《半导体物理与器件》第四版答案第十章
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_______________________________________。
半导体物理与器件第四版课后习题答案6
______________________________________________________________________________________Chapter 66.115105⨯==d o N n cm 3-()4152102105.4105105.1⨯=⨯⨯==d i o N n p cm 3- (a) Minority carrier hole lifetime is a constant.70102-⨯==p pt ττs117401025.2102105.4⨯=⨯⨯==-p opo p R τcm 3-s 1- (b)7144010210105.4-⨯+⨯=+='p o po pp R τδ 20105⨯=cm 3-s 1-_______________________________________ 6.216102⨯==a o N p cm 3-()4162621062.1102108.1-⨯=⨯⨯==o i o p n n cm 3-(a) 21714010105105=⨯⨯=='-n n R τδcm 3-s 1- (b)n ontoptop n n p R τττ===()()()741601051062.1102--⨯⋅⨯⨯=⋅=n o o pt n p ττ131017.6⨯=s_______________________________________ 6.3 (a) Recombination rates are equalpOo nO o pn ττ=1610==d o N n cm 3-()41621021025.210105.1⨯=⨯==o i o n n p cm 3-Then641610201025.210-⨯⨯=nO τ which yields61089.8+⨯=nO τs (b) Generation rate = recombination rate Then96410125.110201025.2⨯=⨯⨯=-G cm 3-s 1- (c)910125.1⨯==G R cm 3-s 1-_______________________________________ 6.4 (a) ()()1083410630010310625.6--⨯⨯⨯===λνhch E or191015.3-⨯=E J; energy of one photon Now1 W = 1 J/s 181017.3⨯⇒photons/s Volume = (1)(0.1) = 0.1 cm 3 Then1.01017.318⨯=g191017.3⨯= e-h pairs/cm 3-s (b)()()61910101017.3-⨯⨯===τδδg p n or141017.3⨯==p n δδcm 3-_______________________________________ 6.5We havepp p p g F t p τ-+∙-∇=∂∂+ andp eD p e J p p p ∇-E =μ The hole particle current density is()p D p e J F p ppp ∇-E =+=+μ Now______________________________________________________________________________________ ()p D p F p p p ∇∙∇-E ∙∇=∙∇+μ We can write()E ∙∇+∇∙E =E ∙∇p p p andp p 2∇=∇∙∇ so()p D p p F p p p 2∇-E ∙∇+∇∙E =∙∇+μThen()E ∙∇+∇∙E -=∂∂p p tpp μpp p p g p D τ-+∇+2 We can then write ()E ∙∇+∇∙E -∇p p p D p p μ2tpp g p p ∂∂=-+τ _______________________________________6.6 From Equation (6.18), pp p pg F t p τ-+∙-∇=∂∂+ For steady-state,0=∂∂tpThenp p p R g F -+∙-∇=+0For a one-dimensional case,192010210⨯-=-=+p p p R g dxdF or19108⨯=+dxdF p cm 3-s 1- _______________________________________ 6.7From Equation (6.18),1910200⨯-+-=+dxdF p or19102⨯-=+dxdF p cm 3-s 1-_______________________________________ 6.8We have the continuity equations (1) ()()[]E ∙∇+∇∙E -∇p p p D p p δμδ2()tp p g p p ∂∂=-+δτ and (2) ()()[]E ∙∇+∇∙E +∇n n n D n n δμδ2()t n n g n n ∂∂=-+δτBy charge neutrality,()()p n n p n δδδδδ∇=∇⇒≡= and()()p n δδ22∇=∇ and ()()t p t n ∂∂=∂∂δδ A lso g g g p n ≡=, R np np ≡=ττ Then we have (1) ()()[]E ∙∇+∇∙E -∇p n n D p p δμδ2()tn R g ∂∂=-+δ and (2) ()()[]E ∙∇+∇∙E +∇n n n D n n δμδ2()tn R g ∂∂=-+δ Multiply Equation (1) by n n μ and Equation(2) by p p μ, and add the two equations. We find()()n pD nD n p p n δμμ2∇+()()n n p p n δμμ∇∙E -+ ()()R g p n p n -++μμ______________________________________________________________________________________ ()()tn pn p n ∂∂+=δμμ Divide by ()p n p n μμ+, then()n p n pD nD p n n p p n δμμμμ2∇⎪⎪⎭⎫⎝⎛+++()()n p n n p p n p n δμμμμ∇∙E ⎥⎥⎦⎤⎢⎢⎣⎡+-()()tn R g ∂∂=-+δ Define()pD n D p n D D p n pD nD D p n p n p n n p p n ++=++='μμμμ and ()pn n p p n p n μμμμμ+-='Then we have()()()R g n n D -+∇∙E '+∇'δμδ2()tn∂∂=δ Q.E.D. _______________________________________6.9 p-type material;minority carriers are electrons(a)n μμ=' From Figure 5.3, 1300≅n μcm 2/V-s (b)()()13000259.0=⋅⎪⎭⎫⎝⎛=='n n e kT D D μ 67.33=cm 2/s (c)7010-==n nt ττs15107⨯==a o N p cm 3-()152102107105.1⨯⨯==a i o N n n41021.3⨯=cm 3-pto nt o pn ττ=pt τ15741071010214.3⨯=⨯- so 41018.2⨯=pt τs_______________________________________6.10 For Ge: 13104.2⨯=i n cm 3- 2222i d d o n N N n +⎪⎪⎭⎫⎝⎛+=()21321313104.221042104⨯+⎪⎪⎭⎫ ⎝⎛⨯+⨯=1310124.5⨯=cm 3-()1313213210124.110124.5104.2⨯=⨯⨯==o i o n n p cm 3- (a) We have:3900=n μcm 2/V-s, 101=n D cm 2/s1900=p μcm 2/V-s, 2.49=p D cm 2/s For very, very low injection, ()p D n D p n D D D p n p n ++='()()()()()()()1313131310124.12.4910124.510110124.110124.52.49101⨯+⨯⨯+⨯= 2.54=cm 2/s and()pn n p p n p n μμμμμ+-='()()()()()()()1313131310124.1190010124.5390010124.510124.119003900⨯+⨯⨯-⨯=1340-=cm 2/V-s (b)For holes, 60102-⨯==p pt ττs For electrons,______________________________________________________________________________________p nt p nττ=6131310210124.110124.5-⨯⨯=⨯ntτ61012.9-⨯=⇒nt τs_______________________________________ 6.11p e n e p n μμσ+= W ith excess carriersn n n o δ+= and p p p o δ+=For an n-type semiconductor, we can write p p n δδδ≡= Then()()p p e p n e o p o n δμδμσ+++= or()()p e p e n e p n o p o n δμμμμσ+++= so()()p e p n δμμσ+=∆In steady-state, pO g p τδ'= So that()()pO p n g e τμμσ'+=∆_______________________________________ 6.12 (a) 1610==a o N p cm 3-()41621021025.210105.1⨯=⨯==o i o p n n cm 3- ()()p p e n n e o p o n δμδμσ+++= ()n e p e p n o p δμμμ++≅ Now ()0/01n t n e g p n ττδδ--'==()()()0/7201105108n t e τ---⨯⨯=()0/141104n t eτ--⨯=cm 3-Then σ()()()161910380106.1-⨯= ()()380900106.119+⨯+-()()0/141104n t e τ--⨯⨯ ()0/10819.0608.0n t e τσ--+= (Ω-cm)1- (b) (i)()608.00=σ(Ω-cm)1-(ii)()690.0=∞σ(Ω-cm)1-_______________________________________ 6.13 (a) For 6100-≤≤t s, ()0/01p t p e g p n ττδδ--'==()()()0/8211105104p t e τ---⨯⨯=()()0/141102p t e τ--⨯= cm 3-At 610-=t s,()()()86105/10146110210--⨯---⨯=e p δ14102⨯=cm 3-Then for 610-≥t s,()()6/1014102p t ep τδ---⨯=cm 3-(b)15105⨯=o n cm 3-()p e n e p n o n δμμμσ++=For 6100-≤≤t s,()()()15191057500106.1⨯⨯=-σ ()()3107500106.119+⨯+-()()0/141102p t eτ--⨯⨯()0/1250.00.6p t eτ--+=(Ω-cm)1-For 610-≥t s,()6/10250.00.6p t eτσ---+=(Ω-cm)1-_______________________________________ 6.14R V I =; ALR σ=______________________________________________________________________________________V L A I ⋅=⇒σFor 1515102108⨯+⨯=+=a d I N N N1610=cm 3- Then, 1300≅n μcm 2/V-s400≅p μcm 2/V-s()p e n e p n o n δμμμσ++≅where 0/0p t p e g p ττδ-'=()()0/720105108p t e τ--⨯⨯= 0/14104p t e τ-⨯=cm 3- ()()()1515191021081300106.1⨯-⨯⨯=-σ()()4001300106.119+⨯+-()0/14104p t e τ-⨯⨯ 0/109.0248.1p t e τσ-+= []()()05.01010109.0248.15/0--+=p t e I τ 0/431018.210496.2p t e τ---⨯+⨯= Aor 0/218.0496.2p t e I τ-+=mA_______________________________________6.151516106102⨯-⨯=-=d a o N N p 16104.1⨯=cm 3-(a)0n g p n τδδ'==021********n τ⨯=⨯ 70105.2-⨯=⇒n τs (b)()0/01n t n eg p n ττδδ--'==()0/141105n t e τ--⨯=()0/71401105.2105n t n e n R ττδ---⨯⨯==' ()not e τ/211102--⨯=cm 3-s 1- (c)(i)()()0/1414110510541n t e τ--⨯=⨯⎪⎭⎫ ⎝⎛()801019.73333.1ln -⨯==n t τs(ii) ()()0/1414110510521n t e τ--⨯=⨯⎪⎭⎫ ⎝⎛ ()701073.12ln -⨯==n t τs (iii)()()0/1414110510543n t e τ--⨯=⨯⎪⎭⎫ ⎝⎛ ()701047.34ln -⨯==n t τs(iv)()()()0/1414110510595.0n t e τ--⨯=⨯()701049.720ln -⨯==n t τs_______________________________________6.161515102108⨯-⨯=-=a d o N N n 15106⨯=cm 3-()415262104.5106108.1-⨯=⨯⨯==o i o n n p cm 3- (a) 0440104.5104p p o o p R ττ-⨯=⨯⇒= so 801035.1-⨯=p τs (b) ()()82101035.1102-⨯⨯='=p g p τδ 13107.2⨯=cm 3- (c) 801035.1-⨯==p ττs _______________________________________6.17 (a)(i)For 71050-⨯≤≤t s()()0/01p t p e g t p ττδ--'=()()()0/7201105105p t e τ---⨯⨯=()0/141105.2p t eτ--⨯= cm3-At 7105-⨯=t s,()1/1141105.2--⨯=e p δ141058.1⨯=cm 3-For 7105-⨯≥t s()()p Ot et p τδ/1051471058.1-⨯--⨯=cm 3-______________________________________________________________________________________ (ii) ()1471058.1105⨯=⨯-p δcm 3- (b) (i) For 61020-⨯≤≤t s()()0/141105.2p t e t p τδ--⨯= cm3-At 6102-⨯=t s,()()()76105/102141105.2--⨯⨯--⨯=e p δ1410454.2⨯=cm 3-For 6102-⨯≥t s,()()p O t et p τδ/10214610454.2-⨯--⨯=cm 3-(ii)()14610454.2102⨯=⨯-p δcm 3- _______________________________________ 6.18 (a) For 61020-⨯≤≤t s ()0/0n t n e g t n ττδ-'=()()0/72110510n t e τ--⨯= 0/14105n t e τ-⨯=cm 3-At 6102-⨯=t s,()()76105/102141105--⨯⨯-⨯=e n δ121016.9⨯= cm 3-For 6102-⨯≥t s ()()0/121411016.9105n t e n τδ--⨯-⨯=121016.9⨯+()12/141016.9110908.40⨯+-⨯=-n t e τcm 3-(b) (i)()141050⨯=n δcm 3-(ii)()1261016.9102⨯=⨯-n δcm 3- (iii)()14105⨯=∞n δcm 3-_______________________________________ 6.19p-type; minority carriers - electrons()()12000259.0=⎪⎭⎫⎝⎛=n n e kT D μ 08.31=cm 2/s()()[]2/1601008.31-==n n n D L τ310575.5-⨯=cm(a) ()()n L x e x p x n /14102-⨯==δδcm 3-(b)()[]n L x n nn e dx deD dx n d eD J /14102-⨯==δ ()n L x nne L eD /14102-⨯-=()()()()n L x e /3141910575.510208.31106.1---⨯⨯⨯-=n L x n e J /1784.0--=A/cm 2Holes diffuse at same rate as minority carrier electrons, son L x p e J /1784.0-+=A/cm 2_______________________________________ 6.20 (a)p-type; 1410=pO p cm 3-and()61421021025.210105.1⨯=⨯==pO i pOp n n cm 3-(b) Excess minority carrier concentrationpO p n n n -=δ At 0=x , 0=p n so that()61025.200⨯-=-=pO n n δcm 3-(c) For the one-dimensional case,()022=-nOnndx n d D τδδ or()0222=-nLndx n d δδ where nO n n D L τ=2The general solution is of the form⎪⎪⎭⎫⎝⎛++⎪⎪⎭⎫ ⎝⎛-=n n L x B L x A n exp exp δ For ∞→x , n δ remains finite, so0=B .Then the solution is⎪⎪⎭⎫⎝⎛--=n pO L x n n exp δ _____________________________________________________________________________________________________________________________ 6.21()n L x e x n /14105-⨯=δcm 3-where ()()[]2/1601025-==n n n D L τ 3105-⨯=cm()()[]n L x n n n e dxdeD dx n d eD J /14105-⨯==δ()n L x nne L eD /14105-⨯-=()()()()nL x e /3141910*********.1---⨯⨯⨯-=n L x n e J /4.0--=A/cm 2(a) For 0=x ,()141050⨯=n δcm 3- ()4.00-=n J A/cm 2 ()4.00+=p J A/cm 2 (b) For 3105-⨯==n L x cm,()()141141084.1105⨯=⨯=-e L n n δcm 3-()147.04.01-=-=-e L J n n A/cm 2()147.04.01+=+=-e L J n p A/cm 2(c)For 31015-⨯=x cm n L 3=()()133141049.21053⨯=⨯=-e L n n δcm 3-()020.04.033-=-=-e L J n n A/cm 2()020.04.033+=+=-e L J n p A/cm 2_______________________________________6.22n-type, so we have()()02=-E -pO o p ppdx p d dx p d D τδδμδ Assume the solution is of the form()sx A p exp =δ Then()()sx As dxp d exp =δ, ()()sx As dx p d exp 222=δ Substituting into the differential equation()()sx As sx As D o p p exp exp 2E -μ ()0exp =-pOsx A τor012=-E -pOo p p s s D τμDividing by p D , we have 0122=-E -ppop L s D s μ The solution for s is⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+⎪⎪⎭⎫ ⎝⎛E ±E =22421p o p p o p pL D D s μμ which can be rewritten as⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+⎪⎪⎭⎫ ⎝⎛E ±E =12212p o p p pop p p D L D L L s μμ Define po p p D L 2E ≡μβThen ⎥⎦⎤⎢⎣⎡+±=211ββpL s In order that p δ0=as +∞→x , use the minus sign for 0>x and the plus sign for0<x . Then the solution is ()x s A p -=exp δ for 0>x ()x s A p +=exp δ for 0<x where⎥⎦⎤⎢⎣⎡+±=±211ββp L s _____________________________________________________________________________________________________________________________ 6.23 Plot_______________________________________6.24 (a) From Equation (6.55)()()022=-E +nO on n ndx n d dx n d D τδδμδ or()()0222=-E +no n n L ndx n d D dx n d δδμδ We have that⎪⎭⎫⎝⎛=ekTD n n μ so we can define ()L e kT D o o nn '≡E =E 1μ Then we can write()()01222=-⋅'+nL ndx n d L dx n d δδδ The solution is of the form()()x n n αδδ-=exp 0 where 0>α Then()()n dxn d δαδ-= and ()()n dxn d δαδ222= Substituting into the differential equation, wefind()()[]0122=--'+nL nn L n δδαδαor0122=-'-nL L ααwhich yields⎥⎥⎦⎤⎢⎢⎣⎡+⎪⎪⎭⎫ ⎝⎛'+'=12212L L L L L n nn α We may note that if 0=E o , then ∞→'Land nL 1=α(b)nO n n D L τ= where ⎪⎭⎫ ⎝⎛=e kT D n n μso ()()1.310259.01200==n D cm 2/s and ()()47104.391051.31--⨯=⨯=n L cmorμ4.39=n L m For 12=E o V/cm, then ()4106.21120259.0-⨯==E ='oe kT L cmand21075.5⨯=αcm 1-(c) Force on the electrons due to the electric field is in the negative x-direction. Therefore, the effective diffusion of the electrons is reduced and the concentration drops off faster with the applied electric field. _______________________________________ 6.25p-type so the minority carriers are electrons and()()()t n n g n n D nO n n ∂∂=-'+∇∙E +∇δτδδμδ2 Uniform illumination means that()()02=∇=∇n n δδ. For ∞=nO τ, we areleft with()g dtn d '=δ which gives 1C t g n +'=δFor 0≤t , 001=⇒=C n δ Thent G n o'=δ for T t ≤≤0 For T t >, 0='g so that()0=dtn d δ AndT G n o'=δ (no recombination) _______________________________________ 6.26______________________________________________________________________________________ n-type, so minority carriers are holes and ()()()t p p g p p D pO p p ∂∂=-'+∇∙E -∇δτδδμδ2 We have ∞=pO τ , 0=E , and()0=∂∂tp δ(steady-state). Then we have ()022='+g dxp d D p δ or ()pD g dx p d '-=22δFor L x L +<<-, oG g '='= constant. Then()1C x D G dx p d p o +'-=δand 2122C x C x D G p po++'-=δ For L x L 3<<, 0='g so we have()022=dx p d δ so that()3C dxp d =δ and 43C x C p +=δFor L x L -<<-3, 0='g so that()022=dx p d δ so that()5C dxp d =δ and 65C x C p +=δThe boundary conditions are: (1) 0=p δ at L x 3+= (2) 0=p δat L x 3-=(3) p δ continuous at L x = (4) p δ continuous at L x -=(5)()dx p d δ continuous at L x = (6) ()dxp d δ continuous at L x -= Applying the boundary conditions, we find()2252x L D G p po -'=δ forL x L +<<-()x L D L G p po-'=3δ for L x L 3<< ()x L D L G p po+'=3δ for L x L -<<-3 _______________________________________6.27 204.080===E L V V/cm ()()60010322025.0-⨯=E =t d p μ 6.390=cm 2/V-s ()()2216t t D pp ∆E =μ()()[]()()62621032161035.9206.390--⨯⨯=42.10=p D cm 2/sWe find02668.06.39042.10==p p D μV This value is very close to 0.0259 for 300=T K._______________________________________ 6.28 (a)Assume that()()⎪⎪⎭⎫ ⎝⎛-=-Dt x Dt t x f 4exp 4,22/1π is the solution to the differential equationt fx f D ∂∂=⎪⎪⎭⎫ ⎝⎛∂∂22 To prove: we can write()⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-=∂∂-Dt x Dt x Dt x f 4exp 42422/1π and()⎪⎪⎭⎫ ⎝⎛-⎢⎣⎡⎪⎭⎫⎝⎛-=∂∂-Dt x Dt x Dt x f 4exp 424222/122π______________________________________________________________________________________ ⎥⎥⎦⎤⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-+Dt x Dt 4exp 422 A lso()⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-=∂∂-Dt x t D x Dt t f 4exp 1442222/1π()⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-+--Dt x t D 4exp 21422/32/1π Substituting the expressions for 22x f∂∂ and t f ∂∂into the differential equation, we find0 = 0.Q.E.D.(b)Considerdx Dt x ⎰+∞∞-⎪⎪⎭⎫ ⎝⎛-4exp 2Let 2x u =, then dx x du ⋅=2 or udu x du dx 22==Let Dta 41= Nowdx Dt x dx Dt x ⎰⎰∞+∞∞-⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛-0224exp 24exp()()du au u du au u -=-=⎰⎰∞∞exp 1exp 2120t D a ππ4==Thent D t D dx Dt x t D πππ444exp 412=⎪⎪⎭⎫ ⎝⎛-⎰+∞∞-1= _______________________________________6.29Plot_______________________________________6.30 (a) ⎪⎪⎭⎫⎝⎛=-i o Fi F n n kT E E ln ()⎪⎪⎭⎫ ⎝⎛⨯⨯=1016105.1104ln 0259.0383225.0=eV(b) ()()7210105102-⨯⨯='==p g p n τδδ 1510=cm 3- ⎪⎪⎭⎫ ⎝⎛+=-i o Fi Fn n n n kT E E δln ()⎪⎪⎭⎫⎝⎛⨯+⨯=101516105.110104ln 0259.0 383865.0=eV ⎪⎪⎭⎫ ⎝⎛+=-i o Fp Fi n p p kT E E δln ()⎪⎪⎭⎫ ⎝⎛⨯≅1015105.110ln 0259.0 28768.0=eV (c) 383225.0383865.0-=-F Fn E E 000640.0=eV or 640.0=meV _______________________________________ 6.31 (a) p-type⎪⎪⎭⎫ ⎝⎛=-i o F Fi n p kT E E ln ()⎪⎪⎭⎫ ⎝⎛⨯⨯=1015105.1105ln 0259.0 or3294.0=-F Fi E E eV (b)______________________________________________________________________________________14105⨯==p n δδcm 3-and ()4152102105.4105105.1⨯=⨯⨯==o i o p n n cm 3-Then⎪⎪⎭⎫⎝⎛+=-i o Fi Fn n n n kT E E δln()⎪⎪⎭⎫⎝⎛⨯⨯+⨯=10144105.1105105.4ln 0259.0 or2697.0=-Fi Fn E E eV and⎪⎪⎭⎫⎝⎛+=-i o Fp Fi n p p kT E E δln()⎪⎪⎭⎫⎝⎛⨯⨯+⨯=101415105.1105105ln 0259.0 or3318.0=-Fp Fi E E eV_______________________________________ 6.32 (a) For n-type,()()Fi F Fi Fn F Fn E E E E E E ---=-⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛+=i o i o n n kT n n n kT ln ln δ ⎪⎪⎭⎫⎝⎛+=o o n n n kT δln So ()⎪⎪⎭⎫⎝⎛⨯+⨯=1515105105ln 0259.000102.0n δ⎪⎭⎫⎝⎛⨯=+⨯0259.000102.0exp 1051051515n δWhich yields 14102⨯≅n δcm 3-(b) ⎪⎪⎭⎫⎝⎛+=-i o Fi Fn n n n kT E E δln ()⎪⎪⎭⎫ ⎝⎛⨯⨯+⨯=101415105.1102105ln 0259.0 33038.0=eV(c) ⎪⎪⎭⎫⎝⎛≅-i Fp Fi n p kT E E δln()⎪⎪⎭⎫⎝⎛⨯⨯=1014105.1102ln 0259.02460.0=eV_______________________________________ 6.33(a) ⎪⎪⎭⎫⎝⎛≅-i Fi Fn n n kT E E δln or ⎥⎦⎤⎢⎣⎡-=kT E E n n Fi Fn i exp δ ()⎥⎦⎤⎢⎣⎡⨯=0259.0270.0exp 105.110141005.5⨯=cm 3-(b)⎪⎪⎭⎫⎝⎛+=-i o Fp Fi n p p kT E E δln()⎪⎪⎭⎫⎝⎛⨯⨯+⨯=101415105.11005.5106ln 0259.0 33618.0=eV (c) (i) ()()F Fi Fp Fi Fp F E E E E E E ---=-⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛+=i o i o n p kT n p p kT ln ln δ ⎪⎪⎭⎫⎝⎛+=o o p p p kT δln (ii) Fp F E E -()⎪⎪⎭⎫⎝⎛⨯⨯+⨯=1514151061005.5106ln 0259.0 310093.2-⨯=eV or 093.2=meV_____________________________________________________________________________________________________________________________ 6.34 (a) (i) ⎪⎪⎭⎫⎝⎛+=-i o Fi Fn n n n kT E E ln ()()()⎪⎪⎭⎫ ⎝⎛⨯=616108.11002.1ln 0259.0 58166.0=eV (ii) ⎪⎪⎭⎫ ⎝⎛≅-i Fp Fi n p kT E E δln ()⎪⎪⎭⎫ ⎝⎛⨯⨯=616108.11002.0ln 0259.047982.0=eV (b) (i) ()⎪⎪⎭⎫⎝⎛⨯⨯=-616108.1101.1ln 0259.0Fi Fn E E58361.0=eV (ii)()⎪⎪⎭⎫⎝⎛⨯⨯=-616108.1101.0ln 0259.0Fp Fi E E52151.0=eV _______________________________________ 6.35Quasi-Fermi level for minority carrier electrons:⎪⎪⎭⎫⎝⎛+=-i o Fi Fn n n n kT E E δln()4162621024.310108.1-⨯=⨯==o i o p n n cm 3-We have()⎪⎭⎫⎝⎛=501014x n δThen()⎥⎦⎤⎢⎣⎡⨯+⨯=--6144108.150101024.3ln x kT E E Fi Fn We findx (μm)(Fi Fn E E -) (eV)0 12 1020 50-0.581 +0.361 +0.379 +0.420 +0.438 +0.462Quasi-Fermi level for holes: we have ⎪⎪⎭⎫ ⎝⎛+=-i o Fp Fi n p p kT E E δln We have 1610=o p cm 3- and p n δδ=. We findx (μm)(Fp Fi E E -) (eV)0 50+0.58115 +0.58140 _______________________________________ 6.36 (a) We can write⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln and⎪⎪⎭⎫⎝⎛+=-i o Fp Fi n p p kT E E δln so that()()Fp F F Fi Fp Fi E E E E E E -=---⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛+=i o i o n p kT n p p kT ln ln δ or()kT p p p kT E E o o Fp F 01.0ln =⎪⎪⎭⎫⎝⎛+=-δ Then______________________________________________________________________________________()010.101.0exp ==+oop p p δ or⇒=010.0op pδlow injection, so that 12105⨯=p δcm 3-(b)⎪⎪⎭⎫ ⎝⎛≅-i Fi Fn n p kT E E δln()⎪⎪⎭⎫⎝⎛⨯⨯=1012105.1105ln 0259.0 or1505.0=-Fi Fn E E eV_______________________________________6.37Plot_______________________________________6.38(a) ⎪⎪⎭⎫⎝⎛≅-i Fp Fi n p kT E E δln ()⎪⎭⎫ ⎝⎛⨯=10105.1ln 0259.0p δ 1110=p δcm 3-,04914.0=-Fp Fi E E eV1210 10877.0 1310 16841.0 1410 0.22805 1510 0.28768 (b) ⎪⎪⎭⎫⎝⎛+=-i o Fi Fn n n n kT E E δln()⎪⎪⎭⎫⎝⎛⨯+⨯=1016105.1102ln 0259.0n δ 1110=n δcm 3-, 365273.0=-Fi Fn E E eV12100.365274 1310 0.365286 1410 0.36540215100.366536_______________________________________ 6.39 (a)()()()p p C n n C n np N C C R p n i t p n '++'+-=2()()()p p n n n np nO pO i '++'+-=ττ2Let i n p n ='='. For 0==p n nO pO i i nO i pO i n n n n R ττττ+-=+-=2 (b) We had defined the net generation rate as()R R g g R g o o '+-'+=-where o o R g =since these are the thermal equilibrium generation and recombination rates.If 0='g , then R R g '-=- andnOpO i n R ττ+-=' so that nOpO in R g ττ++=-Thus a negative recombination rate implies anet positive generation rate._______________________________________6.40 We have that()()()p p C n n C n np N C C R p n i t p n '++'+-=2()()()i nO i pO i n p n n n np +++-=ττ2If n n n o δ+= and n p p o δ+=, then()()()()i o nO i o pO i o o n n p n n n n n p n n R +++++-++=δτδτδδ2______________________________________________________________________________________ ()()()()i o nO i o pO i o o o o n n p n n n n n p n n p n +++++-+++=δτδτδδ22 If i n n <<δ, we can neglect ()2n δ: also 2i o o n p n =Then()()()i o nO i o pO o o n p n n p n n R ++++=ττδ(a) For n-type; O o p n >>, i o n n >> Then 7101+==pOn R τδs 1-(b) For intrinsic, i o o n p n == Then ()()i nO i pO i n n n n R 222ττδ+=or ⇒⨯+=+=--771051011nO pO n R ττδ 61067.1+⨯=nR δs 1- (c) For p-type; o o n p >>, i o n p >>Then6710210511+-⨯=⨯==nO n R τδs 1- _______________________________________ 6.41 (a) From Equation (6.56) ()022=-'+pO p p g dx p d D τδδSolution is of the form⎪⎪⎭⎫⎝⎛++⎪⎪⎭⎫ ⎝⎛-+'=p p pO L x B L x A g p exp exp τδAt +∞=x , pO g p τδ'= so that 0=B , Then⎪⎪⎭⎫⎝⎛-+'=p pO L x A g p exp τδ We have()()00===x x p p s dx p d D δδ We can write()p x L A dx p d -==0δ and ()A g p pO x +'==τδ0Then ()A g s L AD pO pp+'=-τSolving for A , we find s L D g s A pp pO +'-=τThe excess concentration is then()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫⎝⎛-⋅+-'=p p p pO L x s L D s g p exp 1τδ where ()()37101010--===pO p p D L τcm Now()()7211010-=p δ ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⋅+-⨯-p L x s s exp 101013 or ⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⋅+-=p L x s s p exp 10110414δ (i) For 0=s , 1410=p δcm 3- (ii) For 2000=s cm/s,⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--=p L x p exp 167.011014δ (iii) For ∞=s ,⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--=p L x p exp 11014δ______________________________________________________________________________________ (b) (i) For 0=s , ()14100=p δcm 3-(ii) For 2000=s cm/s,()1410833.00⨯=p δcm 3- (iii) For ∞=s , ()00=p δ _______________________________________6.42 ()()710525-⨯==nO n n D L τ4104.35-⨯=cm(a) At 0=x ,()()1572110105102=⨯⨯='-nO g τcm 3- or()15100='=nO g n τδcm 3-For 0>x()()0022222=-⇒=-nnO n L ndx n d n dx n d D δδτδδ The solution is of the form ⎪⎪⎭⎫ ⎝⎛++⎪⎪⎭⎫ ⎝⎛-=n n L x B L x A n exp exp δ At 0=x , ()B A n n +==0δδ At W x =,⎪⎪⎭⎫⎝⎛++⎪⎪⎭⎫ ⎝⎛-==n n L W B L W A n exp exp 0δ Solving these two equations, we find()()()n n L W L W n A 2exp 12exp 0+-+-=δand ()()n L W n B 2exp 10+-=δSubstituting into the general solution, we find()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛+=n n L W L W n n exp exp 0δδ()()⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎦⎤⎢⎣⎡---⎥⎦⎤⎢⎣⎡-+⨯n n L x W L x W exp exp which can be written as()⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡-=n n L W L x W n n sinh sinh 0δδ where ()15100=n δcm 3- and μ4.35=n L m (b) If ∞=nO τ, we have ()022=dx n d δ so the solution is of the form D Cx n +=δ Applying the boundary conditions, we find()⎥⎦⎤⎢⎣⎡-=W x n n 10δδ _______________________________________6.43For ∞=pO τ, we have()022=dxp d δ So the solution is of the formB Ax p +=δ At W x = ()()W x W x p p s dx p d D ===-δδor()B AW s A D p +=-which yields()sW D sAB p +-=At 0=x , the flux of excess holes is()A D dxp d D p x p-=-==01910δso that1819101010-=-=A cm 4-and()⎪⎭⎫⎝⎛+=+=W s sW s B 101010101818 The solution is now______________________________________________________________________________________⎪⎭⎫ ⎝⎛+-=s x W p 101018δ(a)For ∞=s ,()x p -⨯=-418102010δ cm 3- Then()dxp d eD J p p δ-= ()()()181********.1-⨯-=- or6.1=p J A/cm 2 (b)For 3102⨯=s cm/s,()x p -⨯=-418107010δ cm 3- A lso6.1=p J A/cm 2_______________________________________ 6.44For 0<<-x W()022='+on G dxn d D δ so that()1C x D G dx n d no +'-=δ and2122C x C x D G n no++'-=δ For W x <<0,()022=dx n d δ so that43C x C n +=δThe boundary conditions are (1) 0=s at W x -= so that()0=-=Wx dx n d δ (2) ∞=s at W x += so that ()0=W n δ(3) n δ continuous at 0=x(4) ()dxn d δ continuous at 0=x Applying the boundary conditions, we findn o D W G C C '-==31 and noD W G C C 242'+==Then for 0<<-x W()22222W W x x D G n no +--'=δand for W x +<<0()x W D WG n no -'=δ_______________________________________ 6.45Plot_______________________________________ 6.48 (a) GaAs:Ω=⨯==-66101022I V R ()ALR σ∆= and ()p e p n δμμσ+=∆()()13821010510510⨯=⨯='=-p g p τδcm 3-For 1610=d N cm 3-, from Figure 5.3, 7000≅n μcm 2/V-s,310≅p μcm 2/V-s()()()13191053107000106.1⨯+⨯=∆-σ 05848.0=(Ω-cm)1- Let μ20=W mThen ()()441041020--⨯⨯==W d A 81080-⨯=cm 2So ()()86108005848.010-⨯==L RWhich yields 21068.4-⨯=L cm (b) Silicon:Ω=610R , 13105⨯=p δcm 3- For 1610=d N cm 3-, from Figure 5.3,______________________________________________________________________________________ 1300≅n μcm 2/V-s,410≅p μcm 2/V-s()()()13191054101300106.1⨯+⨯=∆-σ01368.0=(Ω-cm)1- Let μ20=W mThen ()()441041020--⨯⨯==W d A 81080-⨯=cm 2So ()()86108001368.010-⨯==LRWhich yields 21009.1-⨯=L cm_______________________________________。
半导体物理与器件习题
第一章 固体晶格结构1．如图是金刚石结构晶胞，若a 是其晶格常数，则其原子密度是 。
2．所有晶体都有的一类缺陷是：原子的热振动，另外晶体中常的缺陷有点缺陷、线缺陷。
3．半导体的电阻率为10-3～109Ωcm 。
4．什么是晶体？晶体主要分几类？5．什么是掺杂？常用的掺杂方法有哪些？答：为了改变导电性而向半导体材料中加入杂质的技术称为掺杂。
常用的掺杂方法有扩散和离子注入。
6．什么是替位杂质？什么是填隙杂质？ 7．什么是晶格？什么是原胞、晶胞？第二章 量子力学初步1．量子力学的三个基本原理是三个基本原理能量量子化原理、波粒二相性原理、不确定原理。
2．什么是概率密度函数？3．描述原子中的电子的四个量子数是： 、 、 、 。
第三章 固体量子理论初步1．能带的基本概念⏹ 能带（energy band ）包括允带和禁带。
⏹ 允带（allowed band ）：允许电子能量存在的能量范围。
⏹ 禁带（forbidden band ）：不允许电子存在的能量范围。
⏹ 允带又分为空带、满带、导带、价带。
⏹ 空带（empty band ）：不被电子占据的允带。
⏹满带（filled band ）：允带中的能量状态（能级）均被电子占据。
导带：有电子能够参与导电的能带，但半导体材料价电子形成的高能级能带通常称为导带。
价带:由价电子形成的能带，但半导体材料价电子形成的低能级能带通常称为价带。
2．什么是漂移电流？漂移电流：漂移是指电子在电场的作用下的定向运动，电子的定向运动所产生的电流。
3．什么是电子的有效质量？晶格中运动的电子，在外力和内力作用下有： Ｆ总＝Ｆ外＋Ｆ内=ma, m 是粒子静止的质量。
Ｆ外=m*n a, m*n 称为电子的有效质量。
4．位于能带底的电子，其有效质量为正，位于能带顶电子，其有效质量为负。
5．在室温T=300K ，Si 的禁带宽度：Eg=1.12eV Ge 的禁带宽度：Eg=0.67eV GaAs 的禁带宽度：Eg=1.43eVEg 具有负温度系数，即T 越大，Eg 越小；Eg 反应了，在相同温度下，Eg 越大，电子跃迁到导带的能力越弱。
半导体物理与器件第十章双极晶体管
而B-C结处于反偏状态，因此在x=xB处过剩载流子浓度的 边界条件为：
nB xB nB x xB nB0 0 nB0 nB0
半导体物理与器件
利用上述边界条件，可以求得上述双极输运方程解得 一般形式中的系数为：
xB eVBE nB 0 nB 0 exp 1 exp kT L B A xB 2 sinh L B
由于B-E结处于正偏状态，因此在发射区中x=0处， 过剩少数载流子空穴浓的边界条件为：
pE 0 pE x' 0 pE 0
eVBE pE 0 exp kT
1
半导体物理与器件
而在发射区表面，复合速度为无穷大，因此在x’=xE处 边界条件为
半导体物理与器件
电子扩散通过基区后，进入反偏的B-C结空间电荷区， 被B-C结电场抽取进入搜集区，能够被拉向收集区的电 子数目取决于由发射区注入到基区中的电子数目（复 合掉的电子数目）。 流入到收集区中的电子数量（构成收集极电流）取决 于发射结上的偏置电压，此即双极型晶体管的放大作 用，即：BJT中流过一个端点的电流取决于另外两个端 点上的外加电压。 其他因素：发射 极空穴电流，基 区复合电流，集 电极反向漏电流
n++ E
P+
n C
半导体物理与器件
定性分析 热平衡和偏置状态
注意这里没有反映出各 个区杂质浓度的区别 正向有源区，电子的输 运过程
半导体物理与器件
B-E结正偏；B-C 结反偏；正向有 源模式
注意基区宽度 回忆：短二极管
半导体物理与器件
发射结正偏，电子扩散注入基区 B-C结反偏，基区中靠近B-C结边界处电子浓度为零。 基区中电子存在着较大的浓度梯度，因此电子可以通 过扩散流过基区，和正偏的PN结二极管类似，少子电 子在通过中性基区的过程中也会与其中的多子空穴发 生一定的复合。
半导体物理与器件第四版课后习题答案2
Chapter 22、1Sketch_______________________________________2、2Sketch_______________________________________2、3Sketch_______________________________________2、4From Problem 2、2, phase t xωλπ-=2= constant Then⎪⎭⎫ ⎝⎛+==⇒=-⋅πλωυωλπ2,02p dt dx dt dx From Problem 2、3, phase t xωλπ+=2 = constantThen ⎪⎭⎫⎝⎛-==⇒=+⋅πλωυωλπ2,02p dt dx dt dx _______________________________________2、5E hchc h E =⇒==λλν Gold: 90.4=E eV ()()19106.190.4-⨯= JSo, ()()()()51910341054.2106.190.410310625.6---⨯=⨯⨯⨯=λcm orμλ254.0=m Cesium: 90.1=E eV ()()19106.190.1-⨯= JSo,()()()()51910341054.6106.190.110310625.6---⨯=⨯⨯⨯=λcm orμλ654.0=m_______________________________________2、6(a) 9341055010625.6--⨯⨯==λhp2710205.1-⨯=kg-m/s 331271032.11011.9102045.1⨯=⨯⨯==--m p υm/s or 51032.1⨯=υcm/s(b) 9341044010625.6--⨯⨯==λh p 2710506.1-⨯=kg-m/s331271065.11011.9105057.1⨯=⨯⨯==--m p υm/s or 51065.1⨯=υcm/s (c) Yes_______________________________________ 2、7(a) (i) ()()()1931106.12.11011.922--⨯⨯==mE p 2510915.5-⨯=kg-m/s925341012.110915.510625.6---⨯=⨯⨯==p h λm or o A 2.11=λ(ii)()()()1931106.1121011.92--⨯⨯=p 241087.1-⨯=kg-m/s1024341054.3108704.110625.6---⨯=⨯⨯=λm or oA 54.3=λ(iii) ()()()1931106.11201011.92--⨯⨯=p 2410915.5-⨯=kg-m/s1024341012.110915.510625.6---⨯=⨯⨯=λm or oA 12.1=λ (b)()()()1927106.12.11067.12--⨯⨯=p 2310532.2-⨯=kg-m/s1123341062.210532.210625.6---⨯=⨯⨯=λm or oA 262.0=λ_______________________________________2、8()03885.00259.02323=⎪⎭⎫⎝⎛==kT E avg eVNowavg avg mE p 2=()()()1931106.103885.01011.92--⨯⨯= or2510064.1-⨯=avg p kg-m/sNow9253410225.610064.110625.6---⨯=⨯⨯==p h λm oroA 25.62=λ_______________________________________2、9pp p hch E λν==Nowmp E ee 22= and221⎪⎪⎭⎫ ⎝⎛=⇒=ee e e h m E hp λλ Set e p E E = and e p λλ10=Then22102121⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛=p ep h m hm hcλλλ which yieldsmchp 2100=λ100221002mc mc h hc hc E E p p =⋅===λ ()()1001031011.922831⨯⨯=-151064.1-⨯=J 25.10=keV _______________________________________2、10(a) 1034108510625.6--⨯⨯==λhp2610794.7-⨯= kg-m/s431261056.81011.910794.7⨯=⨯⨯==--m p υm/s or 61056.8⨯=υcm/s()()243121056.81011.92121⨯⨯==-υm E211033.3-⨯=Jor 219211008.2106.110334.3---⨯=⨯⨯=E eV (b) ()()23311081011.921⨯⨯=-E2310915.2-⨯=Jor 419231082.1106.110915.2---⨯=⨯⨯=E eV ()()3311081011.9⨯⨯==-υm p2710288.7-⨯=kg-m/s827351009.910288.710625.6---⨯-⨯⨯==p h λmor oA 909=λ_______________________________________2、11(a) ()()1083410110310625.6--⨯⨯⨯===λνhch E 151099.1-⨯=J Now1915106.11099.1--⨯⨯==⇒⋅=e E V V e E41024.1⨯=V V 4.12=kV (b)()()15311099.11011.922--⨯⨯==mE p231002.6-⨯=kg-m/s Then1123341010.11002.610625.6---⨯=⨯⨯==p h λm oroA 11.0=λ_______________________________________2、126341010054.1--⨯=∆=∆x p 2810054.1-⨯=kg-m/s_______________________________________2、13(a) (i) =∆∆x p26103410783.8101210054.1---⨯=⨯⨯=∆p kg-m/s (ii)p m p dp d p dp dE E ∆⋅⎪⎪⎭⎫⎝⎛=∆⋅=∆22 mpp p m p ∆=∆⋅=22 Now mE p 2=()()()1931106.1161092--⨯⨯= 2410147.2-⨯=kg-m/s so ()()31262410910783.8101466.2---⨯⨯⨯=∆E1910095.2-⨯=Jor 31.1106.110095.21919=⨯⨯=∆--E eV(b) (i) 2610783.8-⨯=∆p kg-m/s (ii)()()()1928106.1161052--⨯⨯=p 231006.5-⨯=kg-m/s()()28262310510783.81006.5---⨯⨯⨯=∆E 2110888.8-⨯=Jor 219211055.5106.110888.8---⨯=⨯⨯=∆E eV _______________________________________2、143223410054.11010054.1---⨯=⨯=∆=∆x p kg-m/s150010054.132-⨯=∆=∆⇒=m p m p υυ 36107-⨯=∆υm/s_______________________________________2、15(a) =∆∆t E()()1619341023.8106.18.010054.1---⨯=⨯⨯=∆t s (b) 1034105.110054.1--⨯⨯=∆=∆x p 251003.7-⨯=kg-m/s_______________________________________2、16(a) If ()t x ,1ψ and ()t x ,2ψ are solutionstoSchrodinger's wave equation, then()()()()t t x j t x x V x t x m ∂ψ∂=ψ+∂ψ∂⋅-,,,2112122 and()()()()t t x j t x x V x t x m ∂ψ∂=ψ+∂ψ∂⋅-,,,2222222 Adding the two equations, we obtain()()[]t x t x x m ,,221222ψ+ψ∂∂⋅- ()()()[]t x t x x V ,,21ψ+ψ+()()[]t x t x tj ,,21ψ+ψ∂∂=which is Schrodinger's wave equation 、 So ()()t x t x ,,21ψ+ψ is also a solution 、(b) If ()()t x t x ,,21ψ⋅ψ were a solution toSchrodinger's wave equation, then we could write []()[]21212222ψ⋅ψ+ψ⋅ψ∂∂⋅-x V x m[]21ψ⋅ψ∂∂=tjwhich can be written as⎥⎦⎤⎢⎣⎡∂ψ∂⋅∂ψ∂+∂ψ∂ψ+∂ψ∂ψ-x x x x m 2121222221222()[]⎥⎦⎤⎢⎣⎡∂ψ∂ψ+∂ψ∂ψ=ψ⋅ψ+t t j x V 122121 Dividing by 21ψ⋅ψ, we find⎥⎦⎤⎢⎣⎡∂ψ∂∂ψ∂ψψ+∂ψ∂⋅ψ+∂ψ∂⋅ψ-x x x xm21212121222222112 ()⎥⎦⎤⎢⎣⎡∂ψ∂ψ+∂ψ∂ψ=+t t j x V 112211Since 1ψ is a solution, then()tj x V x m ∂ψ∂⋅ψ⋅=+∂ψ∂⋅ψ⋅-1121212112Subtracting these last two equations, we have⎥⎦⎤⎢⎣⎡∂ψ∂∂ψ∂ψψ+∂ψ∂⋅ψ-x x x m 212122222212t j ∂ψ∂⋅ψ⋅=221 Since 2ψ is also a solution, we have()t j x V x m ∂ψ∂⋅ψ⋅=+∂ψ∂⋅ψ⋅-2222222112 Subtracting these last two equations, we obtain()02221212=-∂ψ∂⋅∂ψ∂⋅ψψ⋅-x V xx mThis equation is not necessarily valid, which means that 21ψψ is, in general, not a solutionto Schrodinger's wave equation 、_______________________________________2、1712cos 2312=⎪⎭⎫⎝⎛⎰+-dx x A π()12sin 2312=⎥⎦⎤⎢⎣⎡++-ππx x A 121232=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛--Aso 212=Aor 21=A_______________________________________2、18()1cos 22/12/12=⎰+-dx x n A π()142sin 22/12/12=⎥⎦⎤⎢⎣⎡++-ππn x n x A ⎪⎭⎫⎝⎛==⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛--211414122A Aor 2=A_______________________________________2、19Note that 10*=ψ⋅ψ⎰∞dxFunction has been normalized 、 (a) Nowdx a x a P oa o o 24exp 2⎰⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-=dx a x a oa o o⎰⎪⎪⎭⎫⎝⎛-=42exp 2402exp 22o a o o o a x a a ⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫⎝⎛-=or()⎪⎭⎫ ⎝⎛--=⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎪⎭⎫⎝⎛--=21exp 1142exp 1o oa a P which yields 393.0=P (b)dx a x a P o oa a o o 224exp 2⎰⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-=dx a x a o oa a o o⎰⎪⎪⎭⎫⎝⎛-=242exp 22exp 22o oa a o o o a x a a ⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫⎝⎛-=or()()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛----=21exp 1exp 1Pwhich yields239.0=P (c)dx a x a P oa o o 20exp 2⎰⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-= dx a x a oa o o⎰⎪⎪⎭⎫⎝⎛-=2exp 2o a o o oa x a a 02exp 22⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫⎝⎛-=()()[]12exp 1---= which yields 865.0=P_______________________________________2、20()dx x P 2⎰=ψ(a)dx x a a ⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛⎰2cos 224/0π 4/042sin 22a a a x x a ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛=ππ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=a a a ππ42sin 242 ()()⎥⎦⎤⎢⎣⎡+⎪⎭⎫ ⎝⎛=π4182a a a or 409.0=P(b) dx a x a P a a ⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛=⎰π2/4/2cos 2 2/4/42sin 22a a a a x x a ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛=ππ()⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛=a a a a a ππππ42sin 84sin 42 ⎥⎦⎤⎢⎣⎡--+=π41810412or 0908.0=P(c) dx a x a P a a ⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛=⎰+-π22/2/cos 2 2/2/42sin 22a a a a x x a +-⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛=ππ()()⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛=a a a a a ππππ4sin 44sin 42 or 1=P_______________________________________2、21(a) dx a x a P a ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=⎰π2sin 224/04/0244sin 22a a a x x a ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=ππ()⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=a a a ππ8sin 82or 25.0=P(b) dx a x a P a a ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=⎰π2sin 222/4/ 2/4/244sin 22a a a a x x a ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=ππ()()⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=a a a a a ππππ8sin 882sin 42 or 25.0=P(c) dx a x a P a a ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=⎰+-π2sin 222/2/ 2/2/244sin 22a a a a x x a +-⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=ππ()()⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛-+⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=a a a a a ππππ82sin 482sin 42 or 1=P_______________________________________2、22(a) (i) 481210108108=⨯⨯==k p ωυm/s or 610=p υcm/s9810854.710822-⨯=⨯==ππλk mor oA 54.78=λ (ii)()()431101011.9-⨯==υm p271011.9-⨯=kg-m/s()()24312101011.92121-⨯==υm E2310555.4-⨯=Jor 419231085.2106.110555.4---⨯=⨯⨯=E eV (b) (i) 491310105.1105.1-=⨯-⨯==k p ωυm/s or 610-=p υcm/s991019.4105.122-⨯=⨯==ππλk m or oA 9.41=λ(ii) 271011.9-⨯-=p kg-m/s41085.2-⨯=E eV_______________________________________2、23(a) ()()t kx j Ae t x ω+-=ψ,(b) ()()21921106.1025.0υm E =⨯=- ()2311011.921υ-⨯= so 41037.9⨯=υm/s 61037.9⨯=cm/sFor electron traveling in x -direction, 61037.9⨯-=υcm/s()()4311037.91011.9⨯-⨯==-υm p2610537.8-⨯-=kg-m/s926341076.710537.810625.6---⨯=⨯⨯==p h λm8910097.81076.722⨯=⨯==-πλπk m 1- ()()481037.910097.8⨯⨯=⋅=υωkor 1310586.7⨯=ωrad/s_______________________________________2、24(a) ()()4311051011.9⨯⨯==-υm p 2610555.4-⨯=kg-m/s8263410454.110555.410625.6---⨯=⨯⨯==p h λm881032.410454.122⨯=⨯==-πλπk m 1- ()()481051032.4⨯⨯==υωk131016.2⨯=rad/s (b) ()()631101011.9-⨯=p251011.9-⨯=kg-m/s1025341027.71011.910625.6---⨯=⨯⨯=λm 9101064.810272.72⨯=⨯=-πk m 1- ()()15691064.8101064.8⨯=⨯=ωrad/s _______________________________________2、25()()()2103122342222210751011.9210054.12---⨯⨯⨯==ππn ma n E n()212100698.1-⨯=n E n Jor()19212106.1100698.1--⨯⨯=n E nor ()3210686.6-⨯=n E n eV Then311069.6-⨯=E eV221067.2-⨯=E eV231002.6-⨯=E eV_______________________________________2、26(a) ()()()2103122342222210101011.9210054.12---⨯⨯⨯==ππn ma n E n ()20210018.6-⨯=n J or()()3761.0106.110018.6219202n n E n =⨯⨯=--eV Then376.01=E eV 504.12=E eV 385.33=E eV(b) Ehc ∆=λ ()()19106.1504.1385.3-⨯-=∆E191001.3-⨯=J()()198341001.310310625.6--⨯⨯⨯=λ710604.6-⨯=m or 4.660=λnm_______________________________________2、27(a) 22222ma n E n π =()()()223223423102.11015210054.11015----⨯⨯⨯=⨯πn()622310538.21015--⨯=⨯n or 2910688.7⨯=n (b) 151≅+n E mJ (c) No_______________________________________2、28For a neutron and 1=n :()()()2142722342221101066.1210054.12---⨯⨯==ππma E13103025.3-⨯=Jor6191311006.2106.1103025.3⨯=⨯⨯=--E eV For an electron in the same potential well: ()()()2143122341101011.9210054.1---⨯⨯=πE10100177.6-⨯=J or9191011076.3106.1100177.6⨯=⨯⨯=--E eV _______________________________________2、29Schrodinger's time-independent waveequation()()()()02222=-+∂∂x x V E mx x ψψWe know that()0=x ψ for 2a x ≥ and 2ax -≤We have()0=x V for 22a x a +<<-so in this region()()02222=+∂∂x mEx x ψψThe solution is of the form ()kx B kx A x sin cos +=ψ where22 mEk =Boundary conditions:()0=x ψ at 2,2a x a x +=-= First mode solution: ()x k A x 111cos =ψ where222112ma E a k ππ=⇒=Second mode solution: ()x k B x 222sin =ψ where22222242ma E a k ππ=⇒= Third mode solution: ()x k A x 333cos =ψ where22233293ma E a k ππ=⇒= Fourth mode solution: ()x k B x 444sin =ψ where222442164ma E a k ππ=⇒= _______________________________________2、30The 3-D time-independent wave equation incartesian coordinates for ()0,,=z y x V is:()()()222222,,,,,,z z y x y z y x x z y x ∂∂+∂∂+∂∂ψψψ ()0,,22=+z y x mEψUse separation of variables, so let ()()()()z Z y Y x X z y x =,,ψSubstituting into the wave equation, we obtain222222zZXY y Y XZ x X YZ ∂∂+∂∂+∂∂ 022=+XYZ mEDividing by XYZ and letting222 mEk =, wefind(1) 01112222222=+∂∂⋅+∂∂⋅+∂∂⋅k zZ Z y Y Y x X XWe may set01222222=+∂∂⇒-=∂∂⋅X k xX k x X X x x Solution is of the form()()()x k B x k A x X x x cos sin += Boundary conditions: ()000=⇒=B X and ()an k a x X x x π=⇒==0 where ....3,2,1=x n Similarly, let2221y k y Y Y -=∂∂⋅ and 2221z k zZ Z -=∂∂⋅Applying the boundary conditions, we finda n k y y π=, ....3,2,1=y nan k z z π=, ...3,2,1=z nFrom Equation (1) above, we have02222=+---k k k k z y x or222222 mEk k k k z y x ==++so that()2222222z y x n n n n n n maE E z y x ++=→π _______________________________________2、31 (a)()()()0,2,,22222=⋅+∂∂+∂∂y x mEy y x x y x ψψψSolution is of the form:()y k x k A y x y x sin sin ,⋅=ψWe find()y k x k Ak x y x y x x sin cos ,⋅=∂∂ψ ()y k x k Ak xy x y x x sin sin ,222⋅-=∂∂ψ()y k x k Ak y y x y x y cos sin ,⋅=∂∂ψ()y k x k Ak yy x y x y sin sin ,222⋅-=∂∂ψSubstituting into the original equation, we find:(1) 02222=+--mE k k y xFrom the boundary conditions, 0sin =a k A x , where oA a 40= So an k x x π=, ...,3,2,1=x n Also 0sin =b k A y , where oA b 20= So bn k y y π=, ...,3,2,1=y n Substituting into Eq 、 (1) above⎪⎪⎭⎫ ⎝⎛+=22222222b n an m E y x n n yx ππ (b)Energy is quantized - similar to 1-D result 、There can be more than one quantum stateper given energy - different than 1-D result 、_______________________________________2、32(a) Derivation of energy levels exactly thesame as in the text(b) ()21222222n n maE -=∆π For 1,212==n n Then22223ma E π =∆(i) For oA a 4= ()()()2102722341041067.1210054.13---⨯⨯⨯=∆πE2210155.6-⨯=Jor 319221085.3106.110155.6---⨯=⨯⨯=∆E eV(ii) For 5.0=a cm()()()22272234105.01067.1210054.13---⨯⨯⨯=∆πE3610939.3-⨯=Jor1719361046.2106.110939.3---⨯=⨯⨯=∆E eV _______________________________________2、33(a) For region II, 0>x()()()0222222=-+∂∂x V E mx x O ψψGeneral form of the solution is()()()x jk B x jk A x 22222exp exp -+=ψ where()O V E mk -=222 Term with 2B represents incident wave andterm with 2A represents reflected wave 、 Region I, 0<x()()0212212=+∂∂x mEx x ψψGeneral form of the solution is()()()x jk B x jk A x 11111exp exp -+=ψ where212 mEk =Term involving 1B represents the transmitted wave and the term involving 1A represents reflected wave: but if a particle is transmitted into region I, it will not be reflected so that 01=A 、 Then()()x jk B x 111exp -=ψ()()()x jk B x jk A x 22222exp exp -+=ψ (b)Boundary conditions: (1) ()()0021===x x ψψ(2) 0201==∂∂=∂∂x x x x ψψ Applying the boundary conditions to the solutions, we find221B A B +=112222B k B k A k -=-Combining these two equations, we find212122B k k k k A ⋅⎪⎪⎭⎫⎝⎛+-=212212B k k k B ⋅⎪⎪⎭⎫⎝⎛+=The reflection coefficient is21212*22*22⎪⎪⎭⎫ ⎝⎛+-==k k k k B B A A R The transmission coefficient is()2212141k k k k T R T +=⇒-=_______________________________________2、34()()x k A x 222exp -=ψ()()x k A A x P 2*2222exp -==ψwhere ()222 E V m k o -=()()()34193110054.1106.18.25.31011.92---⨯⨯-⨯=9210286.4⨯=k m 1- (a) For 101055-⨯==oA x m ()x k P 22exp -=()()[]109105102859.42exp -⨯⨯-= 0138.0=(b) For 10101515-⨯==oA x m()()[]1091015102859.42exp -⨯⨯-=P61061.2-⨯= (c) For 10104040-⨯==oA x m()()[]1091040102859.42exp -⨯⨯-=P151029.1-⨯=_______________________________________2、35()a k VE VET o o22exp 116-⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛≅ where ()222 E V m k o -=()()()34193110054.1106.11.00.11011.92---⨯⨯-⨯=or 2k 910860.4⨯=m 1-(a) For 10104-⨯=a m()()[]1091041085976.42exp 0.11.010.11.016-⨯⨯-⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛≅T 0295.0=(b) For 101012-⨯=a m()()[]10910121085976.42exp 0.11.010.11.016-⨯⨯-⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛≅T 51024.1-⨯=(c) υe N J t =, where t N is the density of transmitted electrons 、 1.0=E eV 20106.1-⨯=J ()23121011.92121υυ-⨯==m510874.1⨯=⇒υm/s 710874.1⨯=cm/s()()719310874.1106.1102.1⨯⨯=⨯--t N810002.4⨯=t N electrons/cm 3 Density of incident electrons,10810357.10295.010002.4⨯=⨯=i N cm 3-_______________________________________2、36()a k VEV E T O O 22exp 116-⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛≅ (a) For ()o m m 067.0=()222 E V m k O -=()()()()()2/1234193110054.1106.12.08.01011.9067.02⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⨯⨯-⨯=---or9210027.1⨯=k m 1- Then⎪⎭⎫⎝⎛-⎪⎭⎫ ⎝⎛=8.02.018.02.016T()()[]109101510027.12exp -⨯⨯-⨯ or138.0=T (b) For ()o m m 08.1=2k =()()()()()2/1234193110054.1106.12.08.01011.908.12⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⨯⨯-⨯---or9210124.4⨯=k m 1- Then⎪⎭⎫⎝⎛-⎪⎭⎫ ⎝⎛=8.02.018.02.016T()()[]109101510124.42exp -⨯⨯-⨯or51027.1-⨯=T_______________________________________2、37()a k VE VET o o22exp 116-⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛≅ where ()222 E V m k o -=()()()341962710054.1106.1101121067.12---⨯⨯⨯⨯-⨯=1410274.7⨯=m 1- (a)()()[]14141010274.72exp 121112116-⨯-⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛≅T[]548.14exp 222.1-=710875.5-⨯= (b)()()710875.510-⨯=T()[]a 1410274.72exp 222.1⨯-=()⎪⎭⎫⎝⎛⨯=⨯-61410875.5222.1ln 10274.72a or 1410842.0-⨯=a m_______________________________________2、38Region I ()0<x , 0=V ;Region II ()a x <<0, O V V = Region III ()a x >, 0=V (a) Region I:()()()x jk B x jk A x 11111exp exp -+=ψ(incident) (reflected)where212 mEk =Region II:()()()x k B x k A x 22222exp exp -+=ψwhere ()222 E V m k O -=Region III:()()()x jk B x jk A x 13133exp exp -+=ψ (b)In Region III, the 3B term represents areflected wave 、 However, once a particleis transmitted into Region III, there will not be a reflected wave so that 03=B 、(c) Boundary conditions:At 0=x : ⇒=21ψψ2211B A B A +=+ ⇒=dx d dx d 21ψψ22221111B k A k B jk A jk -=-At a x =: ⇒=32ψψ()()a k B a k A 2222exp exp -+()a jk A 13exp =⇒=dxd dx d 32ψψ()()a k B k a k A k 222222exp exp -- ()a jk A jk 131exp = The transmission coefficient is defined as*11*33A A A A T = so from the boundary conditions, wewant to solve for 3A in terms of 1A 、Solvingfor 1A in terms of 3A , we find(){()()[]a k a k k k k k jA A 2221222131exp exp 4---+= ()()[]}a k a k k jk 2221exp exp 2-+-()a jk 1exp ⨯We then find()(){()[a k k k k k A A A A 22122221*33*11exp 4-=()]22exp a k --()()[]}2222221exp exp 4a k a k k k -++ We have()222 E V m k O -= If we assume that E V O >>, then a k 2 will be large so that ()()a k a k 22exp exp ->> We can then write()(){()[]222122221*33*11exp 4a k k k k k A A A A -= ()[]}222221exp 4a k k k + which becomes()()()a k k k k k A A A A 22122221*33*112exp 4+= Substituting the expressions for 1k and2k , we find222212 O mV k k =+ and()⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡-=22222122 mE E V m k k O ()()E E V m O -⎪⎭⎫ ⎝⎛=222()()E V E V m O O ⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛=1222 Then()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛⎪⎪⎭⎫⎝⎛=E V E V m a k mV A A A A O O O12162exp 222222*33*11()a k V EV E A A O O 2*332exp 116-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛=Finally,()a k V E V E A A A A T O O 2*11*332exp 116-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛== _____________________________________2、39 Region I: 0=V ()()⇒=+∂∂0212212x mE x x ψψ()()()x jk B x jk A x 11111exp exp -+=ψincidentreflectedwhere212 mEk = Region II: 1V V = ()()()⇒=-+∂∂02221222x V E m x x ψψ ()()()x jk B x jk A x 22222exp exp -+=ψ transmitted reflectedwhere()2122V E m k -= Region III: 2V V =()()()⇒=-+∂∂02322232x V E m x x ψψ()()x jk A x 333exp =ψtransmitted where ()2232 V E m k -=There is no reflected wave in Region III 、 The transmission coefficient is defined as:*11*3313*11*3313A A A A k k A A A A T ⋅=⋅=υυ From the boundary conditions, solve for 3Ain terms of 1A 、 The boundary conditions are:At 0=x : ⇒=21ψψ2211B A B A +=+⇒∂∂=∂∂xx 21ψψ22221111B k A k B k A k -=-At a x =: ⇒=32ψψ ()()a jk B a jk A 2222exp exp -+ ()a jk A 33exp =⇒∂∂=∂∂xx 32ψψ()()a jk B k a jk A k 222222exp exp --()a jk A k 333exp = But ⇒=πn a k 22 ()()1exp exp 22=-=a jk a jk Then, eliminating 1B , 2A , 2B from the boundary condition equations, we find()()23131231211344k k k k k k k k k T +=+⋅= _______________________________________2、40 (a) Region I: Since E V O >, we can write()()()0212212=--∂∂x E V m x x O ψψRegion II: 0=V , so()()0222222=+∂∂x mEx x ψψRegion III: 03=⇒∞→ψV The general solutions can be written, keeping in mind that 1ψ must remain finite for 0<x , as()()x k B x 111exp =ψ()()()x k B x k A x 22222cos sin +=ψ ()03=x ψ where()212 E V m k O -= and222 mEk =(b) Boundary conditionsAt 0=x : ⇒=21ψψ21B B = 221121A k B k xx =⇒∂∂=∂∂ψψ At a x =: ⇒=32ψψ ()()0cos sin 2222=+a k B a k A or()a k A B 222tan -= (c)12122211B k k A A k B k ⎪⎪⎭⎫⎝⎛=⇒=and since 21B B =, then2212B k k A ⎪⎪⎭⎫⎝⎛=From ()a k A B 222tan -=, we can write ()a k B k k B 22212tan ⎪⎪⎭⎫⎝⎛-=or()a k k k 221tan 1⎪⎪⎭⎫⎝⎛-=This equation can be written as⎥⎥⎦⎤⎢⎢⎣⎡⋅⋅--=a mE E EV O 22tan 1 or⎥⎥⎦⎤⎢⎢⎣⎡⋅-=-a mE E V EO 22tan This last equation is valid only forspecific values of the total energy E 、 The energy levels are quantized 、_______________________________________2、41 ()222424ne m E o o n ∈-=π(J)()222324n e m o o ∈-=π(eV)()()()[]()22342123193110054.121085.84106.11011.9n----⨯⨯⨯⨯-=πor258.13n E n -= (eV) 58.1311-=⇒=E n eV 395.322-=⇒=E n eV 51.133-=⇒=E n eV 849.044-=⇒=E n eV_______________________________________2、42We have⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫⎝⎛⋅=o oa r a exp 112/3100πψ and*10010024ψψπr P =⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫⎝⎛⋅⋅=o o a r a r 2exp 11432ππ or()⎪⎪⎭⎫ ⎝⎛-⋅=o o a r r a P 2exp 423 To find the maximum probability()0=drr dP()()⎪⎪⎭⎫ ⎝⎛-⎪⎩⎪⎨⎧⎪⎪⎭⎫ ⎝⎛-=o o o a r r a a 2exp 2423 ⎪⎭⎪⎬⎫⎪⎪⎭⎫ ⎝⎛-+o a r r 2exp 2 which giveso oa r a r=⇒+-=10 or o a r = is the radius that gives the greatest probability 、 _______________________________________2、43100ψ is independent of θ and φ, so the wave equation in spherical coordinates reduces to ()()021222=-+⎪⎭⎫ ⎝⎛∂∂∂∂⋅ψψr V E m r r r r o where()r a m r e r V o o o 224 -=∈-=π For ⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛⋅=o o a r a exp 112/3100πψ Then ⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛⋅=∂∂o o o a r a a r exp 1112/3100πψ so⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛⋅-=∂∂o o a r r a r r exp 1122/51002πψ We then obtain2/5100211⎪⎪⎭⎫⎝⎛⋅-=⎪⎪⎭⎫ ⎝⎛∂∂∂∂o a r r r πψ ⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-⨯o o o a r a r a r r exp exp 22 Substituting into the wave equation, we have⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛⋅-o o o o a r a r a r r a r exp exp 21122/52π⎥⎦⎤⎢⎣⎡++r a m E m o o o 2220exp 112/3=⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛⋅⎪⎪⎭⎫ ⎝⎛⨯o o a r a π where ()222241224oo o o a m e m E E -=∈-==π Then the above equation becomes⎪⎩⎪⎨⎧⎥⎦⎤⎢⎣⎡--⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛⋅o o o o a r r a r a r a 222/321exp 11π 022222=⎪⎭⎪⎬⎫⎪⎪⎭⎫ ⎝⎛+-+r a m a m m o o o o oor ⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛⋅o o a r a exp 112/3π0211222=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎪⎭⎫ ⎝⎛+-++-⨯r a a a r a o o o o which gives 0 = 0 and shows that 100ψ is indeed a solution to the wave equation 、 _______________________________________ 2、44All elements are from the Group I column ofthe periodic table 、 All have one valence electron in the outer shell 、 _______________________________________。
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课后习题答案3
Chapter 33.1If o a were to increase, the bandgap energy would decrease and the material would begin to behave less like a semiconductor and more like 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 xt x m ,,2222ψ⋅+∂ψ∂- ()tt x j ∂ψ∂=, Assume the solution is of the form:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Region I: ()0=x V . Substituting theassumed solution into the wave equation, we obtain:()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧∂∂-t E kx j x jku x m exp 22 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⋅⎪⎭⎫ ⎝⎛-=t E kx j x u jE j exp which becomes()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+=t E kx j x Eu exp This equation may be written as()()()()0222222=+∂∂+∂∂+-x u mE x x u x x u jk x u kSetting ()()x u x u 1= for region I, the equation becomes:()()()()021221212=--+x u k dx x du jk dxx u d α where222mE=α Q.E.D.In Region II, ()O V x V =. Assume the same form of the solution:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Substituting into Schrodinger's wave equation, we find:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+t E kx j x u V O exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=t E kx j x Eu exp This equation can be written as:()()()2222x x u x x u jk x u k ∂∂+∂∂+- ()()02222=+-x u mEx u mV OSetting ()()x u x u 2= for region II, this equation becomes()()dx x du jk dxx u d 22222+ ()022222=⎪⎪⎭⎫ ⎝⎛+--x u mV k O α where again222mE=α Q.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 +-+αexp The first derivative is()()()[]x k j A k j dxx du --=ααexp 1 ()()[]x k j B k j +-+-ααexp and 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 the proposed 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 β The 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 and can be written as()[]()[]a k j B a k j A +-+-ααexp exp ()[]b k j C ---βexp()[]0exp =+-b k j D β The 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 =+'αααLet y ka =, x a =α Theny x x xP cos cos sin =+'Consider dy dof this function.()[]{}y x x x P dy d sin cos sin 1-=+⋅'- We find()()()⎭⎬⎫⎩⎨⎧⋅+⋅-'--dy dx x x dy dx x x P cos sin 112y dydxx 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,()()dk d ka d a d dy dxαα===0 And 22 mE=α Sodk dEm mE dk d ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-22/122221 α This implies thatdk dE dk d ==0α for an k π= _______________________________________ 3.8(a) πα=a 1π=⋅a E m o 212()()()()2103123422221102.41011.9210054.12---⨯⨯⨯==ππa m E o19104114.3-⨯=J From Problem 3.5 πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J 12E E E -=∆1918104114.3100198.1--⨯-⨯= 19107868.6-⨯=Jor 24.4106.1107868.61919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J From Problem 3.5, πα617.24=aπ617.2224=⋅a E m o()()()()2103123424102.41011.9210054.1617.2---⨯⨯⨯=πE18103364.2-⨯=J 34E 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 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα859.0=a o ()()()()210312342102.41011.9210054.1859.0---⨯⨯⨯=πoE19105172.2-⨯=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()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka . From Problem 3.5, πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J23E E E -=∆1818100198.1103646.1--⨯-⨯= 19104474.3-⨯=Jor 15.2106.1104474.31919=⨯⨯=∆--E eV_______________________________________3.10(a) πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JFrom Problem 3.6, πα515.12=aπ515.1222=⋅a E m o()()()()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()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JFrom Problem 3.6, πα375.24=aπ375.2224=⋅a E m o()()()()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 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα727.0=a oπ727.022=⋅a E m o o()()()()210312342102.41011.9210054.1727.0---⨯⨯⨯=πo E19108030.1-⨯=Jo 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()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka , From Problem 3.6,πα515.12=aπ515.1222=⋅a E m o()()()()2103423422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J23E E E -=∆191810830.7103646.1--⨯-⨯= 1910816.5-⨯=Jor 635.3106.110816.51919=⨯⨯=∆--E eV_______________________________________3.12For 100=T K, ()()⇒+⨯-=-1006361001073.4170.124gE164.1=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 mWe have()()B curve dkE d 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 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: ⇒<0dk dEvelocity in -x directionPoints C,D: ⇒>0dk dEvelocity 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 J So ()2101211008.0108⨯=⨯-C3811025.1-⨯=⇒CNow ()()38234121025.1210054.12--*⨯⨯==C m 311044.4-⨯=kgor o m m ⋅⨯⨯=--*31311011.9104437.4o 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 321044.4-⨯=kg or o m m ⋅⨯⨯=--*31321011.9104437.4o m m 0488.0=*_______________________________________ 3.17For A: 22k C E E -=-υ()()()2102191008.0106.1025.0⨯-=⨯--C 3921025.6-⨯=⇒C()()39234221025.6210054.12--*⨯⨯-=-=C m31108873.8-⨯-=kgor o m m ⋅⨯⨯-=--*31311011.9108873.8o m m 976.0--=* For B: 22k C E E -=-υ()()()2102191008.0106.13.0⨯-=⨯--C 382105.7-⨯=⇒C()()3823422105.7210054.12--*⨯⨯-=-=C m3210406.7-⨯-=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 ν1410705.2⨯=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 dk E d -=ααcos 2122Then221222*11 αE dk Ed m o k k =⋅== or212*αE m =_______________________________________ 3.21(a) ()[]3/123/24lt dn m m m =*()()[]3/123/264.1082.04oom m =o dn m m 56.0=*(b)o o l t cnm m m m m 64.11082.02123+=+=*oo m m 6098.039.24+=o cn m m 12.0=*_______________________________________ 3.22(a) ()()[]3/22/32/3lh hh dp m m m +=*()()[]3/22/32/3082.045.0o om m +=[]o m ⋅+=3/202348.030187.0o dp m 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 cp m 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 mEψ Use 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 mEDividing by XYZ , we obtain021*********=+∂∂⋅+∂∂⋅+∂∂⋅ mEz Z Z y Y Y x X XLet01222222=+∂∂⇒-=∂∂⋅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=X so 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=+---mE 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 π _______________________________________ 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 where222 mEk =We can then writemEk 2=Taking the differential, we obtaindE Em dE E m dk ⋅⋅=⋅⋅⋅⋅=2112121 Substituting these expressions into the density of states function, we have()dE E mmE a dE E g T ⋅⋅⋅⎪⎭⎫ ⎝⎛=212233 ππ Noting thatπ2h=this density of states function can be simplified and written as()()dE E m h a dE E g T ⋅⋅=2/33324π Dividing by 3a will yield the density of states so that()()E h m E g ⋅=32/324π _______________________________________ 3.25For a one-dimensional infinite potential well,222222k a n E m n ==*π Distance between quantum states()()aa n a n k k n n πππ=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=-+11Now()⎪⎭⎫ ⎝⎛⋅=a dkdk k g T π2NowE m k n *⋅=21dE Em dk n⋅⋅⋅=*2211 Then()dE Em a dE E g n T ⋅⋅⋅=*2212 π Divide by the "volume" a , so ()Em E g n *⋅=21πSo()()()()()EE g 31341011.9067.0210054.11--⨯⋅⨯=π ()EE g 1810055.1⨯=m 3-J 1-_______________________________________ 3.26(a) Silicon, o n m 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 hm 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 mg E kTE p⋅-=⎰-*υυυυπ332/324()()υυυπE kTE pE E hm 32/332/33224-*-⎪⎭⎫ ⎝⎛-=()()[]2/332/333224kT hmp-⎪⎭⎫ ⎝⎛-=*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⨯=m3-or 191012.4⨯=υg cm 3- (ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.5310969.2-⨯⨯=υg2510337.6⨯=m3-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 h m g p-=*υυπ32/324()()[]()E E -⨯⨯=--υπ3342/33110625.61011.956.024E E -⨯=υ55104541.4 For υE E =; 0=υg1.0-=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-⨯=23kTE -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_______________________________________ 3.3622222ma n E n π =For 6=n , Filled state()()()()()2103122234610121011.92610054.1---⨯⨯⨯=πE18105044.1-⨯=Jor 40.9106.1105044.119186=⨯⨯=--E eV For 7=n , Empty state()()()()()2103122234710121011.92710054.1---⨯⨯⨯=πE1810048.2-⨯=Jor 8.12106.110048.219187=⨯⨯=--E eV Therefore 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 π For 5 electrons, the 5th electron occupies the quantum state 1,2,2===z y x n n n ; so()2222252⎪⎭⎫ ⎝⎛++=a n n n m E z y x π()()()()()21031222223410121011.9212210054.1---⨯⨯++⨯=π1910761.3-⨯=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 111or()⎪⎭⎫ ⎝⎛∆+=-kT E E f exp 11122so ()()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 F or()⎪⎪⎭⎫⎝⎛-=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 02.0or 5002.0125.0exp ==⎥⎦⎤⎢⎣⎡+kT ()50ln 25.0=kTor()()⎪⎭⎫⎝⎛===3000259.0063906.050ln 25.0T kT which 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 all temperatures_______________________________________ 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 eVAt 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 fAt 2E E =, 12.13.042.12=-=-E E F eV So()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.012.1expor()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 1or()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 =,()⎪⎪⎭⎫⎝⎛+=⎪⎪⎭⎫ ⎝⎛-+=kT E kTE E E f g F2exp 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 108or ()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⎪⎪⎭⎫ ⎝⎛-+==kT E E f F F 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=eVThen ()2034.010168.02==∆E eV _______________________________________。
(完整版)半导体物理第十章习题答案
第10章 半导体的光学性质和光电与发光现象补充题：对厚度为d 、折射率为n 的均匀半导体薄片，考虑界面对入射光的多次反射，试推导其总透射率T 的表达式，并由此解出用透射率测试结果计算材料对光的吸收系数α的公式。
解：对上图所示的一个夹在空气中的半导体薄片，设其厚度为d ，薄片与空气的两个界面具有相同的反射率R 。
当有波长为λ、强度为I 0的单色光自晶片右侧垂直入射，在界面处反射掉I 0R 部分后，其剩余部分(1－R)I 0进入薄片向左侧传播。
设材料对入射光的吸收系数为α ，则光在薄片中一边传播一边按指数规律exp(－αx )衰减，到达左边边界时其强度业已衰减为（1－R)I 0exp(-αd )。
这个强度的光在这里分为两部分：一部分为反射光，其强度为R(1－R)I 0exp(-αd )；另一部分为透出界面的初级透射光，其强度为(1－R)2I 0exp(-αd )。
左边界的初级反射光经过晶片的吸收返回右边界时，其强度为R(1－R)I 0exp(-2αd )，这部分光在右边界的内侧再次分为反射光和透射光两部分，其反射光强度为R 2(1－R)I 0exp(-2αd )，反射回到左边界时再次被衰减了exp(-αd )倍，即其强度衰减为R 2(1－R)I 0exp(-3αd )。
这部分光在左边界再次分为两部分，其R 2(1－R)2I 0exp(-3αd )部分透出晶片，成为次级透射光。
如此类推，多次反射产生的各级透射光的强度构成了一个以 (1－R)2I 0exp(-αd )为首项，R 2exp(-2αd )为公共比的等比数列。
于是，在左边界外测量到的总透过率可用等比数列求和的公式表示为()22211d id i Re T T R e αα---==-∑由上式可反解出用薄片的透射率测试值求材料吸收吸收的如下计算公式410ln()2A d Tα-+=- 式中，薄片厚度d 的单位为μm ，吸收系数α的单位为cm -1，参数A ，B 分别为21R A R -⎛⎫= ⎪⎝⎭；21R B =空气 薄片 空气入射光I 0 反射光I 0R1.一棒状光电导体长为l ，截面积为S 。
半导体物理与器件第四版课后习题答案1
______________________________________________________________________________________Chapter 1Problem Solutions1.1 (a)fcc: 8 corner atoms 18/1atom6 face atoms32/1atomsTotal of 4 atoms per unit cell (b)bcc: 8 corner atoms 18/1atom1 enclosed atom=1 atom Total of 2 atoms per unit cell(c)Diamond: 8 corner atoms 18/1atom6 faceatoms 32/1atoms4 enclosedatoms= 4 atomsTotal of 8 atoms per unit cell_______________________________________ 1.2 (a)Simple cubic lattice: r a 2Unit cell vol33382rra1 atom per cell, so atom vol 3413r ThenRatio%4.52%10083433rr(b)Face-centered cubic latticerd aa rd22224Unit cell vol 33321622rr a4 atoms per cell, so atom vol3443r ThenRatio%74%10021634433rr (c)Body-centered cubic latticeraa rd3434Unit cell vol 3334ra2 atoms per cell, so atom vol 3423r ThenRatio%68%1003434233r r (d)Diamond lattice Body diagonal raa rd3838Unit cell vol3338r a8 atoms per cell, so atom vol 3483r ThenRatio%34%1003834833rr _______________________________________1.3(a)oA a43.5; From Problem 1.2d,ra38Then oAa r176.18343.583Center of one silicon atom to center ofnearest neighboroAr 35.22______________________________________________________________________________________ (b)Number density22381051043.58cm 3(c)Mass density23221002.609.28105..AN W t At N 33.2grams/cm3_______________________________________1.4(a)4 Ga atoms per unit cell Number density381065.54Density of Ga atoms 221022.2cm34 As atoms per unit cell Density of As atoms 221022.2cm3(b)8 Ge atoms per unit cell Number density381065.58Density of Ge atoms221044.4cm3_______________________________________ 1.5From Figure 1.15 (a)aa d4330.0232oAd 447.265.54330.0(b)aa d7071.022oAd 995.365.57071.0_______________________________________1.674.5423232222sin a a 5.109_______________________________________ 1.7(a) Simple cubic: oAr a 9.32(b)fcc:oAr a515.524(c) bcc:oA r a 503.434(d) diamond:oAra007.9342_______________________________________ 1.8 (a)Br 2035.122035.12oBAr 4287.0(b)oAa 07.2035.12(c)A-atoms: # of atoms1818Density381007.21231013.1cm3B-atoms: # of atoms3216Density381007.23231038.3cm3_______________________________________ 1.9(a)oAr a 5.42# of atoms1818Number density38105.412210097.1cm3______________________________________________________________________________________Mass density AN W t At N ..23221002.65.12100974.1228.0gm/cm3(b)oAr a196.534# of atoms 21818Number density3810196.5222104257.1cm3Mass density23221002.65.12104257.1296.0gm/cm3_______________________________________ 1.10From Problem 1.2, percent volume of fcc atoms is 74%; Therefore after coffee is ground,Volume = 0.74 cm3_______________________________________1.11(b)oAa 8.20.18.1(c)Na: Density38108.22/1221028.2cm3Cl: Density221028.2cm3(d)Na: At. Wt. = 22.99 Cl: At. Wt. = 35.45 So, mass per unit cell23231085.41002.645.352199.2221Then mass density21.2108.21085.43823grams/cm3_______________________________________ 1.12(a)oAa 88.122.223Then oA a 62.4Density of A:22381001.11062.41cm3Density of B:22381001.11062.41cm3(b)Same as (a) (c)Same material_______________________________________ 1.13oAa619.438.122.22(a) For 1.12(a), A-atomsSurface density28210619.411a1410687.4cm2For 1.12(b), B-atoms: oAa 619.4Surface density14210687.41acm2For 1.12(a) and (b), Same material(b) For 1.12(a), A-atoms;oAa 619.4Surface density212a1410315.3cm2B-atoms;Surface density______________________________________________________________________________________14210315.321a cm 2For 1.12(b), A-atoms;oAa 619.4Surface density212a1410315.3cm2B-atoms;Surface density14210315.321acm2For 1.12(a) and (b), Same material_______________________________________ 1.14 (a)Vol. Density31oaSurface Density212oa(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,1,1,21,21Same as (110) plane and [110]direction(iv) (321) plane6,3,211,21,31Intercepts of plane at6,3,2sq p [321] direction is perpendicular to(321) plane_______________________________________1.16(a)31311,31,11(b)12141,21,41_______________________________________ 1.17Intercepts: 2, 4, 331,41,21(634) plane_______________________________________ 1.18(a)oAa d 28.5(b)oAa d734.322(c)oAa d048.333_______________________________________ 1.19(a) Simple cubic(i) (100) plane:Surface density2821073.411a141047.4cm 2(ii) (110) plane:Surface density212a141016.3cm 2(iii) (111) plane: Area of planebh21where oAa b 689.62Now2222243222a a a hSooAh793.573.426______________________________________________________________________________________Area of plane881079304.51068923.62116103755.19cm 2Surface density16103755.19613141058.2cm2(b) bcc(i) (100) plane:Surface density 1421047.41acm2(ii) (110) plane: Surface density222a141032.6cm 2(iii) (111) plane:Surface density16103755.19613141058.2cm2(c) fcc(i) (100) plane:Surface density 1421094.82acm2(ii) (110) plane: Surface density222a141032.6cm 2(iii) (111) plane:Surface density16103755.19213613151003.1cm2_______________________________________ 1.20 (a)(100) plane: - similar to a fcc:Surface density281043.52141078.6cm 2(b)(110) plane:Surface density281043.524141059.9cm2(c)(111) plane: Surface density281043.5232141083.7cm2_______________________________________1.21oAr a703.6237.2424(a)#/cm338310703.64216818a2210328.1cm3(b)#/cm222124142a210703.62281410148.3cm2(c)oA a d74.422703.622(d)# of atoms2213613Area of plane: (see Problem 1.19)oAa b4786.92oAa h2099.826Area88102099.8104786.92121bh______________________________________________________________________________________15108909.3cm2#/cm215108909.32=141014.5cm2oAa d87.333703.633_______________________________________ 1.22Density of silicon atoms 22105cm3and4 valence electrons per atom, soDensity of valence electrons 23102cm3_______________________________________ 1.23Density of GaAs atoms22381044.41065.58cm3An average of 4 valence electrons peratom,SoDensity of valence electrons231077.1cm3_______________________________________ 1.24 (a)%10%10010510532217(b)%104%10010510262215_______________________________________ 1.25 (a)Fraction by weight7221610542.106.2810582.10102(b)Fraction by weight5221810208.206.2810598.3010_______________________________________ 1.26Volume density 1631021dcm3So610684.3dcmoAd 4.368We haveoo Aa 43.5Then85.6743.54.368oa d _______________________________________ 1.27Volume density 1531041dcm 3So61030.6dcmoAd630We have oo Aa 43.5Then11643.5630oa d _______________________________________。
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 kT E 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 , ThenThe ionization energy is or0053.0=E eV_______________________________________ 4.172148.0=eV90518.02148.012.1=-=eV 31090.6⨯=cm 3- (a) Holes338.0=eV_______________________________________ 4.18162.0=eV 958.0162.012.1=-=eV 31041.2⨯=cm 3-365.0=eV_______________________________________ 4.198436.0=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-_______________________________________ 4.23131033.7⨯=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) Holes3294.0=eV _______________________________________4.25()034533.03004000259.0=⎪⎭⎫⎝⎛=kT eV1910601.1⨯=cm 3-19103109.4⨯=cm 3-1210381.2⨯=⇒i n cm 3- 2787.0=eV(a) 84127.027873.012.1=-=-F c E E eV 910134.1⨯=cm 3- (b) Holes2642.0=eV _______________________________________ 4.26141050.4⨯=cm 3-17.125.042.1=-=-F c E E eV 21013.1-⨯=cm 3- (a) 034533.0=kT eV1910078.1⨯=cm 3- 1710236.7⨯=cm 3-3482.0=eV072.13482.042.1=-=-F c E E eV41040.2⨯=cm 3-_____________________________________ 4.27141068.6⨯=cm 3-870.025.012.1=-=-F c E E eV 41023.7⨯=o n cm 3- (a) 034533.0=kT eV1910601.1⨯=cm 3- 1910311.4⨯=cm 3- 3482.0=eV7718.03482.012.1=-=-F c E E eV 91049.8⨯=cm 3-_______________________________________ 4.28(a) ()F c o F N n ηπ2/12=For 2kT E E c F +=, Then ()0.12/1≅F F η 191016.3⨯=cm 3-(b) ()F c o F N n ηπ2/12=171030.5⨯=cm 3-_______________________________________ 4.29So ()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 201090.1⨯=cm 3-(b) ()()0.6107.4217⨯=πo n181018.3⨯=cm 3-_______________________________________ 4.31For the electron concentrationThe Boltzmann approximation applies, so or Define ThenTo find maximum ()()x n E n →, set orwhich yieldsFor the hole concentrationUsing the Boltzmann approximation or Define ThenTo find maximum value of ()()x p E p '→, set()0=''x d x dp Using the results from above,we find the maximum at_______________________________________ 4.32(a) Silicon: We have We can write For045.0=-d c E E eV andkT E E F d 3=-eV we can write or171045.2⨯=o n cm 3-We also haveAgain, we can write ForkT E E a F 3=- and045.0=-υE E a eVThen or161012.9⨯=o p cm 3-(b) GaAs: assume 0058.0=-d c E E eV Then or161087.1⨯=o n cm 3-Assume 0345.0=-υE E a eV Then 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- 1110334.7⨯=⇒i n cm 3-15104⨯==a o N p cm 3-()8152111034.110410334.7⨯=⨯⨯=o n cm 3-1310722.1⨯=⇒i n cm 3- 1410029.1⨯=cm 3-()12142131088.210029.110722.1⨯=⨯⨯=op cm 3-_______________________________________ 4.3515103⨯=cm 3- ()3152621008.1103108.1-⨯=⨯⨯==o i o p n n cm 3-(a) 16103⨯==d o N n cm 3-()416261008.1103108.1-⨯=⨯⨯=op cm 3-(b) 6108.1⨯===i o o n p n cm 3-810580.7⨯=⇒i n cm 3- 15104⨯==a o N p cm 3-()215281044.110410580.7⨯=⨯⨯=on cm 3-1010853.3⨯=⇒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 +⎪⎪⎭⎫⎝⎛+= or15102⨯=≅d o N n cm 3-111088.2⨯= cm 3- (ii)151610710⨯-=-≅d a o N N p 15103⨯=cm 3- 111092.1⨯=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 or(b) We have Now or Then or_______________________________________ 4.38(a) ⇒>d a N N p-type (b) Silicon: or13105.1⨯=o p cm 3- Then()7132102105.1105.1105.1⨯=⨯⨯==o i o p n n cm 3- Germanium: 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-type14108⨯=cm 3-()51421021081.2108105.1⨯=⨯⨯==o i o n n p cm 3-15108.4⨯='⇒aN cm 3-()41521010625.5104105.1⨯=⨯⨯=o n cm 3-_______________________________________ 4.40()155210210125.1102105.1⨯=⨯⨯==o i o p n n cm 3- ⇒>o o p n n-type_______________________________________ 4.411210376.1⨯=⇒i n cm 3-So 111088.6⨯=o n cm 3-, Then 121075.2⨯=o p cm 3-so that 1210064.2⨯=a N cm 3-_______________________________________ 4.42Plot_______________________________________ 4.43Plot_______________________________________ 4.44Plot_______________________________________ 4.45so 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 holes 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 addedSo 16105.3⨯='aN cm 3-()316210105.4105105.1⨯=⨯⨯=on cm 3-_______________________________________ 4.47(a) ⇒<<i o n p n-type on ()16421010125.1102105.1⨯=⨯⨯=cm 3-⇒electrons are majority carriers4102⨯=o p cm 3-⇒holes are minority carriers so 1610825.1⨯=d N cm 3-_______________________________________ 4.482222i a a o n N N p +⎪⎪⎭⎫⎝⎛+= and1510=N cm 3-4.49For 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 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 _______________________________________ 4.50151005.105.1⨯==d o N n cm 3- so 2821025.5⨯=i n NowBy trial and error, 5.536=T K (a) At 300=T K,2652.0=eV At 5.536=T K,()046318.03005.5360259.0=⎪⎭⎫⎝⎛=kT eV1910696.6⨯=cm 3-5124.0=eV then ()2472.0=-∆F c E E eV (b) Closer to the intrinsic energy level._______________________________________ 4.51At 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,410638.7⨯=⇒i n cm 3- At 400=T K,1210381.2⨯=⇒i n cm 3- At 600=T K,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,1510288.3⨯=cm 3-Then, 200=T K, 4212.0=-F Fi E E eV400=T K, 2465.0=-F Fi E E eV600=T K, 0630.0=-F Fi E E eV_______________________________________ 4.52(a)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)For 1410=a N cm 3-, 2889.0=-υE E F eV1510=a N cm 3-,2293.0=-υE E F eV1610=a N cm 3-,1697.0=-υE E F eV1710=a N cm 3-,1100.0=-υE E F eV_______________________________________ 4.53 or0447.0+=-midgap Fi E E eV(a) 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 or131097.1⨯==a o N p cm 3-_______________________________________ 4.54 so 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 eV1610631.1⨯=d N cm 3-15106⨯+'=dN 1610031.1⨯='⇒dN cm 3- Additionaldonor atoms(b) GaAs(i)()⎪⎪⎭⎫ ⎝⎛⨯=-151710107.4ln 0259.0F c E E 15936.0=eV(ii)13346.00259.015936.0=-=-F c E E eV1510718.2⨯=cm 3-1510+'=dN 1510718.1⨯='⇒dN cm 3- Additionaldonor atoms _______________________________________4.56()⎪⎪⎭⎫ ⎝⎛⨯⨯=16191021004.1ln 0259.01620.0=eV ()1876.0102108.2ln 0259.01619=⎪⎪⎭⎫ ⎝⎛⨯⨯=eV(a) For part (a);16102⨯=o p cm 3- 410125.1⨯=cm 3- For part (b):16102⨯=o n cm 3- 410125.1⨯=cm 3-_______________________________________ 4.5715100.3⨯=cm 3- Add additional acceptor impurities 15104⨯=⇒a N cm 3-_______________________________________ 4.58()3161.0105.1103ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=eV()3758.0105.1103ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=eV2786.0=eV 06945.0=eV _______________________________________ 4.59 ()2009.0103100.7ln 0259.01518=⎪⎪⎭⎫ ⎝⎛⨯⨯=eV 360.1=eV7508.0=eV 2526.0=eV 068.1=eV_______________________________________ 4.60n-type ()3504.0105.110125.1ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=eV ______________________________________ 4.61 ()030217.03003500259.0=⎪⎭⎫ ⎝⎛=kT eV ()1921910633.1300350102.1⨯=⎪⎭⎫ ⎝⎛⨯=c N cm 3-()192191045.2300350108.1⨯=⎪⎭⎫ ⎝⎛⨯=υN cm 3- Now So6257.0=⇒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-()5692.0108.1103.6ln 0259.0615=⎪⎪⎭⎫⎝⎛⨯⨯=eV_______________________________________。
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。
半导体物理课后习题答案(1-12章)
＝
(6.62 × 10 − 27 ) 2 ＝0.64eV 48 × 9.1 × 10 − 28 × (3.14 × 10 − 8 ) 2 × 1.6 × 10 − 11
② 77k 时，由（3－46）式得到： Ec － ED ＝ 0.01eV ； T ＝ 77k ； k0 ＝ 1.38×10-23J/K; n0 ＝ 1017 cm − 3 ； Nc ＝ 1.365×1019cm-3； Po 可忽略不计，由于 n 0 = n D ，即
° m0 q 4 ε 0 h2 = 13.6 e V a = = 0.53A ， 0 8ε r2 h2 m0 e2π
∗ 当 ε r = 11.1 ， m p = 0.86m0 时
m∗p E0 13.6 ∆ EA = � 2 � 0.86 m0 ε r 11.12 r1, p = ε r (
9.49 10− 2 eV
( 2mdn ) 4π V
h3
32
(E−
Ec ) 2
1
蝌
E1
E2
dZ = 4π
32
( 2mdn )
h3
32
骣 h2 Ec + 100 琪 ∗ 2 琪8 m L 桫 n Ec 3 2
(E−
Ec ) 2 dE
1
( 2mdn )
h3
2 骣 h2 创 琪100 ∗ 2 3 桫 8mn L
故： Z=1000π 3L3 7. ① 在室温下，锗的有效状态密度 Nc＝1.05×1019cm－3，Nv＝5.7×1018cm－3， 试求锗的载流子有效质量 mn*和 mp*。计算 77k 时的 Nc 和 Nv。已知 300k 时，Eg＝
半导体物理习题答案
第一章半导体中的电子状态例1.证明:对于能带中的电子，K状态和-K状态的电子速度大小相等,方向相反。
即：v(k)= －v(－k)，并解释为什么无外场时，晶体总电流等于零。
解：K状态电子的速度为：（1）同理，－K状态电子的速度则为：（2）从一维情况容易看出：（3）同理有：（4）（5）将式（3）（4）（5）代入式（2）后得：（6）利用（1）式即得：v(－k)= －v(k)因为电子占据某个状态的几率只同该状态的能量有关，即：E(k)=E(－k)故电子占有k状态和-k状态的几率相同，且v(k)=－v(－k)故这两个状态上的电子电流相互抵消，晶体中总电流为零。
例2.已知一维晶体的电子能带可写成：式中，a为晶格常数。
试求：（1）能带的宽度；（2）能带底部和顶部电子的有效质量。
解：（1）由E(k)关系（1）（2）令得：当时，代入（2）得：对应E(k)的极小值。
当时，代入（2）得：对应E(k)的极大值。
根据上述结果，求得和即可求得能带宽度。
故：能带宽度（3）能带底部和顶部电子的有效质量：习题与思考题：1 什么叫本征激发？温度越高，本征激发的载流子越多，为什么？试定性说明之。
2 试定性说明Ge、Si的禁带宽度具有负温度系数的原因。
3 试指出空穴的主要特征。
4 简述Ge、Si和GaAs的能带结构的主要特征。
5 某一维晶体的电子能带为其中E0=3eV，晶格常数a=5×10-11m。
求：（1）能带宽度；（2）能带底和能带顶的有效质量。
6原子中的电子和晶体中电子受势场作用情况以及运动情况有何不同？原子中内层电子和外层电子参与共有化运动有何不同？7晶体体积的大小对能级和能带有什么影响？8描述半导体中电子运动为什么要引入“有效质量”的概念？用电子的惯性质量描述能带中电子运动有何局限性？9 一般来说，对应于高能级的能带较宽，而禁带较窄，是否如此？为什么？10有效质量对能带的宽度有什么影响？有人说：“有效质量愈大，能量密度也愈大，因而能带愈窄。
《半导体物理与器件》第四版答案第十章
《半导体物理与器件》第四版答案第十章-CAL-FENGHAI-(2020YEAR-YICAI)_JINGBIANChapter 1010.1(a) p-type; inversion (b) p-type; depletion (c) p-type; accumulation (d) n-type; inversion_______________________________________ 10.2 (a) (i) ⎪⎪⎭⎫ ⎝⎛=iat fpnN V ln φ ()⎪⎪⎭⎫⎝⎛⨯⨯=1015105.1107ln 0259.03381.0=V 2/14⎥⎦⎤⎢⎣⎡∈=a fp s dTeN x φ()()()()()2/1151914107106.13381.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--51054.3-⨯=cmor μ354.0=dT x m (ii) ()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1103ln 0259.0fpφ3758.0=V ()()()()()2/1161914103106.13758.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx51080.1-⨯=cmor μ180.0=dT x m (b) ()03022.03003500259.0=⎪⎭⎫⎝⎛=kT V ⎪⎪⎭⎫ ⎝⎛-=kTE N N n gc iexp 2υ ()()319193003501004.1108.2⎪⎭⎫⎝⎛⨯⨯=⎪⎭⎫⎝⎛-⨯03022.012.1exp221071.3⨯=so 111093.1⨯=i n cm 3-(i)()⎪⎪⎭⎫⎝⎛⨯⨯=11151093.1107ln 03022.0fpφ3173.0=V ()()()()()2/1151914107106.13173.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx51043.3-⨯=cmor μ343.0=dT x m (ii) ()⎪⎪⎭⎫⎝⎛⨯⨯=11161093.1103ln 03022.0fpφ3613.0=V ()()()()()2/1161914103106.13613.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx51077.1-⨯=cmor μ177.0=dT x m_______________________________________ 10.3(a) ()2/14max ⎥⎦⎤⎢⎣⎡∈=='d fn s d dT d SDeN eN x eN Q φ()()[]2/14fn s d eN φ∈= 1st approximation: Let 30.0=fn φV Then()281025.1-⨯()()()()()()[]30.01085.87.114106.11419--⨯⨯=d N 141086.7⨯=⇒d N cm 3- 2nd approximation:()2814.0105.11086.7ln 0259.01014=⎪⎪⎭⎫⎝⎛⨯⨯=fnφVThen()281025.1-⨯()()()()()()[]2814.01085.87.114106.11419--⨯⨯=d N 141038.8⨯=⇒d N cm 3-(b) ()2831.0105.11038.8ln 0259.01014=⎪⎪⎭⎫⎝⎛⨯⨯=fnφV()566.02831.022===fn s φφV_______________________________________ 10.4p-type silicon(a) Aluminum gate⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫⎝⎛++'-'=fp gmms e E φχφφ2 We have ⎪⎪⎭⎫⎝⎛=i a t fp n N V ln φ ()334.0105.1106ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=VThen()[]334.056.025.320.3++-=ms φ or944.0-=ms φV(b) +n polysilicon gate ⎪⎪⎭⎫⎝⎛+-=fp g mseE φφ2()334.056.0+-= or894.0-=ms φV(c) +p polysilicon gate ()334.056.02-=⎪⎪⎭⎫⎝⎛-=fp gms e E φφ or226.0+=ms φV_______________________________________ 10.5()3832.0105.1104ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=fp φV⎪⎪⎭⎫⎝⎛++'-'=fp g m ms e E φχφφ2 ()3832.056.025.320.3++-=9932.0-=ms φV_______________________________________ 10.6(a) 17102⨯≅d N cm 3-(b) Not possible - ms φ is always positive.(c) 15102⨯≅d N cm 3-_______________________________________ 10.7From Problem 10.5, 9932.0-=ms φV oxssms FB C Q V '-=φ (a) ()()814102001085.89.3--⨯⨯=∈=ox ox oxt C710726.1-⨯=F/cm 2 ()()7191010726.1106.11059932.0--⨯⨯⨯--=FB V 040.1-=V(b) ()()81410801085.89.3--⨯⨯=oxC 710314.4-⨯=F/cm 2()()7191010314.4106.11059932.0--⨯⨯⨯--=FB V 012.1-=V_______________________________________ 10.8(a) 42.0-≅ms φV42.0-==ms FB V φV (b)()()781410726.1102001085.89.3---⨯=⨯⨯=oxC F/cm2(i)()()7191010726.1106.1104--⨯⨯⨯-='-=∆ox ss FBC Q V0371.0-=V (ii)()()7191110726.1106.110--⨯⨯-=∆FBV 0927.0-=V (c) 42.0-==ms FB V φV()()781410876.2101201085.89.3---⨯=⨯⨯=ox C F/cm 2(i)()()7191010876.2106.1104--⨯⨯⨯-=∆FBV 0223.0-=V(ii)()()7191110876.2106.110--⨯⨯-=∆FB V0556.0-=V_______________________________________ 10.9⎪⎪⎭⎫⎝⎛++'-'=fp gmms e E φχφφ2 where ()365.0105.1102ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=fpφVThen()365.056.025.320.3++-=ms φ or975.0-=ms φV Now oxssms FB C Q V '-=φ or()ox FB ms ssC V Q -='φ We have()()814104501085.89.3--⨯⨯=∈=ox ox oxt Cor81067.7-⨯=ox C F/cm 2 So now()[]()81067.71975.0-⨯⋅---='ssQ 91092.1-⨯=C/cm 2 or10102.1⨯='eQ sscm 2- _______________________________________ 10.10 ()3653.0105.1102ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=fp φV()()()()()2/1161914102106.13653.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx510174.2-⨯=cm()dT a SDx eN Q ='max()()()5161910174.2102106.1--⨯⨯⨯=810958.6-⨯=C/cm 2()()781410301.2101501085.89.3---⨯=⨯⨯=oxC F/cm 2()fp ms oxss SDTN C Q Q V φφ2max ++'-'=()()71910810301.2106.110710958.6---⨯⨯⨯-⨯=()3653.02++ms φ ms φ+=9843.0(a) n + poly gate on p-type:12.1-≅ms φV136.012.19843.0-=-=TN V V (b) p + poly gate on p-type:28.0+≅ms φV26.128.09843.0+=+=TN V V (c) Al gate on p-type: 95.0-≅ms φV 0343.095.09843.0+=-=TN V V _______________________________________10.11 ()3161.0105.1103ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=fn φV()()()()()2/1151914103106.13161.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx510223.5-⨯=cm()dT d SDx eN Q ='max ()()()5151910223.5103106.1--⨯⨯⨯= 810507.2-⨯=C/cm 2()()781410301.2101501085.89.3---⨯=⨯⨯=oxC F/cm 2()fn ms ox ss SDTPC Q Q V φφ2max -+⎥⎥⎦⎤⎢⎢⎣⎡'+'-=()()⎥⎦⎤⎢⎣⎡⨯⨯⨯+⨯-=---71019810301.2107106.110507.2()3161.02-+ms φms TP V φ+-=7898.0(a) n + poly gate on n-type:41.0-≅ms φV20.141.07898.0-=--=TP V V (b) p + poly gate on n-type:0.1+≅ms φV210.00.17898.0+=+-=TP V V (c) Al gate on n-type: 29.0-≅ms φV 08.129.07898.0-=--=TP V V _______________________________________10.12 ()3294.0105.1105ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=fpφVThe surface potential is()659.03294.022===fp s φφV We have 90.0-='-=oxssms FB C Q V φV Now ()FB s oxSDT V C Q V ++'=φmaxWe obtain 2/14⎥⎦⎤⎢⎣⎡∈=a fp s dTeN x φ()()()()()2/1151914105106.13294.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--or410413.0-⨯=dT x cm Then()()()()4151910413.0105106.1max --⨯⨯⨯='SDQ or()810304.3max -⨯='SDQ C/cm 2 We also find ()()814104001085.89.3--⨯⨯=∈=ox ox oxt Cor810629.8-⨯=ox C F/cm 2 Then90.0659.010629.810304.388-+⨯⨯=--T Vor142.0+=T V V_______________________________________10.13 ()()814102201085.89.3--⨯⨯=∈=ox ox oxt C710569.1-⨯=F/cm 2 ()()1019104106.1⨯⨯='-ssQ9104.6-⨯=C/cm 2 By trial and error, let 16104⨯=a N cm 3-.Now ()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1104ln 0259.0fpφ3832.0=V ()()()()()2/1161914104106.13832.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx510575.1-⨯=cm()max SDQ ' ()()()5161910575.1104106.1--⨯⨯⨯= 710008.1-⨯=C/cm 2 94.0-≅ms φV Then ()fp ms oxss SDTN C Q Q V φφ2max ++'-'=79710569.1104.610008.1---⨯⨯-⨯= ()3832.0294.0+-Then 428.0=TN V V 45.0≅V_______________________________________10.14 ()()814101801085.89.3--⨯⨯=∈=ox ox oxt C7109175.1-⨯=F/cm 3-()()1019104106.1⨯⨯='-ssQ 9104.6-⨯=C/cm 2 By trial and error, let 16105⨯=d N cm 3- Now ()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1105ln 0259.0fnφ3890.0=V()()()()()2/1161914105106.13890.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx510419.1-⨯=cm()max SDQ ' ()()()5161910419.1105106.1--⨯⨯⨯= 710135.1-⨯=C/cm 3- 10.1+≅ms φV Then()()fn ms oxssSDTP C Q Q V φφ2max -+'+'-=()797109175.1104.610135.1---⨯⨯+⨯-=()3890.0210.1-+Then 303.0-=TP V V, which is within the specified value._______________________________________10.15We have 710569.1-⨯=ox C F/cm 29104.6-⨯='ssQ C/cm 2 By trial and error, let 14105⨯=d N cm 3- Now ()⎪⎪⎭⎫⎝⎛⨯⨯=1014105.1105ln 0259.0fnφ2697.0=V ()()()()()2/1141914105106.12697.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx410182.1-⨯=cm()max SDQ ' ()()()4141910182.1105106.1--⨯⨯⨯= 910456.9-⨯=C/cm 2 33.0-≅ms φV Then()()fn ms oxssSDTP C Q Q V φφ2max -+'+'-=⎪⎪⎭⎫⎝⎛⨯⨯+⨯-=---79910569.1104.610456.9 ()2697.0233.0--970.0=VThen 970.0-=TP V V 975.0-≅ V which meets the specification._______________________________________10.16(a) 03.1-≅ms φV ()()814101801085.89.3--⨯⨯=oxC 7109175.1-⨯=F/cm 2Now oxssms FB C Q V '-=φ ()()71019109175.1106106.103.1--⨯⨯⨯--=08.1-=FB V V (b) ()⎪⎪⎭⎫⎝⎛⨯=1015105.110ln 0259.0fpφ2877.0=V ()()()()()2/115191410106.12877.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dTx510630.8-⨯=cm()max SDQ ' ()()()5151910630.810106.1--⨯⨯= 810381.1-⨯=C/cm 2 Now ()fp FB oxSDTN V C Q V φ2max ++'=()2877.0208.1109175.110381.178+-⨯⨯=-- or 433.0-=TN V V_______________________________________10.17(a) We have n-type material under the gate, so 2/14⎥⎦⎤⎢⎣⎡∈==d fn s C dTeN t x φwhere ()288.0105.110ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯=fnφVThen ()()()()()2/115191410106.1288.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dTxor410863.0-⨯==C dT t x cm μ863.0=m (b)()()fn ms oxoxss SDT t Q Q V φφ2max -+⎪⎪⎭⎫⎝⎛∈'+'-= For an +n polysilicon gate,()288.056.02--=⎪⎪⎭⎫⎝⎛--=fn gms e E φφ or272.0-=ms φV Now()()()()4151910863.010106.1max --⨯⨯='SDQor()81038.1max -⨯='SDQ C/cm 2 We have()()91019106.110106.1--⨯=⨯='ssQ C/cm 2We now find()()()()81498105001085.89.3106.11038.1----⨯⨯⨯+⨯-=T V()288.02272.0-- or07.1-=T V V_______________________________________10.18(b) ⎪⎪⎭⎫⎝⎛++'-'=fp gmms e E φχφφ2 where20.0-='-'χφmV and ()3473.0105.110ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯=fpφVThen()3473.056.020.0+--=ms φ or107.1-=ms φV(c) For 0='ssQ ()fp ms oxox SDTN t Q V φφ2max ++⎪⎪⎭⎫⎝⎛∈'= We find ()()()()()2/116191410106.13473.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dTxor41030.0-⨯=dT x cm μ30.0=m Now()()()()416191030.010106.1max --⨯⨯='SDQ or()810797.4max -⨯='SDQ C/cm 2 Then ()()()()14881085.89.31030010797.4---⨯⨯⨯=TV()3473.02107.1+- or00455.0+=T V V 0≅V_______________________________________10.19Plot_______________________________________10.20 Plot_______________________________________10.21 Plot_______________________________________10.22 Plot_______________________________________10.23(a) For 1=f Hz (low freq), ()()814101201085.89.3--⨯⨯=∈=ox ox oxt C710876.2-⨯=F/cm 2a st s ox ox oxFBeNV t C ∈⎪⎪⎭⎫ ⎝⎛∈∈+∈='()()()()()()()16191481410106.11085.87.110259.07.119.3101201085.89.3----⨯⨯⎪⎭⎫ ⎝⎛+⨯⨯= 710346.1-⨯='FBC F/cm 2 dT soxox oxx t C ⋅⎪⎪⎭⎫ ⎝⎛∈∈+∈='minNow()3473.0105.110ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯=fpφV()()()()()2/116191410106.13473.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dTx51000.3-⨯=cmThen()()()5814min1000.37.119.3101201085.89.3---⨯⎪⎭⎫ ⎝⎛+⨯⨯='C810083.3-⨯=F/cm 2C '(inv)710876.2-⨯==ox C F/cm 2 (b) 1=f MHz (high freq), 710876.2-⨯=ox C F/cm 2 (unchanged)710346.1-⨯='FBC F/cm 2 (unchanged)8min10083.3-⨯='C F/cm 2 (unchanged)C '(inv)8min10083.3-⨯='=C F/cm 2 (c) 10.1-≅==ms FB V φV ()fp FB oxSDTN V C Q V φ2max ++'=Now()dT a SDx eN Q ='max ()()()516191000.310106.1--⨯⨯= 81080.4-⨯=C/cm 2 ()3473.0210.110876.21080.478+-⨯⨯=--TNV 2385.0-=TNV V_______________________________________10.24(a) 1=f Hz (low freq), ()()814101201085.89.3--⨯⨯=∈=ox ox oxt C710876.2-⨯=F/cm 2a st s ox ox oxFBeNV t C ∈⎪⎪⎭⎫ ⎝⎛∈∈+∈='()()()()()()()141914814105106.11085.87.110259.07.119.3101201085.89.3⨯⨯⨯⎪⎭⎫⎝⎛+⨯⨯=---- 810726.4-⨯='FBC F/cm 2 dT soxox oxx t C ⋅⎪⎪⎭⎫ ⎝⎛∈∈+∈='minNow()2697.0105.1105ln 0259.01014=⎪⎪⎭⎫⎝⎛⨯⨯=fnφV()()()()()2/1141914105106.12697.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx410182.1-⨯=cmThen()()()4814min10182.17.119.3101201085.89.3---⨯⎪⎭⎫ ⎝⎛+⨯⨯='C910504.8-⨯=F/cm 2C '(inv)710876.2-⨯==ox C F/cm 2 (b) 1=f MHz (high freq), 710876.2-⨯=ox C F/cm 2 (unchanged)810726.4-⨯='FBC F/cm 2 (unchanged)9min10504.8-⨯='C F/cm 2 (unchanged)C '(inv)9min10504.8-⨯='=C F/cm 2 (c) 95.0≅=ms FB V φV ()fn FB oxSDTP V C Q V φ2max -+'-=Now()dT d SDx eN Q ='max ()()()4141910182.1105106.1--⨯⨯⨯= 910456.9-⨯=C/cm 2 Then ()2697.0295.010876.210456.979-+⨯⨯-=--TPV 378.0+=TPV V_______________________________________10.25The amount of fixed oxide charge at x is()x x ∆ρ C/cm 2By lever action, the effect of this oxide chargeon the flatband voltage is ()x x t x C V oxoxFB ∆⎪⎪⎭⎫⎝⎛-=∆ρ1 If we add the effect at each point, wemustintegrate so that ()dx t x x C V oxt oxoxFB ⎰-=∆01ρ _______________________________________10.26(a) We have ρx Q tSS ()='∆Then ∆V C x x t dx FB oxoxoxt =-()z10ρ ≈-'F H G I K J F H I K -z 1C t t Q t dx oxox oxox oxSSt tt ∆∆bg =-'--=-'F H I K 1C Q t t t t Q CoxSS ox oxSS ox∆∆a f or∆V Q t FB SSox ox=-'∈F H G I KJ =-⨯⨯⨯⨯---()16108102001039885101910814...b gb gb g b gor∆V FB =-00742.V(b)We haveρx Q t SS ox()='=⨯⨯⨯--16108102001019108.b gb g =⨯=-64103.ρONow∆V C x x t dx C t xdx FB oxox ox O ox oxoxt t =-=-()zz10ρρor ∆V t FB O oxox=-∈ρ22=-⨯⨯⨯---()6410200102398851038214...bgbg bgor∆V FB =-00371.V (c) ρρx x t O ox()F H G I KJ =We find12216108102001019108t Q ox O SSO ρρ='⇒=⨯⨯⨯--.bgb gor ρO =⨯-128102. Now ∆V C t x x t dx FB oxoxOoxt ox=-⋅⋅F H G I KJ z110ρ=-⋅z122C tx dx oxOoxoxt ρa fwhich becomes ∆V t t xt FB oxoxOoxoxO ox oxt =-∈⋅⋅=-∈F H G I KJ 133232ρρa fThen∆V FB =-⨯⨯⨯---()12810200103398851028214...bgb g b gor 0494.0-=∆FB V V_______________________________________10.27 Sketch_______________________________________10.28 Sketch_______________________________________10.29 (b)⎪⎪⎭⎫⎝⎛-=-=2ln i d a t bi FB n N N V V V ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯-=2101616105.11010ln 0259.0 or695.0-=FB V V(c) Apply 3-=G V V, 3≅ox V VFor 3+=G V V,sdx d ∈-=Eρ n-side: d eN =ρ1C x eN eN dx d sd s d +∈-=E ⇒∈-=E0=E at n x x -=, then sn d xeN C ∈-=1so ()n sdx x eN +∈-=E for 0≤≤-x x n In the oxide, 0=ρ, so=E ⇒=E0dxd constant. From the boundary conditions, in the oxidesnd x eN ∈-=E In the p-region,2C x eN eN dx d sa s a s +∈=E ⇒∈+=∈-=Eρ 0=E at ()p ox x t x +=, then ()[]x x t eN p ox sa-+∈-=E At ox t x =, sn d s p a xeN x eN ∈-=∈-=ESo that n d p a x N x N =Since d a N N =, then p n x x = The potential is ⎰E -=dx φFor zero bias, we can write bi p ox n V V V V =++where p ox n V V V ,, are the voltage drops acrossthe n-region, the oxide, and the p-region,respectively. For the oxide: soxn d ox ox t x eN t V ∈=⋅E = For the n-region:()C x x x eNx V n sdn '+⎪⎪⎭⎫ ⎝⎛⋅+∈=22 Arbitrarily, set 0=n V at n x x -=, thensnd x eN C ∈='22so that()()22n sdn x x eN x V +∈=At 0=x , snd n x eN V ∈=22which is thevoltagedrop across the n-region. Because of symmetry, p n V V =. Then for zero bias, we havebi ox n V V V =+2which can be written asbi soxn d s n d V t x eN x eN =∈+∈2or02=∈-+dsbi ox n n eN V t x x Solving for n x , we obtain dbis ox ox n eN V t t x ∈+⎪⎪⎭⎫ ⎝⎛+-=222 If we apply a voltage G V , then replace bi V byG bi V V +, so()dG bi s ox ox p n eN V V t t x x +∈+⎪⎪⎭⎫ ⎝⎛+-==222 We find2105008-⨯-==p n x x()()()()()1619142810106.1695.31085.87.11210500---⨯⨯+⎪⎪⎭⎫ ⎝⎛⨯+ which yields510646.4-⨯==p n x x cm Now soxn d ox t x eN V ∈=()()()()()()148516191085.87.111050010646.410106.1----⨯⨯⨯⨯=or359.0=ox V V We also findsnd p n x eN V V ∈==22()()()()()142516191085.87.11210646.410106.1---⨯⨯⨯=or67.1==p n V V V_______________________________________10.30(a) n-type(b) We have 731210110210200---⨯=⨯⨯=ox C F/cm 2 Also()()7141011085.89.3--⨯⨯=∈=⇒∈=ox ox ox ox ox oxC t t Cor61045.3-⨯=ox t cm 5.34=nm oA 345= (c) oxss ms FB C Q V '-=φ or71050.080.0-'--=-ss Qwhich yields8103-⨯='ssQ C/cm 21110875.1⨯=cm 2- (d)⎪⎪⎭⎫ ⎝⎛∈⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛∈∈+∈='dss ox ox oxFBeN e kT t C()()[][6141045.31085.89.3--⨯÷⨯=()()()()()⎥⎥⎦⎤⨯⨯⨯⎪⎭⎫ ⎝⎛+--161914102106.11085.87.110259.07.119.3which yields81082.7-⨯='FBC F/cm 2 or156=FB C pF_______________________________________10.31(a) Point 1: Inversion 2: Threshold 3: Depletion 4: Flat-band 5: Accumulation_______________________________________10.32We have()()[]fp ms x GS ox nV V C Q φφ2+---='()()max SD ssQ Q '+'- Now let DS x V V =, so()⎩⎨⎧--='DS GS ox nV V C Q()()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡+-'+'+fpms ox ss SDC Q Q φφ2maxFor a p-type substrate, ()max SD Q ' is a negative value, so we can write()⎩⎨⎧--='DS GS ox nV V C Q()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡++'-'-fp ms ox ss SD C Q Q φφ2maxUsing the definition of thresholdvoltage T V , we have()[]T DS GS ox nV V V C Q ---=' At saturation()T GS DS DS V V sat V V -==which then makes nQ 'equal to zero at thedrain terminal._______________________________________10.33 (a) ()[]222DS DS T GS n D V V V V LW k I --⋅'=()()()()[]22.02.04.08.028218.0--⎪⎭⎫ ⎝⎛=0864.0=mA(b) ()22T GS n D V V L W k I -⋅'=()()24.08.08218.0-⎪⎭⎫ ⎝⎛=1152.0=mA(c) Same as (b), 1152.0=D I mA(d) ()22T GS n D V V L W k I -⋅'=()()24.02.18218.0-⎪⎭⎫ ⎝⎛=4608.0=mA_______________________________________10.34 (a) ()[]222SDSD T SG p D V V V V LW k I -+⋅'=()()()()[]225.025.04.08.0215210.0--⎪⎭⎫ ⎝⎛=103.0=D I mA(b) ()22T SG p D V V L Wk I +⋅'=()()24.08.015210.0-⎪⎭⎫ ⎝⎛=12.0=mA(c) ()22T SG p D V V L W k I +⋅'=()()24.02.115210.0-⎪⎭⎫ ⎝⎛=48.0=mA(d) Same as (c), 48.0=D I mA_______________________________________10.35(a) ()22T GS n D V V LW k I -⋅'=()28.04.126.00.1-⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=L W26.9=⇒LW(b) ()()28.085.126.926.0-⎪⎭⎫ ⎝⎛=DI 06.3=mA(c) ()[]222DS DS T GS n D V V V V LW k I --⋅'=()()()()[]215.015.08.02.1226.926.0--⎪⎭⎫ ⎝⎛=271.0=mA_______________________________________10.36(a) Assume biased in saturation region()22T SG p D V V L Wk I +⋅'=()()2020212.010.0T V +⎪⎭⎫ ⎝⎛=289.0+=⇒T V VNote: 0.1=SD V V 289.00+=+>T SG V V V So the transistor is biased in the saturation region.(b) ()()2289.04.020212.0+⎪⎭⎫ ⎝⎛=D I570.0=mA(c) ()()[()15.0289.06.0220212.0+⎪⎭⎫ ⎝⎛=D I()]215.0- or293.0=D I mA_______________________________________10.37()()781410138.3101101085.89.3---⨯=⨯⨯=oxC F/cm 2()()()()2.122010138.342527-⨯==L W C K ox n n μ 310111.1-⨯=A/V 2=1.111 mA/V 2 (a) 0=GS V , 0=D I6.0=GS V V, ()15.0=sat V DS V, ()()()245.06.0111.1-=sat I D025.0=mA2.1=GS V V, ()75.0=sat V DS V, ()()()245.02.1111.1-=sat I D 625.0=mA 8.1=GS V V, ()35.1=sat V DS V, ()()()245.08.1111.1-=sat I D 025.2=mA 4.2=GS V V, ()95.1=sat V DS V, ()()()245.04.2111.1-=sat I D 225.4=mA (c)0=D I for 45.0≤GS V V 6.0=GS V V,()()()()[]21.01.045.06.02111.1--=D I 0222.0=mA 2.1=GS V V,()()()()[]21.01.045.02.12111.1--=D I 156.0=mA 8.1=GS V V,()()()()[]21.01.045.08.12111.1--=D I 289.0=mA 4.2=GS V V,()()()()[]21.01.045.04.22111.1--=D I 422.0=mA_______________________________________10.38 ()()814101101085.89.3--⨯⨯=∈=ox ox oxt C710138.3-⨯=F/cm 2 L WC K ox p p 2μ=()()()()2.123510138.32107-⨯=41061.9-⨯=A/V 2=0.961 mA/V 2(a) 0=SG V , 0=D I6.0=SG V V, ()25.0=sat V SD V ()()()235.06.0961.0-=sat I D060.0=mA2.1=SG V V, ()85.0=sat V SD V ()()()235.02.1961.0-=sat I D 694.0=mA8.1=SG V V, ()45.1=sat V SD V ()()()235.08.1961.0-=sat I D 02.2=mA4.2=SG V V, ()05.2=sat V SD V ()()()235.04.2961.0-=sat I D 04.4=mA (c)0=D I for 35.0≤SG V V6.0=SG V V()()()()[]21.01.035.06.02961.0--=D I 0384.0=mA 2.1=SG V V()()()()[]21.01.035.02.12961.0--=D I 154.0=mA 8.1=SG V V()()()()[]21.01.035.08.12961.0--=D I 269.0=mA 4.2=SG V V()()()()[]21.01.035.04.22961.0--=D I 384.0=mA_______________________________________10.39(a) From Problem10.37,111.1=n K mA/V 2 For 8.0-=GS V V, 0=D I0=GS V , ()8.0=sat V DS V ()()()28.00111.1+=sat I D 711.0=mA8.0+=GS V V, ()6.1=sat V DS V ()()()28.08.0111.1+=sat I D 84.2=mA6.1=GS V V, ()4.2=sat V DS V ()()()28.06.1111.1+=sat I D 40.6=mA_______________________________________10.40 Sketch_______________________________________10.41 Sketch_______________________________________10.42We have()T DS T GS DS V V V V sat V -=-= so that()T DS DS V sat V V +=Since ()sat V V DS DS >, the transistor is alwaysbiased in the saturation region. Then ()2T GS n D V V K I -=where, from Problem 10.37,111.1=n K mA/V 2and 45.0=T V V______10.43From Problem 10.38, 961.0=p K mA/V 2()()[]22SD SD T SG p D V V V V K I -+= ()T SG p V SDDd V V K V I g SD +=∂∂=→20For 35.0≤SG V V, 0=d gFor 35.0>SG V V,()()35.0961.02-=SG d V g For 4.2=SG V V,()()35.04.2961.02-=d g 94.3=mA/V_______________________________________10.44 (a) GSDm V I g ∂∂=()()[]{}22DSDS T GS n GSV V V V K V --∂∂=()DS n V K 2= ()()05.0225.1n K = 5.12=⇒n K mA/V 2(b) ()()()[()]205.005.03.08.025.12--=D I 594.0=mA (c) ()()23.08.05.12-=D I 125.3=mA_______________________________________10.45We find that 2.0≅T V V Now ()()T GS oxn D V V LC W sat I -⋅=2μ where ()()814104251085.89.3--⨯⨯=∈=ox ox oxt Cor81012.8-⨯=ox C F/cm 2We are given 10=L W . From the graph, for3=GS V V, we have ()033.0≅sat I D , then ()2.032033.0-⋅=LC W oxn μ or310139.02-⨯=L C W oxn μ or()()3810139.01012.81021--⨯=⨯n μ which yields342=n μcm 2/V-s_______________________________________10.46 (a)()T GS DS V V sat V -= or8.48.04=⇒-=GS GS V V V(b)()()()sat V K V V K sat I DSn T GS n D 22=-= so()244102n K =⨯- which yieldsμ5.12=n K A/V 2 (c)()2.18.02=-=-=T GS DS V V sat V V so ()sat V V DS DS >()()()258.021025.1-⨯=-sat I D or()μ18=sat I D A (d)()sat V V DS DS <()[]22DSDS T GS n D V V V V K I --= ()()()()[]25118.0321025.1--⨯=- orμ5.42=D I A_______________________________________10.47 (a) ()()814101801085.89.3--⨯⨯=oxC7109175.1-⨯=F/cm 2(i)()()7109175.1450-⨯=='ox n nC k μ 510629.8-⨯=A/V 2or μ29.86='nk A/V 2 (ii)()()22T GS n D V V L W k sat I -⎪⎭⎫ ⎝⎛⎪⎪⎭⎫⎝⎛'= ()24.02208629.08.0-⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=L W24.7=⇒LW(b) (i) ()()7109175.1210-⨯=='ox p p C k μ510027.4-⨯=A/V 2 or μ27.40='p k A/V 2(ii) ()()22T SG p D V V L W k sat I +⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛'= ()24.02204027.08.0-⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=L W5.15=⇒LW_______________________________________10.48From Problem 10.37, 111.1=n K mA/V 2(a) ()()[]{}22DSDS T GS n GSmL V V V V K V g --∂∂=()()()()1.02111.12==DS n V Kso 222.0=mL g mA/V (b) (){}2T GS n GSms V V K V g -∂∂=()()()45.05.1111.122-=-=T GS n V V Kso 33.2=ms g mA/V_______________________________________10.49From Problem 10.38, 961.0=p K mA/V 2(a) ()()[]{}22SDSD T SG p SGmL V V V V K V g -+∂∂=()()()()1.02961.02==SD p V Kor 192.0=mL g mA/V (b) ()[]2T SG p SGms V V K V g +∂∂=()()()35.05.1961.022-=+=T SG p V V Kor 21.2=ms g mA/V_______________________________________10.50 (a) oxa s C N e ∈=2γNow ()()814101501085.89.3--⨯⨯=ox C 710301.2-⨯=F/cm 2Then()()()()716141910301.21051085.87.11106.12---⨯⨯⨯⨯=γ 5594.0=γV 2/1 (b) ()3890.0105.1105ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=fp φV(i)()()()()()2/1161914105106.13890.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx510419.1-⨯=cm()max SDQ ' ()()()5161910419.1105106.1--⨯⨯⨯= 710135.1-⨯=C/cm 2()fp FB oxSDTO V C Q V φ2max ++'=()3890.025.010301.210135.177+-⨯⨯=--7713.0=VLWC K ox n n 2μ=()()()()2.12810301.24507-⨯=410452.3-⨯=A/V 2 or 3452.0=n K mA/V 2For 0=D I , 7713.0==TO GS V V V For 5.0=D I ()()27713.03452.0-=GS V 975.1=⇒GS V V(c) (i) For 0=SB V , 7713.0==TO T V V V (ii) 1=SB V V,()()[1389.025594.0+=∆T V ()]389.02-2525.0=V024.12525.07713.0=+=T V V (iii) 2=SB V V,()()[2389.025594.0+=∆T V ()]389.02-4390.0=V210.14390.07713.0=+=T V V (iv) 4=SB V V,()()[4389.025594.0+=∆T V ()]389.02-7294.0=V501.17294.07713.0=+=T V V _______________________________________10.51()3473.0105.110ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯=fpφ V[]fp SB fp T V V φφγ22-+=∆ ()()[5.23473.0212.0+=()]3473.02- or114.0=∆T V VNow T TO T V V V ∆+= 114.05.0+=TO V 386.0=⇒TO V V_______________________________________10.52 (a) ()()814102001085.89.3--⨯⨯=oxC 710726.1-⨯=F/cm 2oxd s C Ne ∈=2γ()()()()715141910726.11051085.87.11106.12---⨯⨯⨯⨯=2358.0=γV 2/1 (b) ()3294.0105.1105ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=fnφV[]fn BS fn T V V φφγ22-+-=∆ ()()[BS V +-=-3294.022358.022.0()]3294.02-39.2=⇒BS V V_______________________________________10.53(a) +n poly-to-p-type 0.1-=⇒ms φV()288.0105.110ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯=fp φValso 2/14⎥⎦⎤⎢⎣⎡∈=a fp s dTeN x φ()()()()()2/115191410106.1288.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--or410863.0-⨯=dT x cm Now()()()()4151910863.010106.1max --⨯⨯='SDQor()81038.1max -⨯='SDQ C/cm 2 Also()()814104001085.89.3--⨯⨯=∈=ox ox oxt Cor81063.8-⨯=ox C F/cm 2 We find()()91019108105106.1--⨯=⨯⨯='ssQ C/cm 2 Then ()fp ms oxss SDT C Q Q V φφ2max ++'-'=()288.020.11063.81081038.1898+-⎪⎪⎭⎫⎝⎛⨯⨯-⨯=--- or357.0-=T V V(b) For NMOS, apply SB V and T V shifts in apositive direction, so for 0=T V , we want 357.0+=∆T V V. So []fpSB fpoxa s T V C N e V φφ222-+∈=∆or()()()()81514191063.8101085.87.11106.12357.0---⨯⨯⨯=+ ()()[]288.02288.02-+⨯SB V or[]576.0576.0211.0357.0-+=SB V which yields 43.5=SB V V_______________________________________10.54 Plot_______________________________________10.55 (a)()T GS oxn m V V L C W g -=μ ()T GS oxox n V V t L W -∈=μ()()()()()65.0510*******.89.340010814-⨯⨯=-- or26.1=m g mS Nowsm m m s m m mr g g g r g g g +=='⇒+='118.01 which yields⎪⎭⎫⎝⎛-=⎪⎭⎫ ⎝⎛-=18.0126.1118.011ms g r or198.0=s r k Ω(b) For 3=GS V V, 683.0=m g mS Then()()602.0198.0683.01683.0=+='mg mSor88.0683.0602.0=='m m g g which is a 12% reduction._______________________________________10.56(a) The ideal cutoff frequency for no overlapcapacitance is, ()222LV V C g f T GS n gs m T πμπ-==()()()24102275.04400-⨯-=πor17.5=T f GHz (b) Now。
半导体物理_第十章
从上页图中可见，虽然发射结和收集结均处于正 向偏臵状态，但是发射结上的正偏电压还是略高于收 集结上的正偏电压，因此在基区内部仍然存在着过剩 少子电子的浓度梯度，由此形成BJT器件的收集极电 流。下图所示为工作在饱和状态时BJT器件中各区的 能带情况示意图。
最后，我们再给出BJT处于反向放大状态时不同 区域的少数载流子浓度分布，如下图所示，从图中可 见，由于器件发射结处于反偏状态，而收集结处于正 偏状态，电子由收集区注入到基区，最后扩散到发射 结附近并被发射结电场拉向发射区，基区中的过剩少 子电子的浓度梯度也与正向放大状态正好相反。
如果BJT器件的发射区厚度xE与发射区中少数载 流子空穴的扩散长度LE相当或可以比拟的话，则发射 区中过剩少数载流子空穴的浓度为指数分布。 最后我们再来看BJT器件中收集区内部的少子浓 度分布。在BJT器件的收集区中，稳态条件下过剩少 子空穴的双极输运方程为：
由此即可得出BJT器件中收集区内部过剩少数载流子 空穴的浓度分布为：
C. 基区中的少子电子存在着比较大的浓度梯度，因 此电子可以通过扩散流过基区，和正偏的PN结二极管 类似，少子电子在通过中性基区的过程中也会与其中 的多子空穴发生一定的复合。
D. 电子扩散通过基区之后，将进入反偏的收集结空 间电荷区中，收集结中的电场将把扩散过来的电子拉 向收集区，能够被拉向收集区的电子数目取决于由发 射区注入到基区中的电子数目。 E. 流入到收集区中的电子数量（构成收集极电流） 取决于发射结上的偏臵电压，此即双极型晶体管的放 大作用，即：BJT中流过一个端点的电流取决于另外 两个端点上的外加电压。 发生在正向放大模式BJT中的其它次要的物理过程还有： F. 基区中的少子电子将与基区中的多子空穴相复 合，因此基区中的多子空穴必须不断地得到补充，这 个过程构成了基极空穴电流的一个组成部分。
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《半导体物理与器件》第四版答案第十章-CAL-FENGHAI-(2020YEAR-YICAI)_JINGBIANChapter 1010.1(a) p-type; inversion (b) p-type; depletion (c) p-type; accumulation (d) n-type; inversion_______________________________________ 10.2 (a) (i) ⎪⎪⎭⎫ ⎝⎛=iat fpnN V ln φ ()⎪⎪⎭⎫⎝⎛⨯⨯=1015105.1107ln 0259.03381.0=V 2/14⎥⎦⎤⎢⎣⎡∈=a fp s dTeN x φ()()()()()2/1151914107106.13381.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--51054.3-⨯=cmor μ354.0=dT x m (ii) ()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1103ln 0259.0fpφ3758.0=V ()()()()()2/1161914103106.13758.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx51080.1-⨯=cmor μ180.0=dT x m (b) ()03022.03003500259.0=⎪⎭⎫⎝⎛=kT V ⎪⎪⎭⎫ ⎝⎛-=kTE N N n gc iexp 2υ ()()319193003501004.1108.2⎪⎭⎫⎝⎛⨯⨯=⎪⎭⎫⎝⎛-⨯03022.012.1exp221071.3⨯=so 111093.1⨯=i n cm 3-(i)()⎪⎪⎭⎫⎝⎛⨯⨯=11151093.1107ln 03022.0fpφ3173.0=V ()()()()()2/1151914107106.13173.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx51043.3-⨯=cmor μ343.0=dT x m (ii) ()⎪⎪⎭⎫⎝⎛⨯⨯=11161093.1103ln 03022.0fpφ3613.0=V ()()()()()2/1161914103106.13613.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx51077.1-⨯=cmor μ177.0=dT x m_______________________________________ 10.3(a) ()2/14max ⎥⎦⎤⎢⎣⎡∈=='d fn s d dT d SDeN eN x eN Q φ()()[]2/14fn s d eN φ∈= 1st approximation: Let 30.0=fn φV Then()281025.1-⨯()()()()()()[]30.01085.87.114106.11419--⨯⨯=d N 141086.7⨯=⇒d N cm 3- 2nd approximation:()2814.0105.11086.7ln 0259.01014=⎪⎪⎭⎫⎝⎛⨯⨯=fnφVThen()281025.1-⨯()()()()()()[]2814.01085.87.114106.11419--⨯⨯=d N 141038.8⨯=⇒d N cm 3-(b) ()2831.0105.11038.8ln 0259.01014=⎪⎪⎭⎫⎝⎛⨯⨯=fnφV()566.02831.022===fn s φφV_______________________________________ 10.4p-type silicon(a) Aluminum gate⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫⎝⎛++'-'=fp gmms e E φχφφ2 We have ⎪⎪⎭⎫⎝⎛=i a t fp n N V ln φ ()334.0105.1106ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=VThen()[]334.056.025.320.3++-=ms φ or944.0-=ms φV(b) +n polysilicon gate ⎪⎪⎭⎫⎝⎛+-=fp g mseE φφ2()334.056.0+-= or894.0-=ms φV(c) +p polysilicon gate ()334.056.02-=⎪⎪⎭⎫⎝⎛-=fp gms e E φφ or226.0+=ms φV_______________________________________ 10.5()3832.0105.1104ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=fp φV⎪⎪⎭⎫⎝⎛++'-'=fp g m ms e E φχφφ2 ()3832.056.025.320.3++-=9932.0-=ms φV_______________________________________ 10.6(a) 17102⨯≅d N cm 3-(b) Not possible - ms φ is always positive.(c) 15102⨯≅d N cm 3-_______________________________________ 10.7From Problem 10.5, 9932.0-=ms φV oxssms FB C Q V '-=φ (a) ()()814102001085.89.3--⨯⨯=∈=ox ox oxt C710726.1-⨯=F/cm 2 ()()7191010726.1106.11059932.0--⨯⨯⨯--=FB V 040.1-=V(b) ()()81410801085.89.3--⨯⨯=oxC 710314.4-⨯=F/cm 2()()7191010314.4106.11059932.0--⨯⨯⨯--=FB V 012.1-=V_______________________________________ 10.8(a) 42.0-≅ms φV42.0-==ms FB V φV (b)()()781410726.1102001085.89.3---⨯=⨯⨯=oxC F/cm2(i)()()7191010726.1106.1104--⨯⨯⨯-='-=∆ox ss FBC Q V0371.0-=V (ii)()()7191110726.1106.110--⨯⨯-=∆FBV 0927.0-=V (c) 42.0-==ms FB V φV()()781410876.2101201085.89.3---⨯=⨯⨯=ox C F/cm 2(i)()()7191010876.2106.1104--⨯⨯⨯-=∆FBV 0223.0-=V(ii)()()7191110876.2106.110--⨯⨯-=∆FB V0556.0-=V_______________________________________ 10.9⎪⎪⎭⎫⎝⎛++'-'=fp gmms e E φχφφ2 where ()365.0105.1102ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=fpφVThen()365.056.025.320.3++-=ms φ or975.0-=ms φV Now oxssms FB C Q V '-=φ or()ox FB ms ssC V Q -='φ We have()()814104501085.89.3--⨯⨯=∈=ox ox oxt Cor81067.7-⨯=ox C F/cm 2 So now()[]()81067.71975.0-⨯⋅---='ssQ 91092.1-⨯=C/cm 2 or10102.1⨯='eQ sscm 2- _______________________________________ 10.10 ()3653.0105.1102ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=fp φV()()()()()2/1161914102106.13653.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx510174.2-⨯=cm()dT a SDx eN Q ='max()()()5161910174.2102106.1--⨯⨯⨯=810958.6-⨯=C/cm 2()()781410301.2101501085.89.3---⨯=⨯⨯=oxC F/cm 2()fp ms oxss SDTN C Q Q V φφ2max ++'-'=()()71910810301.2106.110710958.6---⨯⨯⨯-⨯=()3653.02++ms φ ms φ+=9843.0(a) n + poly gate on p-type:12.1-≅ms φV136.012.19843.0-=-=TN V V (b) p + poly gate on p-type:28.0+≅ms φV26.128.09843.0+=+=TN V V (c) Al gate on p-type: 95.0-≅ms φV 0343.095.09843.0+=-=TN V V _______________________________________10.11 ()3161.0105.1103ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=fn φV()()()()()2/1151914103106.13161.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx510223.5-⨯=cm()dT d SDx eN Q ='max ()()()5151910223.5103106.1--⨯⨯⨯= 810507.2-⨯=C/cm 2()()781410301.2101501085.89.3---⨯=⨯⨯=oxC F/cm 2()fn ms ox ss SDTPC Q Q V φφ2max -+⎥⎥⎦⎤⎢⎢⎣⎡'+'-=()()⎥⎦⎤⎢⎣⎡⨯⨯⨯+⨯-=---71019810301.2107106.110507.2()3161.02-+ms φms TP V φ+-=7898.0(a) n + poly gate on n-type:41.0-≅ms φV20.141.07898.0-=--=TP V V (b) p + poly gate on n-type:0.1+≅ms φV210.00.17898.0+=+-=TP V V (c) Al gate on n-type: 29.0-≅ms φV 08.129.07898.0-=--=TP V V _______________________________________10.12 ()3294.0105.1105ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=fpφVThe surface potential is()659.03294.022===fp s φφV We have 90.0-='-=oxssms FB C Q V φV Now ()FB s oxSDT V C Q V ++'=φmaxWe obtain 2/14⎥⎦⎤⎢⎣⎡∈=a fp s dTeN x φ()()()()()2/1151914105106.13294.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--or410413.0-⨯=dT x cm Then()()()()4151910413.0105106.1max --⨯⨯⨯='SDQ or()810304.3max -⨯='SDQ C/cm 2 We also find ()()814104001085.89.3--⨯⨯=∈=ox ox oxt Cor810629.8-⨯=ox C F/cm 2 Then90.0659.010629.810304.388-+⨯⨯=--T Vor142.0+=T V V_______________________________________10.13 ()()814102201085.89.3--⨯⨯=∈=ox ox oxt C710569.1-⨯=F/cm 2 ()()1019104106.1⨯⨯='-ssQ9104.6-⨯=C/cm 2 By trial and error, let 16104⨯=a N cm 3-.Now ()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1104ln 0259.0fpφ3832.0=V ()()()()()2/1161914104106.13832.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx510575.1-⨯=cm()max SDQ ' ()()()5161910575.1104106.1--⨯⨯⨯= 710008.1-⨯=C/cm 2 94.0-≅ms φV Then ()fp ms oxss SDTN C Q Q V φφ2max ++'-'=79710569.1104.610008.1---⨯⨯-⨯= ()3832.0294.0+-Then 428.0=TN V V 45.0≅V_______________________________________10.14 ()()814101801085.89.3--⨯⨯=∈=ox ox oxt C7109175.1-⨯=F/cm 3-()()1019104106.1⨯⨯='-ssQ 9104.6-⨯=C/cm 2 By trial and error, let 16105⨯=d N cm 3- Now ()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1105ln 0259.0fnφ3890.0=V()()()()()2/1161914105106.13890.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx510419.1-⨯=cm()max SDQ ' ()()()5161910419.1105106.1--⨯⨯⨯= 710135.1-⨯=C/cm 3- 10.1+≅ms φV Then()()fn ms oxssSDTP C Q Q V φφ2max -+'+'-=()797109175.1104.610135.1---⨯⨯+⨯-=()3890.0210.1-+Then 303.0-=TP V V, which is within the specified value._______________________________________10.15We have 710569.1-⨯=ox C F/cm 29104.6-⨯='ssQ C/cm 2 By trial and error, let 14105⨯=d N cm 3- Now ()⎪⎪⎭⎫⎝⎛⨯⨯=1014105.1105ln 0259.0fnφ2697.0=V ()()()()()2/1141914105106.12697.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx410182.1-⨯=cm()max SDQ ' ()()()4141910182.1105106.1--⨯⨯⨯= 910456.9-⨯=C/cm 2 33.0-≅ms φV Then()()fn ms oxssSDTP C Q Q V φφ2max -+'+'-=⎪⎪⎭⎫⎝⎛⨯⨯+⨯-=---79910569.1104.610456.9 ()2697.0233.0--970.0=VThen 970.0-=TP V V 975.0-≅ V which meets the specification._______________________________________10.16(a) 03.1-≅ms φV ()()814101801085.89.3--⨯⨯=oxC 7109175.1-⨯=F/cm 2Now oxssms FB C Q V '-=φ ()()71019109175.1106106.103.1--⨯⨯⨯--=08.1-=FB V V (b) ()⎪⎪⎭⎫⎝⎛⨯=1015105.110ln 0259.0fpφ2877.0=V ()()()()()2/115191410106.12877.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dTx510630.8-⨯=cm()max SDQ ' ()()()5151910630.810106.1--⨯⨯= 810381.1-⨯=C/cm 2 Now ()fp FB oxSDTN V C Q V φ2max ++'=()2877.0208.1109175.110381.178+-⨯⨯=-- or 433.0-=TN V V_______________________________________10.17(a) We have n-type material under the gate, so 2/14⎥⎦⎤⎢⎣⎡∈==d fn s C dTeN t x φwhere ()288.0105.110ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯=fnφVThen ()()()()()2/115191410106.1288.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dTxor410863.0-⨯==C dT t x cm μ863.0=m (b)()()fn ms oxoxss SDT t Q Q V φφ2max -+⎪⎪⎭⎫⎝⎛∈'+'-= For an +n polysilicon gate,()288.056.02--=⎪⎪⎭⎫⎝⎛--=fn gms e E φφ or272.0-=ms φV Now()()()()4151910863.010106.1max --⨯⨯='SDQor()81038.1max -⨯='SDQ C/cm 2 We have()()91019106.110106.1--⨯=⨯='ssQ C/cm 2We now find()()()()81498105001085.89.3106.11038.1----⨯⨯⨯+⨯-=T V()288.02272.0-- or07.1-=T V V_______________________________________10.18(b) ⎪⎪⎭⎫⎝⎛++'-'=fp gmms e E φχφφ2 where20.0-='-'χφmV and ()3473.0105.110ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯=fpφVThen()3473.056.020.0+--=ms φ or107.1-=ms φV(c) For 0='ssQ ()fp ms oxox SDTN t Q V φφ2max ++⎪⎪⎭⎫⎝⎛∈'= We find ()()()()()2/116191410106.13473.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dTxor41030.0-⨯=dT x cm μ30.0=m Now()()()()416191030.010106.1max --⨯⨯='SDQ or()810797.4max -⨯='SDQ C/cm 2 Then ()()()()14881085.89.31030010797.4---⨯⨯⨯=TV()3473.02107.1+- or00455.0+=T V V 0≅V_______________________________________10.19Plot_______________________________________10.20 Plot_______________________________________10.21 Plot_______________________________________10.22 Plot_______________________________________10.23(a) For 1=f Hz (low freq), ()()814101201085.89.3--⨯⨯=∈=ox ox oxt C710876.2-⨯=F/cm 2a st s ox ox oxFBeNV t C ∈⎪⎪⎭⎫ ⎝⎛∈∈+∈='()()()()()()()16191481410106.11085.87.110259.07.119.3101201085.89.3----⨯⨯⎪⎭⎫ ⎝⎛+⨯⨯= 710346.1-⨯='FBC F/cm 2 dT soxox oxx t C ⋅⎪⎪⎭⎫ ⎝⎛∈∈+∈='minNow()3473.0105.110ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯=fpφV()()()()()2/116191410106.13473.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dTx51000.3-⨯=cmThen()()()5814min1000.37.119.3101201085.89.3---⨯⎪⎭⎫ ⎝⎛+⨯⨯='C810083.3-⨯=F/cm 2C '(inv)710876.2-⨯==ox C F/cm 2 (b) 1=f MHz (high freq), 710876.2-⨯=ox C F/cm 2 (unchanged)710346.1-⨯='FBC F/cm 2 (unchanged)8min10083.3-⨯='C F/cm 2 (unchanged)C '(inv)8min10083.3-⨯='=C F/cm 2 (c) 10.1-≅==ms FB V φV ()fp FB oxSDTN V C Q V φ2max ++'=Now()dT a SDx eN Q ='max ()()()516191000.310106.1--⨯⨯= 81080.4-⨯=C/cm 2 ()3473.0210.110876.21080.478+-⨯⨯=--TNV 2385.0-=TNV V_______________________________________10.24(a) 1=f Hz (low freq), ()()814101201085.89.3--⨯⨯=∈=ox ox oxt C710876.2-⨯=F/cm 2a st s ox ox oxFBeNV t C ∈⎪⎪⎭⎫ ⎝⎛∈∈+∈='()()()()()()()141914814105106.11085.87.110259.07.119.3101201085.89.3⨯⨯⨯⎪⎭⎫⎝⎛+⨯⨯=---- 810726.4-⨯='FBC F/cm 2 dT soxox oxx t C ⋅⎪⎪⎭⎫ ⎝⎛∈∈+∈='minNow()2697.0105.1105ln 0259.01014=⎪⎪⎭⎫⎝⎛⨯⨯=fnφV()()()()()2/1141914105106.12697.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx410182.1-⨯=cmThen()()()4814min10182.17.119.3101201085.89.3---⨯⎪⎭⎫ ⎝⎛+⨯⨯='C910504.8-⨯=F/cm 2C '(inv)710876.2-⨯==ox C F/cm 2 (b) 1=f MHz (high freq), 710876.2-⨯=ox C F/cm 2 (unchanged)810726.4-⨯='FBC F/cm 2 (unchanged)9min10504.8-⨯='C F/cm 2 (unchanged)C '(inv)9min10504.8-⨯='=C F/cm 2 (c) 95.0≅=ms FB V φV ()fn FB oxSDTP V C Q V φ2max -+'-=Now()dT d SDx eN Q ='max ()()()4141910182.1105106.1--⨯⨯⨯= 910456.9-⨯=C/cm 2 Then ()2697.0295.010876.210456.979-+⨯⨯-=--TPV 378.0+=TPV V_______________________________________10.25The amount of fixed oxide charge at x is()x x ∆ρ C/cm 2By lever action, the effect of this oxide chargeon the flatband voltage is ()x x t x C V oxoxFB ∆⎪⎪⎭⎫⎝⎛-=∆ρ1 If we add the effect at each point, wemustintegrate so that ()dx t x x C V oxt oxoxFB ⎰-=∆01ρ _______________________________________10.26(a) We have ρx Q tSS ()='∆Then ∆V C x x t dx FB oxoxoxt =-()z10ρ ≈-'F H G I K J F H I K -z 1C t t Q t dx oxox oxox oxSSt tt ∆∆bg =-'--=-'F H I K 1C Q t t t t Q CoxSS ox oxSS ox∆∆a f or∆V Q t FB SSox ox=-'∈F H G I KJ =-⨯⨯⨯⨯---()16108102001039885101910814...b gb gb g b gor∆V FB =-00742.V(b)We haveρx Q t SS ox()='=⨯⨯⨯--16108102001019108.b gb g =⨯=-64103.ρONow∆V C x x t dx C t xdx FB oxox ox O ox oxoxt t =-=-()zz10ρρor ∆V t FB O oxox=-∈ρ22=-⨯⨯⨯---()6410200102398851038214...bgbg bgor∆V FB =-00371.V (c) ρρx x t O ox()F H G I KJ =We find12216108102001019108t Q ox O SSO ρρ='⇒=⨯⨯⨯--.bgb gor ρO =⨯-128102. Now ∆V C t x x t dx FB oxoxOoxt ox=-⋅⋅F H G I KJ z110ρ=-⋅z122C tx dx oxOoxoxt ρa fwhich becomes ∆V t t xt FB oxoxOoxoxO ox oxt =-∈⋅⋅=-∈F H G I KJ 133232ρρa fThen∆V FB =-⨯⨯⨯---()12810200103398851028214...bgb g b gor 0494.0-=∆FB V V_______________________________________10.27 Sketch_______________________________________10.28 Sketch_______________________________________10.29 (b)⎪⎪⎭⎫⎝⎛-=-=2ln i d a t bi FB n N N V V V ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯-=2101616105.11010ln 0259.0 or695.0-=FB V V(c) Apply 3-=G V V, 3≅ox V VFor 3+=G V V,sdx d ∈-=Eρ n-side: d eN =ρ1C x eN eN dx d sd s d +∈-=E ⇒∈-=E0=E at n x x -=, then sn d xeN C ∈-=1so ()n sdx x eN +∈-=E for 0≤≤-x x n In the oxide, 0=ρ, so=E ⇒=E0dxd constant. From the boundary conditions, in the oxidesnd x eN ∈-=E In the p-region,2C x eN eN dx d sa s a s +∈=E ⇒∈+=∈-=Eρ 0=E at ()p ox x t x +=, then ()[]x x t eN p ox sa-+∈-=E At ox t x =, sn d s p a xeN x eN ∈-=∈-=ESo that n d p a x N x N =Since d a N N =, then p n x x = The potential is ⎰E -=dx φFor zero bias, we can write bi p ox n V V V V =++where p ox n V V V ,, are the voltage drops acrossthe n-region, the oxide, and the p-region,respectively. For the oxide: soxn d ox ox t x eN t V ∈=⋅E = For the n-region:()C x x x eNx V n sdn '+⎪⎪⎭⎫ ⎝⎛⋅+∈=22 Arbitrarily, set 0=n V at n x x -=, thensnd x eN C ∈='22so that()()22n sdn x x eN x V +∈=At 0=x , snd n x eN V ∈=22which is thevoltagedrop across the n-region. Because of symmetry, p n V V =. Then for zero bias, we havebi ox n V V V =+2which can be written asbi soxn d s n d V t x eN x eN =∈+∈2or02=∈-+dsbi ox n n eN V t x x Solving for n x , we obtain dbis ox ox n eN V t t x ∈+⎪⎪⎭⎫ ⎝⎛+-=222 If we apply a voltage G V , then replace bi V byG bi V V +, so()dG bi s ox ox p n eN V V t t x x +∈+⎪⎪⎭⎫ ⎝⎛+-==222 We find2105008-⨯-==p n x x()()()()()1619142810106.1695.31085.87.11210500---⨯⨯+⎪⎪⎭⎫ ⎝⎛⨯+ which yields510646.4-⨯==p n x x cm Now soxn d ox t x eN V ∈=()()()()()()148516191085.87.111050010646.410106.1----⨯⨯⨯⨯=or359.0=ox V V We also findsnd p n x eN V V ∈==22()()()()()142516191085.87.11210646.410106.1---⨯⨯⨯=or67.1==p n V V V_______________________________________10.30(a) n-type(b) We have 731210110210200---⨯=⨯⨯=ox C F/cm 2 Also()()7141011085.89.3--⨯⨯=∈=⇒∈=ox ox ox ox ox oxC t t Cor61045.3-⨯=ox t cm 5.34=nm oA 345= (c) oxss ms FB C Q V '-=φ or71050.080.0-'--=-ss Qwhich yields8103-⨯='ssQ C/cm 21110875.1⨯=cm 2- (d)⎪⎪⎭⎫ ⎝⎛∈⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛∈∈+∈='dss ox ox oxFBeN e kT t C()()[][6141045.31085.89.3--⨯÷⨯=()()()()()⎥⎥⎦⎤⨯⨯⨯⎪⎭⎫ ⎝⎛+--161914102106.11085.87.110259.07.119.3which yields81082.7-⨯='FBC F/cm 2 or156=FB C pF_______________________________________10.31(a) Point 1: Inversion 2: Threshold 3: Depletion 4: Flat-band 5: Accumulation_______________________________________10.32We have()()[]fp ms x GS ox nV V C Q φφ2+---='()()max SD ssQ Q '+'- Now let DS x V V =, so()⎩⎨⎧--='DS GS ox nV V C Q()()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡+-'+'+fpms ox ss SDC Q Q φφ2maxFor a p-type substrate, ()max SD Q ' is a negative value, so we can write()⎩⎨⎧--='DS GS ox nV V C Q()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡++'-'-fp ms ox ss SD C Q Q φφ2maxUsing the definition of thresholdvoltage T V , we have()[]T DS GS ox nV V V C Q ---=' At saturation()T GS DS DS V V sat V V -==which then makes nQ 'equal to zero at thedrain terminal._______________________________________10.33 (a) ()[]222DS DS T GS n D V V V V LW k I --⋅'=()()()()[]22.02.04.08.028218.0--⎪⎭⎫ ⎝⎛=0864.0=mA(b) ()22T GS n D V V L W k I -⋅'=()()24.08.08218.0-⎪⎭⎫ ⎝⎛=1152.0=mA(c) Same as (b), 1152.0=D I mA(d) ()22T GS n D V V L W k I -⋅'=()()24.02.18218.0-⎪⎭⎫ ⎝⎛=4608.0=mA_______________________________________10.34 (a) ()[]222SDSD T SG p D V V V V LW k I -+⋅'=()()()()[]225.025.04.08.0215210.0--⎪⎭⎫ ⎝⎛=103.0=D I mA(b) ()22T SG p D V V L Wk I +⋅'=()()24.08.015210.0-⎪⎭⎫ ⎝⎛=12.0=mA(c) ()22T SG p D V V L W k I +⋅'=()()24.02.115210.0-⎪⎭⎫ ⎝⎛=48.0=mA(d) Same as (c), 48.0=D I mA_______________________________________10.35(a) ()22T GS n D V V LW k I -⋅'=()28.04.126.00.1-⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=L W26.9=⇒LW(b) ()()28.085.126.926.0-⎪⎭⎫ ⎝⎛=DI 06.3=mA(c) ()[]222DS DS T GS n D V V V V LW k I --⋅'=()()()()[]215.015.08.02.1226.926.0--⎪⎭⎫ ⎝⎛=271.0=mA_______________________________________10.36(a) Assume biased in saturation region()22T SG p D V V L Wk I +⋅'=()()2020212.010.0T V +⎪⎭⎫ ⎝⎛=289.0+=⇒T V VNote: 0.1=SD V V 289.00+=+>T SG V V V So the transistor is biased in the saturation region.(b) ()()2289.04.020212.0+⎪⎭⎫ ⎝⎛=D I570.0=mA(c) ()()[()15.0289.06.0220212.0+⎪⎭⎫ ⎝⎛=D I()]215.0- or293.0=D I mA_______________________________________10.37()()781410138.3101101085.89.3---⨯=⨯⨯=oxC F/cm 2()()()()2.122010138.342527-⨯==L W C K ox n n μ 310111.1-⨯=A/V 2=1.111 mA/V 2 (a) 0=GS V , 0=D I6.0=GS V V, ()15.0=sat V DS V, ()()()245.06.0111.1-=sat I D025.0=mA2.1=GS V V, ()75.0=sat V DS V, ()()()245.02.1111.1-=sat I D 625.0=mA 8.1=GS V V, ()35.1=sat V DS V, ()()()245.08.1111.1-=sat I D 025.2=mA 4.2=GS V V, ()95.1=sat V DS V, ()()()245.04.2111.1-=sat I D 225.4=mA (c)0=D I for 45.0≤GS V V 6.0=GS V V,()()()()[]21.01.045.06.02111.1--=D I 0222.0=mA 2.1=GS V V,()()()()[]21.01.045.02.12111.1--=D I 156.0=mA 8.1=GS V V,()()()()[]21.01.045.08.12111.1--=D I 289.0=mA 4.2=GS V V,()()()()[]21.01.045.04.22111.1--=D I 422.0=mA_______________________________________10.38 ()()814101101085.89.3--⨯⨯=∈=ox ox oxt C710138.3-⨯=F/cm 2 L WC K ox p p 2μ=()()()()2.123510138.32107-⨯=41061.9-⨯=A/V 2=0.961 mA/V 2(a) 0=SG V , 0=D I6.0=SG V V, ()25.0=sat V SD V ()()()235.06.0961.0-=sat I D060.0=mA2.1=SG V V, ()85.0=sat V SD V ()()()235.02.1961.0-=sat I D 694.0=mA8.1=SG V V, ()45.1=sat V SD V ()()()235.08.1961.0-=sat I D 02.2=mA4.2=SG V V, ()05.2=sat V SD V ()()()235.04.2961.0-=sat I D 04.4=mA (c)0=D I for 35.0≤SG V V6.0=SG V V()()()()[]21.01.035.06.02961.0--=D I 0384.0=mA 2.1=SG V V()()()()[]21.01.035.02.12961.0--=D I 154.0=mA 8.1=SG V V()()()()[]21.01.035.08.12961.0--=D I 269.0=mA 4.2=SG V V()()()()[]21.01.035.04.22961.0--=D I 384.0=mA_______________________________________10.39(a) From Problem10.37,111.1=n K mA/V 2 For 8.0-=GS V V, 0=D I0=GS V , ()8.0=sat V DS V ()()()28.00111.1+=sat I D 711.0=mA8.0+=GS V V, ()6.1=sat V DS V ()()()28.08.0111.1+=sat I D 84.2=mA6.1=GS V V, ()4.2=sat V DS V ()()()28.06.1111.1+=sat I D 40.6=mA_______________________________________10.40 Sketch_______________________________________10.41 Sketch_______________________________________10.42We have()T DS T GS DS V V V V sat V -=-= so that()T DS DS V sat V V +=Since ()sat V V DS DS >, the transistor is alwaysbiased in the saturation region. Then ()2T GS n D V V K I -=where, from Problem 10.37,111.1=n K mA/V 2and 45.0=T V V______10.43From Problem 10.38, 961.0=p K mA/V 2()()[]22SD SD T SG p D V V V V K I -+= ()T SG p V SDDd V V K V I g SD +=∂∂=→20For 35.0≤SG V V, 0=d gFor 35.0>SG V V,()()35.0961.02-=SG d V g For 4.2=SG V V,()()35.04.2961.02-=d g 94.3=mA/V_______________________________________10.44 (a) GSDm V I g ∂∂=()()[]{}22DSDS T GS n GSV V V V K V --∂∂=()DS n V K 2= ()()05.0225.1n K = 5.12=⇒n K mA/V 2(b) ()()()[()]205.005.03.08.025.12--=D I 594.0=mA (c) ()()23.08.05.12-=D I 125.3=mA_______________________________________10.45We find that 2.0≅T V V Now ()()T GS oxn D V V LC W sat I -⋅=2μ where ()()814104251085.89.3--⨯⨯=∈=ox ox oxt Cor81012.8-⨯=ox C F/cm 2We are given 10=L W . From the graph, for3=GS V V, we have ()033.0≅sat I D , then ()2.032033.0-⋅=LC W oxn μ or310139.02-⨯=L C W oxn μ or()()3810139.01012.81021--⨯=⨯n μ which yields342=n μcm 2/V-s_______________________________________10.46 (a)()T GS DS V V sat V -= or8.48.04=⇒-=GS GS V V V(b)()()()sat V K V V K sat I DSn T GS n D 22=-= so()244102n K =⨯- which yieldsμ5.12=n K A/V 2 (c)()2.18.02=-=-=T GS DS V V sat V V so ()sat V V DS DS >()()()258.021025.1-⨯=-sat I D or()μ18=sat I D A (d)()sat V V DS DS <()[]22DSDS T GS n D V V V V K I --= ()()()()[]25118.0321025.1--⨯=- orμ5.42=D I A_______________________________________10.47 (a) ()()814101801085.89.3--⨯⨯=oxC7109175.1-⨯=F/cm 2(i)()()7109175.1450-⨯=='ox n nC k μ 510629.8-⨯=A/V 2or μ29.86='nk A/V 2 (ii)()()22T GS n D V V L W k sat I -⎪⎭⎫ ⎝⎛⎪⎪⎭⎫⎝⎛'= ()24.02208629.08.0-⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=L W24.7=⇒LW(b) (i) ()()7109175.1210-⨯=='ox p p C k μ510027.4-⨯=A/V 2 or μ27.40='p k A/V 2(ii) ()()22T SG p D V V L W k sat I +⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛'= ()24.02204027.08.0-⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=L W5.15=⇒LW_______________________________________10.48From Problem 10.37, 111.1=n K mA/V 2(a) ()()[]{}22DSDS T GS n GSmL V V V V K V g --∂∂=()()()()1.02111.12==DS n V Kso 222.0=mL g mA/V (b) (){}2T GS n GSms V V K V g -∂∂=()()()45.05.1111.122-=-=T GS n V V Kso 33.2=ms g mA/V_______________________________________10.49From Problem 10.38, 961.0=p K mA/V 2(a) ()()[]{}22SDSD T SG p SGmL V V V V K V g -+∂∂=()()()()1.02961.02==SD p V Kor 192.0=mL g mA/V (b) ()[]2T SG p SGms V V K V g +∂∂=()()()35.05.1961.022-=+=T SG p V V Kor 21.2=ms g mA/V_______________________________________10.50 (a) oxa s C N e ∈=2γNow ()()814101501085.89.3--⨯⨯=ox C 710301.2-⨯=F/cm 2Then()()()()716141910301.21051085.87.11106.12---⨯⨯⨯⨯=γ 5594.0=γV 2/1 (b) ()3890.0105.1105ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=fp φV(i)()()()()()2/1161914105106.13890.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx510419.1-⨯=cm()max SDQ ' ()()()5161910419.1105106.1--⨯⨯⨯= 710135.1-⨯=C/cm 2()fp FB oxSDTO V C Q V φ2max ++'=()3890.025.010301.210135.177+-⨯⨯=--7713.0=VLWC K ox n n 2μ=()()()()2.12810301.24507-⨯=410452.3-⨯=A/V 2 or 3452.0=n K mA/V 2For 0=D I , 7713.0==TO GS V V V For 5.0=D I ()()27713.03452.0-=GS V 975.1=⇒GS V V(c) (i) For 0=SB V , 7713.0==TO T V V V (ii) 1=SB V V,()()[1389.025594.0+=∆T V ()]389.02-2525.0=V024.12525.07713.0=+=T V V (iii) 2=SB V V,()()[2389.025594.0+=∆T V ()]389.02-4390.0=V210.14390.07713.0=+=T V V (iv) 4=SB V V,()()[4389.025594.0+=∆T V ()]389.02-7294.0=V501.17294.07713.0=+=T V V _______________________________________10.51()3473.0105.110ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯=fpφ V[]fp SB fp T V V φφγ22-+=∆ ()()[5.23473.0212.0+=()]3473.02- or114.0=∆T V VNow T TO T V V V ∆+= 114.05.0+=TO V 386.0=⇒TO V V_______________________________________10.52 (a) ()()814102001085.89.3--⨯⨯=oxC 710726.1-⨯=F/cm 2oxd s C Ne ∈=2γ()()()()715141910726.11051085.87.11106.12---⨯⨯⨯⨯=2358.0=γV 2/1 (b) ()3294.0105.1105ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=fnφV[]fn BS fn T V V φφγ22-+-=∆ ()()[BS V +-=-3294.022358.022.0()]3294.02-39.2=⇒BS V V_______________________________________10.53(a) +n poly-to-p-type 0.1-=⇒ms φV()288.0105.110ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯=fp φValso 2/14⎥⎦⎤⎢⎣⎡∈=a fp s dTeN x φ()()()()()2/115191410106.1288.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--or410863.0-⨯=dT x cm Now()()()()4151910863.010106.1max --⨯⨯='SDQor()81038.1max -⨯='SDQ C/cm 2 Also()()814104001085.89.3--⨯⨯=∈=ox ox oxt Cor81063.8-⨯=ox C F/cm 2 We find()()91019108105106.1--⨯=⨯⨯='ssQ C/cm 2 Then ()fp ms oxss SDT C Q Q V φφ2max ++'-'=()288.020.11063.81081038.1898+-⎪⎪⎭⎫⎝⎛⨯⨯-⨯=--- or357.0-=T V V(b) For NMOS, apply SB V and T V shifts in apositive direction, so for 0=T V , we want 357.0+=∆T V V. So []fpSB fpoxa s T V C N e V φφ222-+∈=∆or()()()()81514191063.8101085.87.11106.12357.0---⨯⨯⨯=+ ()()[]288.02288.02-+⨯SB V or[]576.0576.0211.0357.0-+=SB V which yields 43.5=SB V V_______________________________________10.54 Plot_______________________________________10.55 (a)()T GS oxn m V V L C W g -=μ ()T GS oxox n V V t L W -∈=μ()()()()()65.0510*******.89.340010814-⨯⨯=-- or26.1=m g mS Nowsm m m s m m mr g g g r g g g +=='⇒+='118.01 which yields⎪⎭⎫⎝⎛-=⎪⎭⎫ ⎝⎛-=18.0126.1118.011ms g r or198.0=s r k Ω(b) For 3=GS V V, 683.0=m g mS Then()()602.0198.0683.01683.0=+='mg mSor88.0683.0602.0=='m m g g which is a 12% reduction._______________________________________10.56(a) The ideal cutoff frequency for no overlapcapacitance is, ()222LV V C g f T GS n gs m T πμπ-==()()()24102275.04400-⨯-=πor17.5=T f GHz (b) Now。