being empty is()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F 222exp 1111⎪⎭⎫ ⎝⎛∆-+⎪⎭⎫ ⎝⎛∆-=⎪⎭⎫ ⎝⎛∆-+-=kT E kT E kT E exp 1exp exp 111 or()⎪⎭⎫ ⎝⎛∆+=-kT E E f exp 11122 so ()()22111E f E f -= Q.E.D._______________________________________ 3.39(a) At energy 1E , we want01.0exp 11exp 11exp 1111=⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-+-⎪⎪⎭⎫ ⎝⎛-kT E E kT E E kT E E F F FThis expression can be written as01.01exp exp 111=-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-+kT E E kT E E F For()⎪⎪⎭⎫⎝⎛-=kT E E F 1exp 01.01Then()100ln 1kT E E F += orkT E E F 6.41+= (b)At kT E E F 6.4+=,()()6.4exp 11exp 1111+=⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F which yields()01.000990.01≅=E f_______________________________________ 3.40 (a)()()⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--=0259.050.580.5exp exp kT E E f F F 61032.9-⨯=(b) ()060433.03007000259.0=⎪⎭⎫⎝⎛=kT eV31098.6060433.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f (c) ()⎥⎦⎤⎢⎣⎡--≅-kT E E f F F exp 1 ⎥⎦⎤⎢⎣⎡-=kT 25.0exp 02.0______________________________________________________________________________________or 5002.0125.0exp ==⎥⎦⎤⎢⎣⎡+kT ()50ln 25.0=kTor()()⎪⎭⎫⎝⎛===3000259.0063906.050ln 25.0T kTwhich yields 740=T K_______________________________________ 3.41 (a)()00304.00259.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 0.304%(b) At 1000=T K, 08633.0=kT eV Then()1496.008633.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 14.96%(c) ()997.00259.00.785.6exp 11=⎪⎭⎫⎝⎛-+=E for 99.7% (d)At F E E =, ()21=E f for alltemperatures_______________________________________ 3.42(a) For 1E E =()()⎥⎦⎤⎢⎣⎡--≅⎪⎪⎭⎫ ⎝⎛-+=kT E E kTE E E fF F11exp exp 11Then()611032.90259.030.0exp -⨯=⎪⎭⎫ ⎝⎛-=E fFor 2E E =,82.030.012.12=-=-E E F eV Then()⎪⎭⎫ ⎝⎛-+-=-0259.082.0exp 1111E for()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛---≅-0259.082.0exp 111E f141078.10259.082.0exp -⨯=⎪⎭⎫ ⎝⎛-=(b) For 4.02=-E E F eV, 72.01=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.072.0exp exp 1kT E E E f F or()131045.8-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1⎪⎭⎫ ⎝⎛-=0259.04.0expor()71096.11-⨯=-E f_______________________________________ 3.43(a) At 1E E =()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.030.0exp exp 1kT E E E f F or()61032.9-⨯=E f At 2E E =, 12.13.042.12=-=-E E F eV So()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1⎪⎭⎫ ⎝⎛-=0259.012.1exp______________________________________________________________________________________or()191066.11-⨯=-E f(b) For 4.02=-E E F ,02.11=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.002.1exp exp 1kT E E E f F or()181088.7-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1⎪⎭⎫ ⎝⎛-=0259.04.0expor ()71096.11-⨯=-E f_______________________________________ 3.44()1exp 1-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+=kTE E E f Fso()()2exp 11-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+-=kT E E dE E df F⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛⨯kT E E kT F exp 1 or()2exp 1exp 1⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛-=kT E E kT E E kT dE E df F F (a) At 0=T K, For()00exp =⇒=∞-⇒<dE dfE E F()0exp =⇒+∞=∞+⇒>dEdfE E FAt -∞=⇒=dEdfE E F(b) At 300=T K, 0259.0=kT eVFor F E E <<,0=dE dfFor F E E >>, 0=dEdfAt F E E =,()()65.91110259.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1-(c) At 500=T K, 04317.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =, ()()79.511104317.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1- _______________________________________ 3.45(a) At midgap E E =,()⎪⎪⎭⎫ ⎝⎛+=⎪⎪⎭⎫ ⎝⎛-+=kTE kTE E E f gF2exp 11exp 11Si: 12.1=g E eV, ()()⎥⎦⎤⎢⎣⎡+=0259.0212.1exp 11E for()101007.4-⨯=E fGe: 66.0=g E eV______________________________________________________________________________________()()⎥⎦⎤⎢⎣⎡+=0259.0266.0exp 11E for()61093.2-⨯=E f GaAs: 42.1=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0242.1exp 11E for()121024.1-⨯=E f(b) Using the results of Problem 3.38, the answers to part (b) are exactly the same as those given in part (a)._______________________________________ 3.46(a) ()⎥⎦⎤⎢⎣⎡--=kT E E f F F exp ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 108 or ()810ln 60.0+=kT()032572.010ln 60.08==kT eV()⎪⎭⎫⎝⎛=3000259.0032572.0Tso 377=T K(b) ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 106()610ln 60.0+=kT()043429.010ln 60.06==kT ()⎪⎭⎫⎝⎛=3000259.0043429.0Tor 503=T K_______________________________________ 3.47(a) At 200=T K,()017267.03002000259.0=⎪⎭⎫⎝⎛=kT eV⎪⎪⎭⎫ ⎝⎛-+==kTE E f FF exp 1105.019105.01exp =-=⎪⎪⎭⎫⎝⎛-kT E E F()()()19ln 017267.019ln ==-kT E E F 05084.0=eV By symmetry, for 95.0=F f , 05084.0-=-F E E eVThen ()1017.005084.02==∆E eV (b) 400=T K, 034533.0=kT eV For 05.0=F f , from part (a),()()()19ln 034533.019ln ==-kT E E F 10168.0=eV Then ()2034.010168.02==∆E eV_______________________________________。
《半导体物理与器件》习题库
《半导体物理与器件》习题库目录《半导体物理与器件》习题库 (1)第1章思考题和习题 (2)第2章思考题和习题 (3)第3章思考题和习题 (6)第4章思考题和习题 (10)第5章半导体器件制备技术 (12)第6章Ga在SiO2/Si结构下的开管掺杂 (13)第1章思考题和习题1. 300K时硅的晶格常数a=5.43Å，求每个晶胞内所含的完整原子数和原子密度为多少？2. 综述半导体材料的基本特性及Si、GaAs的晶格结构和特征。
3. 画出绝缘体、半导体、导体的简化能带图，并对它们的导电性能作出定性解释。
4. 以硅为例，简述半导体能带的形成过程。
5. 证明本征半导体的本征费米能级E i位于禁带中央。
6. 简述迁移率、扩散长度的物理意义。
7. 室温下硅的有效态密度Nc=2.8×1019cm-3,κT=0.026eV,禁带宽度Eg=1.12eV，如果忽略禁带宽度随温度的变化，求：（a）计算77K、300K、473K 3个温度下的本征载流子浓度。
（b） 300K本征硅电子和空穴的迁移率分别为1450cm2/V·s和500cm2/V·s，计算本征硅的电阻率是多少？8. 某硅棒掺有浓度分别为1016/cm3和1018/cm3的磷，求室温下的载流子浓度及费米能级E FN的位置（分别从导带底和本征费米能级算起）。
9. 某硅棒掺有浓度分别为1015/cm3和1017/cm3的硼，求室温下的载流子浓度及费米能级E FP的位置（分别从价带顶和本征费米能级算起）。
10. 求室温下掺磷为1017/cm3的N+型硅的电阻率与电导率。
11. 掺有浓度为3×1016cm-3的硼原子的硅，室温下计算：（a）光注入△n=△p=3×1012cm-3的非平衡载流子，是否为小注入？为什么？（b）附加光电导率△σ为多少？（c）画出光注入下的准费米能级E’FN和E’FP(E i为参考)的位置示意图。
半导体物理与器件课后练习题含答案
半导体物理与器件课后练习题含答案1. 简答题1.1 什么是p型半导体？答案： p型半导体是指通过加入掺杂物（如硼、铝等）使得原本的n型半导体中含有空穴，从而形成的半导体材料。
具有p型性质的半导体材料被称为p型半导体。
1.2 什么是n型半导体？答案： n型半导体是指通过加入掺杂物（如磷、锑等）使得原本的p型半导体中含有更多的自由电子，从而形成的半导体材料。
具有n型性质的半导体材料被称为n型半导体。
1.3 什么是pn结？答案： pn结是指将p型半导体和n型半导体直接接触形成的结构。
在pn结的界面处，p型半导体中的空穴和n型半导体中的自由电子会相互扩散，形成空间电荷区，从而形成一定的电场。
当外加正向电压时，电子和空穴在空间电荷区中相遇，从而发生复合并产生少量电流；而当外加反向电压时，电场反向，空间电荷区扩大，从而形成一个高电阻的结，电流几乎无法通过。
2. 计算题2.1 若硅片的掺杂浓度为1e16/cm³，电子迁移率为1350 cm²/Vs，电离能为1.12 eV，则硅片的载流子浓度为多少？解题过程：根据硅片的掺杂浓度为1e16/cm³，可以判断硅片的类型为n型半导体。
因此易知载流子为自由电子。
根据电离能为1.12 eV，可以推算出自由电子的有效密度为：n = N * exp(-Eg / (2kT)) = 6.23e9/cm³其中，N为硅的密度，k为玻尔兹曼常数（1.38e-23 J/K），T为温度（假定为室温300K），Eg为硅的带隙（1.12 eV）。
因此，载流子浓度为1e16 + 6.23e9 ≈ 1e16 /cm³。
2.2 假设有一n+/p结的二极管，其中n+区的掺杂浓度为1e19/cm³，p区的掺杂浓度为1e16/cm³，假设该二极管在正向电压下的漏电流为1nA，求该二极管的有效面积。
解题过程：由于该二极管的正向电压下漏电流为1nA，因此可以利用肖特基方程计算出它的开启电压：I = I0 * (exp(qV / (nkT)) - 1)其中，I0为饱和漏电流（假定为0），q为电子电荷量，V为电压，n为调制系数（一般为1），k为玻尔兹曼常数，T为温度。
半导体物理与器件第四版课后习题答案(供参考)
Chapter 44.1⎪⎪⎭⎫ ⎝⎛-=kTE N N n gc i exp 2υ ⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=kT E T N N g O cO exp 3003υwhere cO N and O N υ are the values at 300 K.(b) Germanium_______________________________________ 4.2Plot_______________________________________ 4.3(a) ⎪⎪⎭⎫⎝⎛-=kT E N N n g c i exp 2υ ()()()319192113001004.1108.2105⎪⎭⎫⎝⎛⨯⨯=⨯T()()⎥⎦⎤⎢⎣⎡-⨯3000259.012.1exp T()3382330010912.2105.2⎪⎭⎫⎝⎛⨯=⨯T()()()()⎥⎦⎤⎢⎣⎡-⨯T 0259.030012.1expBy trial and error, 5.367≅T K(b)()252122105.2105⨯=⨯=i n()()()()()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=T T 0259.030012.1exp 30010912.2338By trial and error, 5.417≅T K_______________________________________ 4.4At 200=T K, ()⎪⎭⎫⎝⎛=3002000259.0kT017267.0=eVAt 400=T K, ()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV()()()()17222102210025.31040.11070.7200400⨯=⨯⨯=ii nn⎥⎦⎤⎢⎣⎡-⎥⎦⎤⎢⎣⎡-⨯⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛=017267.0exp 034533.0exp 30020030040033g g E E⎥⎦⎤⎢⎣⎡-=034533.0017267.0exp 8g g E E()[]9578.289139.57exp 810025.317-=⨯g Eor()1714.38810025.3ln 9561.2817=⎪⎪⎭⎫⎝⎛⨯=g E or 318.1=g E eVNow ()32103004001070.7⎪⎭⎫⎝⎛=⨯o co N N υ⎪⎭⎫ ⎝⎛-⨯034533.0318.1exp ()()172110658.2370.210929.5-⨯=⨯o co N N υ so 371041.9⨯=o co N N υcm 6-_______________________________________ 4.5()()⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-=kT kT kT A n B n i i 20.0exp 90.0exp 10.1exp For 200=T K, 017267.0=kT eV For 300=T K, 0259.0=kT eV For 400=T K, 034533.0=kT eV(a) For 200=T K, ()()610325.9017267.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i (b) For 300=T K, ()()41043.40259.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i (c) For 400=T K, ()()31005.3034533.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i _______________________________________ 4.6(a) ()⎥⎦⎤⎢⎣⎡---∝kT E E E E f g F c F c exp()⎥⎦⎤⎢⎣⎡---∝kT E E E E c c exp()⎥⎦⎤⎢⎣⎡--⨯kT E E F c exp Let x E E c =-Then ⎪⎭⎫⎝⎛-∝kT x x f g F c expTo find the maximum value: ()⎪⎭⎫⎝⎛-∝-kT x x dx f g d F c exp 212/10exp 12/1=⎪⎭⎫ ⎝⎛-⋅-kT x x kT which yields2212/12/1kTx kT x x =⇒= The maximum value occurs at2kTE E c +=(b)()()⎥⎦⎤⎢⎣⎡---∝-kT E E E E f g F F exp 1υυ()⎥⎦⎤⎢⎣⎡---∝kT E E E E υυexp()⎥⎦⎤⎢⎣⎡--⨯kT E E F υexp Let x E E =-υThen ()⎪⎭⎫ ⎝⎛-∝-kT x x f g F exp 1υTo find the maximum value()[]0exp 1=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛-∝-kT x x dx d dx f g d F υ Same as part (a). Maximum occurs at2kTx =or2kTE E -=υ_______________________________________ 4.7()()()()⎥⎦⎤⎢⎣⎡---⎥⎦⎤⎢⎣⎡---=kT E E E E kT E E E E E n E n c c c c 221121exp expwherekT E E c 41+= and 22kTE E c += Then()()()⎥⎦⎤⎢⎣⎡--=kT E E kT kTE n E n 2121exp 24()5.3exp 22214exp 22-=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛--=or()()0854.021=E n E n_______________________________________ 4.8Plot_______________________________________ 4.9Plot_______________________________________ 4.10⎪⎪⎭⎫ ⎝⎛=-**ln 43n p midgap Fi m m kT E E Silicon: o p m m 56.0*=, o n m m 08.1*=0128.0-=-midgap Fi E E eV Germanium: o pm m 37.0*=,o n m m 55.0*=0077.0-=-midgap Fi E E eV Gallium Arsenide: o p m m 48.0*=,o n m m 067.0*= 0382.0+=-midgap Fi E E eV _______________________________________ 4.11 ()⎪⎪⎭⎫⎝⎛=-c midgap Fi N N kT E E υln 21()()kT kT 4952.0108.21004.1ln 211919-=⎪⎪⎭⎫ ⎝⎛⨯⨯=_______________________________________ 4.12(a) ⎪⎪⎭⎫ ⎝⎛=-**ln 43n pmidgap Fi m m kT E E()⎪⎭⎫⎝⎛=21.170.0ln 0259.04363.10-⇒meV(b) ()⎪⎭⎫⎝⎛=-080.075.0ln 0259.043midgap Fi E E47.43+⇒meV_______________________________________ 4.13Let ()==K E g c constant Then()()dE E fE g n FE co c⎰∞=dE kT E E Kc E F⎰∞⎪⎪⎭⎫⎝⎛-+=exp 11()dE kT E E K cE F ⎰∞⎥⎦⎤⎢⎣⎡--≅exp Let kT E E c-=η so that ηd kT dE ⋅=We can write ()()c F c F E E E E E E -+-=-so that()()()η-⋅⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--exp exp exp kT E E kT E E F c F The integral can then be written as()()ηηd kT E E kT K n F c o ⎰∞-⎥⎦⎤⎢⎣⎡--⋅⋅=0exp exp which becomes()⎥⎦⎤⎢⎣⎡--⋅⋅=kT E E kT K n F c o exp _______________________________________ 4.14Let ()()c c E E C E g -=1 for c E E ≥ Then()()dE E fE g n FE co c⎰∞=()dE kT E E E E C c E Fc ⎰∞⎪⎪⎭⎫⎝⎛-+-=exp 11()()dE kT E E E E C F E C c⎥⎦⎤⎢⎣⎡---≅⎰∞exp 1LetkTE E c-=η so that ηd kT dE ⋅= We can write()()F c c F E E E E E E -+-=-Then()⎥⎦⎤⎢⎣⎡--=kT E E C n F c o exp 1()()dE kT E E E E c E c c⎥⎦⎤⎢⎣⎡---⨯⎰∞exp or()⎥⎦⎤⎢⎣⎡--=kT E E C n F c o exp 1 ()()()[]()ηηηd kT kT -⨯⎰∞exp 0We find that()()()11exp exp 0+=---=-∞∞⎰ηηηηηdSo()()⎥⎦⎤⎢⎣⎡--=kT E E kT C n F c o exp 21 _______________________________________ 4.15We have ⎪⎪⎭⎫⎝⎛=∈*1m m a r o r o For germanium, 16=∈r , o m m 55.0*= Then()()()53.02955.01161=⎪⎭⎫⎝⎛=o a roroA r 4.151=The ionization energy can be written as ()6.132*⎪⎪⎭⎫⎝⎛∈∈⎪⎪⎭⎫ ⎝⎛=s o o m m E eV ()()029.06.131655.02=⇒=E eV_______________________________________ 4.16We have ⎪⎪⎭⎫⎝⎛=∈*1m m a r o r o For gallium arsenide, 1.13=∈r ,o m m 067.0*= Then()()oA r 10453.0067.011.131=⎪⎭⎫⎝⎛=The ionization energy is()()()6.131.13067.06.1322*=⎪⎪⎭⎫ ⎝⎛∈∈⎪⎪⎭⎫ ⎝⎛=s o o m m E or0053.0=E eV_______________________________________ 4.17(a) ⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1519107108.2ln 0259.02148.0=eV(b) ()F c g F E E E E E --=-υ90518.02148.012.1=-=eV(c) ()⎥⎦⎤⎢⎣⎡--=kT E E N p F o υυexp()⎥⎦⎤⎢⎣⎡-⨯=0259.090518.0exp 1004.119 31090.6⨯=cm 3- (d) Holes(e) ⎪⎪⎭⎫⎝⎛=-i o Fi F n n kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1015105.1107ln 0259.0338.0=eV_______________________________________ 4.18(a) ⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=16191021004.1ln 0259.0162.0=eV (b) ()υE E E E E F g F c --=- 958.0162.012.1=-=eV(c) ()⎪⎭⎫⎝⎛-⨯=0259.0958.0exp 108.219o n31041.2⨯=cm 3-(d) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1102ln 0259.0365.0=eV_______________________________________ 4.19(a) ⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=519102108.2ln 0259.08436.0=eV ()F c g F E E E E E --=-υ 8436.012.1-= 2764.0=-υE E F eV(b) ()⎪⎭⎫⎝⎛-⨯=0259.027637.0exp 1004.119o p1410414.2⨯=cm 3-(c) p-type_______________________________________ 4.20(a) ()032375.03003750259.0=⎪⎭⎫⎝⎛=kT eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫ ⎝⎛⨯=032375.028.0exp 300375107.42/317o n 141015.1⨯=cm 3-()28.042.1-=--=-F c g F E E E E E υ 14.1=eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=032375.014.1exp 3003751072/318o p 31099.4⨯=cm 3-(b) ()⎪⎪⎭⎫⎝⎛⨯⨯=-14171015.1107.4ln 0259.0F c E E2154.0=eV()2154.042.1-=--=-F c g F E E E E E υ 2046.1=eV()⎥⎦⎤⎢⎣⎡-⨯=0259.02046.1exp 10718o p21042.4-⨯=cm 3-_______________________________________ 4.21(a) ()032375.03003750259.0=⎪⎭⎫⎝⎛=kT eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=032375.028.0exp 300375108.22/319o n 151086.6⨯= cm 3-()28.012.1-=--=-F c g F E E E E E υ 840.0=eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=032375.0840.0exp 3003751004.12/319o p 71084.7⨯=cm 3-(b) ⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=151910862.6108.2ln 0259.02153.0=eV9047.02153.012.1=-=-υE E F eV()⎥⎦⎤⎢⎣⎡-⨯=0259.0904668.0exp 1004.119o p 31004.7⨯=cm 3-_______________________________________ 4.22(a) p-type(b) 28.0412.14===-g F E E E υeV()⎥⎦⎤⎢⎣⎡--=kT E E N p F o υυexp ()⎥⎦⎤⎢⎣⎡-⨯=0259.028.0exp 1004.119141010.2⨯=cm 3- ()υE E E E E F g F c --=- 84.028.012.1=-=eV()⎥⎦⎤⎢⎣⎡--=kT E E N n F c c o exp ()⎥⎦⎤⎢⎣⎡-⨯=0259.084.0exp 108.219 51030.2⨯=cm 3-_______________________________________ 4.23(a) ⎥⎦⎤⎢⎣⎡-=kT E E n n Fi Fi o exp ()⎥⎦⎤⎢⎣⎡⨯=0259.022.0exp 105.110131033.7⨯=cm 3-⎥⎦⎤⎢⎣⎡-=kT E E n p F Fii o exp ()⎥⎦⎤⎢⎣⎡-⨯=0259.022.0exp 105.11061007.3⨯=cm 3-(b) ⎥⎦⎤⎢⎣⎡-=kT E E n n Fi Fi o exp()⎥⎦⎤⎢⎣⎡⨯=0259.022.0exp 108.16 91080.8⨯=cm 3-⎥⎦⎤⎢⎣⎡-=kT E E n p F Fii o exp()⎥⎦⎤⎢⎣⎡-⨯=0259.022.0exp 108.16 21068.3⨯=cm 3-_______________________________________ 4.24(a) ⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=151********.1ln 0259.01979.0=eV (b) ()υE E E E E F g F c --=- 92212.019788.012.1=-=eV(c) ()⎥⎦⎤⎢⎣⎡-⨯=0259.092212.0exp 108.219o n31066.9⨯=cm 3- (d) Holes(e) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1015105.1105ln 0259.03294.0=eV _______________________________________ 4.25()034533.03004000259.0=⎪⎭⎫⎝⎛=kT eV()2/3193004001004.1⎪⎭⎫⎝⎛⨯=υN1910601.1⨯=cm 3-()2/319300400108.2⎪⎭⎫⎝⎛⨯=c N19103109.4⨯=cm 3-()()1919210601.1103109.4⨯⨯=i n⎥⎦⎤⎢⎣⎡-⨯034533.012.1exp 24106702.5⨯=1210381.2⨯=⇒i n cm 3- (a) ⎪⎪⎭⎫ ⎝⎛=-oF pN kT E E υυln ()⎪⎪⎭⎫⎝⎛⨯⨯=151910510601.1ln 034533.02787.0=eV(b) 84127.027873.012.1=-=-F c E E eV(c) ()⎥⎦⎤⎢⎣⎡-⨯=034533.084127.0exp 103109.419o n910134.1⨯=cm 3- (d) Holes(e) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=121510381.2105ln 034533.02642.0=eV _______________________________________ 4.26(a) ()⎥⎦⎤⎢⎣⎡-⨯=0259.025.0exp 10718o p141050.4⨯=cm 3-17.125.042.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=0259.017.1exp 107.417o n21013.1-⨯=cm 3-(b) 034533.0=kT eV ()2/318300400107⎪⎭⎫⎝⎛⨯=υN1910078.1⨯=cm 3- ()2/317300400107.4⎪⎭⎫⎝⎛⨯=c N1710236.7⨯=cm 3- ⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=14191050.410078.1ln 034533.03482.0=eV072.13482.042.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=034533.007177.1exp 10236.717o n 41040.2⨯=cm 3-_____________________________________ 4.27(a) ()⎥⎦⎤⎢⎣⎡-⨯=0259.025.0exp 1004.119o p141068.6⨯=cm 3-870.025.012.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=0259.0870.0exp 108.219o n41023.7⨯=o n cm 3- (b) 034533.0=kT eV ()2/3193004001004.1⎪⎭⎫⎝⎛⨯=υN1910601.1⨯=cm 3- ()2/319300400108.2⎪⎭⎫ ⎝⎛⨯=c N1910311.4⨯=cm 3-⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=14191068.610601.1ln 034533.03482.0=eV7718.03482.012.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=034533.077175.0exp 10311.419o n91049.8⨯=cm 3-_______________________________________ 4.28(a) ()F c o F N n ηπ2/12=For 2kT E E c F +=,5.02==-=kTkT kT E E c F F η Then ()0.12/1≅F F η()()0.1108.2219⨯=πo n191016.3⨯=cm 3-(b) ()F c o F N n ηπ2/12=()()0.1107.4217⨯=π171030.5⨯=cm 3-_______________________________________ 4.29()F o F N p ηπυ'=2/12()()FF ηπ'⨯=⨯2/119191004.12105So ()26.42/1='FF η We find kTE E FF-=≅'υη0.3()()0777.00259.00.3==-F E E υeV_______________________________________ 4.30(a) 44==-=kTkTkT E E c F F ηThen ()0.62/1≅F F η ()F c o F N n ηπ2/12=()()0.6108.2219⨯=π201090.1⨯=cm 3-(b) ()()0.6107.4217⨯=πo n181018.3⨯=cm 3-_______________________________________ 4.31For the electron concentration ()()()E f E g E n F c =The Boltzmann approximation applies, so ()()c nE E hm E n -=32/3*24π()⎥⎦⎤⎢⎣⎡--⨯kT E E F exp or()()()⎥⎦⎤⎢⎣⎡--=kT E E h m E n F c nexp 2432/3*π()⎥⎦⎤⎢⎣⎡---⨯kT E E kT E E kTc c exp DefinekTE E x c-= Then()()()x x K x n E n -=→exp To find maximum ()()x n E n →, set()()x x K dx x dn -⎢⎣⎡==-exp 2102/1 +()()⎥⎦⎤--x x exp 12/1or()⎥⎦⎤⎢⎣⎡--=-x x Kx 21exp 02/1which yieldskT E E kT E E x c c 2121+=⇒-==For the hole concentration ()()()[]E f E g E p F -=1υUsing the Boltzmann approximation ()()E E h m E p p-=υπ32/3*24()⎥⎦⎤⎢⎣⎡--⨯kT E E F exp or()()()⎥⎦⎤⎢⎣⎡--=kT E E h mE pF pυπexp 2432/3*()⎥⎦⎤⎢⎣⎡---⨯kT E E kT E E kTυυexp DefinekTEE x -='υThen()()x x K x p '-''='exp To find maximum value of ()()x p E p '→,set()0=''x d x dp Using the results from above,we find the maximum atkT E E 21-=υ_______________________________________4.32(a) Silicon: We have()⎥⎦⎤⎢⎣⎡--=kT E E N n F c c o exp We can write()()F d d c F c E E E E E E -+-=- For045.0=-d c E E eV andkT E E F d 3=-eV we can write()⎥⎦⎤⎢⎣⎡--⨯=30259.0045.0exp 108.219o n()()737.4exp 108.219-⨯= or171045.2⨯=o n cm 3- We also have()⎥⎦⎤⎢⎣⎡--=kT E E N p F o υυexp Again, we can write()()υυE E E E E E a a F F -+-=- ForkT E E a F 3=- and045.0=-υE E a eV Then()⎥⎦⎤⎢⎣⎡--⨯=0259.0045.03exp 1004.119o p ()()737.4exp 1004.119-⨯= or161012.9⨯=o p cm 3-(b) GaAs: assume 0058.0=-d c E E eV Then()⎥⎦⎤⎢⎣⎡--⨯=30259.00058.0exp 107.417o n ()()224.3exp 107.417-⨯= or161087.1⨯=o n cm 3-Assume 0345.0=-υE E a eV Then()⎥⎦⎤⎢⎣⎡--⨯=30259.00345.0exp 10718o p ()()332.4exp 10718-⨯=or161020.9⨯=o p cm 3-_______________________________________ 4.33Plot_______________________________________ 4.34(a) 151510310154⨯=-⨯=o p cm 3- ()415210105.7103105.1⨯=⨯⨯=o n cm 3-(b) 16103⨯==d o N n cm 3- ()316210105.7103105.1⨯=⨯⨯=o p cm 3-(c) 10105.1⨯===i o o n p n cm 3-(d) ()()3191923003751004.1108.2⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯3750259.030012.1exp1110334.7⨯=⇒i n cm 3- 15104⨯==a o N p cm 3-()8152111034.110410334.7⨯=⨯⨯=on cm 3-(e) ()()3191923004501004.1108.2⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯4500259.030012.1exp1310722.1⨯=⇒i n cm 3-()2132141410722.1210210⨯+⎪⎪⎭⎫ ⎝⎛+=o n1410029.1⨯=cm 3-()12142131088.210029.110722.1⨯=⨯⨯=o p cm 3-_______________________________________ 4.35(a) 151510104-⨯=-=d a o N N p 15103⨯=cm 3-()3152621008.1103108.1-⨯=⨯⨯==o i o p n n cm 3-(b) 16103⨯==d o N n cm 3- ()416261008.1103108.1-⨯=⨯⨯=op cm 3-(c) 6108.1⨯===i o o n p n cm 3-(d) ()()318172300375100.7107.4⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯3750259.030042.1exp810580.7⨯=⇒i n cm 3- 15104⨯==a o N p cm 3-()215281044.110410580.7⨯=⨯⨯=on cm 3-(e) ()()318172300450100.7107.4⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯4500259.030042.1exp1010853.3⨯=⇒i n cm 3- 1410==d o N n cm 3-()7142101048.11010853.3⨯=⨯=op cm 3-_______________________________________ 4.36(a) Ge: 13104.2⨯=i n cm 3-(i)2222i dd o n N N n +⎪⎪⎭⎫ ⎝⎛+=()21321515104.221022102⨯+⎪⎪⎭⎫ ⎝⎛⨯+⨯=or15102⨯=≅d o N n cm 3-()152132102104.2⨯⨯==o i o n n p111088.2⨯= cm 3- (ii)151610710⨯-=-≅d a o N N p 15103⨯=cm 3- ()152132103104.2⨯⨯==o i o p n n111092.1⨯=cm 3-(b) GaAs: 6108.1⨯=i n cm 3- (i)15102⨯=≅d o N n cm()315261062.1102108.1-⨯=⨯⨯=op cm 3-(ii)15103⨯=-≅d a o N N p cm 3-()315261008.1103108.1-⨯=⨯⨯=on cm 3-(c) The result implies that there is only one minority carrier in a volume of 310cm 3. _______________________________________ 4.37(a) For the donor level⎪⎪⎭⎫ ⎝⎛-+=kT E E N n F d d d exp 2111⎪⎭⎫ ⎝⎛+=0259.020.0exp 2111or41085.8-⨯=dd N n(b) We have()⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F F exp 11Now()()F c c F E E E E E E -+-=- or245.0+=-kT E E FThen()⎪⎭⎫ ⎝⎛++=0259.0245.01exp 11E f For()51087.2-⨯=E f F_______________________________________ 4.38(a) ⇒>d a N N p-type (b) Silicon:1313101105.2⨯-⨯=-=d a o N N p or13105.1⨯=o p cm 3- Then()7132102105.1105.1105.1⨯=⨯⨯==o i o p n n cm 3- Germanium:2222i da d a o n N N N N p +⎪⎪⎭⎫ ⎝⎛-+-=()21321313104.22105.12105.1⨯+⎪⎪⎭⎫⎝⎛⨯+⎪⎪⎭⎫ ⎝⎛⨯=or131026.3⨯=o p cm 3- Then()131321321076.110264.3104.2⨯=⨯⨯==o i o p n n cm 3-Gallium Arsenide:13105.1⨯=-=d a o N N p cm 3- and()216.0105.1108.113262=⨯⨯==o i o p n n cm 3- _______________________________________ 4.39(a) ⇒>a d N N n-type(b) 1515102.1102⨯-⨯=-≅a d o N N n 14108⨯=cm 3-()51421021081.2108105.1⨯=⨯⨯==o i o n n p cm 3-(c) ()d a ao N N N p -+'≅ 151515102102.1104⨯-⨯+'=⨯aN 15108.4⨯='⇒aN cm 3-()41521010625.5104105.1⨯=⨯⨯=on cm 3-_______________________________________ 4.40()155210210125.1102105.1⨯=⨯⨯==o i o p n n cm 3- ⇒>o o p n n-type_______________________________________ 4.41()()318192300250100.61004.1⎪⎭⎫⎝⎛⨯⨯=i n()()⎥⎦⎤⎢⎣⎡-⨯3002500259.066.0exp24108936.1⨯=1210376.1⨯=⇒i n cm 3- 2222414i o o i o i o n n n n p n n =⇒==i o n n 21=⇒So 111088.6⨯=o n cm 3-,Then 121075.2⨯=o p cm 3-2222i aa o n N N p +⎪⎪⎭⎫ ⎝⎛+= 212210752.2⎪⎪⎭⎫ ⎝⎛-⨯a N242108936.12⨯+⎪⎪⎭⎫ ⎝⎛=a N()21224210752.2105735.7⎪⎪⎭⎫ ⎝⎛+⨯-⨯aa N N 242108936.12⨯+⎪⎪⎭⎫ ⎝⎛=a N so that 1210064.2⨯=a N cm 3-_______________________________________ 4.42Plot_______________________________________ 4.43Plot_______________________________________ 4.44Plot_______________________________________ 4.452222i ad a d o n N N N N n +⎪⎪⎭⎫ ⎝⎛-+-= 2102.1102101.1141414⨯-⨯=⨯2214142102.1102i n +⎪⎪⎭⎫⎝⎛⨯-⨯+()()221321314104104101.1i n +⨯=⨯-⨯22727106.1109.4i n +⨯=⨯ so 131074.5⨯=i n cm 3-1314272103101.1103.3⨯=⨯⨯==o i o n n p cm 3- _______________________________________ 4.46(a) ⇒>d a N N p-typeMajority carriers are holes1616105.1103⨯-⨯=-=d a o N N p16105.1⨯=cm 3-Minority carriers are electrons()4162102105.1105.1105.1⨯=⨯⨯==o i o p n n cm 3- (b) Boron atoms must be addedd a ao N N N p -+'=161616105.1103105⨯-⨯+'=⨯aN So 16105.3⨯='aN cm 3-()316210105.4105105.1⨯=⨯⨯=o n cm 3-_______________________________________ 4.47(a) ⇒<<i o n p n-type (b) oi o o i o p n n n n p 22=⇒=on ()16421010125.1102105.1⨯=⨯⨯=cm 3-⇒electrons are majority carriers4102⨯=o p cm 3-⇒holes are minority carriers (c) a d o N N n -=151610710125.1⨯-=⨯d N so 1610825.1⨯=d N cm 3-_______________________________________ 4.48⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln2222i a a o n N N p +⎪⎪⎭⎫ ⎝⎛+= and1510=N cm 3-_______________________________________ 4.49(a) ⎪⎪⎭⎫⎝⎛=-d c F c N N kT E E ln()⎪⎪⎭⎫⎝⎛⨯=d N 19108.2ln 0259.0 For 1410cm 3-, 3249.0=-F c E E eV 1510cm 3-, 2652.0=-F c E E eV 1610cm 3-, 2056.0=-F c E E eV 1710cm 3-, 1459.0=-F c E E eV(b) ⎪⎪⎭⎫⎝⎛=-i d Fi F n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯=10105.1ln 0259.0d N For 1410cm 3-, 2280.0=-Fi F E E eV 1510cm 3-, 2877.0=-Fi F E E eV 1610cm 3-, 3473.0=-Fi F E E eV 1710cm 3-, 4070.0=-Fi F E E eV _______________________________________ 4.50(a) 2222i dd o n N N n +⎪⎪⎭⎫ ⎝⎛+= 151005.105.1⨯==d o N n cm 3- ()21515105.01005.1⨯-⨯()2215105.0i n +⨯=so 2821025.5⨯=i nNow()()3191923001004.1108.2⎪⎭⎫ ⎝⎛⨯⨯=T n i()()⎥⎦⎤⎢⎣⎡-⨯3000259.012.1exp T()3382830010912.21025.5⎪⎭⎫ ⎝⎛⨯=⨯T⎥⎦⎤⎢⎣⎡-⨯T 973.12972exp By trial and error, 5.536=T K (b) At 300=T K,⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯=-151910108.2ln 0259.0F c E E2652.0=eV At 5.536=T K,()046318.03005.5360259.0=⎪⎭⎫⎝⎛=kT eV()2/3193005.536108.2⎪⎭⎫⎝⎛⨯=c N1910696.6⨯=cm 3-⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=-15191005.110696.6ln 046318.0F c E E5124.0=eV then ()2472.0=-∆F c E E eV (c) Closer to the intrinsic energy level._______________________________________ 4.51⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln At 200=T K, 017267.0=kT eV 400=T K, 034533.0=kT eV 600=T K, 0518.0=kT eVAt 200=T K,()()3191923002001004.1108.2⎪⎭⎫⎝⎛⨯⨯=in⎥⎦⎤⎢⎣⎡-⨯017267.012.1exp410638.7⨯=⇒i n cm 3- At 400=T K,()()3191923004001004.1108.2⎪⎭⎫⎝⎛⨯⨯=in⎥⎦⎤⎢⎣⎡-⨯034533.012.1exp 1210381.2⨯=⇒i n cm 3- At 600=T K,()()3191923006001004.1108.2⎪⎭⎫⎝⎛⨯⨯=in⎥⎦⎤⎢⎣⎡-⨯0518.012.1exp1410740.9⨯=⇒i n cm 3- At 200=T K and 400=T K, 15103⨯==a o N p cm 3- At 600=T K,2222i a a o n N N p +⎪⎪⎭⎫⎝⎛+=()2142151510740.921032103⨯+⎪⎪⎭⎫ ⎝⎛⨯+⨯=1510288.3⨯=cm 3-Then, 200=T K, 4212.0=-F Fi E E eV 400=T K, 2465.0=-F Fi E E eV600=T K, 0630.0=-F Fi E E eV_______________________________________ 4.52(a)()⎪⎪⎭⎫⎝⎛⨯=⎪⎪⎭⎫ ⎝⎛=-6108.1ln 0259.0ln a i a F Fi N n N kT E EFor 1410=a N cm 3-,4619.0=-F Fi E E eV1510=a N cm 3-,5215.0=-F Fi E E eV1610=a N cm 3-,5811.0=-F Fi E E eV1710=a N cm 3-,6408.0=-F Fi E E eV (b)()⎪⎪⎭⎫⎝⎛⨯=⎪⎪⎭⎫ ⎝⎛=-a a F N N N kT E E 18100.7ln 0259.0ln υυ For 1410=a N cm 3-,2889.0=-υE E F eV1510=a N cm 3-,2293.0=-υE E F eV1610=a N cm 3-,1697.0=-υE E F eV1710=a N cm 3-,1100.0=-υE E F eV_______________________________________ 4.53(a) ⎪⎪⎭⎫ ⎝⎛=-**ln 43n p midgap Fi m m kT E E ()()10ln 0259.043= or0447.0+=-midgap Fi E E eV(b) Impurity atoms to be added so 45.0=-F midgap E E eV (i) p-type, so add acceptor atoms (ii)4947.045.00447.0=+=-F Fi E E eV Then⎪⎪⎭⎫⎝⎛-=kT E E n p F Fi i o exp()⎪⎭⎫⎝⎛=0259.04947.0exp 105or131097.1⨯==a o N p cm 3-_______________________________________4.54()⎥⎦⎤⎢⎣⎡--=-=kT E E N N N n F c c a d o exp so()⎪⎭⎫⎝⎛-⨯+⨯=0259.0215.0exp 108.21051915d N15151095.6105⨯+⨯=or16102.1⨯=d N cm 3-_______________________________________ 4.55(a) Silicon(i)⎪⎪⎭⎫⎝⎛=-d c F c N N kT E E ln()2188.0106108.2ln 0259.01519=⎪⎪⎭⎫⎝⎛⨯⨯=eV(ii)1929.00259.02188.0=-=-F c E E eV()⎥⎦⎤⎢⎣⎡--=kT E E N N F c c d exp()⎥⎦⎤⎢⎣⎡-⨯=0259.01929.0exp 108.2191610631.1⨯=d N cm 3-15106⨯+'=dN 1610031.1⨯='⇒dN cm 3- Additionaldonor atoms (b) GaAs(i)()⎪⎪⎭⎫⎝⎛⨯=-151710107.4ln 0259.0F c E E15936.0=eV(ii)13346.00259.015936.0=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=0259.013346.0exp 107.417d N1510718.2⨯=cm 3-1510+'=dN 1510718.1⨯='⇒dN cm 3- Additionaldonor atoms _______________________________________4.56 (a) ⎪⎪⎭⎫ ⎝⎛=-a F Fi N N kT E E υln()⎪⎪⎭⎫⎝⎛⨯⨯=16191021004.1ln 0259.01620.0=eV(b) ⎪⎪⎭⎫⎝⎛=-d c Fi F N N kT E E ln ()1876.0102108.2ln 0259.01619=⎪⎪⎭⎫⎝⎛⨯⨯=eV (c) For part (a); 16102⨯=o p cm 3-()162102102105.1⨯⨯==o i o p n n410125.1⨯=cm 3-For part (b): 16102⨯=o n cm 3-()162102102105.1⨯⨯==o i o n n p410125.1⨯=cm 3-_______________________________________4.57⎥⎦⎤⎢⎣⎡-=kT E E n n Fi F i o exp ()⎥⎦⎤⎢⎣⎡⨯=0259.055.0exp 108.1615100.3⨯=cm 3-Add additional acceptor impurities a d o N N n -= a N -⨯=⨯151510710315104⨯=⇒a N cm 3-_______________________________________ 4.58(a) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln ()3161.0105.1103ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=eV (b) ⎪⎪⎭⎫⎝⎛=-i o Fi F n n kT E E ln ()3758.0105.1103ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=eV(c) Fi F E E =(d) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln ()⎪⎪⎭⎫ ⎝⎛⨯⨯⎪⎭⎫ ⎝⎛=111510334.7104ln 3003750259.0 2786.0=eV(e) ⎪⎪⎭⎫ ⎝⎛=-i o Fi F n n kT E E ln()⎪⎪⎭⎫ ⎝⎛⨯⨯⎪⎭⎫ ⎝⎛=131410722.110029.1ln 3004500259.0 06945.0=eV _______________________________________ 4.59 (a) ⎪⎪⎭⎫ ⎝⎛=-o F p N kT E E υυln()2009.0103100.7ln 0259.01518=⎪⎪⎭⎫ ⎝⎛⨯⨯=eV(b) ()⎪⎪⎭⎫ ⎝⎛⨯⨯=--4181008.1100.7ln 0259.0υE E F 360.1=eV(c) ()⎪⎪⎭⎫⎝⎛⨯⨯=-618108.1100.7ln 0259.0υE E F7508.0=eV(d) ()⎪⎭⎫⎝⎛=-3003750259.0υE E F()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯152/318104300375100.7ln 2526.0=eV(e) ()⎪⎭⎫⎝⎛=-3004500259.0υE E F()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯72/3181048.1300450100.7ln 068.1=eV_______________________________________ 4.60n-type⎪⎪⎭⎫⎝⎛=-i o Fi F n n kT E E ln()3504.0105.110125.1ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=eV ______________________________________ 4.612222i aa o n N N p +⎪⎪⎭⎫ ⎝⎛+= 21051008.51515⨯=⨯22152105i n +⎪⎪⎭⎫ ⎝⎛⨯+()21515105.21008.5⨯-⨯()2215105.2i n +⨯=230301025.6106564.6i n +⨯=⨯29210064.4⨯=⇒i n⎥⎦⎤⎢⎣⎡-=kT E N N n g c i exp 2υ()030217.03003500259.0=⎪⎭⎫⎝⎛=kT eV()1921910633.1300350102.1⨯=⎪⎭⎫ ⎝⎛⨯=c N cm 3-()192191045.2300350108.1⨯=⎪⎭⎫ ⎝⎛⨯=υN cm 3- Now()()1919291045.210633.110064.4⨯⨯=⨯⎥⎦⎤⎢⎣⎡-⨯030217.0exp g ESo()()()⎥⎦⎤⎢⎣⎡⨯⨯⨯=29191910064.41045.210633.1ln 030217.0g E 6257.0=⇒g E eV_______________________________________ 4.62(a) Replace Ga atoms ⇒Silicon acts as adonor()()1415105.310705.0⨯=⨯=d N cm 3-Replace As atoms ⇒Silicon acts as anacceptor()()15151065.610795.0⨯=⨯=a N cm 3-(b) ⇒>d a N N p-type(c) 1415105.31065.6⨯-⨯=-=d a o N N p 15103.6⨯=cm 3-()4152621014.5103.6108.1-⨯=⨯⨯==o i o p n n cm 3- (d) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()5692.0108.1103.6ln 0259.0615=⎪⎪⎭⎫⎝⎛⨯⨯=eV_______________________________________。
半导体器件物理课后习题(施敏)
8. 给定一个未知掺杂的硅晶样品，霍耳测量提供了以下的 信息：W = 0.05 cm，A = 1.610-3 cm2（参考图8），I = 2.5 mA，且磁场为30T（1特斯拉（T）= 10-4 Wb/cm2）。若测 量出的霍耳电压为 +10 mV，求半导体样品的霍耳系数、导 体型态、多数载流子浓度、电阻率及迁移率。
3 5.43 2.35Å 4
(b)计算硅中（100），（110），（111）三平面 上每平方厘米的原子数。
1 (1) 从(100)面上看，每个单胞侧面上有 4 1 2 个原子 4
2 2 14 6 . 78 10 所以，每平方厘米的原子数= 2 a (5.43 108 ) 2
答：因为镓为III族元素，最外层有3个电子；锡为IV族元 素，最外层有4个电子，所以锡替换镓后作为施主提供电 子，此时电子为多子，所以该半导体为n型。
12. 求出在300K时一非简并n型半导体导带中电
子的动能。
解：在能量为dE范围内单位体积的电子数 N(E)F(E)dE, 而导带中每个电子的动能为E-Ec 所以导带中单位体积电子总动能为
因为热平衡时，样品内部没有载流子的净流动，所以有
J n漂移 J n扩散 J n 0
根据欧姆定律的微分形式
J n漂移 E ( x)
(a) q
E
J n扩散 ( x)
Dn N 0 exp( ax)
a q kT n N 0 exp( ax) q a kT n N 0 exp( ax) a kT n N D q n N D a kT q
20. 对一掺杂1016 cm-3磷施主原子，且施主能级ED= 0.045 eV的n型硅样品而言，找出在77K时中性施主 浓度对电离施主浓度的比例；此时费米能级低于导 带底部0.0459eV（电离施主的表示式可见问题19）。
半导体物理与器件习题
第一章 固体晶格结构1．如图是金刚石结构晶胞，若a 是其晶格常数，则其原子密度是 。
2．所有晶体都有的一类缺陷是：原子的热振动，另外晶体中常的缺陷有点缺陷、线缺陷。
3．半导体的电阻率为10-3～109Ωcm 。
4．什么是晶体？晶体主要分几类？5．什么是掺杂？常用的掺杂方法有哪些？答：为了改变导电性而向半导体材料中加入杂质的技术称为掺杂。
常用的掺杂方法有扩散和离子注入。
6．什么是替位杂质？什么是填隙杂质？ 7．什么是晶格？什么是原胞、晶胞？第二章 量子力学初步1．量子力学的三个基本原理是三个基本原理能量量子化原理、波粒二相性原理、不确定原理。
2．什么是概率密度函数？3．描述原子中的电子的四个量子数是： 、 、 、 。
第三章 固体量子理论初步1．能带的基本概念⏹ 能带（energy band ）包括允带和禁带。
⏹ 允带（allowed band ）：允许电子能量存在的能量范围。
⏹ 禁带（forbidden band ）：不允许电子存在的能量范围。
⏹ 允带又分为空带、满带、导带、价带。
⏹ 空带（empty band ）：不被电子占据的允带。
⏹满带（filled band ）：允带中的能量状态（能级）均被电子占据。
导带：有电子能够参与导电的能带，但半导体材料价电子形成的高能级能带通常称为导带。
价带:由价电子形成的能带，但半导体材料价电子形成的低能级能带通常称为价带。
2．什么是漂移电流？漂移电流：漂移是指电子在电场的作用下的定向运动，电子的定向运动所产生的电流。
3．什么是电子的有效质量？晶格中运动的电子，在外力和内力作用下有： Ｆ总＝Ｆ外＋Ｆ内=ma, m 是粒子静止的质量。
Ｆ外=m*n a, m*n 称为电子的有效质量。
4．位于能带底的电子，其有效质量为正，位于能带顶电子，其有效质量为负。
5．在室温T=300K ，Si 的禁带宽度：Eg=1.12eV Ge 的禁带宽度：Eg=0.67eV GaAs 的禁带宽度：Eg=1.43eVEg 具有负温度系数，即T 越大，Eg 越小；Eg 反应了，在相同温度下，Eg 越大，电子跃迁到导带的能力越弱。
半导体物理与器件习题答案
半导体物理与器件习题答案【篇一：半导体物理与器件课后习题2】图3.35所示色e-k关系曲线表示了两种可能的价带。
说明其中哪一种对应的空穴有效质量较大。
为什么？解：图中b曲线对应的空穴有效质量较大空穴的有效质量： m*p?1 21de?222?dk?图中曲线a的弯曲程度大于曲线bd2e 故 22dkd2e?22dkba?m*p?a??m*p?b?3.16 图3.37所示为两种不同半导体材料导带中电子的e-k关系抛物线，试确定两种电子的有效质量（以自由电子质量为单位）。
解：e-k关系曲线k=0附近的图形 ?k2近似于抛物线故有：e?ec? *2mn由图可知 ec?0①对于a曲线1??1.055?10?0.1????2k210-10??*?31? ?4.97?10kg?0.55me 有mn（a）?-192e0.07?1.06?10 ?-342?2②对于b曲线有1??1.055?10??0.1?-10?22?k10?32m*??4.97?10kg?0.055men （b）?-192e0.7?1.06?10 ?-342?23.20 硅的能带图3.23b所示导带的最小能量出现在[100]方向上。
最小值附近一维方向上的能量可以近似为s(k?k0) e?e0?e1co?其中k0是最小能量的k值。
是确定k?k0时的粒子的有效质量。
解：导带能量最小值附近一维方向上的能量e?e0?e1cos?(k?k0) d2e ?22??2e1cos?(k?k0) dkd2e当k?k0时 cos?(k?k0)?1；22??2e1dk 11d2e?*?222又mn?dk?2?k?k0时粒子的有效质量为：m?2?e1 *n3.24 试确定t=300k时gaas中ev和ev-kt之间的总量子态数量。
h3?3*2pev?e当t=300k时 gaas中ev和ev?kt之间总量子态数量：h3h36.6262?10?3432?1.38?103?23?30032?3.28?10?7cm?33.37 某种材料t=300k时的费米能级为6.25ev。
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1.1 确定晶胞中的原子数：（a ）面心立方；（b ）体心立方；(c)金刚石晶格。
解：（a ）面心立方： 8个拐角原子×81=1个原子6个面原子×21=3个原子∴ 面心立方中共含4个原子（b ）体心立方：8个拐角原子×81=1个原子1个中心原子 =1个原子 ∴ 体心立方中共含2个原子（c ）金刚石晶格：8个拐角原子×81=1个原子6个面原子×21 =3个原子4个中心原子 =4个原子 ∴ 金刚是晶格中共含8个原子1.15 计算如下平面硅原子的面密度：（a ）（100），（b ）（110），（c ）（111）。
解：(a):（100）平面面密度，通过把晶格原子数与表面面积相除得：面密度=()28-1043.52⨯个原子=214/1078.6cm 个原子⨯(b):(110)表面面密度=()28-1043.524⨯个原子=214/1059.9cm 个原子⨯(c):(111)表面面密度=()28-1043.534⨯个原子=214/1083.7cm 个原子⨯1.19（a ）如果硅中加入浓度为2×1610/3cm 的替位硼杂质原子，计算单晶中硅原子替位的百分率。
（b ）对于浓度为1510/3cm 的硼杂质原子，重新计算（a ） 解：（a ）：硅原子的体密度()32238-/1000.51043.58cm 个原子个原子⨯≈⨯=∴ 硅原子替位百分率=005-0022161041001000.5102⨯=⨯⨯⨯ （b ）同理：硅原子替位百分率=006-0022161021001000.5101⨯=⨯⨯⨯3.14 图3.35所示色E-k 关系曲线表示了两种可能的价带。
说明其中哪一种对应的空穴有效质量较大。
为什么？解：图中B 曲线对应的空穴有效质量较大空穴的有效质量： 2222*11m k d E d p ⨯=图中曲线A 的弯曲程度大于曲线B故 BAkd E d kd Ed 222222>∴()()**m m B p A p <3.16 图3.37所示为两种不同半导体材料导带中电子的E-k 关系抛物线，试确定两种电子的有效质量（以自由电子质量为单位）。
解：E-k 关系曲线k=0附近的图形近似于抛物线故有：*2m 2k n C E E =-由图可知 0=C E①对于A 曲线 有()0.55me kg 104.97 1006.107.01011.010055.12k m 3119-210-234-22*≈⨯=⨯⨯⎪⎭⎫ ⎝⎛⨯⨯⨯==-E A n ）（②对于B 曲线有()0.055me kg 104.97 1006.17.01011.010055.12k m 3219-210-234-22*≈⨯=⨯⨯⎪⎭⎫ ⎝⎛⨯⨯⨯==-E B n ）（3.20 硅的能带图3.23b 所示导带的最小能量出现在[100]方向上。
最小值附近一维方向上的能量可以近似为)(cos 010k k E E E --=α其中0k 是最小能量的k 值。
是确定0k k =时的粒子的有效质量。
解：导带能量最小值附近一维方向上的能量)(cos 010k k E E E --=α)(cos 012222k k E kd Ed -=∴αα当0k k =时 1)(cos 0=-k k α； 12222E k d Ed α=又2222*1m 1k d E d n =∴0k k =时粒子的有效质量为：122*m E nα =3.24 试确定T=300K 时GaAs 中T E E k -v v 和之间的总量子态数量。
解：根据()E E h E V V -=323*p2m π4)(g当T=300K 时 GaAs 中T E E V V k -和之间总量子态数量：()()()()()()()()3723233342331-23323*pk 23323*pk 323*p1028.33001038.132106262.6109.1090.672π4322m π4322m π42m π4)(g ------⨯≈⨯⨯⨯⨯⨯⨯==-⎪⎭⎫⎝⎛-=-=⎰cm kT hE E h dEE E hE V V VV E TE V E TE V T3.37 某种材料T=300K 时的费米能级为6.25eV 。
该材料中的电子符合费米-狄拉克函数。
（a ）求6.50eV 处能级被电子占据的概率。
（b ）如果温度上升为T=950K ，重复前面的计算（假设F E 不变）.(c)如果比费米能级低0.03eV 处能级为空的概率是1%。
此时温度为多少？解：根据费米-狄拉克分布函数：⎪⎭⎫⎝⎛-+=kTE E E FF ex p 11)(f（a ）在6.50eV 处能级被电子占据的概率：%1037.61038.1300106.125.6-50.6exp 11)(f 323-19--⨯≈⎥⎦⎤⎢⎣⎡⨯⨯⨯⨯+=）（E F（b ）温度上升为950K 时 6.50eV 能级被占据概率：%1052.41038.1950106.125.6-50.6exp 11)(f 323-19--⨯≈⎥⎦⎤⎢⎣⎡⨯⨯⨯⨯+=）（E F（c ）有题意可知比费米能级低0.3eV 处能级为空的概率为1%，即被占据的概率为99%KT In T kT e kT e kT 7570101.011038.11060.13.00101.013.0exp 99.013.0exp 199.0e 3.0-exp 112319=⎪⎭⎫⎝⎛=⨯÷⨯=⎪⎭⎫⎝⎛=⎪⎭⎫⎝⎛-+=⎪⎭⎫ ⎝⎛+∴--解得：故此时温度为757K习题44.14 假设某种半导体材料的导带状态密度为一常量K,且假设费米-狄拉克统计分布和波尔兹曼近似有效。
试推导热平衡状态下导带内电子浓度的表达式。
解：令常数())(g 常数K E c =，则：()()E d KT E E K Ed KT E E K Ed E f E EcF Ec F EcF c ⎰⎰⎰∞∞∞⎪⎭⎫ ⎝⎛-≈⎪⎭⎫ ⎝⎛-+==exp exp 11g n 0 设,KTE E F-=η则)exp(exp exp )()(d ηη-⎪⎭⎫⎝⎛-=⎪⎭⎫⎝⎛----=-∴=KTE E KT E E E E E E E E KTd E CF F C F C F上式可写为⎪⎭⎫⎝⎛-⋅⋅=-⎪⎭⎫ ⎝⎛-⋅⋅=⎰∞KT E E kT K d KT E E kT K C F C F exp n )exp(exp n 000即ηη4.22 (a)考虑T=300K 时的硅。
若ev E E F F 35.0i =-求0p (b)假设(a)中的0p 保持不变，求T=400K 时F F E E -i 的值(c)求出(a)与(b)中的0n解：当T=300K 时，硅的ev kT cm 0259.0,105.1n 310i=⨯=- 则)m (1011.10259.035.0exp 105.1exp n p 3-16100c kT E E F Fi i ⨯=⎪⎭⎫ ⎝⎛⨯=⎪⎭⎫ ⎝⎛-=(b)当T=300K 时，硅中318317100.7,cm 107.4--⨯=⨯=cm N N v C 当T=400K 时)(1038.2n )03454.01.12exp(-)300400)(100.7)(107.4(n 03454.0)300400)(0259.0()kT Egexp(-n 312i 3181722-⨯=⨯⨯====cm evKT N N iV C i则：()ev n p kT E E iF F 292.01038.21011.1ln 03453.0)ln(12160i =⎪⎪⎭⎫⎝⎛⨯⨯==-(c)由(a)得:)cm 1003.21011.1105.1(n 3416210020-⨯=⨯⨯==（）p n i对（b ）有：)cm 1010.51011.11038.2(n 3816212020-⨯=⨯⨯==（）p n i习题四（2）4.34 已知T450K 时的一块硅样品，掺杂了浓度为的1531.510cm -⨯硼和浓度为的143810cm -⨯砷。
（a ）该材料时n 型半导体还是p 型半导体？(b)计算电子的浓度和空穴的浓度。
(c)计算已电离的杂质浓度。
解：T=450K 时 对于硅： 1.12g E ev =2191919323133Eg n exp(-)kT450 1.12 1.6010(2.810)(1.0410)()exp(-)300450 1.38102.9610()iC V N N cm ----=⨯⨯=⨯⨯⨯⨯⨯⨯⨯≈⨯ (a),da N N P ∴>故为型半导体(b)空穴浓度：a 01514143-p 21.510-81027.010()d N N cm -=+⨯⨯=+≈⨯ 电子浓度:()21321430140 1.7210 4.2310()710in n cm p -⨯===⨯⨯(c)d d d N N n +=- ;0a a N N p -=-450K 时为强电离区故00d N p == 从而已电离的杂质浓度为1415153810 1.510 2.310()d a d a N N N N cm +--+=+=⨯+⨯=⨯4.51(a)T300K 时硅中掺杂了浓度为15310cm -的磷原子，确定硅的费米能级相对于本征费米能级的位置。
(b)假如加入的杂质换为浓度为15310cm -的硼原子重复（a ）.（c ）分别计算与中的电子子浓度。
解：（a ）:i 1510ln()2100.0259ln 1.5100.2877d F F i N E E kT n ev-=⎛⎫⨯=⨯ ⎪⨯⎝⎭=即硅的费米能级高于本征费米能级0.2877ev 处； （b ）1512ln()100.0259ln 1.5100.2877dFi F i N E E kT n ev-=⎛⎫=⨯ ⎪⨯⎝⎭=即硅的费米能级低于本征费米能级0.2877ev 处；（c):（a ）2000n ;a a i p N n p n =+=得：0n 2d d N N =+≈ 故：电子浓度1530n 10dN cm -== （b)153010a p N cm-==()2102530150 1.5102.251010i n n cm p -⨯===⨯习题55.9 在一块特殊的半导体材料中s -/1000u 2nv cm =，s -/600u 2p v cm =,31910-==cm N N v C ,且这些参数不随温度变化。
测得T=300K 时的本征电导率为。
求T=500K 时的电导率? 解： 电导率)(p u n u en p n i +=σT=300K 时本征电导率为16-)cm -10-Ω（故 )(p n i i u u en +=σ即 391961091.3)1600(106.110)300(---⨯≈⨯⨯=cm K n ievN N N N i V C V C i122.1)1091.3()10(ln0259.0n kTln Eg )kT Egexp(-n 2921922=⨯=⎪⎪⎭⎫⎝⎛==故又313i232623-1921921024.2500n )(1002.5)500101.38106.1122.1exp(-)10()kTEg exp(-500n---⨯=⨯≈⨯⨯⨯⨯==cm K cm n N N K V C i）（）（所以 从而有131319)-(1074.5)6001000(1024.2106.1)()500(---Ω⨯≈+⨯⨯⨯⨯=+=cm u u en K p n i σ5.29半导体中总电流恒定，由电子漂移电流和空穴扩散电流组成。