半导体物理与器件第四版课后习题答案4
半导体物理与器件(尼曼第四版)答案
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4复习进程
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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 习题4答案
3 2
3 2
3
3
77K时有：
Nc77 K 77 =( ) Nc300 K 300
18
3 2
Nv77 K 77 =( ) Nv300 K 300
−3
3 2
Nc = 1.365 ×10 cm
Nv = 7.41×10 cm
17
−3
根据公式：
n = n0 p0 = NcNve
2 i
Eg − k0T
分别解得300K时：
600K时，ni与ND-NA相比不能忽略 ： ni ≈ 2 ×1017 cm −3
1 n0 = N D − N A + ( N D − N A ) 2 + 4ni2 2
p0 = 1 −( N D − N A ) + ( N D − N A ) 2 + 4ni2 2
{
}
500K时，本征载流子数据可查表。但是也可 以计算通过公式计算：
ni = 2 × 1016cm −3
本征激发和Nd相比不能忽略
n0 = 1 N D − N A + ( N D − N A ) 2 + 4ni2 2
{
}
n 0 = 2.2 × 1016cm −3
p 0 = 1.7 × 10 cm
1 2V (m ) dZ ( E ) = = ( E − Ec) 2 dE π h
3 * 2 n 2 3
m = mdn = s (ml m )
* n
2 3
1 2 3 t
第三题
根据Nc和Nv的定义：
Nc = (m k T ) 2π h
3 2 * n 0 3 2
Nv =
(m* k0T ) p 2π h
半导体物理学第四版答案
半导体物理学第四版答案【篇一：半导体物理学第四章答案】. 300k时，ge的本征电阻率为47?cm，如电子和空穴迁移率分别为3900cm2/( v.s)和1900cm/( v.s)。
试求ge 的载流子浓度。
解：在本征情况下，n?p?ni,由??1/??211知 ?nqun?pqupniq(un?up)ni?1113?3??2.29?10cm?19?q(un?up)47?1.602?10?(3900?1900)2. 试计算本征si在室温时的电导率，设电子和空穴迁移率分别为1350cm2/( v.s)和500cm2/( v.s)。
当掺入百万分之一的as后，设杂质全部电离，试计算其电导率。
比本征si的电导率增大了多少倍？解：300k时，un?1350cm2/(v?s),up?500cm2/(v?s)，查表3-2或图3-7可知，室温下si的本征载流子浓度约为ni?1.0?1010cm?3。
本征情况下，??nqun?pqup?niq(un?up)?1?1010?1.602?10-19?(1350+500)?3.0?10?6s/cm11金钢石结构一个原胞内的等效原子个数为8??6??4?8个，查看附录b知si的晶格常数为820.543102nm，则其原子密度为822?3。
?5?10cm?73(0.543102?10)1?5?1016cm?3，杂质全部电离后，nd??ni，1000000掺入百万分之一的as,杂质的浓度为nd?5?1022?这种情况下，查图4-14（a）可知其多子的迁移率为800 cm2/( v.s) ??ndqun?5?1016?1.602?10-19?800?6.4s/cm?6.4??2.1?106倍比本征情况下增大了?6?3?103. 电阻率为10?.m的p型si样品，试计算室温时多数载流子和少数载流子浓度。
解：查表4-15(b)可知，室温下，10?.m的p型si样品的掺杂浓度na约为1.5?1015cm?3,查表3-2或图3-7可知，室温下si的本征载流子浓度约为ni?1.0?1010cm?3,na??nip?na?1.5?1015cm?3ni(1.0?1010)24?3n???6.7?10cm15p1.5?104. 0.1kg的ge单晶，掺有3.2?10-9kg的sb，设杂质全部电离，试求该材料的电阻率??n=0.38m2/( v.s),ge的单晶密度为5.32g/cm3,sb原子量为121.8?。
半导体物理与器件英文版第四版课后练习题含答案
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.。
半导体物理习题参考答案第四章
第4章 半导体的导电性2.试计算本征Si 在室温时的电导率，设电子和空穴迁移率分别为1350cm 2/V ⋅s 和500 cm 2/V ⋅s 。
当掺入百万分之一的As 后，设杂质全部电离，试计算其电导率。
掺杂后的电导率比本征Si 的电导率增大了多少倍？解：将室温下Si 的本征载流子密度1.5⨯1010/cm 3及题设电子和空穴的迁移率代入电导率公式()i i n p n q σμμ=+g 算得500克Si 单晶的体积为3214.6 cm 2.33V ==，于是知B 的浓度 ∴1816-32.510 1.1610 cm 214.6A Z N V ⨯===⨯ 室温下硅中此等浓度的B 杂质应已完全电离，查表4-14知相应的空穴迁移率为400 cm 2/V ⋅s 。
故161911 1.35cm 1.1610 1.610400A p N q ρμ-===Ω⋅⨯⨯⨯⨯ 6. 设Si 中电子的迁移率为0.1 m 2/(V .s)，电导有效质量m C =0.26m 0，加以强度为104V/m 的电场，试求平均自由时间和平均自由程。
解：由迁移率的定义式*n c cq m τμ=知平均自由时间 *c c n m qμτ⋅= 代入相关数据，得3113190.269.1100.1 1.48101.610n s τ---⨯⨯⨯==⨯⨯8. 0.1A 的。
为5.3⨯10 cm 的施主。
10. 试求本征Si 在473K 时的电阻率。
解：由图4-13查出T=473K 时本征硅中电子和空穴的迁移率分别是2440 cm /V s n μ=⋅，2140 cm /V s p μ=⋅在温度变化不大时可忽略禁带宽度随温度的变化，则任意温度下的本征载流子密度可用室温下的等效态密度N C (300)和N V (300)、禁带宽度E g (300)和室温kT=0.026eV 表示为3/23(300)300()(300)(300)(exp() cm 3000.026g i C V E T n T N N T⋅=－代入相关数据，得193/2133473 1.12300(473)10()exp() =4.110 cm 30020.026473i n ⨯=-⨯⨯⨯－ 该值与图3-7中T=200℃（473K ）所对应之值低大约一个数量级，这里有忽略禁带变窄的因素，也有其他因素（参见表3-2，计算值普遍比实测值低）。
半导体物理与器件第四版课后习题4
______________________________________________________________________________________Chapter 44.1⎪⎪⎭⎫ ⎝⎛-=kTE N N n gc iexp 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 eV______________________________________________________________________________________ Now ()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 exp To find the maximum value:()⎪⎭⎫⎝⎛-∝-kT x x dx f g d F c exp 212/10exp 12/1=⎪⎭⎫⎝⎛-⋅-kT x x kT which yields2212/12/1kT x kT x x =⇒= The maximum value occurs at 2kTE 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 1υ To find the maximum value ()[]0exp 1=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛-∝-kT x x dx d dx f g d F υ Same as part (a). Maximum occurs at 2kT x =or______________________________________________________________________________________2kTE E -=υ _______________________________________4.7()()()()⎥⎦⎤⎢⎣⎡---⎥⎦⎤⎢⎣⎡---=kT E E E E kT E E E E E n E n c c c c 221121exp exp where kT E E c 41+= and 22kT E E c += Then()()()⎥⎦⎤⎢⎣⎡--=kT E E kT kT E n E n 2121exp 24()5.3exp 22214exp 22-=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛--=or()()0854.021=E n E n_______________________________________ 4.8Plot_______________________________________ 4.9Plot_______________________________________ 4.10⎪⎪⎭⎫ ⎝⎛=-**ln 43n pmidgap Fi m m kT E ESilicon: o p m m 56.0*=, o n m m 08.1*=0128.0-=-midgap Fi E E eV Germanium: o p m 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 0259.043 63.10-⇒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 c E F ⎰∞⎥⎦⎤⎢⎣⎡--≅exp LetkT 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 _______________________________________4.14 Let ()()c c E E C E g -=1 for cE 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 00+=---=-∞∞⎰ηηηηηdSo ()()⎥⎦⎤⎢⎣⎡--=kT E E kT C n F c o exp 21 _______________________________________ 4.15 We 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 1004.11931090.6⨯=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 n 31041.2⨯=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 E 2154.0=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 1004.119 141010.2⨯=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 108.21951030.2⨯=cm 3-_______________________________________ 4.23(a) ⎥⎦⎤⎢⎣⎡-=kT E E n n Fi F i 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 F i o exp ()⎥⎦⎤⎢⎣⎡⨯=0259.022.0exp 108.1691080.8⨯=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 0259.0 1979.0=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) ⎪⎪⎭⎫ ⎝⎛=-o F p N 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 n 910134.1⨯=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 n 21013.1-⨯=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 034533.0______________________________________________________________________________________ 3482.0=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 p 141068.6⨯=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==-=kT kT kT E E c F Fη Then ()0.12/1≅F F η()()0.1108.2219⨯=πo n 191016.3⨯=cm 3-(b) ()F c oF N n ηπ2/12= ()()0.1107.4217⨯=π171030.5⨯=cm 3-_______________________________________ 4.29()Fo 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 n 181018.3⨯=cm 3-______________________________________________________________________________________ _______________________________________ 4.31 For the electron concentration ()()()E f E g E n F c =The Boltzmann approximation applies, so()()c n E E hm E n -=32/3*24π()⎥⎦⎤⎢⎣⎡--⨯kT E E F exp or()()()⎥⎦⎤⎢⎣⎡--=kT E E h m E n F c n exp 2432/3*π()⎥⎦⎤⎢⎣⎡---⨯kT E E kT E E kT c c exp Define kTE E x c -= Then()()()x x K x n E n -=→expTo 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 02/1 which yields kT 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 m E p F pυπexp 2432/3* ()⎥⎦⎤⎢⎣⎡---⨯kT E E kT E E kT υυexp Define kT E E x -='υ Then ()()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-⨯=or 171045.2⨯=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 -+-=- For kT E E a F 3=- and 045.0=-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 cm3-(b)16103⨯==d o N n cm 3-()316210105.7103105.1⨯=⨯⨯=o p cm3-(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 cm3-()8152111034.110410334.7⨯=⨯⨯=o n 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⨯=⨯⨯=op 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-⨯=⨯⨯=o p 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⎪⎭⎫ ⎝⎛⨯⨯=i n ()()()()⎥⎦⎤⎢⎣⎡-⨯4500259.030042.1exp 1010853.3⨯=⇒i n cm 3- 1410==d o N n cm 3- ()7142101048.11010853.3⨯=⨯=o p cm 3- _______________________________________4.36 (a) Ge: 13104.2⨯=i n cm 3- (i)2222i d d o n N N n +⎪⎪⎭⎫ ⎝⎛+=()21321515104.221022102⨯+⎪⎪⎭⎫ ⎝⎛⨯+⨯=or15102⨯=≅d o N n cm 3- ()152132102104.2⨯⨯==o i o n n p 111088.2⨯= cm 3-(ii)151610710⨯-=-≅d a o N N p15103⨯=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-⨯=⨯⨯=o p cm 3- (ii)15103⨯=-≅d a o N N p cm 3-()315261008.1103108.1-⨯=⨯⨯=o n 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 2111or 41085.8-⨯=d d N n(b) We have ()⎪⎪⎭⎫⎝⎛-+=kT E E E f F F exp 11Now ()()F c c F E E E E E E -+-=-or 245.0+=-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 por 13105.1⨯=o p cm 3- Then()7132102105.1105.1105.1⨯=⨯⨯==o i o p n n cm 3- Germanium:______________________________________________________________________________________2222i d a d a o n N N N N p +⎪⎪⎭⎫ ⎝⎛-+-= ()21321313104.22105.12105.1⨯+⎪⎪⎭⎫ ⎝⎛⨯+⎪⎪⎭⎫ ⎝⎛⨯= or 131026.3⨯=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 n14108⨯=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 a a o n N N p +⎪⎪⎭⎫⎝⎛+= 212210752.2⎪⎪⎭⎫ ⎝⎛-⨯a N 242108936.12⨯+⎪⎪⎭⎫ ⎝⎛=a N()21224210752.2105735.7⎪⎪⎭⎫⎝⎛+⨯-⨯a a N N242108936.12⨯+⎪⎪⎭⎫ ⎝⎛=aN so that 1210064.2⨯=a N cm 3-_______________________________________ 4.42Plot_______________________________________ 4.43Plot_______________________________________ 4.44Plot_____________________________________________________________________________________________________________________________ 4.45 2222i a d 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-type Majority carriers are holes 1616105.1103⨯-⨯=-=d a o N N p 16105.1⨯=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⨯=⨯⨯=on cm 3-_______________________________________ 4.47(a) ⇒<<i o n p n-type (b) oi o o i o p n n n n p 22=⇒=o n ()16421010125.1102105.1⨯=⨯⨯=cm 3- ⇒electrons are majority carriers 4102⨯=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 +⎪⎪⎭⎫⎝⎛+= and 1510=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 n Now()()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⎪⎭⎫⎝⎛⨯⨯=i n⎥⎦⎤⎢⎣⎡-⨯0518.012.1exp 1410740.9⨯=⇒i n cm 3- At 200=T K and 400=T K, 15103⨯==a o N p cm 3- At 600=T K,2222i aa 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 E For 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 eV 1610=a N cm 3-, 1697.0=-υE E F eV 1710=a N cm 3-, 1100.0=-υE E F eV_______________________________________ 4.53(a) ⎪⎪⎭⎫ ⎝⎛=-**ln 43n pmidgap 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 N______________________________________________________________________________________15151095.6105⨯+⨯=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⨯='⇒d N cm 3- Additional donor 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⨯=op 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 p 410125.1⨯=cm 3- _______________________________________4.57 ⎥⎦⎤⎢⎣⎡-=kT E E n n Fi F i o exp ()⎥⎦⎤⎢⎣⎡⨯=0259.055.0exp 108.16 15100.3⨯=cm 3- Add additional acceptor impuritiesa d o N N n -=a N -⨯=⨯1515107103 15104⨯=⇒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 F 7508.0=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 a a 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_______________________________________。
半导体物理与器件第四版课后习题答案(供参考).doc
Chapter 44.1n i 2E gN c N expkTT 3E gexpN cO N O300kTwhere N cO and N Oare the values at 300 K.(a) SiliconT (K) kT (eV) n i (cm 3) 200 0.01727 7.68 104 400 0.03453 2.38 1210 6000.05189.74 1014(c) GaAs(b) GermaniumT (K)n i (cm 3 ) n i (cm 3 ) 200 2.16 10101.38 4008.60 1014 3.28 109 6003.82 10165.72 1012_______________________________________ 4.2Plot_______________________________________4.3(a) n i 2 N c NexpE gkT31121919T5 2.8 1.04 101010300exp1.120.0259 T 300T 32.5 10 232.912 10 38300exp1.12 3000.0259 TBy trial and error, T 367.5 K(b)n i25 10 1222.5 10 2532.912 10 38T exp 1.12 300300 0.0259 TBy trial and error,T 417.5 K _______________________________________4.4At T200 K, kT0.02592003000. 017267eVAt T400 K, kT0.02594003000. 034533eVn i 2400 7.70 101023.025 10 17n i 2 2001.40 10 2 23400expE g3000.0345333200Egexp300 0.017267E gE g8 exp0.0345330.0172673.025 10178 exp E g 57 .9139 28.9578orE g 28.9561ln 3.025 1017 38.17148 or E g 1.318 eVNow7.70 1010N co N o340023001.318 exp0.03453321N co N o 2.370 175.929 10 2.658 10so N co N o 9.41 10 37 cm 6_______________________________________4.5exp 1.10n i kT 0.20Bexpn i A 0.90 kTexp kTFor T 200 K, kT 0.017267 eVFor T 300 K, kT 0.0259 eVFor T 400 K, kT 0.034533 eV(a) For T 200K,n i B exp 0.20 9.325 10 6n i A 0.017267(b) For T 300K,n i Bexp 0.204.43 10 4n i A 0.0259 (c) For T 400K,n i Bexp 0.203.05 10 3n i A 0.034533_______________________________________ 4.6(a) g c f FE E FE E c expkTThen g c f F x expxkTTo find the maximum value:d g c f F 1 x1 / 2 exp xdx 2 kT1 x1 /2 exp x 0kT kTwhich yields1/ 21 x kT2x1/ 2 x 2kTThe maximum value occurs atEkTE c2(b)g 1 f FE F EE E expkTE EE E expkTexpE F EkTLet E E xThen g 1 f F x expxkTTo find the maximum valued g 1 f F d xdx dxx expkTSame as part (a). Maximum occurs atxkT2E E c exp E E ckTorkTE E2E c EF expkTLet E E c x _______________________________________ 4.7E1 E c exp E1 E cn E1 kTn E2E2 E c exp E2 E c kTwhereE1 E c 4kT and E 2 E c kT 2Thenn E1 4kTexp E1 E2n E2 kT kT22 2 exp 4 12 exp 3.522orn E10.0854n E 2_______________________________________ 4.8Plot_______________________________________4.9Plot_______________________________________ 4.10E Fi E midgap 3kT ln m*pm n* 4Silicon: m*p 0.56 m o , m n* 1.08m oE Fi E midgap 0.0128 eVGermanium: m*p 0. 37m o ,*0.55m om nE Fi E midgap 0 .0077 eVGallium Arsenide: m*p 0.48m o ,m n* 0.067m oE Fi E midgap 0 .0382 eV_______________________________________ 4.11E Fi E midgap 1 kT ln N2 N c1kT ln 1.04 1019 0.4952 kT2 2.8 1019T (K) kT (eV) ( E Fi E midgap )(eV) 200 0.01727 0.0086 400 0.03453 0.0171 600 0.0518 0.0257_______________________________________4.12(a) E Fi E midgapm*p3 kT ln4 m n*3 0.0259 ln0.704 1.2110.63 meV(b) E Fi E midgap 3 0.0259 ln0.754 0.08043.47 meV_______________________________________4.13Let g c E K constantThenn o g c E f F E dEE cK1dEE E FEc 1 expkTK expE E FdEkTE cLetE E cso that dE kT dkTWe can writeE EF E c E F E E cso thatE E Fexp E c E FexpexpkTkTThe integral can then be written asn o K kT exp E c E Fexp d kTwhich becomesn o K kTE c EF expkT_______________________________________4.14Let g c E C1E E c for E E cThenn o g c E f F E dEE cC1 E E cdEE c 1exp E EF kTC1 EE E FdE E C expE ckTLetE E cdE kT dso thatkTWe can writeE EF E E c E c E FThenE c E Fn o C1 expkTE E cE E cdE expE ckT orn oE c EF C1 expkTkT exp kT d 0We find thatexp d exp 1 1So2 E c E Fn o C1 kT expkT_______________________________________4.15r1 m oWe have rm*a oFor germanium, r 16 , m* 0.55m oThenr1 16 1 a o 29 0.530.55oror1 15.4 AThe ionization energy can be written asm*2E o 13.6 eVm o s0.552 13.6 E 0.029 eV16_______________________________________ 4.16We have r1 m orm*a oFor gallium arsenide, r 13.1 , *m0.067 m o1or1 13.1 104 A0.530.067The ionization energy ism*20.067E o 13.6 13.6m o s 13.1 2orE0.0053 eV_______________________________________4.17Nc(a) E c E F kT ln2.8 10190.0259 ln 157 100.2148 eV(b) E F E E g E c E F1.12 0.2148 0.90518eV(c) p o NE F E expkT1.04 19 0.9051810 exp0.02596.90 103cm 3(d) Holesn o(e) E F E Fi kT lnn i710 150.0259 ln1.5 10100.338 eV_______________________________________4.18N(a) E F E kT lnp o190.0259 ln 1.0410210160.162 eV(b) E c E F E g E F E1.12 0.162 0.958 eV(c) n o 2.8 19 0.95810 exp0.02592.41 103cm3p o(d) E Fi E F kT lnn i2 10 160.0259 ln 101.5 100.365eV_______________________________________4.19Nc(a) E c E F kT ln0.0259 ln 2.810192 1050.8436 eVE F E E g E c E F1.12 0.8436E F E 0.2764 eV(b) p o 1.04 1019 exp 0.276370.02592.414 1014cm3(c)p-type_______________________________________4.20(a) kT3750.032375 eV0.02593003 / 2n o 4.7 10 17 375 exp 0.28300 0.0323751.15 1014cm3E F E E g E c E F 1.42 0.281.14 eV375 3 / 2 1.14 p o 7 18 exp10300 0.0323754.99 103cm 3(b) E c E F 0.0259 ln 4.7 10171.15 10 140.2154 eVE F E E g E c E F 1.42 0.21541.2046 eVp o 7 10 18 exp 1.20460.02594.42 10 2cm 3_______________________________________ 4.21(a) kT 0.0259 3750.032375 eV 300375 3 / 2 0.28n o 2.8 19 exp10300 0.0323756.86 1015cm 3E F E E g E c E F 1.12 0.280.840 eV375 3 / 20.840p o 1.04 1019 exp300 0.0323757.84 107cm 3(b) E c E F kT ln N cn o0.0259 ln2.8 10196.862 10 150.2153 eVE F E 1.12 0.2153 0.9047 eVp o 1.04 10 19 exp 0.9046680.02597.04 103 cm 3_______________________________________4.22(a) p-typeE g(b) E F E1.124 0.28 eV4p o N exp E F EkT1.04 10 19 exp 0.280.02592.10 1014cm 3E c EF E g E F E1.12 0.28 0.84 eVn o N c exp E c E FkT2.8 1019exp0.840.02592.30 105cm 3_______________________________________4.23(a) n o n iE F E FiexpkT1.5 1010 exp 0.220.02597.3313cm310p oE Fi E Fn i expkT1.5 1010 exp 0.220.02593.07 106cm 3(b) n o n iE F E FiexpkT1.8 10 6 exp 0.220.02598.80 109cm 3p o n i expE Fi E FkT1.8 106 exp 0.220.02593.68 102cm 3_______________________________________4.24(a) E F ENkT lnp o0.0259 ln1.04 10 195 10 150.1979 eV(b) E c E F E g E F E1.12 0.19788 0.92212 eV(c) n o 2.8 1019 exp 0.922120.02599.66 103cm 3(d) Holesp o(e) E Fi E F kT lnn i510 150.0259 ln1.5 10100.3294 eV _______________________________________4.25kT 0.0259 4000.034533 eV 3003 / 2N 1.04 10 19400300 1.601 1019cm 33 / 2N c 2.8 1019400300 4.3109 1019cm 30.2642 eV _______________________________________4.26(a) p o 7 1018 exp 0.250.02594.50 1014cm 3E c EF 1.42 0.25 1.17 eVn o 4.7 10 17 exp 1.170.02591.13 10 2cm 3(b)kT 0.034533eV3 / 2N 7 10184003001.078 1019cm 33 / 217 400N c 4.7 103007.236 1017cm3expn i 2 4.3109 10 19 1.601 10191.12NE F E kT lnp o19 0.0345335.67022410n i 2.381 1012 cm 3(a) E F ENkT lnp o0.034533 ln 1.601 10195 1015 0.2787 eV(b) E c E F 1.12 0.27873 0.84127 eV(c) n o 4.3109 10 19 exp 0.841270.0345331.134 109cm3(d) Holes(e) E Fi E F kT ln p on i510150.034533 ln2.381 10120.034533 ln1.078104.50 10 140.3482 eVE c EF 1.42 0.3482 1.072 eVn o 7.236 1017 exp 1 .071770. 0345332.40 104cm 3_____________________________________4.27(a) p o 1.04 1019 exp 0.250.02596.68 1014cm 3E c EF 1.12 0.25 0.870 eVn o 2.8 10 19 exp 0.8700.0259n o7.2310 4 cm 3(b)kT0.034533 eV3 / 2N 1.04 10194003001.601 1019cm 33 / 2N c 2.8 1019 4003004.311 1019cm 3NE F E kT lnp o1.60110 190.034533ln6.6810140.3482 eVE c EF 1.12 0.34820.7718 eVn o 4.311 1019 exp 0.771750.0345338.49 109cm 3_______________________________________4.282(a) n o N c F1 / 2 FFor E F E c kT 2 ,E F E c kT 2 FkT 0.5kTThen F1/ 2 F 1.0n o 2 2.8 1019 1.03.16 1019cm 3(b) n o 2 N c F1 / 2 F24.7 1017 1.05.30 1017cm 3_______________________________________ 4.29p o 2 N F1/2 F5 1019 2 1.04 1019 F1/2 FSo F1/ 2 F 4.26We find F 3.0E E FkTE EF 3.0 0.0259 0.0777 eV_______________________________________4.30E F E c 4kT(a) F 4kT kTThen F1 / 2 F 6.02N c F1 / 2n o F2 2.8 1019 6.01.90 10 20 cm 3(b) n o 2 4.7 1017 6.03.18 1018cm 3_______________________________________ 4.31For the electron concentrationn E g c E f F EThe Boltzmann approximation applies, so4 * 3 / 22m nE E cn Eh3E E FexpkTor4 2m n* 3 / 2 E c E Fexpn E h3kTE E c E E ckT expkTkTDefinexEE ckTThenn E n x K x exp xTo find maximumn E n x , setdn x 0 K 1 x 1 / 2 exp xdx 2x 1 / 21 expxorKx 1 / 2 expx1 x2which yieldsx1 E E cE E c12kTkT2For the hole concentrationp Eg E 1f F EUsing the Boltzmann approximation4 2m p * 3 / 2p EEEh 3E F EexpkT or3 / 242m *p E F Ep Eh 3expkTE E E EkTexpkTkTDefinexE EkTThenp xK x exp xTo find maximum value ofp Ep x ,setdp xUsing the results from0 dxabove,we find the maximum at1E E kT2_______________________________________4.32 (a) Silicon:We haven oN c expE cE FkTWe can writeE c E FE c E d E d E FForE c E d 0.045 eV andE dE F3kT eVwe can writen o2.8 1019 exp 0.04530.02592.8 1019exp 4.737or10 17 cm3n o2.45 We also havep oN expE F EkTAgain, we can writeE FEE FE aE aEForE FE a3kTandE aE0.045eVThenp o1.04 1019 exp 3 0.0450.02591.04 1019 exp4.737orp o9.12 10 16 cm 3(b) GaAs: assume E c E d0.0058eVThenn o4.7 1017 exp0.0058 30.025917exp 3.2244.7 10orn o1.87 1016 cm3Assume E a E 0.0345 eVThenp o71018 exp0.0345 30.02597 1018 exp 4.332orp o9.20 1016 cm 3_______________________________________ 4.33Plot_______________________________________4.34 10 151015 cm 3(a)p o415 31.5 10 10 2n o7.5 10 4 cm33 10153(b) n oN d316cm1010 2p o1.5 107.5 10 3cm 33 1016 (c)n op on i 1.5 10 10cm33(d) n i 22.8 10 19 1.041019 375300 exp1.12 3000.0259 375n i7.334 1011 cm3p o N a4 10 15 cm 37.334 10 11 2n o1.34 10 8 cm34 10 153(e) n i 22.8 10 19 1.04 10 19 4503001.12 300exp0.0259 450133n i1.722 10 cm14142n o1.722 10 1310102221.029 1014 cm 31.722 1013 2p o2.88 1012 cm 31.029 1014_______________________________________(a) p oN aN d4 101510153 1015 cm 3n i 2 1.8 10 6 2n o1.08 10 3cm 3p o3 1015(b) n oN d 3 10 16 cm 3p o1.8 10 6 2 1.08 10 4 cm33 10163(c) n o p on i1.8 10 6cm375 3(d) n i 24.7 1017 7.0 10 18300 exp1.42 3000.0259 375n i 7.580 10 8 cm 3p o N a4 1015 cm 38 2n o7.580 10 1.44 10 2 cm 34 10 153 (e) 2 4.7 10 17 7.0 18450 n i 10 300 exp1.42 3000.0259 450n i 3.853 1010 cm3n oN d10 14 cm 33.853 1010 2p o1.48 10 7 cm 310 14_______________________________________4.3610 13 cm 3(a) Ge: n i2.42(i) n oN dN dn i 22 22 10152 210152.4 13 22210or2 1015 cm 3n oN d4.35n i 2 2.4 1013 2p o2 1015n o2.88 1011 cm 3(ii) p o N a N d 10167 10153 1015 cm 32n i22.4 10 13n op o310 151.92 1011cm3(b) GaAs: n i 1.8 10 6cm3(i) n o N d2 1015 cm62p o1.8 10 1.62 10 3cm32 10 15(ii) p oN aN d3 10 15 cm 362n o1.8 101.08 10 3cm 33 1015 (c) The result implies that there is only one 33minority carrier in a volume of 10 cm ._______________________________________4.37(a) For the donor leveln d 1N d1 1exp EdE F2kT11 1 exp 0.2020.0259orn d8.85 10 4N d (b) We havef F E1E E F1expkTNowE E FE E cE c E ForE EF kT 0.245Thenf F E10.2451 exp 1 0.0259orf F E 2.87 10 5_______________________________________4.38N aN d(a) p-type(b) Silicon:10131013p oN aN d 2.5 1 or1013 cm 3p o1.5Thenn i 21.5 10 10 210 7cm 3n o1.5p o 1.5 1013 Germanium:N aN d N a N d 2p o2n i 221.5131.5 10 1322.4 101310222or3.26 10 13 cm 3p oThen2n i 22.4 10 13n o1.76 10 13p o3.264 1013cm 3Gallium Arsenide:p oN a N d1.5 10 13 cm 3and2n i 21.8 10 6n o0.216 cm 3p o1.5 1013_______________________________________4.39 (a) N d N an-type(b) n oN d N a 2 10151.2 10158 1014 cm 3n i 21.5 101022.81 10 5cm 3p o8 14n o10(c)p o N aN a N d4 1015N a 1.2 10 152 1015N a 4.8 10 15 cm31.5 10 102n o5.625 10 4cm 3 4 1015_______________________________________4.40n i21.5 101021. 153n o2 10 5 125 10cmp on o p on-type_______________________________________4.413n i 21.04 10196.0 10 18 250300 exp0.660.0259250 3001.8936 102412n i 1.376cm310 n on i 2 n i 2n o 21n i 2p o4n o 4n o1n i2Son o 6.88 1011 cm 3 ,Then p o2.75 1012cm3N a N a 2p on i 222N a22.752 10122N a21.8936 10 24227.5735 10 242.752 10 12 N aN a2N a 21.8936 10 242so that N a 2.064 1012cm 3_______________________________________4.42Plot_______________________________________4.43Plot_______________________________________4.44Plot_______________________________________ 4.45N d N aN dN a 2n o2n i 2214141.1 1014 2 10 1.2 102 2 10141.2 1014 2n i 221.1 10144 10 1324 10132n i 24.9 10 271.6 10 27n i2so n i5.74 10 13 cm 3p on i 23.3 10 273 133n o 1.1 10 1410 cm_______________________________________4.46(a)N a N d p-typeMajority carriers are holesp o N a N d16163 101.5 101.5 1016 cm 3Minority carriers are electrons210 10 2n on i 1.5 1.5 10 4 cm 3p o 1.5 1016(b) Boron atoms must be addedp o N a N aN d5 1016N a 3 10161.5 1016So N a3.5 10 16 cm 31.5 10 102n o4.5 10 3cm 35 10 16_______________________________________4.47p on i (a)n-type(b) p on i 2 n on i 2n op o1.5 10 1021016 cm3n o4 1.125 2 10electrons are majoritycarriersp o2 10 4cm3holes are minority carriers(c) n oN d N a1.125 101615N d 7 10so N d1.825 1016 cm3_______________________________________4.48E Fi E FkT lnp on iFor GermaniumT (K)kT (eV)n i (cm 3)200 0.01727 2.16 1010400 0.03453 8.60 1410 6000.05183.82 1016N aN a 2p o n i 2and22N a10 15 cm 3T (K)p o (cm3)E Fi EF (eV)200 1.0 1015 0.1855 4001.49 1015 0.01898 6003.87 10160.000674_______________________________________4.49(a) E c E FkT lnN cN d0.0259 ln 2.8 1019N dFor 1014cm 3 , E cE F 0.3249eV15 cm 3 ,E cE F0.2652eV1016cm 3, E c E F 0.2056eV 101017 cm 3 , E c E F0.1459eV(b) E F E FikT lnN dn i0.0259 lnN d1.51010For 1014cm 3 , E FE Fi 0.2280 eV15cm 3, E F E Fi 0.2877 eV10 1016 cm 3 , E F E Fi 0.3473 eV 1017 cm 3 ,E F E Fi0.4070 eV_______________________________________ 4.50N d N d 2(a) n on i 222n o1.05N d1.05 10 15 cm 31.05 10150.5 10 1520.5 10152n i2son i 25.25 10 28Now3n i 22.8 1019 1.04 1019T300exp1.120.0259 T 30035.25 10 28 2.912 10 38 T300exp 12972.973TBy trial and error, T 536.5K(b) At T 300 K,E c EF kT ln N cn oE c EF 0.0259 ln 2.8 1019 1015T 536.5 K, 0.2652 eVAt536.5kT0.02590.046318 eV3003 / 2N c 2.8 1019 536.53006.696 1019cm 3E c E FN c kT lnn oE c E F6.696 10 19 0.046318 ln10151.050.5124 eVthen E c E F 0.2472 eV(c)Closer to the intrinsic energy level._______________________________________4.51p oE Fi EF kT lnn iAt T 200K, kT 0.017267 eVT 400 K, kT 0.034533 eVT 600 K, kT 0.0518 eV At T 200K,22.8 10191019 200n i 1.04300exp1.120.017267n i 7.638 10 4 cm 3At T 400 K,3n i 2 2.8 1019 1.04 10 19 4003001.12exp0.034533n i 2.381 1012 cm 3At T 600 K,322.8 1019 19 600n i 1.04 10300exp 1.120.0518n i 9.740 1014 cm 3At T 200 K and T 400 K,p o N a 3 1015 cm 3At T 600 K,N a N a2p o n i22 23 15 3 10 15 2 9.740 10 1410 22 23.288 1015cm3Then, T 200K, E Fi E F 0.4212eVT 400K,E Fi EF 0.2465 eVT600K,E Fi EF 0.0630 eV_______________________________________4.52(a)N a N aE Fi EF kT ln 0.0259 ln6n i 1.8 10For N a10 14 cm 3 ,E FiE F0.4619 eVN a 10 15 cm 3,E FiE F0.5215 eV163,N a 10 cmE FiE F0.5811 eVN a 10 17cm 3,E FiE F 0.6408 eV(b)E FEN7.0 1018kT ln0.0259 lnN aN aFor N a10 14 cm 3 ,E F E0.2889 eVN a 10 15 cm 3 ,E FE0.2293 eV163,N a 10 cmE F E0.1697 eVN a 10 17 cm3,E F E 0.1100 eV_______________________________________ 4.53(a) E Fi3 m *p E midgapkT ln4m n *3 0.0259 ln 104 orE Fi E midgap 0.0447 eV(b) Impurity atoms to be added soE midgap EF 0.45 eV(i) p-type, so add acceptor atoms(ii)E Fi EF 0.0447 0.45 0.4947 eVThenp oE FiE Fn i expkT10 5exp 0.49470.0259 or10 13 cm3p o N a1.97_______________________________________4.54n oN d N aN c expE c E FkTsoN d 5 10 15 2.8 10 19 exp0.2150.025951015 6.95 1015orcm 3N d 1.2 1016_______________________________________4.55(a) Silicon(i) E cE F N ckT lnN d0.0259 ln 2.8 10 190.2188 eV6 1015(ii) E cE F0.2188 0.0259 0.1929 eVN dN c expE c E FkT2.8 10 19 exp0.19290.0259N d1.631 1016 cm3N d 6 1015N d1.031 10 16 cm 3Additional donor atoms(b) GaAs(i) E c E F0.0259 ln4.7101710150.15936eV(ii) E cE F0.15936 0.0259 0.13346 eVN d4.7 1017 exp0.133460.02592.718 1015 cm 3N d 1015N d1.718 10 15 cm3Additionaldonor atoms_______________________________________ 4.56(a) E Fi E FN kT lnN a0.0259 ln 1.04 10190.1620 eV2 1016(b) E F E Fi kT ln N c N d0.0259 ln 2.8 1019 0.1876 eV2 10 16(c) For part (a);p o 2 1016 cm 3n i2 1.5 1010 2n op o 2 10161.125 104cm3For part (b):3n o 2 1016 cmn i 2 1.5 1010 2p on o 2 10 161.125 104cm3_______________________________________ 4.57n oE F E Fin i expkT1.8 10 6 exp 0.550.02593.0 1015cm 3Add additional acceptor impuritiesn o N d N a3 10 15 7 10 15 N aN a 4 10 15 cm 3_______________________________________(a) E Fi E F kT lnpon i0.02593 10 150.3161 eVln10 101.5(b) E F E Fin okT lnn i0.02593 10160.3758 eVln10 101.5(c) E F E Fi(d) E Fi E Fp okT lnn i0.0259 375 ln 4 1015300 7.334 10 110.2786 eV(e) E F E Fi kT lnnon i140.0259 450 ln 1.029 10300 1.722 10 130.06945eV_______________________________________4.59(a) E F ENkT lnp o0.0259 ln7.0 10180.2009 eV3 1015(b) E F E 0.0259 l n7.0 10 181.08 10 41.360 eV(c) E F E 0.0259 l n 7.0 10181.8 10 60.7508 eV4.58(d) E F E 0.0259 375300ln 7.0 10 18 375 300 3 / 24 10 150.2526 eV(e) E F E 0.0259 450 300ln 7.0 10 18 450 300 3/ 21.48 10 71.068 eV_______________________________________4.60n-typeE F E Fi kT ln n o n i0.02591.125 10 16ln100.3504 eV1.5 10______________________________________ 4.61N a N a 2 p o 22 2 n i5.08 1015 5 101525 10 15 2n i225.08 10 15 2.5 10 15 22.5 1015 2n i26.6564 10 30 6.25 10 30 n i2n i 2 4.064 10 29n i2 N c N expE gkTkT 0.02593500.030217 eV3003502N c 1.2 10 19 1.633 1019 cm 33003502N 1.8 1019 2.45 10 19 cm 3300Now4.064 10 29 1.633 1019 2.45 1019E gexp0.030217SoE g 0.030217 ln 1.633 10 19 2.45 10 194.064 10 29E g 0.6257 eV_______________________________________4.62(a) Replace Ga atoms Silicon acts as adonorN d0.05 7 1015 3.5 10 14 cm 3Replace As atoms Silicon acts asanacceptorN a 0.95 7 1015 6.65 10 15 cm 3(b) N a N d p-type(c) p o N a N d 6.65 1015 3.5 10146.3 1015cm 3n i 2 1. 810 6 2n o 5.14 10 4 cm 3 p o 6 .3 1015(d) E Fi E F kT ln p o n i0.0259 ln 6.3 10 150.5692 eV1.8 10 6_______________________________________。
半导体物理与器件第四版答案
半导体物理与器件第四版答案半导体物理与器件第四版答案【篇一：半导体物理第五章习题答案】>1. 一个n型半导体样品的额外空穴密度为1013cm-3，已知空穴寿命为100?s，计算空穴的复合率。
解：复合率为单位时间单位体积内因复合而消失的电子-空穴对数，因此1013u1017cm?3?s ?6100?102. 用强光照射n型样品，假定光被均匀吸收，产生额外载流子，产生率为gp，空穴寿命为?，请①写出光照开始阶段额外载流子密度随时间变化所满足的方程；②求出光照下达到稳定状态时的额外载流子密度。
解：⑴光照下，额外载流子密度?n=?p，其值在光照的开始阶段随时间的变化决定于产生和复合两种过程，因此，额外载流子密度随时间变化所满足的方程由产生率gp和复合率u的代数和构成，即 d(?p)?p gp? dt?d(?p)0，于是由上式得⑵稳定时额外载流子密度不再随时间变化，即dtp?p?p0?gp?3. 有一块n型硅样品，额外载流子寿命是1?s，无光照时的电阻率是10??cm。
今用光照射该样品，光被半导体均匀吸收，电子-空穴对的产生率是1022/cm3?s，试计算光照下样品的电阻率，并求电导中少数载流子的贡献占多大比例？解：光照被均匀吸收后产生的稳定额外载流子密度p??n?gp??1022?10?6?1016 cm－3取?n?1350cm2/(v?s)，?p?500cm/(v?s)，则额外载流子对电导率的贡献2pq(?n??p)?1016?1.6?10?19?(1350?500)?2.96 s/cm无光照时?0?10.1s/cm，因而光照下的电导率02.96?0.1?3.06s/cm相应的电阻率 ??110.33??cm 3.06少数载流子对电导的贡献为：?p?pq?p??pq?p?gp?q?p代入数据：?p?(p0??p)q?p??pq?p?1016?1.6?10?19?500?0.8s/cm∴p?00.80.26?26﹪ 3.06即光电导中少数载流子的贡献为26﹪4.一块半导体样品的额外载流子寿命? =10?s，今用光照在其中产生非平衡载流子，问光照突然停止后的20?s时刻其额外载流子密度衰减到原来的百分之几？解：已知光照停止后额外载流子密度的衰减规律为p(t)??p0e?因此光照停止后任意时刻额外载流子密度与光照停止时的初始密度之比即为t??p(t)e? ?p0t当t?20?s?2?10?5s时20??p(20)e10?e?2?0.135?13.5﹪ ?p05. 光照在掺杂浓度为1016cm-3的n型硅中产生的额外载流子密度为?n=?p= 1016cm-3。
半导体物理习题答案第四章
第4章 半导体的导电性2.试计算本征Si 在室温时的电导率，设电子和空穴迁移率分别为1350cm 2/V ⋅s 和500 cm 2/V ⋅s 。
当掺入百万分之一的As 后，设杂质全部电离，试计算其电导率。
掺杂后的电导率比本征Si 的电导率增大了多少倍？解：将室温下Si 的本征载流子密度1.5⨯1010/cm 3及题设电子和空穴的迁移率代入电导率公式()i i n p n q σμμ=+即得：101961.510 1.610(1350500) 4.4410 s/cm i σ--=⨯⨯⨯⨯+=⨯；已知室温硅的原子密度为5⨯1022/cm 3，掺入1ppm 的砷，则砷浓度22616351010510 cm D N --=⨯⨯=⨯在此等掺杂情况下可忽略少子对材料电导率的贡献，只考虑多子的贡献。
这时，电子密度n 0因杂质全部电离而等于N D ；电子迁移率考虑到电离杂质的散射而有所下降，查表4-14知n-Si 中电子迁移率在施主浓度为5⨯1016/cm 3时已下降为800 cm 2/V ⋅s 。
于是得1619510 1.610800 6.4 s cm n nq σμ-==⨯⨯⨯⨯=/该掺杂硅与本征硅电导率之比866.4 1.44104.4410i σσ-==⨯⨯ 即百万分之一的砷杂质使硅的电导率增大了1.44亿倍5. 500g 的Si 单晶中掺有4.5⨯10-5g 的B ，设杂质全部电离，求其电阻率。
（硅单晶的密度为2.33g/cm 3，B 原子量为10.8）。
解：为求电阻率须先求杂质浓度。
设掺入Si 中的B 原子总数为Z ，则由1原子质量单位=1.66⨯10-24g算得618244.510 2.51010.8 1.6610Z --⨯==⨯⨯⨯个 500克Si 单晶的体积为3500214.6 cm 2.33V ==，于是知B 的浓度 ∴1816-32.510 1.1610 cm 214.6A Z N V ⨯===⨯ 室温下硅中此等浓度的B 杂质应已完全电离，查表4-14知相应的空穴迁移率为400 cm 2/V ⋅s 。
半导体物理与器件第四版课后习题答案(完整教资)
Chapter 1Problem Solutions1.1 (a) fcc: 8 corner atoms 18/1=⨯atom 6 face atoms 32/1=⨯atomsTotal of 4 atoms per unit cell (b) bcc: 8 corner atoms 18/1=⨯atom1 enclosed atom =1 atomTotal of 2 atoms per unit cell (c) Diamond: 8 corner atoms 18/1=⨯atom 6 face atoms 32/1=⨯atoms4 enclosed atoms = 4 atomsTotal of 8 atoms per unit cell_______________________________________ 1.2 (a) Simple cubic lattice: r a 2=Unit cell vol ()33382r r a === 1 atom per cell, so atom vol ()⎪⎪⎭⎫⎝⎛=3413r π ThenRatio %4.52%10083433=⨯⎪⎪⎭⎫ ⎝⎛=rr π (b) Face-centered cubic latticer da a r d ⋅==⇒==22224Unit cell vol ()33321622r r a ⋅=⋅==4 atoms per cell, so atom vol()⎪⎪⎭⎫⎝⎛=3443r π ThenRatio ()%74%10021634433=⨯⋅⎪⎪⎭⎫⎝⎛=r r π (c) Body-centered cubic latticer a a r d ⋅=⇒==3434 Unit cell vol 3334⎪⎪⎭⎫⎝⎛⋅==r a 2 atoms per cell, so atom vol()⎪⎪⎭⎫⎝⎛=3423r πThenRatio ()%68%1003434233=⨯⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫⎝⎛=r r π (d) Diamond lattice Body diagonalr a a r d ⋅=⇒===3838Unit cell vol 3338⎪⎪⎭⎫⎝⎛==r a 8 atoms per cell, so atom vol ()⎪⎪⎭⎫⎝⎛=3483r π ThenRatio ()%34%1003834833=⨯⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫⎝⎛=r r π _______________________________________ 1.3(a)oA a 43.5=; From Problem 1.2d,r a ⋅=38Then ()o A a r 176.18343.583===Center of one silicon atom to center ofnearest neighbor oA r 35.22== (b) Number density()22381051043.58⨯=⨯=-cm 3-(c)Mass density()()()23221002.609.28105..⨯⨯===A N W t At N ρ 33.2=⇒ρ grams/cm 3_______________________________________ 1.4 (a) 4 Ga atoms per unit cellNumber density ()381065.54-⨯=⇒Density of Ga atoms221022.2⨯=cm 3-4 As atoms per unit cell ⇒Density of As atoms 221022.2⨯=cm 3- (b) 8 Ge atoms per unit cellNumber density ()381065.58-⨯=⇒Density of Ge atoms 221044.4⨯=cm 3-_______________________________________ 1.5From Figure 1.15(a)()a a d 4330.0232=⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛= =()()oA d 447.265.54330.0=⇒ (b)()a a d 7071.022=⎪⎭⎫⎝⎛=()()oA d 995.365.57071.0=⇒= _______________________________________ 1.6︒=⇒==⎪⎭⎫ ⎝⎛74.5423232222sin θθa a︒=⇒5.109θ_______________________________________ 1.7(a) Simple cubic: oA r a 9.32== (b) fcc: oA r a 515.524==(c) bcc: oA ra 503.434==(d) diamond: ()oA r a 007.9342==_______________________________________ 1.8(a)()()B r 2035.122035.12+= oB A r 4287.0= (b) ()oA a 07.2035.12==(c)A-atoms: # of atoms 1818=⨯= Density ()381007.21-⨯=231013.1⨯=cm 3-B-atoms: # of atoms 3216=⨯=Density ()381007.23-⨯=231038.3⨯= cm 3- _______________________________________ 1.9 (a)oA r a 5.42==# of atoms 1818=⨯= Number density ()38105.41-⨯=2210097.1⨯=cm 3-Mass density ()AN W t At N ..==ρ ()()23221002.65.12100974.1⨯⨯==228.0gm/cm 3(b)o A ra 196.534==# of atoms 21818=+⨯Number density ()3810196.52-⨯=22104257.1⨯=cm 3-Mass density ()()23221002.65.12104257.1⨯⨯==ρ296.0=gm/cm 3_______________________________________ 1.10From Problem 1.2, percent volume of fcc atoms is 74%; Therefore after coffee is ground,Volume = 0.74 cm 3_______________________________________ 1.11(b)oA a 8.20.18.1=+= (c)Na: Density ()()38108.22/1-⨯=221028.2⨯=cm 3-Cl: Density 221028.2⨯=cm 3- (d) Na: At. Wt. = 22.99Cl: At. Wt. = 35.45So, mass per unit cell ()()23231085.41002.645.352199.2221-⨯=⨯⎪⎭⎫⎝⎛+⎪⎭⎫ ⎝⎛= Then mass density()21.2108.21085.43823=⨯⨯=--ρ grams/cm 3_______________________________________ 1.12 (a)()()oA a 88.122.223=+=Then oA a 62.4= Density of A:()22381001.11062.41⨯=⨯=-cm 3-Density of B: ()22381001.11062.41⨯=⨯=-cm 3-(b) Same as (a) (c) Same material_______________________________________ 1.13()()o A a 619.438.122.22=+=(a) For 1.12(a), A-atoms Surface density ()28210619.411-⨯==a 1410687.4⨯=cm 2-For 1.12(b), B-atoms: oA a 619.4= Surface density 14210687.41⨯==acm 2- For 1.12(a) and (b), Same material(b) For 1.12(a), A-atoms; o A a 619.4= Surface density212a =1410315.3⨯=cm 2-B-atoms;Surface density14210315.321⨯==a cm 2- For 1.12(b), A-atoms; o A a 619.4= Surface density212a =1410315.3⨯=cm 2-B-atoms;Surface density14210315.321⨯==a cm 2- For 1.12(a) and (b), Same material_______________________________________ 1.14(a) Vol. Density 31oa =Surface Density 212oa=(b) Same as (a)_______________________________________ 1.15 (i) (110) plane(see Figure 1.10(b))(ii) (111) plane(see Figure 1.10(c))(iii) (220) plane ⇒()0,1,1,21,21⇒⎪⎭⎫⎝⎛∞Same as (110) plane and [110] direction(iv) (321) plane ()6,3,211,21,31⇒⎪⎭⎫⎝⎛⇒Intercepts of plane at 6,3,2===s q p[321] direction is perpendicular to (321) plane_______________________________________ 1.16(a)()31311,31,11⇒⎪⎭⎫⎝⎛(b)()12141,21,41⇒⎪⎭⎫⎝⎛_______________________________________ 1.17Intercepts: 2, 4, 3 ⇒⎪⎭⎫⎝⎛⇒31,41,21(634) plane_______________________________________ 1.18(a) oA a d 28.5==(b) o A a d 734.322==(c) o A a d 048.333==_______________________________________ 1.19 (a) Simple cubic(i) (100) plane:Surface density ()2821073.411-⨯==a141047.4⨯=cm 2- (ii) (110) plane:Surface density 212a =141016.3⨯=cm 2- (iii) (111) plane:Area of plane bh 21=where oA a b 689.62== Now ()()2222243222a a a h =⎪⎪⎭⎫⎝⎛-= So ()o A h 793.573.426==Area of plane ()()881079304.51068923.621--⨯⨯= 16103755.19-⨯=cm 2Surface density 16103755.19613-⨯⨯=141058.2⨯=cm 2- (b) bcc(i) (100) plane:Surface density 1421047.41⨯==a cm 2- (ii) (110) plane:Surface density 222a =141032.6⨯=cm 2- (iii) (111) plane:Surface density 16103755.19613-⨯⨯=141058.2⨯=cm 2- (c) fcc(i) (100) plane:Surface density 1421094.82⨯==acm 2-(ii) (110) plane:Surface density 222a =141032.6⨯=cm 2- (iii) (111) plane:Surface density 16103755.19213613-⨯⨯+⨯=151003.1⨯=cm 2-_______________________________________ 1.20 (a) (100) plane: - similar to a fcc:Surface density ()281043.52-⨯=141078.6⨯=cm 2- (b) (110) plane:Surface density ()281043.524-⨯=141059.9⨯=cm 2- (c) (111) plane:Surface density ()()281043.5232-⨯= 141083.7⨯=cm 2-_______________________________________ 1.21()o A r a 703.6237.2424===(a) #/cm 3()38310703.64216818-⨯=⨯+⨯=a 2210328.1⨯=cm 3-(b) #/cm 222124142a ⨯+⨯= ()210703.6228-⨯=1410148.3⨯=cm 2- (c) ()o A a d 74.422703.622===(d)# of atoms 2213613=⨯+⨯=Area of plane: (see Problem 1.19)oA a b 4786.92==o A ah 2099.826==Area ()()88102099.8104786.92121--⨯⨯==bh 15108909.3-⨯=cm 2#/cm 215108909.32-⨯= =141014.5⨯ cm 2-()o A a d 87.333703.633===_______________________________________ 1.22Density of silicon atoms 22105⨯=cm 3- and4 valence electrons per atom, so Density of valence electrons 23102⨯=cm 3-_______________________________________ 1.23Density of GaAs atoms()22381044.41065.58⨯=⨯=-cm 3- An average of 4 valence electrons per atom,SoDensity of valence electrons231077.1⨯=cm 3-_______________________________________ 1.24(a) %10%10010510532217-=⨯⨯⨯ (b) %104%10010510262215-⨯=⨯⨯⨯ _______________________________________ 1.25 (a) Fraction by weight()()()()7221610542.106.2810582.10102-⨯=⨯⨯≅ (b) Fraction by weight()()()()5221810208.206.2810598.3010-⨯=⨯≅ _______________________________________ 1.26Volume density 1631021⨯==dcm 3-So 610684.3-⨯=d cm oA d 4.368=⇒ We have oo A a 43.5=Then85.6743.54.368==o a d _______________________________________ 1.27Volume density 1531041⨯==dcm 3-So 61030.6-⨯=d cm oA d 630=⇒ We have oo A a 43.5= Then11643.5630==o a d _______________________________________。
半导体物理与器件第四版课后习题答案
半导体物理与器件第四版课后习题答案第一章半导体材料基础知识1.1 小题一根据题目描述，当n=5时，半导体材料的载流子浓度为’n=2.5×1015cm(-3)’，求势垒能为多少？解答：根据势垒能公式E_g = E_c - E_v其中E_g为势垒能，E_c为导带底，E_v为价带顶。
根据载流子浓度和温度的关系n = 2 * （2 * pi * m_e * k * T / h^2）^(3/2) * e^(-E_g / (2 * k * T))其中m_e为载流子质量，k为玻尔兹曼常数，T为绝对温度。
可以得到E_g = -2 * k * T * ln(n / (2 * （2 * pi * m_e * k * T / h^2）^(3/2)))代入已知条件，计算得到势垒能为E_g = -2 * 1.38 * 10^(-23) * 300 * ln(2.5 * 10^15 / (2 * （2 * pi * 9.1 * 10^(-31) * 1.38 * 10^ (-23) * 300 / (6.63 * 10^(-34))^2）^(3/2)))1.1 小题二根据题目描述，当势垒能E_g=1.21eV时，求温度为多少时，载流子浓度为’n=5.0×1015cm(-3)’？解答：按照1.1 小题一的公式，可以求出温度TT = E_g / (2 * k * ln(n / (2 * （2 * pi * m_e * k * T / h^2）^(3/2))))将已知数据代入公式，计算得到温度T = 1.21 / (2 * 1.38 * 10^(-23) * ln(5 * 10^15/ (2 * （2 * pi * 9.1 * 10^(-31) * 1.38 * 10^(-2 3) * T / (6.63 * 10^(-34))^2）^(3/2))))第二章半导体材料与器件基本特性2.1 小题一根据题目描述，当Si掺杂浓度[N_b]为5×10^15 cm(-3)和[P_e]为2×1017 cm^(-3)，求Si中的载流子浓度和导电类型。
半导体物理与器件第四版课后习题答案（供参考）
半导体物理与器件第四版课后习题答案（供参考）Chapter 44.1 where cO N and O N υ are the values at 300 K.4.2Plot_______________________________________4.3By trial and error, 5.367?T K(b)By trial and error, 5.417?T K_______________________________________4.4At 200=T K, ()=3002000259.0kT017267.0=eVAt 400=T K, ()?=3004000259.0kT 034533.0=eVoror 318.1=g E eV Now so 371041.9?=o co N N υcm 6-_______________________________________4.5 For 200=T K, 017267.0=kT eV For 300=T K, 0259.0=kT eVFor 400=T K, 034533.0=kT eV(a) For 200=T K,(b) For 300=T K,(c) For 400=T K,_______________________________________4.6Let x E E c =-Then ??-∝kT x x f g F c exp To find the maximum value: which yields The maximum value occurs at (b)Let x E E =-υ Then ()??-∝-kT x x f g F exp 1υ To find the maximum value Same as part (a). Maximum occurs at or_______________________________________4.7 wherekT E E c 41+= and 22kT E E c += Then or _______________________________________ 4.8 Plot_______________________________________ 4.9 Plot _______________________________________4.10 Silicon: o p m m 56.0*=, o n m m 08.1*=0128.0-=-midgap Fi E E eV Germanium: o p m m 37.0*=,o n m m 55.0*=0077.0-=-midgap Fi EE eVGallium Arsenide: o p m m 48.0*=, 0382.0+=-midgapFi E E eV _______________________________________4.12 63.10-?meV 47.43+?meV _______________________________________ 4.13 Let ()==K E g c constant Then Let kTE E c-=η so that ηd kT dE ?= We can writeso that The integral can then be written as which becomes_______________________________________ 4.14Let ()()c c E E C E g -=1 for c E E ≥ Then LetkTE E c-=η so that ηd kT dE ?=We can write Then orWe find that So_______________________________________ 4.15We have=∈*1m m a r o r o For germanium, 16=∈r , o m m 55.0*= Then orThe 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 , Then The ionization energy is or0053.0=E eV_______________________________________ 4.17 2148.0=eV90518.02148.012.1=-=eV 31090.6?=cm 3- (a) Holes 338.0=eV_______________________________________ 4.18 162.0=eV 958.0162.012.1=-=eV 31041.2?=cm 3-365.0=eV_______________________________________ 4.19 8436.0=eV2764.0=-υE E F eV1410414.2?=cm 3- (a) p-type_______________________________________ 4.20 (a) ()032375.03003750259.0==kT eV141015.1?=cm 3-14.1=eV31099.4?=cm 3-2154.0=eV 2046.1=eV 21042.4-?=cm 3-_______________________________________ 4.21 (a) ()032375.03003750259.0==kT eV151086.6?= cm 3-840.0=eV 71084.7?=cm 3-2153.0=eV9047.02153.012.1=-=-υE E F eV 31004.7?=cm 3-_______________________________________ 4.22(a) p-type(b) 28.0412.14===-g F E E E υeV141010.2?=cm 3- 84.028.012.1=-=eV 51030.2?=cm 3-_______________________________________ 4.23 131033.7?=cm 3- 61007.3?=cm 3- 91080.8?=cm 3-21068.3?=cm 3-_______________________________________ 4.241979.0=eV 92212.019788.012.1=-=eV 31066.9?=cm 3- (a) Holes 3294.0=eV _______________________________________。
半导体物理课后习题解答(刘恩科第四版)
1－1． （P32）设晶格常数为 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 E v（k）分 别为： Ec(k)=
h 2 k 2 h 2 (k k1) 2 h2k 2 3h 2 k 2 + 和 Ev(k)= － ； 3m0 m0 6m0 m0
m0 为电子惯性质量，k1＝ 1/2a； 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禁带宽度 Eg 根据
h2 (
的 Nc 和 Nv：
2(2 m k T ' ) 3 3 3 N T' 2 ' T' 2 77 2 h ( ) ; Nc ( ) Nc ( ) 1.05 1019 1.365 1019 3 Nc T T 300 * 2(2 mn k 0T ) 2 h3
2）T＝300k 时：
T 2 4.774 10 4 500 2 Eg (500) Eg (0) 0.7437 0.58132eV ； T 500 235
查图 3-7(P 61)可得： ni 2.2 10 ，属于过渡区，
16
1
( N D N A ) [( N D N A ) 2 4ni2 ] 2 n0 2.464 1016 ； 2
∴ exp(
即： E F E D k 0T ln 2
n0 Nc exp(
取对数后得：
Ec E F N ) D k 0T 2
整理得下式：
EC E D k 0T ln 2 N ln D k 0T 2 Nc
E D N ln D k 0T Nc
半导体物理分章答案第四章-36页文档资料
L [111] Γ
X[100]
• Si的能带结构
L [111]
Γ
X[100]
• Ge的能带结构
L [111]
Γ
X[100]
2、高场畴及耿氏振荡
更多精品资源请访问
docin/sanshengshiyuan doc88/sanshenglu
著） • 能谷间散射：等同能谷间散射高温下较易发生；不同
能谷间散射一般在强电场下发生。
（1）电离杂质散射（即库仑散射）
载流子的散射几率P 散射几单率位Pi∝时N间iT内-3/2 一个载流子受到散射的平均 （次N数i：。为主杂要质浓用度于总描和述）散射的强弱。
（2）晶格振动散射
晶格振动表现为格波
N个原胞组成的晶体→格波波矢有N个。格波的总数 等于原子自由度总数
§4.7 多能谷散射 耿氏效应
1、双能谷模型和砷化镓的能带结构
（1）负微分电导、负微分迁移率 半导体材料的载流子运动速度随电场的增加而减
小称为负微分电导。
（2）双能谷模型 半导体有两个能谷，它们之间有能量间隔△E。在外
电场为零时，导带电子按晶格温度和各自的状态密度所决
定的分布规律分布于两能谷之中。外电场增加时载流子将
（A）声学波散射：
在长声学波中，纵波对散射起主要作用（通过体变 产生附加势场）。
对于单一极值，球形等能面的半导体，理论推导得
到
Ps 16c2kh04Tu(2mn*)2 v
Ec
c
V V0
其中u纵弹性波波速。 由上式可知
3
Ps T 2
半导体物理与器件第四版课后习题标准答案4
Chapter 44.1⎪⎪⎭⎫ ⎝⎛-=kTE N N n gc i exp 2υ ⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=kT E T N N g O cO exp 3003υ where cO N and O N υ are the values at 300 K.4.2Plot_______________________________________ 4.3(a) ⎪⎪⎭⎫⎝⎛-=kT E N N n g c i exp 2υ()()()319192113001004.1108.2105⎪⎭⎫⎝⎛⨯⨯=⨯T()()⎥⎦⎤⎢⎣⎡-⨯0259.012.1exp 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⨯=⨯⨯=i i n n⎥⎦⎤⎢⎣⎡-⎥⎦⎤⎢⎣⎡-⨯⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛=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 υso 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 212/1 0exp 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 exp wherekT E E c 41+= and 22kTE E c +=Then()()()⎥⎦⎤⎢⎣⎡--=kT E E kT kT E n E n 2121exp 24()5.3exp 22214exp 22-=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛--=or()()0854.021=E n E n _______________________________________ 4.8Plot_______________________________________ 4.9Plot_______________________________________ 4.10⎪⎪⎭⎫ ⎝⎛=-**ln 43n pmidgap Fi m m kT E ESilicon: o pm m 56.0*=, o nm 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 eVGallium 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⎥⎦⎤⎢⎣⎡---⨯⎰∞expor()⎥⎦⎤⎢⎣⎡--=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 1004.11931090.6⨯=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 n 31041.2⨯=cm 3- (d) ⎪⎪⎭⎫ ⎝⎛=-i o F Fi n p kT E E ln()⎪⎪⎭⎫ ⎝⎛⨯⨯=1016105.1102ln 0259.0 365.0=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 --=-υ8436.012.1-=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 p 21042.4-⨯=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 p31004.7⨯=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 1004.119 141010.2⨯=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 108.21951030.2⨯=cm 3-_______________________________________ 4.23(a) ⎥⎦⎤⎢⎣⎡-=kT E E n n Fi Fi o exp()⎥⎦⎤⎢⎣⎡⨯=0259.022.0exp 105.110 131033.7⨯=cm 3-⎥⎦⎤⎢⎣⎡-=kT E E n p F Fii o exp()⎥⎦⎤⎢⎣⎡-⨯=0259.022.0exp 105.110 61007.3⨯=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 108.1621068.3⨯=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⎥⎦⎤⎢⎣⎡-⨯034533.012.1exp24106702.5⨯=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 p 141050.4⨯=cm 3-17.125.042.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=0259.017.1exp 107.417o n 21013.1-⨯=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 n41040.2⨯=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()Fo 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 h m E n -=32/3*24π()⎥⎦⎤⎢⎣⎡--⨯kT E E F expor ()()()⎥⎦⎤⎢⎣⎡--=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 m E p F 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 fromabove,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⨯=⨯⨯=on cm 3-(b) 16103⨯==d o N n cm 3- ()316210105.7103105.1⨯=⨯⨯=op cm 3- (c) 10105.1⨯===i o o n p n cm 3- (d) ()()3191923003751004.1108.2⎪⎭⎫⎝⎛⨯⨯=i n()()()()⎥⎦⎤⎢⎣⎡-⨯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⎪⎭⎫⎝⎛⨯⨯=i n()()()()⎥⎦⎤⎢⎣⎡-⨯4500259.030012.1exp1310722.1⨯=⇒i n cm 3-()2132141410722.1210210⨯+⎪⎪⎭⎫ ⎝⎛+=o n1410029.1⨯=cm 3-()12142131088.210029.110722.1⨯=⨯⨯=op cm 3-_______________________________________ 4.35(a) 151510104-⨯=-=d a o N N p15103⨯=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 d d 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 F Then()⎪⎭⎫ ⎝⎛++=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⎪⎭⎫⎝⎛⨯⨯=in()()⎥⎦⎤⎢⎣⎡-⨯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 a a o n N N p +⎪⎪⎭⎫⎝⎛+= 212210752.2⎪⎪⎭⎫ ⎝⎛-⨯a N 242108936.12⨯+⎪⎪⎭⎫ ⎝⎛=a N()21224210752.2105735.7⎪⎪⎭⎫⎝⎛+⨯-⨯a a N N242108936.12⨯+⎪⎪⎭⎫ ⎝⎛=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⨯=⨯⨯=on 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 Nso 1610825.1⨯=d N cm 3-_______________________________________ 4.48⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln1510=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 d d o n N N n +⎪⎪⎭⎫⎝⎛+= 151005.105.1⨯==d o N n cm 3- ()21515105.01005.1⨯-⨯()2215105.0i n +⨯=so 2821025.5⨯=i n Now()()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 E 2652.0=eVAt 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 eV At 200=T K,()()3191923002001004.1108.2⎪⎭⎫⎝⎛⨯⨯=i n⎥⎦⎤⎢⎣⎡-⨯017267.012.1exp410638.7⨯=⇒i n cm 3- At 400=T K,()()3191923004001004.1108.2⎪⎭⎫⎝⎛⨯⨯=in⎥⎦⎤⎢⎣⎡-⨯034533.012.1exp1210381.2⨯=⇒i n cm 3-At 600=T K,()()3191923006001004.1108.2⎪⎭⎫ ⎝⎛⨯⨯=in⎥⎦⎤⎢⎣⎡-⨯0518.012.1exp 1410740.9⨯=⇒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 eV400=T K,2465.0=-F Fi E E eV 600=T K,0630.0=-F Fi E E eV _______________________________________4.52 (a)()⎪⎪⎭⎫⎝⎛⨯=⎪⎪⎭⎫ ⎝⎛=-6108.1ln 0259.0ln a iaF Fi N nN kT E E For 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 eV 1710=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= or 0447.0+=-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 eVThen⎪⎪⎭⎫⎝⎛-=kT E E n p F Fi i o exp()⎪⎭⎫ ⎝⎛=0259.04947.0exp 105 or 131097.1⨯==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- Additional donor 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⨯='⇒d N 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 p 410125.1⨯=cm 3-_______________________________________ 4.57⎥⎦⎤⎢⎣⎡-=kT E E n n Fi Fi o exp ()⎥⎦⎤⎢⎣⎡⨯=0259.055.0exp 108.1615100.3⨯=cm 3- Add additional acceptor impurities a d o N N n -= a N -⨯=⨯1515107103 15104⨯=⇒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 F360.1=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_______________________________________。
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Plot
_______________________________________
4.45
so cm
cm
_______________________________________
4.46
(a) p-type
Majority carriers are holes
cm
Minority carriers are electrons
(c)
so cm
_______________________________________
4.48
For Germanium
(K)
(eV)
(cm )
and
cm
(K)
(cm )
(eV)
_______________________________________
4.49
(a)
For cm , eV
4.29
So
We find
eV
_______________________________________
4.30
(a)
Then
cm
(b)
cm
_______________________________________
4.31
For the electron concentration
The Boltzmann approximation applies, so
4.22
(a)p-type
(b) eV
cm
eV
cm
_______________________________________
4.23
(a)
cm
cm
(b)
cm
cm
_______________________________________
4.24
(a)
eV
(b)
eV
(c)
cm
(d)Holes
(e)
Chapter 4
4.1
where and are the values at 300 K.
(a) Silicon
(K)
(eV)
(cm )
(b) Germanium (c) GaAs
(K)
(cm )
(cm )
_______________________________________
4.2
Plot
cm
and
cm
_______________________________________
4.39
(a) n-type
(b)
cm
cm
(c)
cm
cm
_______________________________________
4.40
cm
n-type
_______________________________________
meV
(b)
meV
_______________________________________
4.13
Let constant
Then
Let
so that
We can write
so that
The integral can then be written as
which becomes
_______________________________________
4.55
(a)Silicon
(i)
eV
(ii) eV
cm
cm Additional
donor atoms
(b)GaAs
(i)
eV
(ii) eV
cm
cm Additional
donor atoms
_______________________________________
4.56
(a)
eV
_______________________________________
4.20
(a) eV
cm
eV
cm
(b)
eV
eV
cm
_______________________________________
4.21
(a) eV
cm
eV
cm
(b)
eV
eV
cm
_______________________________________
4.7
where
and
Then
or
_______________________________________
4.8
Plot
_______________________________________
4.9
Plot
_______________________________________
4.10
4.37
(a) For the donor level
or
(b) We have
Now
or
Then
or
_______________________________________
4.38
(a) p-type
(b)Silicon:
or
cm
Then
cm
Germanium:
or
cm
Then
cm
Gallium Arsenide:
eV
eV
_______________________________________
4.16
We have
For gallium arsenide, ,
Then
The ionization energy is
or
eV
_______________________________________
4.17
4.41
cm
So cm ,
Then cm
so that cm
_______________________________________
4.42
Plot
_______________________________________
4.43
Plot
_______________________________________
we find the maximum at
_______________________________________
4.32
(a)Silicon: We have
We can write
For
eV and eV
we can write
or
cm
We also have
Again, we can write
4.36
(a)Ge: cm
(i)
or
cm
cm
(ii)
cm
cm
(b)GaAs: cm
(i) cm
cm
(ii) cm
cm
(c) The result implies that there is only one minority carrier in a volume of cm .
_______________________________________
Then
To find the maximum value:
which yields
The maximum value occurs at
nd the maximum value
Same as part (a). Maximum occurs at
or
_______________________________________
cm
(b)Boron atoms must be added
So cm
cm
_______________________________________
4.47
(a) n-type
(b)
cm
electrons are majority carriers
cm
holes are minority carriers
_______________________________________
4.3
(a)
By trial and error, K
(b)
By trial and error, K
_______________________________________
4.4
At K,
eV
At K,
eV
or
or eV
For
and eV
Then
or
cm
(b)GaAs: assume eV
Then
or
cm
Assume eV
Then
or
cm
_______________________________________
4.33
Plot
_______________________________________
4.34
_____________________________________
4.27
(a)
cm
eV
cm
(b) eV
cm
cm
eV
eV
cm
_______________________________________