下载文档原格式
半导体物理与器件(尼曼第四版)答案第一章：半导体材料与晶体1.1 半导体材料的基本特性半导体材料是一种介于导体和绝缘体之间的材料。
它的基本特性包括：1.带隙：半导体材料的价带与导带之间存在一个禁带或带隙，是电子在能量上所能占据的禁止区域。
2.拉伸系统：半导体材料的结构是由原子或分子构成的晶格结构，其中的原子或分子以确定的方式排列。
3.载流子：在半导体中，存在两种载流子，即自由电子和空穴。
自由电子是在导带上的，在外加电场存在的情况下能够自由移动的电子。
空穴是在价带上的，当一个价带上的电子从该位置离开时，会留下一个类似电子的空位，空穴可以看作电子离开后的痕迹。
4.掺杂：为了改变半导体材料的导电性能，通常会对其进行掺杂。
掺杂是将少量元素添加到半导体材料中，以改变载流子浓度和导电性质。
1.2 半导体材料的结构与晶体缺陷半导体材料的结构包括晶体结构和非晶态结构。
晶体结构是指材料具有有序的周期性排列的结构，而非晶态结构是指无序排列的结构。
晶体结构的特点包括：1.晶体结构的基本单位是晶胞，晶胞在三维空间中重复排列。
2.晶格常数是晶胞边长的倍数，用于描述晶格的大小。
3.晶体结构可分为离子晶体、共价晶体和金属晶体等不同类型。
晶体结构中可能存在各种晶体缺陷，包括：1.点缺陷：晶体中原子位置的缺陷，主要包括实际缺陷和自间隙缺陷两种类型。
2.线缺陷：晶体中存在的晶面上或晶内的线状缺陷，主要包括位错和脆性断裂两种类型。
3.面缺陷：晶体中存在的晶面上的缺陷，主要包括晶面位错和穿孔两种类型。
1.3 半导体制备与加工半导体制备与加工是指将半导体材料制备成具有特定电性能的器件的过程。
它包括晶体生长、掺杂、薄膜制备和微电子加工等步骤。
晶体生长是将半导体材料从溶液或气相中生长出来的过程。
常用的晶体生长方法包括液相外延法、分子束外延法和气相外延法等。
掺杂是为了改变半导体材料的导电性能，通常会对其进行掺杂。
常用的掺杂方法包括扩散法、离子注入和分子束外延法等。
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_______________________________________。
半导体物理第四版答案【篇一：(考试范围)半导体物理学课后题答案】格常数为a的一维晶格，导带极小值附近能量ec(k)和价带极大值附近能量ev(k)分别为：h2(k?k1)2h2k2h2k213h2k2,ev(k) ec（k）= 3m0m06m0m0m0为电子惯性质量，k1?a,a?0.314nm。
试求：（1）禁带宽度;（2）导带底电子有效质量; （3）价带顶电子有效质量;（4）价带顶电子跃迁到导带底时准动量的变化解：（1）导带：2?2k2?2(k?k1)由??03m0m03k14d2ec2?22?28?2203m0m03m0dk得：k?所以：在k?价带：dev6?2k0得k?0dkm0d2ev6?2又因为0,所以k?0处，ev取极大值2m0dk?2k123因此：eg?ec(k1)?ev(0)??0.64ev412m03k处，ec取极小值4(2)m*nc22decdk23?m0 83k?k141(3)m*nv22devdk2k?01m06(4)准动量的定义：p??k所以：?p?(?k)3k?k143(k)k0k107.951025n/s42. 晶格常数为0.25nm的一维晶格，当外加102v/m，107 v/m的电场时，试分别计算电子自能带底运动到能带顶所需的时间。
解：根据：f?qe?h (0t1k?k得?t?qet)8.27108s1.61019102(0a8.271013s)?107t21.61019半导体物理第2章习题5. 举例说明杂质补偿作用。
当半导体中同时存在施主和受主杂质时，若（1） ndna因为受主能级低于施主能级，所以施主杂质的电子首先跃迁到na个受主能级上，还有nd-na个电子在施主能级上，杂质全部电离时，跃迁到导带中的导电电子的浓度为n= nd-na。
即则有效受主浓度为naeff≈ nd-na （2）nand施主能级上的全部电子跃迁到受主能级上，受主能级上还有na-nd个空穴，它们可接受价带上的na-nd个电子，在价带中形成的空穴浓度p= na-nd. 即有效受主浓度为naeff≈ na-nd （3）na?nd时，不能向导带和价带提供电子和空穴，称为杂质的高度补偿 6. 说明类氢模型的优点和不足。
半导体器件物理课后作业第二章对发光二极管（LED）、光电二极管（PD）、隧道二极管、齐纳二极管、变容管、快恢复二极管和电荷存储二极管这7个二端器件，请选择其中的4个器件，简述它们的工作原理和应用场合。
解：发光二极管它是半导体二极管的一种，是一种固态的半导体器件，可以把电能转化成光能；常简写为LED。
工作原理：发光二极管与普通二极管一样是由一个PN结组成，也具有单向导电性。
当给发光二极管加上正向电压后，从P区注入到N区的空穴和由N区注入到P区的电子，在PN结附近数微米内分别与N区的电子和P区的空穴复合，产生自发辐射的荧光。
不同的半导体材料中电子和空穴所处的能量状态不同。
当电子和空穴复合时释放出的能量多少是不同的，释放出的能量越多，则发出的光的波长越短；反之，则发出的光的波长越长。
应用场合：常用的是发红光、绿光或黄光的二极管，它们主要用于各种LED显示屏、彩灯、工作（交通）指示灯以及居家LED节能灯。
光电二极管光电二极管（Photo-Diode）和普通二极管一样，也是由一个PN结组成的半导体器件，也具有单方向导电特性，但在电路中它不是作整流元件，而是把光信号转换成电信号的光电传感器件。
工作原理：普通二极管在反向电压作用时处于截止状态，只能流过微弱的反向电流，光电二极管在设计和制作时尽量使PN结的面积相对较大，以便接收入射光，而电极面积尽量小些，而且PN结的结深很浅，一般小于1微米。
光电二极管是在反向电压作用下工作的，没有光照时，反向电流极其微弱，叫暗电流；当有光照时，携带能量的光子进入PN结后，把能量传给共价键上的束缚电子，使部分电子挣脱共价键，从而产生电子—空穴对，称为光生载流子。
它们在反向电压作用下参加漂移运动，使反向电流迅速增大到几十微安，光的强度越大，反向电流也越大。
这种特性称为“光电导”。
光电二极管在一般照度的光线照射下，所产生的电流叫光电流。
如果在外电路上接上负载，负载上就获得了电信号，而且这个电信号随着光的变化而相应变化。
第一章习题1．设晶格常数为a 的一维晶格，导带极小值附近能量E c (k)和价带极大值附近能量E V (k)分别为：E c =0220122021202236)(,)(3m k h m k h k E m k k h m k h V -=-+ 0m 。
试求：为电子惯性质量，nm a ak 314.0,1==π（1）禁带宽度;（2） 导带底电子有效质量; （3）价带顶电子有效质量;（4）价带顶电子跃迁到导带底时准动量的变化 解：（1）eV m k E k E E E k m dk E d k m kdk dE Ec k k m m m dk E d k k m k k m k V C g V V V c 64.012)0()43(0,060064338232430)(2320212102220202020222101202==-==<-===-==>=+===-+ 因此：取极大值处，所以又因为得价带：取极小值处，所以：在又因为：得：由导带：043222*83)2(1m dk E d mk k C nC===sN k k k p k p m dk E d mk k k k V nV/1095.7043)()()4(6)3(25104300222*11-===⨯=-=-=∆=-== 所以：准动量的定义：2. 晶格常数为0.25nm 的一维晶格，当外加102V/m ，107 V/m 的电场时，试分别计算电子自能带底运动到能带顶所需的时间。
解：根据：tkhqE f ∆∆== 得qE k t -∆=∆sat sat 137192821911027.810106.1)0(1027.810106.1)0(----⨯=⨯⨯--=∆⨯=⨯⨯--=∆ππ第三章习题和答案1. 计算能量在E=E c 到2*n 2C L 2m 100E E π+= 之间单位体积中的量子态数。
解322233*28100E 21233*22100E 0021233*231000L 8100)(3222)(22)(1Z VZZ )(Z )(22)(2322C22C L E m h E E E m V dE E E m V dE E g V d dEE g d E E m V E g c nc C n l m h E C n l m E C n n c n c πππππ=+-=-====-=*++⎰⎰**）（）（单位体积内的量子态数）（2. 试证明实际硅、锗中导带底附近状态密度公式为式（3-6）。
半导体物理与器件英文版第四版课后练习题含答案Chapter 1: Crystal PropertiesMultiple Choice Questions1.Which of the following statements is correct? A. The latticestructure of a crystal can be described by three crystal axes that are normal to each other. B. For a crystal with a primitive cubic unit cell, the coordination number is 8. C. In a crystal lattice with a face-centered cubic (FCC) unit cell, each atom has only six nearest neighbors. D. The Miller indices of a crystal planeperpendicular to the x-axis and passing through point (1, 2, 3) are (1, 2, 3).Answer: A2.Which of the following statements is correct? A. The crystalstructure of diamond is face-centered cubic (FCC). B. The density of silicon is smaller than that of germanium. C. The coordination number of germanium is 4. D. The Miller indices of a crystal plane parallel to the x-axis and passing through point (1, 2, 3) are (1, 0, 0).Answer: DShort Answer Questions1.What is the difference between a lattice and a unit cell?2.Define the concept of coordination number and give anexample of a coordination number 6 crystal structure.3.Define the concept of a crystal plane and expln how Millerindices are used to describe crystal planes.Answers:1.A lattice is an infinitely repeating arrangement of pointsin space that defines the basic symmetry of a crystal, while aunit cell is the smallest repeating unit of a crystal lattice that can be used to reconstruct the entire crystal by translation.2.Coordination number is the number of nearest neighbors of anatom in a crystal lattice. An example of a coordination number 6 crystal structure is the hexagonal close-packed (HCP) structure.3.A crystal plane is an imaginary flat surface in a crystalthat can be used to define the orientation of the crystal in space.Miller indices are a set of integers that describe the orientation of a crystal plane relative to the crystal axes. The Millerindices of a plane are determined by finding the reciprocals of the intercepts of the plane with the crystal axes and thenreducing these reciprocals to the smallest set of integers that give a unique designation of the plane.。
半导体物理与器件(尼曼第四版)答案之第一部分-半导体属性
1. 导电性：
半导体材料是指在电声信号强度及温度变化范围内，具有显著能量带隙、静电屏蔽能力和较强导电性的半导体物质。
其导电性取决于半导体物质的原子结构和物理性质。
值得注意的是，半导体材料具有非常高的电阻率，其电阻率取决于半导体材料中存在的空穴和电子的数量及相应的电子移动速率。
在常温下，半导体物质的电阻率可以达到106到1012欧姆之间的数字，而在低温和高温下，电阻率几乎可以忽略不计。
2. 光电效应：
半导体物质具有光电效应，即半导体物质可以在受到光照时发生微小变化。
由于半导体物质具有光电效应，因此，当光照在半导体物质上时，可以产生电压，从而使半导体物质的电阻率发生变化，产生静电效应。
这种光电效应可以被用于光电器件的研制中，例如太阳能电池，光敏电阻等等，具有十分广阔的应用范围。
3. 热敏性：
半导体物质具有高的热敏性，当温度发生变化时，半导体物质的性质也会发生变化。
当温度提高时，半导体物质开始呈现出热电效应，其电阻率会随着温度提高而减小，而当温度降低时，会出现负热效应，其电阻会随着温度降低而增加。
因此，半导体物质的热敏性可以被利用于研制热敏电阻、热敏电容等等的器件中。
第一章常用半导体器件自测题一、判断下列说法是否正确，用“√”和“×”表示判断结果填入空内。
（1）在N型半导体中如果掺入足够量的三价元素，可将其改型为P型半导体。
（）（2）因为N型半导体的多子是自由电子，所以它带负电。
（）（3）PN结在无光照、无外加电压时，结电流为零。
（）（4）处于放大状态的晶体管，集电极电流是多子漂移运动形成的。
（）（5）结型场效应管外加的栅-源电压应使栅-源间的耗尽层承受反向电压，才能保证其R G S大的特点。
（）（6）若耗尽型N沟道MOS管的U G S大于零，则其输入电阻会明显变小。
（）解：（1）√（2）×（3）√（4）×（5）√（6）×二、选择正确答案填入空内。
（1）PN结加正向电压时，空间电荷区将。
A. 变窄B. 基本不变C. 变宽（2）稳压管的稳压区是其工作在。
A. 正向导通B.反向截止C.反向击穿（3）当晶体管工作在放大区时，发射结电压和集电结电压应为。
A. 前者反偏、后者也反偏B. 前者正偏、后者反偏C. 前者正偏、后者也正偏（4）U G S＝0V时，能够工作在恒流区的场效应管有。
A. 结型管B. 增强型MOS管C. 耗尽型MOS管解：（1）A （2）C （3）B （4）A C三、写出图T1.3所示各电路的输出电压值，设二极管导通电压U D＝0.7V。
图T1.3解：U O1≈1.3V，U O2＝0，U O3≈－1.3V，U O4≈2V，U O5≈1.3V，U O6≈－2V。
四、已知稳压管的稳压值U Z＝6V，稳定电流的最小值I Z m i n＝5mA。
求图T1.4所示电路中U O1和U O2各为多少伏。
图T1.4解：U O1＝6V，U O2＝5V。
五、电路如图T1.6所示，V C C＝15V，β＝100，U B E＝0.7V。
试问：（1）R b ＝50k Ω时，u O ＝？ （2）若T 临界饱和，则R b ≈？ 解：（1）R b ＝50k Ω时，基极电流、集电极电流和管压降分别为26b BE BB B =-=R U V I μ AV2mA 6.2 C C CC CE B C =-===R I V U I I β所以输出电压U O ＝U C E ＝2V 。
半导体物理与器件答案半导体物理与器件答案篇一：半导体物理习题及答案复习思索题与自测题第一章1. 原子中的电子和晶体中电子受势场作用状况以及运动状况有何不同, 原子中内层电子和外层电子参加共有化运动有何不同。
答：原子中的电子是在原子核与电子库伦互相作用势的束缚作用下以电子XX的形式存在，没有一个固定的轨道；而晶体中的电子是在整个晶体内运动的共有化电子，在晶体周期性势场中运动。
当原子相互靠近结成固体时，各个原子的内层电子仍旧组成围绕各原子核的封闭壳层,和孤立原子一样;然而，外层价电子则参加原子间的互相作用，应当把它们看成是属于整个固体的一种新的运动状态。
组成晶体原子的外层电子共有化运动较强，其行为与自由电子相像，称为准自由电子，而内层电子共有化运动较弱，其行为与孤立原子的电子相像。
2.描述半导体中电子运动为什么要引入有效质量的概念, 用电子的惯性质量描述能带中电子运动有何局限性。
答：引进有效质量的意义在于它概括了半导体内部势场的作用，使得在解决半导体中电子在外力作用下的运动规律时，可以不涉及半导体内部势场的作用。
惯性质量描述的是真空中的自由电子质量，而不能描述能带中不自由电子的运动，通常在晶体周期性势场作用下的电子惯性运动，成为有效质量3.一般来说, 对应于高能级的能带较宽,而禁带较窄,是否如此，为什么？答：不是，能级的宽窄取决于能带的疏密程度，能级越高能带越密，也就是越窄；而禁带的宽窄取决于掺杂的浓度，掺杂浓度高，禁带就会变窄，掺杂浓度低，禁带就比较宽。
4.有效质量对能带的宽度有什么影响，有人说:有效质量愈大,能量密度也愈大,因此能带愈窄.是否如此，为什么？答：有效质量与能量函数对于K的二次微商成反比，对宽窄不同的各个能带，1〔k〕随k的改变状况不同，能带越窄，二次微商越小，有效质量越大，内层电子的能带窄，有效质量大；外层电子的能带宽，有效质量小。
5.简述有效质量与能带结构的关系；答：能带越窄，有效质量越大，能带越宽，有效质量越小。
______________________________________________________________________________________Chapter 55.1 (a)()()()1519101300106.111-⨯==d n Ne μρ Ω=808.4-cm(b) 208.08077.411===ρσ(Ω-cm)1-_______________________________________ 5.2a p N e μσ=or ()()380106.180.119-⨯==p a e N μσ 161096.2⨯=cm 3-_______________________________________ 5.3 (a) d n N e μσ= ()d n N μ19106.110-⨯=From Figure 5.3, for 16106⨯=d N cm 3- we find 1050≅n μcm 2/V-s which gives ()()()16191061050106.1⨯⨯=-σ 08.10=(Ω-cm)1-(b) ap N e μρ1=()ap N μ19106.1120.0-⨯=From Figure 5.3, for 1710=a N cm 3- we find 320≅p μcm 2/V-s which gives()()()195.010320106.111719=⨯=-ρΩ-cm_______________________________________ 5.4(a) ap N e μρ1=()ap N μ19106.1135.0-⨯=From Figure 5.3, for 16108⨯=a N cm 3- we find 220≅p μcm 2/V-s which gives()()()1619108220106.11⨯⨯=-ρ355.0=Ω-cm (b) d n N e μσ=()d n N μ19106.1120-⨯= From Figure 5.3, for17102⨯=d N cm 3-, then 3800≅n μcm 2/V-s which gives()()()17191023800106.1⨯⨯=-σ6.121=(Ω-cm)1-_______________________________________ 5.5()A N e LA L A L R d n μσρ=== or ()RAeN Ld n =μ()()()()1.070102106.15.21519⨯⨯=- 1116=cm 2/V-s_______________________________________ 5.6(a) 1610==d o N n cm 3- and()4162621024.310108.1-⨯=⨯==o i o n n p cm 3- (b)E =o n n e J μFor GaAs doped at 1610=d N cm 3-, 7500≅n μcm 2/V-s Then()()()()10107500106.11619-⨯=J or120=J A/cm 2 (b) (i) 1610==a o N p cm 3-______________________________________________________________________________________421024.3-⨯==o i o p n n cm 3- (ii) For GaAs doped at 1610=a N cm 3-, 310≅p μcm 2/V-sE =o p p e J μ()()()()1010310106.11619-⨯= or96.4=J A/cm 2_______________________________________ 5.7 (a) ()R IR V 1.010=⇒= orΩ=100R (b)RALA L R =⇒=σσ or()()01.0101001033==--σ(Ω-cm)1- (c) d n N e μσ≅ or()()d N 1350106.101.019-⨯= Then131063.4⨯=d N cm 3-(d)o p p e μσ≅or()()o p 480106.101.019-⨯= Then141030.1⨯=o p cm 3-d a N N -= So1515141013.1101030.1⨯=+⨯=a N cm 3- Note: For the doping concentrations obtained, the assumed mobility values are valid._______________________________________ 5.8(a) ()AN e LA L R a p μσ== For 16102⨯=a N cm 3-, then 400≅p μcm 2/V-s()()()()()41619105.8102400106.1075.0--⨯⨯⨯=R Ω=93.68 0290.093.682===R V I A or 0.29=I mA (b) ()()Ω==⇒∝79.206393.68R L R00967.079.2062===R V I A or 67.9=I mA (c) d o ep J υ=For (a), 12.34105.8100.2943=⨯⨯=--J A/cm 2Then ()()1619102106.112.34⨯⨯==-o d ep J υ 410066.1⨯=cm/s For (b), 38.11105.81067.943=⨯⨯=--J A/cm 2 ()()1619102106.138.11⨯⨯=-d υ31055.3⨯=cm/s_______________________________________ 5.9 (a) For 15102⨯=d N cm 3-, then8000≅n μcm 2/V-sΩ=⨯==-200102553I V R ()AN e LR d n μ=or ()RA N e L d n μ=()()()()()515191052001028000106.1--⨯⨯⨯= 0256.0=cm(b)d o en AIJ υ==______________________________________________________________________________________or ()o d en A I =υ ()()()151953102106.11051025⨯⨯⨯⨯=--- 61056.1⨯=cm/s(c) ()A en I d o υ=()()()()561519105105102106.1--⨯⨯⨯⨯= 080.0= A or 80=I mA _______________________________________5.10 (a)313===E L V V/cm ⇒E =n d μυ3104=E =d n υμ or 3333=n μcm 2/V-s (b) ()()3800=E =n d μυor3104.2⨯=d υcm/s_______________________________________5.11 (a) Silicon: For 1=E kV/cm, 6102.1⨯=d υcm/s Then11641033.8102.110--⨯=⨯==d t d t υs For GaAs: 6105.7⨯=d υcm/s Then 11641033.1105.710--⨯=⨯==d t d t υs (b) Silicon: For 50=E kV/cm, 6105.9⨯=d υcm/sThen11641005.1105.910--⨯=⨯=t t s For GaAs: 6107⨯=d υcm/sThen 11641043.110710--⨯=⨯=t t s _______________________________________ 5.12()ip n o p o n n e p e n e μμμμρ+=+=11 (a) 1410==d a N N cm 3- 1350≅⇒n μcm 2/V-s 480≅p μcm 2/V-s ()()()1019105.14801350106.11⨯+⨯=-ρ Ω⨯=51028.2-cm(b) 1610==d a N N cm 3- 1250≅⇒n μcm 2/V-s410≅p μcm 2/V-s()()()1019105.14101250106.11⨯+⨯=-ρ Ω⨯=51051.2-cm (c) 1810==d a N N cm 3- 290≅⇒n μcm 2/V-s130≅p μcm 2/V-s()()()1019105.1130290106.11⨯+⨯=-ρ Ω⨯=51092.9-cm_____________________________________________________________________________________________________________________________ 5.13 (a) GaAs: ()o p o p p p e μμσ19106.15-⨯=⇒≅ From Figure 5.3, and using trial and error,wefind 17103.1⨯≅o p cm 3- and240≅p μcm 2/V-s Then ()5172621049.2103.1108.1-⨯=⨯⨯==o io p n n cm 3- (b) Silicon:o n n e μρσ≅=1or()()()1350106.181119-⨯==n o e n μρ which gives141079.5⨯=o n cm 3- and()51421021089.31079.5105.1⨯=⨯⨯==o i o n n p cm 3- Note: For the doping concentrations obtainedin part (b), the assumed mobility values arevalid._______________________________________ 5.14()p n i i en μμσ+=Then()()i n 6001000106.110196+⨯=-- ori n (300 K)91091.3⨯=cm 3- Now⎪⎪⎭⎫⎝⎛-=kT E N N n g c i exp 2υ or ⎪⎪⎭⎫⎝⎛=2ln i c g n N N kT E υ()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯=292191091.310ln 0259.0which gives122.1=gE eVNow2i n (500K)()()()⎥⎦⎤⎢⎣⎡-=3005000259.0122.1exp 10219261015.5⨯= ori n (500 K)131027.2⨯=cm 3- Then()()()60010001027.2106.11319+⨯⨯=-i σ which givesi σ(500 K)31081.5-⨯= (Ω-cm)1- _______________________________________ 5.15 (a) (i) Silicon: ()p n i i en μμσ+=()()()4801350105.1106.11019+⨯⨯=-i σ or61039.4-⨯=i σ(Ω-cm)1- (ii) Ge:()()()19003900104.2106.11319+⨯⨯=-i σ or21023.2-⨯=i σ(Ω-cm)1- (iii) GaAs:()()()4008500108.1106.1619+⨯⨯=-i σ or91056.2-⨯=i σ(Ω-cm)1-(b) ALR σ=(i) Si:______________________________________________________________________________________ ()()Ω⨯=⨯⨯⨯=---98641036.510851039.410200R (ii) Ge:()()Ω⨯=⨯⨯⨯=---68241006.110851023.210200R (iii) GaAs:()()Ω⨯=⨯⨯⨯=---128941019.910851056.210200R_______________________________________5.16 (a) d n N e μσ=()d n N μ19106.125.0-⨯=From Figure 5.3, for 15102.1⨯=d N cm 3-, then1300≅n μcm 2/V-sSo ()()()1519102.11300106.1⨯⨯=-σ 2496.0=(Ω-cm)1- (b) Using Figure 5.2,(i) For 250=T K (︒-23C),1800≅⇒n μcm 2/V-s()()()1519102.11800106.1⨯⨯=-σ 346.0=(Ω-cm)1-(ii) For 400=T K (︒127C), 670≅⇒n μcm 2/V-s()()()1519102.1670106.1⨯⨯=-σ129.0=(Ω-cm)1- _______________________________________ 5.17()dx x t t avg⎰=01σσdx d x t to ⎪⎭⎫ ⎝⎛-=⎰exp 10σ ()tod x d t 0exp ⎪⎭⎫ ⎝⎛--=σ ⎥⎦⎤⎢⎣⎡-⎪⎭⎫ ⎝⎛--=1exp d t td o σ ()()()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛--=3.05.1exp 15.13.02097.3=(Ω-cm)1-_______________________________________5.18 (a) 3.1331015024=⨯==E -L V V/cm(b) ()()x N e x d n μσ= ()dx T x T e Tn avg ⎪⎭⎫ ⎝⎛-⨯⋅=⎰111.111021016μσ()()T n T x x T e 0216111.12102⎥⎦⎤⎢⎣⎡-⨯=μ()()⎥⎦⎤⎢⎣⎡-⨯=T T T T e n 111.12102216μ ()()55.010216⨯=n e μ()()()()55.010*******.11619⨯⨯=- 32.1=avg σ(Ω-cm)1- (c) ()()()21015010105.732.1444⋅⨯⨯=⋅=---V L A I avg σ 51032.1-⨯= A or μ2.13=I A (d) Top surface;()()()1619102750106.1⨯⨯=-σ4.2=(Ω-cm)1-()()3203.1334.2==E =σJ A/cm 2 Bottom surface:()()()1519102750106.1⨯⨯=-σ24.0=(Ω-cm)1-()()323.13324.0==E =σJ A/cm 2_______________________________________ 5.19Plot_____________________________________________________________________________________________________________________________ 5.20 (a) 10=E V/cm so ()()41035.1101350⨯==E =n d μυcm/sor21035.1⨯=d υ m/sThen2*21d n m T υ=()()()22311035.11011.908.121⨯⨯=- or271097.8-⨯=T J 81060.5-⨯⇒ eV (b) 1=E kV/cm ()()61035.110001350⨯==d υcm/s or41035.1⨯=d υ m/sThen ()()()24311035.11011.908.121⨯⨯=-Tor 231097.8-⨯=T J 41060.5-⨯⇒ eV_______________________________________5.21(a) ⎪⎪⎭⎫ ⎝⎛-=kT E N N n gc i exp 2υ ()()⎪⎭⎫ ⎝⎛-⨯⨯=0259.010.1exp 1011021919 191018.7⨯= or 91047.8⨯=i n cm 3- For1410=d N cm 3->>1410=⇒o i n n cm 3-ThenE =σJ E =o n n e μ ()()()()100101000106.11419-⨯=or60.1=J A/cm 2(b) A 5% increase is due to a 5% increase inelectron concentration, so 2214221005.1i d d o n N N n +⎪⎪⎭⎫ ⎝⎛+=⨯= which becomes ()()2213213141051051005.1i n +⨯=⨯-⨯ and yields 2621025.5⨯=i n ()()⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛⨯⨯=kT E T g exp 30010110231919 or()()⎥⎦⎤⎢⎣⎡-⎪⎭⎫ ⎝⎛=⨯-3000259.010.1exp 30010625.2312T T By trial and error, we find456=T K _______________________________________5.22 (a) o p o n p e n e μμσ+= and o i o p n n 2=Theno p oi n p e p n e μμσ+=2 To find the minimum conductivity, set()p o i n o e p n e dp d μμσ+-==2210 which yields 2/1⎪⎪⎭⎫⎝⎛=p n i o n p μμ (Answer to part (b)) Substituting into the conductivity expression()[]2/12min p n i i n n n e μμμσσ== ()[]2/1p n i p n e μμμ+ which simplifies to pn i en μμσ2min =The intrinsic conductivity is defined as______________________________________________________________________________________()pn ii p n i i en en μμσμμσ+=⇒+=The minimum conductivity can then bewritten as pn pn i μμμμσσ+=2min_______________________________________5.23 (a) n-type: 16105⨯==d o N n cm 3- ()162102105105.1⨯⨯==o i o n n p 3105.4⨯=cm 3- p-type: 16102⨯==a o N p cm 3- ()41621010125.1102105.1⨯=⨯⨯=o n cm 3- compensated: a d o N N n -=1616102105⨯-⨯=16103⨯=cm 3-()316210105.7103105.1⨯=⨯⨯=o p cm 3- (b) From Figure 5.3, n-type: 1100≅n μcm 2/V-s p-type: 400≅p μcm 2/V-scompensated: 1000≅n μcm 2/V-s (c) n-type: o n n e μσ=()()()16191051100106.1⨯⨯=- 8.8=(Ω-cm)1-p-type: o p p e μσ=()()()1619102400106.1⨯⨯=-28.1=(Ω-cm)1-compensated: o n n e μσ=()()()16191031000106.1⨯⨯=- 8.4=(Ω-cm)1-(d) σσJJ =E ⇒E =n-type: 6.138.8120==E V/cm p-type: 75.9328.1120==E V/cmcompensated: 258.4120==E V/cm_______________________________________5.24 3211111μμμμ++= 50011500120001++= 0020.0000667.000050.0++= or 003167.01=μThen 316=μcm 2/V-s _______________________________________5.25()()2/32/330013003001300+-⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛=T T n μ (a) At 200=T K, ()238820030013002/3=⎪⎭⎫ ⎝⎛=n μcm 2/V-s (b) At 400=T K, 844=n μcm 2/V-s_______________________________________5.26 006.05001250111121=+=+=μμμ Then 167=μcm 2/V-s_______________________________________5.27Plot _____________________________________________________________________________________________________________________________ 5.28 Plot _______________________________________5.29 ()⎪⎪⎭⎫ ⎝⎛--⨯==001.0010514n eD dx dneD J n n n ()()()⎪⎪⎭⎫ ⎝⎛-⨯⨯=-010.0010525106.119.01419n Then ()()()()()010525106.1010.019.01419n -⨯=⨯-which yields()141025.00⨯=n cm 3-_______________________________________5.30xn eD dx dn eD J nn n ∆∆== ()()⎥⎦⎤⎢⎣⎡-⨯-⨯⨯=-012.0010510227106.1151619n J4.5-=n J A/cm 2 _______________________________________5.31(a) xn eD dx dn eD J nn n ∆∆== ()()()⎥⎦⎤⎢⎣⎡⨯--⨯=---411519102001030106.12x n()11833108.4108.4104x n ---⨯-⨯=⨯ which yields ()1411067.1⨯=x n cm 3-(b) ()()()⎥⎦⎤⎢⎣⎡⨯--⨯=---4115191020010230106.12x n()117231068.31068.3104x n ---⨯-⨯=⨯ ()1411091.8⨯=x n cm 3-_______________________________________ 5.32⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛+-=-=216110L x dx d eD dx dp eD J p p p⎪⎭⎫ ⎝⎛+⋅⋅-=L x L eD p 121016(a) For 0=x , ()()()()41619101221010106.1--⨯⨯-=p J 7.26-=A/cm 2 (b) For μ6-=x m,()()()()416191012126121010106.1--⨯⎪⎭⎫ ⎝⎛-⨯-=p J 3.13-=A/cm 2(c) For μ12-=x m, 0=p J_______________________________________ 5.33For electrons: []n L x n n n e dxdeD dx dn eD J /1510-== ()n L x n L e eD n/1510--= At 0=x , ()()()21021025106.131519-=⨯⨯-=--n J A/cm 2 For holes:[]pL x p p p e dxd eD dx dp eD J /15105+⨯-=-= ()p L x p L eeD p /15105+⨯-=For 0=x ,()()()4151910*********.1--⨯⨯⨯-=p J 16-=A/cm 2 ()()00=+==x J x J J p n Total()18162-=-+-=A/cm 2 _____________________________________________________________________________________________________________________________ 5.34[]pL x p p p edx d eD dx dp eD J /15105-⨯-=-= ()pL x p L eeD p /15105-⨯= (a) (i) ()()()41519105010510106.1--⨯⨯⨯=p J6.1=A/cm 2(ii) ()()()41519105.2210548106.1--⨯⨯⨯=p J 07.17=A/cm 2(b) (i) ()()()411519105010510106.1---⨯⨯⨯=e J p 589.0=A/cm 2(ii) ()()()411519105.2210548106.1---⨯⨯⨯=e J p28.6=A/cm 2_______________________________________ 5.35dxdneD n e J n n n +E =μ or()()E ⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛-⨯=--18exp 10960106.1401619x()()()16191025106.1-⨯+⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛⨯-⨯-18exp 101814x Then()⎪⎭⎫ ⎝⎛--E ⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛-=-18exp 22.2218exp 536.140x xWe find ()()⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛-=E 18exp 536.14018exp 22.22x x or()⎪⎭⎫⎝⎛+-=E 18exp 0.265.14x _______________________________________5.36(a) []L x n n n e dx d eD dx dn eD J /15102-⨯==()L e eD L x n /15102-⨯-= ()()()4/1519101510227106.1---⨯⨯⨯-=L x e L x e /76.5--= (b) ()L x n Total p e J J J /76.510----=-= []1076.5/-=-L x e A/cm 2 (c) We have ()E =E =o p p p e J μσ ()()()E ⨯=---1619/10420106.11076.5L x e So []88.1457.8/-=E -L x e V/cm _______________________________________5.37(a) ()()dxx dn eD x n e J nn +E =μ We have 8000=n μcm 2/V-s, so that()()20780000259.0==n D cm 2/s Then()()()()x n 128000106.110019-⨯=()()()dx x dn 207106.119-⨯+ which yields()()()()dxx dn x n 171410312.310536.1100--⨯+⨯= Solution is of the form ()⎪⎭⎫ ⎝⎛-+=d x B A x n exp so that______________________________________________________________________________________()⎪⎭⎫⎝⎛--=d x d B dx x dn exp Substituting into the differential equation,we have ()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛-+⨯=-d x B A exp 10536.110014()⎪⎭⎫ ⎝⎛-⨯--d x B d exp 10312.317This equation is valid for all x , so ()A 1410536.1100-⨯= or151051.6⨯=AAlso⎪⎭⎫⎝⎛-⨯-d x B exp 10536.114()0exp 10312.317=⎪⎭⎫ ⎝⎛-⨯--d x B dwhich yields310156.2-⨯=d cm At 0=x , ()500=E n e n μ so that()()()()B A +⨯=-128000106.15019 which yields1510255.3⨯-=B Then()⎪⎭⎫⎝⎛-⨯-⨯=d x x n exp 10255.31051.61515cm 3-(b)At 0=x , ()151510255.31051.60⨯-⨯=n Or()151026.30⨯=n cm 3- At μ50=x m,()⎪⎭⎫⎝⎛-⨯-⨯=56.2150exp 10255.31051.6501515nor()151019.650⨯=n cm 3- (c)At μ50=x m,()E =50n e J n drf μ ()()()()121019.68000106.11519⨯⨯=- or ()08.9550==x J drf A/cm 2Then ()08.9510050-==x J diffor()92.450==x J diff A/cm 2_______________________________________ 5.38⎪⎪⎭⎫⎝⎛-=kT E E n n Fi F i exp(a)b ax E E Fi F +=-, 4.0=b()4.01015.03+=-awhich yields2105.2⨯-=aThenx E E Fi F 2105.24.0⨯-=- so⎪⎪⎭⎫⎝⎛⨯-=kT x n n i 2105.24.0exp (b) dxdneD J nn =⎪⎪⎭⎫⎝⎛⨯-⎪⎪⎭⎫ ⎝⎛⨯-=kT x kT n eD i n 22105.24.0exp 105.2 Assume 300=T K, so 0259.0=kT eVand10105.1⨯=i n cm 3- Then()()()()()0259.0105.2105.125106.121019⨯⨯⨯-=-n J⎪⎪⎭⎫⎝⎛⨯-⨯0259.0105.24.0exp 2x or______________________________________________________________________________________ ⎪⎪⎭⎫ ⎝⎛⨯-⨯-=-0259.0105.24.0exp 1079.524x J n (i) At 0=x , 31095.2⨯-=n J A/cm 2 (ii) At μ5=x m, 7.23-=n J A/cm 2_______________________________________ 5.39(a) dxdneD n e J nn n +E =μ()()()E ⎪⎭⎫⎝⎛-⨯=--L x 1101000106.1801619()()⎪⎪⎭⎫ ⎝⎛-⨯+-L 1619109.25106.1where 34101010--=⨯=L cm We find()()44.41106.16.1803-E ⎪⎭⎫⎝⎛-E =--xor()44.4116.180+E ⎪⎭⎫⎝⎛-=L xSolving for the electric field, we find⎪⎭⎫ ⎝⎛-=E 11.24L x V/cm(b) For 20-=n J A/cm 2()44.4116.120+E ⎪⎭⎫⎝⎛-=L xThen⎪⎭⎫ ⎝⎛-=E L x 13.13 V/cm_______________________________________ 5.40(a) ()()dx x dN x N e kT d dX ⋅⋅⎪⎭⎫⎝⎛-=E 1 ()[]L x do L x do e N dx de N //0259.0--⋅-= ()L x do Lx do e N L e N //10259.0--⎪⎭⎫⎝⎛-⋅-=410100259.00259.0-⨯==Lor 9.25=E X V/cm(b)()()09.250--=E -=⎰L dx LX φ()()0259.010109.254-=⨯-=-V or 9.25-=φmV_______________________________________ 5.41From Example 5.6()()()()()()xx x 33191619101100259.010********.0-=-=E dx V x ⎰-E -=4100()()()⎰---=410033101100259.0x dx()()[]4100333101ln 101100259.0--⎪⎭⎫⎝⎛--=x()()()[]1ln 1.01ln 0259.0--=or73.2-=V mV_______________________________________ 5.42()()dx x dN x N e kT d d x ⋅⋅⎪⎭⎫⎝⎛-=E 1 For ()L x do d e N x N /-=So 5000259.0==E LX V/cm Which yields 51018.5-⨯=L cm_____________________________________________________________________________________________________________________________ 5.43(a) We have()dxx dN eD dx dneD J d nn diff == ()⎪⎭⎫ ⎝⎛-⋅-=L x N L eD do n expWe have()()0259.06000=⎪⎭⎫⎝⎛=e kT D n n μor4.155=n D cm 2/s Then ()()()()⎪⎭⎫ ⎝⎛-⨯⨯⨯-=--L x J diff exp 101.01054.155106.141619or⎪⎭⎫⎝⎛-⨯-=L x J diff exp 10243.15 A/cm 2(b) diff drf J J +=0 Now E =n e J n drf μ()()()E ⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛-⨯⨯=-L x exp 1056000106.11619 or()E ⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛-=L x J drf exp 48We have diff drf J J -= so()⎪⎭⎫ ⎝⎛-⨯=E ⎥⎦⎤⎢⎣⎡⎪⎭⎫⎝⎛-L x L x exp 10243.1exp 485 which yields 31059.2⨯=E V/cm _______________________________________5.44 Plot _______________________________________5.45(a) (i) ()()8.2911500259.0==n D cm 2/s (ii) ()()6.16062000259.0==n D cm 2/s (b) (i) 9.3080259.08==p μcm 2/V-s (ii) 13510259.035==p μcm 2/V-s _______________________________________5.46 110-=L cm, 210-=W cm,310-=d cm (a) ()()()()()519222310106.1102105102.1----⨯⨯⨯⨯-=-=ned B I V Z X H 310875.1-⨯-=Vor 875.1-=H V mV(b)1875.01010875.123-=⨯-==E --W V H H V/cm _______________________________________5.47 (a) ned B I V zx H -=()()()()()5192126105106.110510510250----⨯⨯⨯⨯⨯-= or3125.0-=H V mV (b) 23102103125.0--⨯⨯-==E W V H H or 21056.1-⨯-=E H V/cm(c)W denV LI x x n =μ()()()()()()()542119361051021.0105106.11010250-----⨯⨯⨯⨯⨯= or 3125.0=n μm 2/V-s 3125=cm 2/V-s_____________________________________________________________________________________________________________________________ 5.48 (a) ⇒<0H V n-type(b) ()()()()()35193102.510106.110.01050.0----⨯-⨯⨯-=-=H Z X edV B I n 211001.6⨯=m 3-or 151001.6⨯=n cm 3-(c) W denV L I X X n =μ()()()()()()()542119331010151001.6106.110105.0-----⨯⨯⨯= 03466.0=m 2/V-sor 6.346=n μcm 2/V-s_______________________________________5.49(a)()()23105105.16--⨯⨯-=E =W V H H or 825.0-=H V mV (b) =H V negative ⇒n-type (c) H z x edV B I n -=()()()()()35192310825.0105106.1105.6105.0-----⨯-⨯⨯⨯⨯-= or 2110924.4⨯=n m 3-1510924.4⨯=cm 3- (d) W d enV LI x x n =μ()()()()()()()5421192310510525.110924.4106.1105.0105.0-----⨯⨯⨯⨯⨯⨯= or1015.0=n μm 2/V-s 1015=cm 2/V-s _______________________________________5.50 (a) =H V negative ⇒n-type(b) H zx edV B I n -= ()()()()()321923105.41001.0106.1105.2105.2-----⨯-⨯⨯⨯⨯-= or201068.8⨯=n m 3-141068.8⨯=cm 3- (c) W d enV LI x x n =μ ()()()()()⎥⎦⎤⎢⎣⎡⨯⨯⨯⨯=---2.21068.8106.1105.0105.2201923 ()()⎥⎦⎤⎢⎣⎡⨯⨯⨯--221001.01005.01or8182.0=n μm 2/V-s 8182=cm 2/V-s(d) n e n μρσ==1 ()()()14191068.88182106.1⨯⨯=- or 88.0=ρ(Ω-cm)_______________________________________。
