半导体物理与器件第四版课后习题答案.doc

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半导体物理与器件(尼曼第四版)答案

半导体物理与器件(尼曼第四版)答案第一章：半导体材料与晶体1.1 半导体材料的基本特性半导体材料是一种介于导体和绝缘体之间的材料。

它的基本特性包括：1.带隙：半导体材料的价带与导带之间存在一个禁带或带隙，是电子在能量上所能占据的禁止区域。

2.拉伸系统：半导体材料的结构是由原子或分子构成的晶格结构，其中的原子或分子以确定的方式排列。

3.载流子：在半导体中，存在两种载流子，即自由电子和空穴。

自由电子是在导带上的，在外加电场存在的情况下能够自由移动的电子。

空穴是在价带上的，当一个价带上的电子从该位置离开时，会留下一个类似电子的空位，空穴可以看作电子离开后的痕迹。

4.掺杂：为了改变半导体材料的导电性能，通常会对其进行掺杂。

掺杂是将少量元素添加到半导体材料中，以改变载流子浓度和导电性质。

1.2 半导体材料的结构与晶体缺陷半导体材料的结构包括晶体结构和非晶态结构。

晶体结构是指材料具有有序的周期性排列的结构，而非晶态结构是指无序排列的结构。

晶体结构的特点包括：1.晶体结构的基本单位是晶胞，晶胞在三维空间中重复排列。

2.晶格常数是晶胞边长的倍数，用于描述晶格的大小。

3.晶体结构可分为离子晶体、共价晶体和金属晶体等不同类型。

晶体结构中可能存在各种晶体缺陷，包括：1.点缺陷：晶体中原子位置的缺陷，主要包括实际缺陷和自间隙缺陷两种类型。

2.线缺陷：晶体中存在的晶面上或晶内的线状缺陷，主要包括位错和脆性断裂两种类型。

3.面缺陷：晶体中存在的晶面上的缺陷，主要包括晶面位错和穿孔两种类型。

1.3 半导体制备与加工半导体制备与加工是指将半导体材料制备成具有特定电性能的器件的过程。

它包括晶体生长、掺杂、薄膜制备和微电子加工等步骤。

晶体生长是将半导体材料从溶液或气相中生长出来的过程。

常用的晶体生长方法包括液相外延法、分子束外延法和气相外延法等。

掺杂是为了改变半导体材料的导电性能，通常会对其进行掺杂。

常用的掺杂方法包括扩散法、离子注入和分子束外延法等。

半导体物理与器件第四版课后习题答案4复习进程

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半导体物理与器件第四版课后习题标准答案

半导体物理与器件第四版课后习题答案————————————————————————————————作者：————————————————————————————————日期：2______________________________________________________________________________________3Chapter 33.1If o a were to increase, the bandgap energy would decrease and the material would beginto behave less like a semiconductor and morelike a metal. If o a were to decrease, the bandgap energy would increase and thematerial would begin to behave more like an insulator._______________________________________ 3.2Schrodinger's wave equation is:()()()t x x V x t x m ,,2222ψ⋅+∂ψ∂-η()tt x j ∂ψ∂=,ηAssume the solution is of the form:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x ηexp , Region I: ()0=x V . Substituting theassumed solution into the wave equation, we obtain:()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧∂∂-t E kx j x jku x m ηηexp 22 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u ηexp()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⋅⎪⎭⎫ ⎝⎛-=t E kx j x u jE j ηηηexp which becomes()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m ηηexp 222()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jk ηexp 2()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u ηexp 22()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+=t E kx j x Eu ηexp This equation may be written as()()()()0222222=+∂∂+∂∂+-x u mE x x u x x u jk x u k ηSetting ()()x u x u 1= for region I, the equation becomes: ()()()()021221212=--+x u k dx x du jk dxx u d α where222ηmE=αQ.E.D.In Region II, ()O V x V =. Assume the same form of the solution:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x ηexp , Substituting into Schrodinger's wave equation, we find:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m ηηexp 222()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jk ηexp 2()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u ηexp 22()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+t E kx j x u V O ηexp()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=t E kx j x Eu ηexp This equation can be written as:______________________________________________________________________________________4()()()2222xx u x x u jk x u k ∂∂+∂∂+- ()()02222=+-x u mEx u mV O ηη Setting ()()x u x u 2= for region II, this equation becomes()()dx x du jkdx x u d 22222+()022222=⎪⎪⎭⎫ ⎝⎛+--x u mV k O ηα where again222ηmE=αQ.E.D._______________________________________ 3.3We have ()()()()021221212=--+x u k dx x du jk dxx u d α Assume the solution is of the form: ()()[]x k j A x u -=αexp 1 ()[]x k j B +-+αexpThe first derivative is()()()[]x k j A k j dxx du --=ααexp 1 ()()[]x k j B k j +-+-ααexpand the second derivative becomes()()[]()[]x k j A k j dxx u d --=ααexp 2212()[]()[]x k j B k j +-++ααexp 2Substituting these equations into the differential equation, we find()()[]x k j A k ---ααexp 2()()[]x k j B k +-+-ααexp 2(){()[]x k j A k j jk --+ααexp 2()()[]}x k j B k j +-+-ααexp ()()[]{x k j A k ---ααexp 22 ()[]}0exp =+-+x k j B α Combining terms, we obtain()()()[]222222αααα----+--k k k k k ()[]x k j A -⨯αexp()()()[]222222αααα--++++-+k k k k k()[]0exp =+-⨯x k j B α We find that00= Q.E.D.For the differential equation in ()x u 2 and theproposed solution, the procedure is exactly the same as above._______________________________________ 3.4We have the solutions()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp for a x <<0 and()()[]x k j C x u -=βexp 2()[]x k j D +-+βexp for 0<<-x b .The first boundary condition is ()()0021u u =which yields0=--+D C B AThe second boundary condition is201===x x dx dudx du which yields()()()C k B k A k --+--βαα ()0=++D k βThe third boundary condition is ()()b u a u -=21 which yields()[]()[]a k j B a k j A +-+-ααexp exp ()()[]b k j C --=βexp ()()[]b k j D -+-+βexp______________________________________________________________________________________5and can be written as()[]()[]a k j B a k j A +-+-ααexp exp ()[]b k j C ---βexp ()[]0exp =+-b k j D βThe fourth boundary condition isbx a x dx dudx du -===21 which yields()()[]a k j A k j --ααexp()()[]a k j B k j +-+-ααexp()()()[]b k j C k j ---=ββexp()()()[]b k j D k j -+-+-ββexp and can be written as()()[]a k j A k --ααexp()()[]a k j B k +-+-ααexp ()()[]b k j C k ----ββexp()()[]0exp =+++b k j D k ββ_______________________________________ 3.5(b) (i) First point: πα=aSecond point: By trial and error, πα729.1=a (ii) First point: πα2=aSecond point: By trial and error, πα617.2=a _______________________________________ 3.6(b) (i) First point: πα=aSecond point: By trial and error, πα515.1=a (ii) First point: πα2=aSecond point: By trial and error, πα375.2=a _______________________________________ 3.7ka a aaP cos cos sin =+'ααα Let y ka =, x a =α Theny x x xP cos cos sin =+'Consider dydof this function.()[]{}y x x x P dyd sin cos sin 1-=+⋅'- We find()()()⎭⎬⎫⎩⎨⎧⋅+⋅-'--dy dx x x dy dx x x P cos sin 112y dydx x sin sin -=-Theny x x x x x P dy dx sin sin cos sin 12-=⎭⎬⎫⎩⎨⎧-⎥⎦⎤⎢⎣⎡+-' For πn ka y ==,...,2,1,0=n 0sin =⇒y So that, in general,()()dkd ka d a d dy dxαα===0 And22ηmE=α SodkdEm mE dk d ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-22/122221ηηα This implies thatdk dE dk d ==0α for an k π= _______________________________________ 3.8(a) πα=a 1π=⋅a E m o 212η______________________________________________________________________________________6()()()()2103123422221102.41011.9210054.12---⨯⨯⨯==ππa m E o η19104114.3-⨯=J From Problem 3.5 πα729.12=aπ729.1222=⋅a E m o η()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J12E E E -=∆1918104114.3100198.1--⨯-⨯= 19107868.6-⨯=Jor 24.4106.1107868.61919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J From Problem 3.5, πα617.24=aπ617.2224=⋅a E m o η()()()()2103123424102.41011.9210054.1617.2---⨯⨯⨯=πE18103364.2-⨯=J34E E E -=∆1818103646.1103364.2--⨯-⨯= 1910718.9-⨯=Jor 07.6106.110718.91919=⨯⨯=∆--E eV_______________________________________ 3.9(a) At π=ka , πα=a 1π=⋅a E m o 212η()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα859.0=a o()()()()210312342102.41011.9210054.1859.0---⨯⨯⨯=πo E19105172.2-⨯=J o E E E -=∆11919105172.2104114.3--⨯-⨯= 2010942.8-⨯=Jor 559.0106.110942.81920=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J At π=ka . From Problem 3.5,πα729.12=aπ729.1222=⋅a E m o η()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J 23E E E -=∆1818100198.1103646.1--⨯-⨯= 19104474.3-⨯=Jor 15.2106.1104474.31919=⨯⨯=∆--E eV_____________________________________________________________________________________________________________________________73.10(a) πα=a 1π=⋅a E m o 212η()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JFrom Problem 3.6, πα515.12=aπ515.1222=⋅a E m o η()()()()2103123422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J 12E E E -=∆1919104114.310830.7--⨯-⨯= 19104186.4-⨯=Jor 76.2106.1104186.41919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JFrom Problem 3.6, πα375.24=aπ375.2224=⋅a E m o η()()()()2103123424102.41011.9210054.1375.2---⨯⨯⨯=πE18109242.1-⨯=J 34E E E -=∆1818103646.1109242.1--⨯-⨯=1910597.5-⨯=Jor 50.3106.110597.51919=⨯⨯=∆--E eV _____________________________________3.11(a) At π=ka , πα=a 1π=⋅a E m o 212η()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=J At 0=ka , By trial and error, πα727.0=a o π727.022=⋅a E m o o η()()()()210312342102.41011.9210054.1727.0---⨯⨯⨯=πo E19108030.1-⨯=J o E E E -=∆11919108030.1104114.3--⨯-⨯= 19106084.1-⨯=Jor 005.1106.1106084.11919=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka , From Problem 3.6, πα515.12=aπ515.1222=⋅a E m o η()()()()2103423422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J 23E E E -=∆191810830.7103646.1--⨯-⨯=______________________________________________________________________________________81910816.5-⨯=Jor 635.3106.110816.51919=⨯⨯=∆--E eV_______________________________________ 3.12For 100=T K,()()⇒+⨯-=-1006361001073.4170.124g E164.1=g E eV200=T K, 147.1=g E eV 300=T K, 125.1=g E eV 400=T K, 097.1=g E eV 500=T K, 066.1=g E eV 600=T K, 032.1=g E eV _______________________________________ 3.13The effective mass is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m ηWe have()()B curve dkEd A curve dk E d 2222>so that ()()B curve m A curve m **<_______________________________________ 3.14The effective mass for a hole is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m p η We have that()()B curve dkEd A curve dk E d 2222> so that ()()B curve m A curve m p p **<_______________________________________ 3.15Points A,B: ⇒<0dkdEvelocity in -x direction Points C,D: ⇒>0dkdEvelocity in +x directionPoints A,D: ⇒<022dk Ednegative effective massPoints B,C: ⇒>022dkEd positive effective mass_______________________________________ 3.16For A: 2k C E i =At 101008.0+⨯=k m 1-, 05.0=E eV Or()()2119108106.105.0--⨯=⨯=E JSo ()2101211008.0108⨯=⨯-C3811025.1-⨯=⇒CNow ()()38234121025.1210054.12--*⨯⨯==C m η 311044.4-⨯=kgor o m m ⋅⨯⨯=--*31311011.9104437.4 o m m 488.0=*For B: 2k C E i =At 101008.0+⨯=k m 1-, 5.0=E eV Or ()()2019108106.15.0--⨯=⨯=E JSo ()2101201008.0108⨯=⨯-C 3711025.1-⨯=⇒CNow ()()37234121025.1210054.12--*⨯⨯==C m η321044.4-⨯=kgor o m m ⋅⨯⨯=--*31321011.9104437.4o m m 0488.0=*_____________________________________________________________________________________________________________________________93.17For A: 22k C E E -=-υ()()()2102191008.0106.1025.0⨯-=⨯--C3921025.6-⨯=⇒C ()()39234221025.6210054.12--*⨯⨯-=-=C m η31108873.8-⨯-=kgor o m m ⋅⨯⨯-=--*31311011.9108873.8o m m 976.0--=* For B: 22k C E E -=-υ()()()2102191008.0106.13.0⨯-=⨯--C382105.7-⨯=⇒C()()3823422105.7210054.12--*⨯⨯-=-=C m η3210406.7-⨯-=kgor o m m ⋅⨯⨯-=--*31321011.910406.7o m m 0813.0-=*_______________________________________ 3.18(a) (i) νh E =or ()()341910625.6106.142.1--⨯⨯==h E ν1410429.3⨯=Hz(ii) 141010429.3103⨯⨯===νλc E hc 51075.8-⨯=cm 875=nm(b) (i) ()()341910625.6106.112.1--⨯⨯==h E ν 1410705.2⨯=Hz(ii) 141010705.2103⨯⨯==νλc410109.1-⨯=cm 1109=nm _______________________________________ 3.19(c) Curve A: Effective mass is a constantCurve B: Effective mass is positive around 0=k , and is negativearound 2π±=k ._______________________________________ 3.20()[]O O k k E E E --=αcos 1 Then()()()[]O k k E dkdE---=ααsin 1()[]O k k E -+=ααsin 1 and()[]O k k E dkEd -=ααcos 2122Then221222*11ηηαE dk Ed m o k k =⋅== or212*αE m η=_______________________________________ 3.21(a) ()[]3/123/24l t dnm m m =*()()[]3/123/264.1082.04o o m m =o dnm m 56.0=*(b)oo l t cn m m m m m 64.11082.02123+=+=*oo m m 6098.039.24+=o cnm m 12.0=*_______________________________________ 3.22(a) ()()[]3/22/32/3lh hh dp m m m +=*______________________________________________________________________________________10()()[]3/22/32/3082.045.0o o m m +=[]o m ⋅+=3/202348.030187.0o dpm m 473.0=*(b) ()()()()2/12/12/32/3lh hh lh hh cpm m m m m ++=* ()()()()om ⋅++=2/12/12/32/3082.045.0082.045.0 o cpm m 34.0=*_______________________________________3.23For the 3-dimensional infinite potential well, ()0=x V when a x <<0, a y <<0, and a z <<0. In this region, the wave equation is:()()()222222,,,,,,z z y x y z y x x z y x ∂∂+∂∂+∂∂ψψψ()0,,22=+z y x mEψηUse separation of variables technique, so let ()()()()z Z y Y x X z y x =,,ψSubstituting into the wave equation, we have222222zZXY y Y XZ x X YZ ∂∂+∂∂+∂∂ 022=⋅+XYZ mEηDividing by XYZ , we obtain021*********=+∂∂⋅+∂∂⋅+∂∂⋅ηmE z Z Z y Y Y x X X Let01222222=+∂∂⇒-=∂∂⋅X k x X k x X X xx The solution is of the form: ()x k B x k A x X x x cos sin += Since ()0,,=z y x ψ at 0=x , then ()00=Xso that 0=B .Also, ()0,,=z y x ψ at a x =, so that()0=a X . Then πx x n a k = where...,3,2,1=x n Similarly, we have2221y k y Y Y -=∂∂⋅ and 2221z k zZ Z -=∂∂⋅From the boundary conditions, we find πy y n a k = and πz z n a k = where...,3,2,1=y n and ...,3,2,1=z n From the wave equation, we can write022222=+---ηmE k k k z y xThe energy can be written as()222222⎪⎭⎫ ⎝⎛++==a n n n m E E z y x n n n z y x πη _______________________________________ 3.24The total number of quantum states in the 3-dimensional potential well is given (in k-space) by()332a dk k dk k g T ⋅=ππ where222ηmEk =We can then writeηmEk 2=Taking the differential, we obtaindE Em dE E m dk ⋅⋅=⋅⋅⋅⋅=2112121ηηSubstituting these expressions into the densityof states function, we have()dE EmmE a dE E g T ⋅⋅⋅⎪⎭⎫ ⎝⎛=212233ηηππ Noting thatπ2h=ηthis density of states function can be simplified and written as______________________________________________________________________________________()()dE E m h a dE E g T ⋅⋅=2/33324πDividing by 3a will yield the density of states so that()()E hm E g ⋅=32/324π _______________________________________ 3.25For a one-dimensional infinite potential well,222222k an E m n ==*πη Distance between quantum states()()aa n a n k k n n πππ=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=-+11Now()⎪⎭⎫ ⎝⎛⋅=a dkdk k g T π2NowE m k n*⋅=21ηdE Em dk n⋅⋅⋅=*2211η Then()dE Em a dE E g n T ⋅⋅⋅=*2212ηπDivide by the "volume" a , so()Em E g n *⋅=21πηSo()()()()()EE g 31341011.9067.0210054.11--⨯⋅⨯=π ()EE g 1810055.1⨯=m 3-J 1-_______________________________________ 3.26(a) Silicon, o nm m 08.1=*()()c nc E E h m E g -=*32/324π()dE E E h m g kTE E c nc c c⋅-=⎰+*232/324π()()kT E E c nc cE E h m 22/332/33224+*-⋅⋅=π()()2/332/323224kT h m n⋅⋅=*π ()()[]()()2/33342/33123210625.61011.908.124kT ⋅⋅⨯⨯=--π ()()2/355210953.7kT ⨯=(i) At 300=T K, 0259.0=kT eV ()()19106.10259.0-⨯=2110144.4-⨯=J Then()()[]2/3215510144.4210953.7-⨯⨯=c g25100.6⨯=m 3- or 19100.6⨯=c g cm 3-(ii) At 400=T K, ()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV()()19106.1034533.0-⨯=21105253.5-⨯=J Then()()[]2/32155105253.5210953.7-⨯⨯=c g2510239.9⨯=m 3-or 191024.9⨯=c g cm 3-(b) GaAs, o nm m 067.0=*()()[]()()2/33342/33123210625.61011.9067.024kT g c ⋅⋅⨯⨯=--π ()()2/3542102288.1kT ⨯=______________________________________________________________________________________(i) At 300=T K, 2110144.4-⨯=kT J()()[]2/3215410144.42102288.1-⨯⨯=c g2310272.9⨯=m 3- or 171027.9⨯=c g cm 3-(ii) At 400=T K, 21105253.5-⨯=kT J()()[]2/32154105253.52102288.1-⨯⨯=c g2410427.1⨯=m 3- 181043.1⨯=c g cm 3-_______________________________________ 3.27(a) Silicon, o p m m 56.0=* ()()E E h mE g p-=*υυπ32/324()dE E E h m g E kTE p⋅-=⎰-*υυυυπ332/324()()υυυπE kTE pE E h m 32/332/33224-*-⎪⎭⎫ ⎝⎛-=()()[]2/332/333224kT h mp-⎪⎭⎫ ⎝⎛-=*π ()()[]()()2/33342/33133210625.61011.956.024kT ⎪⎭⎫ ⎝⎛⨯⨯=--π ()()2/355310969.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215510144.4310969.2-⨯⨯=υg2510116.4⨯=m 3- or 191012.4⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J ()()[]2/32155105253.5310969.2-⨯⨯=υg2510337.6⨯=m 3-or 191034.6⨯=υg cm 3- (b) GaAs, o p m m 48.0=*()()[]()()2/33342/33133210625.61011.948.024kT g ⎪⎭⎫ ⎝⎛⨯⨯=--πυ ()()2/3553103564.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215510144.43103564.2-⨯⨯=υg2510266.3⨯=m 3- or 191027.3⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.53103564.2-⨯⨯=υg2510029.5⨯=m 3-or 191003.5⨯=υg cm 3-_______________________________________ 3.28(a) ()()c nc E E h m E g -=*32/324π()()[]()c E E -⨯⨯=--3342/33110625.61011.908.124πc E E -⨯=56101929.1 For c E E =; 0=c g1.0+=c E E eV; 4610509.1⨯=c g m 3-J 1-2.0+=c E E eV;4610134.2⨯=m 3-J 1- 3.0+=c E E eV; 4610614.2⨯=m 3-J 1- 4.0+=c E E eV; 4610018.3⨯=m 3-J 1-(b) ()E E hm g p-=*υυπ32/324()()[]()E E -⨯⨯=--υπ3342/33110625.61011.956.024E E -⨯=υ55104541.4______________________________________________________________________________________For υE E =; 0=υg 1.0-=υE E eV; 4510634.5⨯=υg m 3-J 1-2.0-=υE E eV;4510968.7⨯=m 3-J 1-3.0-=υE E eV; 4510758.9⨯=m 3-J 1-4.0-=υE E eV;4610127.1⨯=m 3-J 1-_______________________________________ 3.29(a) ()()68.256.008.12/32/32/3=⎪⎭⎫ ⎝⎛==**pnc m m g g υ(b) ()()0521.048.0067.02/32/32/3=⎪⎭⎫ ⎝⎛==**pncmm g g υ_______________________________________3.30 Plot_______________________________________ 3.31(a) ()()()!710!7!10!!!-=-=i i i i i N g N g W()()()()()()()()()()()()1201238910!3!7!78910===(b) (i) ()()()()()()()()12!10!101112!1012!10!12=-=i W 66= (ii)()()()()()()()()()()()()1234!8!89101112!812!8!12=-=i W 495=_______________________________________ 3.32()⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F exp 11(a) kT E E F =-, ()()⇒+=1exp 11E f ()269.0=E f(b) kT E E F 5=-, ()()⇒+=5exp 11E f()31069.6-⨯=E f (c) kT E E F 10=-, ()()⇒+=10exp 11E f()51054.4-⨯=E f_______________________________________ 3.33()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F exp 1111or()⎪⎪⎭⎫⎝⎛-+=-kT E E E f F exp 111(a) kT E E F =-, ()269.01=-E f (b) kT E E F 5=-, ()31069.61-⨯=-E f (c) kT E E F 10=-, ()51054.41-⨯=-E f_______________________________________ 3.34(a) ()⎥⎦⎤⎢⎣⎡--≅kT E E f F F exp c E E =; 61032.90259.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f 2kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.020259.030.0exp F f 61066.5-⨯=kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.00259.030.0exp F f 61043.3-⨯=23kT E c +; ()()⎥⎦⎤⎢⎣⎡+-=0259.020259.0330.0exp F f 61008.2-⨯=kT E c 2+; ()()⎥⎦⎤⎢⎣⎡+-=0259.00259.0230.0exp F f 61026.1-⨯=______________________________________________________________________________________(b) ⎥⎦⎤⎢⎣⎡-+-=-kT E E f F F exp 1111()⎥⎦⎤⎢⎣⎡--≅kT E E F exp υE E =; ⎥⎦⎤⎢⎣⎡-=-0259.025.0exp 1F f 51043.6-⨯= 2kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.020259.025.0exp 1F f 51090.3-⨯=kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.00259.025.0exp 1F f 51036.2-⨯=23kT E -υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.020259.0325.0exp 1F f 51043.1-⨯= kT E 2-υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.00259.0225.0exp 1F f 61070.8-⨯=_______________________________________ 3.35()()⎥⎦⎤⎢⎣⎡-+-=⎥⎦⎤⎢⎣⎡--=kT E kT E kT E E f F c F F exp exp and()⎥⎦⎤⎢⎣⎡--=-kT E E f F F exp 1()()⎥⎦⎤⎢⎣⎡---=kT kT E E F υexp So ()⎥⎦⎤⎢⎣⎡-+-kT E kT E F c exp()⎥⎦⎤⎢⎣⎡+--=kT kT E E F υexp Then kT E E E kT E F F c +-=-+υOr midgap c F E E E E =+=2υ_______________________________________ 3.3622222man E n πη= For 6=n , Filled state()()()()()2103122234610121011.92610054.1---⨯⨯⨯=πE18105044.1-⨯=Jor 40.9106.1105044.119186=⨯⨯=--E eVFor 7=n , Empty state()()()()()2103122234710121011.92710054.1---⨯⨯⨯=πE1810048.2-⨯=Jor 8.12106.110048.219187=⨯⨯=--E eVTherefore 8.1240.9<<F E eV_______________________________________ 3.37(a) For a 3-D infinite potential well ()222222⎪⎭⎫⎝⎛++=a n n n mE z y x πη For 5 electrons, the 5thelectron occupies the quantum state 1,2,2===z y x n n n ; so()2222252⎪⎭⎫ ⎝⎛++=a n n n m E z y x πη()()()()()21031222223410121011.9212210054.1---⨯⨯++⨯=π 1910761.3-⨯=Jor 35.2106.110761.319195=⨯⨯=--E eV For the next quantum state, which is empty, the quantum state is 2,2,1===z y x n n n . This quantum state is at the same energy, so 35.2=F E eV(b) For 13 electrons, the 13th electronoccupies the quantum state______________________________________________________________________________________3,2,3===z y x n n n ; so ()()()()()2103122222341310121011.9232310054.1---⨯⨯++⨯=πE 1910194.9-⨯=Jor 746.5106.110194.9191913=⨯⨯=--E eVThe 14th electron would occupy the quantum state 3,3,2===z y x n n n . This state is at the same energy, so 746.5=F E eV_______________________________________ 3.38The probability of a state at E E E F ∆+=1 being occupied is()⎪⎭⎫ ⎝⎛∆+=⎪⎪⎭⎫ ⎝⎛-+=kT E kT E E E f F exp 11exp 11111 The probability of a state at E E E F ∆-=2being empty is()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F 222exp 1111⎪⎭⎫ ⎝⎛∆-+⎪⎭⎫ ⎝⎛∆-=⎪⎭⎫ ⎝⎛∆-+-=kT E kT E kT E exp 1exp exp 111 or()⎪⎭⎫ ⎝⎛∆+=-kT E E f exp 11122 so ()()22111E f E f -= Q.E.D._______________________________________ 3.39(a) At energy 1E , we want01.0exp 11exp 11exp 1111=⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-+-⎪⎪⎭⎫ ⎝⎛-kT E E kT E E kT E E F F FThis expression can be written as01.01exp exp 111=-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-+kT E E kT E E F For()⎪⎪⎭⎫⎝⎛-=kT E E F 1exp 01.01Then()100ln 1kT E E F += orkT E E F 6.41+= (b)At kT E E F 6.4+=,()()6.4exp 11exp 1111+=⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F which yields()01.000990.01≅=E f_______________________________________ 3.40 (a)()()⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--=0259.050.580.5exp exp kT E E f F F 61032.9-⨯=(b) ()060433.03007000259.0=⎪⎭⎫⎝⎛=kT eV31098.6060433.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f (c) ()⎥⎦⎤⎢⎣⎡--≅-kT E E f F F exp 1 ⎥⎦⎤⎢⎣⎡-=kT 25.0exp 02.0______________________________________________________________________________________or 5002.0125.0exp ==⎥⎦⎤⎢⎣⎡+kT ()50ln 25.0=kTor()()⎪⎭⎫⎝⎛===3000259.0063906.050ln 25.0T kTwhich yields 740=T K_______________________________________ 3.41 (a)()00304.00259.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 0.304%(b) At 1000=T K, 08633.0=kT eV Then()1496.008633.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 14.96%(c) ()997.00259.00.785.6exp 11=⎪⎭⎫⎝⎛-+=E for 99.7% (d)At F E E =, ()21=E f for alltemperatures_______________________________________ 3.42(a) For 1E E =()()⎥⎦⎤⎢⎣⎡--≅⎪⎪⎭⎫ ⎝⎛-+=kT E E kTE E E fF F11exp exp 11Then()611032.90259.030.0exp -⨯=⎪⎭⎫ ⎝⎛-=E fFor 2E E =,82.030.012.12=-=-E E F eV Then()⎪⎭⎫ ⎝⎛-+-=-0259.082.0exp 1111E for()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛---≅-0259.082.0exp 111E f141078.10259.082.0exp -⨯=⎪⎭⎫ ⎝⎛-=(b) For 4.02=-E E F eV, 72.01=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.072.0exp exp 1kT E E E f F or()131045.8-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1⎪⎭⎫ ⎝⎛-=0259.04.0expor()71096.11-⨯=-E f_______________________________________ 3.43(a) At 1E E =()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.030.0exp exp 1kT E E E f F or()61032.9-⨯=E f At 2E E =, 12.13.042.12=-=-E E F eV So()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1⎪⎭⎫ ⎝⎛-=0259.012.1exp______________________________________________________________________________________or()191066.11-⨯=-E f(b) For 4.02=-E E F ,02.11=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.002.1exp exp 1kT E E E f F or()181088.7-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1⎪⎭⎫ ⎝⎛-=0259.04.0expor ()71096.11-⨯=-E f_______________________________________ 3.44()1exp 1-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+=kTE E E f Fso()()2exp 11-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+-=kT E E dE E df F⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛⨯kT E E kT F exp 1 or()2exp 1exp 1⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛-=kT E E kT E E kT dE E df F F (a) At 0=T K, For()00exp =⇒=∞-⇒<dE dfE E F()0exp =⇒+∞=∞+⇒>dEdfE E FAt -∞=⇒=dEdfE E F(b) At 300=T K, 0259.0=kT eVFor F E E <<,0=dE dfFor F E E >>, 0=dEdfAt F E E =,()()65.91110259.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1-(c) At 500=T K, 04317.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =, ()()79.511104317.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1- _______________________________________ 3.45(a) At midgap E E =,()⎪⎪⎭⎫ ⎝⎛+=⎪⎪⎭⎫ ⎝⎛-+=kTE kTE E E f gF2exp 11exp 11Si: 12.1=g E eV, ()()⎥⎦⎤⎢⎣⎡+=0259.0212.1exp 11E for()101007.4-⨯=E fGe: 66.0=g E eV______________________________________________________________________________________()()⎥⎦⎤⎢⎣⎡+=0259.0266.0exp 11E for()61093.2-⨯=E f GaAs: 42.1=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0242.1exp 11E for()121024.1-⨯=E f(b) Using the results of Problem 3.38, the answers to part (b) are exactly the same as those given in part (a)._______________________________________ 3.46(a) ()⎥⎦⎤⎢⎣⎡--=kT E E f F F exp ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 108 or ()810ln 60.0+=kT()032572.010ln 60.08==kT eV()⎪⎭⎫⎝⎛=3000259.0032572.0Tso 377=T K(b) ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 106()610ln 60.0+=kT()043429.010ln 60.06==kT ()⎪⎭⎫⎝⎛=3000259.0043429.0Tor 503=T K_______________________________________ 3.47(a) At 200=T K,()017267.03002000259.0=⎪⎭⎫⎝⎛=kT eV⎪⎪⎭⎫ ⎝⎛-+==kTE E f FF exp 1105.019105.01exp =-=⎪⎪⎭⎫⎝⎛-kT E E F()()()19ln 017267.019ln ==-kT E E F 05084.0=eV By symmetry, for 95.0=F f , 05084.0-=-F E E eVThen ()1017.005084.02==∆E eV (b) 400=T K, 034533.0=kT eV For 05.0=F f , from part (a),()()()19ln 034533.019ln ==-kT E E F 10168.0=eV Then ()2034.010168.02==∆E eV_______________________________________。

半导体物理与器件第四课后习题答案3.doc

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=αIn 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_______________________________________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 that 00=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 dyd 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 And22 mE=αSodk 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 -=_______________________________________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 _______________________________________。

《半导体物理与器件》第四版答案第13章

Chapter 1313.1Sketch_______________________________________ 13.2Sketch_______________________________________ 13.3(a) sdpO N ea V ∈=22(i)()()()()()141624191085.81.1321031040.0106.1---⨯⨯⨯⨯=pOV 312.3=V(ii) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=261816108.1102103ln 0259.0bi V328.1=V312.3328.1-=-=pO bi p V V V984.1-=V(b) ()2/122⎥⎦⎤⎢⎣⎡-+∈=d GS DS bi s eN V V V h(i) 2h()()()()()()2/1161914103106.15.00328.11085.81.132⎥⎦⎤⎢⎣⎡⨯⨯--+⨯=-- 521097.2-⨯=h cm μ297.0=m μ103.0297.040.02=-=-h a m (ii) 2h()()()()()()2/1161914103106.15.05.0328.11085.81.132⎥⎦⎤⎢⎣⎡⨯⨯--+⨯=-- 521035.3-⨯=h cm μ335.0=m μ065.0335.040.02=-=-h a m (iii) 2h()()()()()()2/1161914103106.15.05.2328.11085.81.132⎥⎦⎤⎢⎣⎡⨯⨯--+⨯=-- 521057.4-⨯=h cm μ457.0=m 022=-⇒>h a a h(c) ()()GS bi pO DS V V V sat V --=(i)()()0328.1312.3--=sat V DS 984.1=V(ii) ()()()0.1328.1312.3---=sat V DS 984.0=V_______________________________________ 13.4(a) sdpO N ea V ∈=22(i)()()()()()141624191085.87.1121031040.0106.1---⨯⨯⨯⨯=pOV 709.3=V(ii) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101816105.1102103ln 0259.0bi V860.0=V709.3860.0-=-=pO bi p V V V849.2-=V(b) ()2/122⎥⎦⎤⎢⎣⎡-+∈=d GS DS bi s eN V V V h(i) 2h()()()()()()2/1161914103106.15.00860.01085.87.112⎥⎦⎤⎢⎣⎡⨯⨯--+⨯=-- 521042.2-⨯=h cm μ242.0=m μ158.0242.040.02=-=-h a m (ii) 2h()()()()()()2/1161914103106.15.05.0860.01085.87.112⎥⎦⎤⎢⎣⎡⨯⨯--+⨯=-- 521083.2-⨯=h cm μ283.0=mμ117.0283.040.02=-=-h a m(iii) 2h()()()()()()2/1161914103106.15.05.2860.01085.87.112⎥⎦⎤⎢⎣⎡⨯⨯--+⨯=-- 521008.4-⨯=h cm μ408.0=m 022=-⇒>h a a h(c) ()()GS bi pO DS V V V sat V --=(i)()()0860.0705.3--=sat V DS845.2=V(ii) ()()()0.1860.0705.3---=sat V DS 845.1=V_______________________________________ 13.5(a) 2222ea V N N ea V pO s a s apO ∈=⇒∈= ()()()()()2419141065.0106.175.21085.81.132---⨯⨯⨯=a N1510433.9⨯=cm 3-(b) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯=261815108.11010433.9ln 0259.0bi V 280.1=V280.175.2-=-=bi pO p V V V47.1=V(c) 2265.015.0h h a -==- μ50.02=h m()2/122⎥⎦⎤⎢⎣⎡++∈=a GS SD bi s eN V V V h()241050.0-⨯()()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯++⨯=--15191410433.9106.1028.11085.81.132GSV()()GSV +⨯=⨯--28.1105363.1105.299347.0=⇒GS V V(d) ()2/122⎥⎦⎤⎢⎣⎡++∈=a GS SD bi s eN V V V h()241065.0-⨯()()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯+⨯=--15191410433.9106.128.11085.81.132SD V ()()()SD V +⨯=⨯--28.1105363.11065.092447.1=⇒SD V V_______________________________________ 13.6(a) 22ea V N pOs a ∈=()()()()()2419141065.0106.175.21085.87.112---⨯⨯⨯=or a N 1510425.8⨯=cm 3- (b) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯=2101815105.11010425.8ln 0259.0bi V 8095.0=V8095.075.2-=-=bi pO p V V V9405.1=V(c) 2265.015.0h h a -==- μ50.02=⇒h m()2/122⎥⎦⎤⎢⎣⎡++∈=a GS SD bi s eN V V V h()241050.0-⨯()()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯++⨯=--15191410425.8106.108095.01085.87.112GS V ()()GS V +⨯=⨯--8095.010536.1105.299 8178.0=⇒GS V V(d) ()2/122⎥⎦⎤⎢⎣⎡++∈=a GS SD bi s eN V V V h()241065.0-⨯()()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯+⨯=--15191410425.8106.18095.01085.87.112SDV()()()SDV +⨯=⨯--8095.010536.11065.092494.1=⇒SD V V_______________________________________ 13.7(a) ⎪⎪⎭⎫⎝⎛=2ln i d a t bi n N N V V ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101816105.1103102ln 0259.0860.0=V bi pO p V V V -=86.3860.00.3=⇒-=pO pO V V VNow 2/12⎥⎦⎤⎢⎣⎡∈=a pO s eN V a()()()()()2/1161914102106.186.31085.87.112⎥⎦⎤⎢⎣⎡⨯⨯⨯=--5100.5-⨯≅cm μ50.0=m(b) 86.3=pO V V(c) ()()GS bi pO SD V V V sat V +-=(i) ()0.386.086.3=-=sat V SD V (ii) ()()5.15.186.086.3=+-=sat V SD V _______________________________________13.8(a) ⎪⎪⎭⎫⎝⎛=2ln i d a t bi n N N V V()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=261816108.1103102ln 0259.0 328.1=Vbi pO p V V V -=328.4328.10.3=⇒-=pO pO V V V2/12⎥⎦⎤⎢⎣⎡∈=a pO s eN V a()()()()()2/1161914102106.1328.41085.81.132⎥⎦⎤⎢⎣⎡⨯⨯⨯=-- 51060.5-⨯=cm μ560.0=m(b) 328.4=pO V V(c) ()()GS bi pO SD V V V sat V +-=(i) ()()0.30328.1328.4=+-=sat V SD V (ii) ()()5.15.1328.1328.4=+-=sat V SD V _______________________________________ 13.9(a) ()()GS bi pO DS V V V sat V --= Now()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101816105.1104104ln 0259.0bi V886.0=V We find886.5886.05=⇒-=pO pO V V V2/12⎥⎦⎤⎢⎣⎡∈=a pO s eN V a()()()()()2/1161914104106.1886.51085.87.112⎥⎦⎤⎢⎣⎡⨯⨯⨯=--51036.4-⨯=cm μ436.0=m(b) (i) 886.5=pO V V(ii) 0.5886.5886.0-=-=-=pO bi p V V V V _______________________________________13.10 (a) ()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯=261815108.110105ln 0259.0bi V 264.1=V()()GS bi pO SD V V V sat V +-= ()0.1264.15.3+-=pO V 764.5=⇒pO V V 2/12⎥⎦⎤⎢⎣⎡∈=a pO s eN V a ()()()()()2/1151914105106.1764.51085.81.132⎥⎦⎤⎢⎣⎡⨯⨯⨯=-- 410293.1-⨯=cm μ293.1=m (b) (i) 764.5=pO V V(ii) 264.1764.5-=-=bi pO p V V V 5.4=V _______________________________________13.11 (a) ()L W a eN I s d n P ∈=6321μ()()()[]()()14216191085.87.11610106.11000--⨯⨯=()()43441020105.010400---⨯⨯⨯⨯or03.11=P I mA (b)sdPO N ea V ∈=22()()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=---141624191085.87.11210105.0106.1 or93.1=PO V V Also()()()()874.0105.11010ln 0259.02101619=⎥⎥⎦⎤⎢⎢⎣⎡⨯=bi V VNow()()GS bi PO DS V V V sat V --= GS V +-=874.093.1 or()GS DS V sat V +=056.1We have 056.193.1874.0-=-=-=PO bi P V V V V Then (i) For 0=GS V , ()06.1=sat V DS V (ii) For 264.041-==P GS V V V, ()792.0=sat V DS V (iii) For 528.021-==P GS V V V,()528.0=sat V DS V(iv) For 792.043-==P P V V V, ()264.0=sat V DS V (c)()⎢⎢⎣⎡⎪⎪⎭⎫⎝⎛--=PO GS bi P D V V V I sat I 3111 ⎥⎥⎦⎤⎪⎪⎭⎫ ⎝⎛--⨯PO GS bi V V V 321 ()⎢⎣⎡⎪⎪⎭⎫⎝⎛--=93.1874.03103.1GS V ⎥⎥⎦⎤⎪⎪⎭⎫ ⎝⎛--⨯93.1874.0321GS V (i) For 0=GS V , ()258.01=sat I D mA(ii) For 264.0-=GS V V,()141.01=sat I D mA(iii) For 528.0-=GS V V, ()0608.01=sat I D mA (iv) For 792.0-=GS V V,()0148.01=sat I D mA_______________________________________13.12⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--=2/111PO GSbi O d V V V G g where()93.11003.133311-⨯==PO P O V I Gor 311060.1-⨯=O G S 60.1=mS Then13.13n-channel JFET - GaAs (a) LW aN e G d n O μ=1()()()4161910101028000106.1--⨯⨯⨯= ()()441035.01030--⨯⨯⨯ or 311069.2-⨯=O G S(b) ()()GS bi PO DS V V V sat V --= We havesd PO N ea V ∈=22 ()()()()()141624191085.81.1321021035.0106.1---⨯⨯⨯⨯= or 69.1=PO V V We find()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=261618108.1102105ln 0259.0bi Vor 34.1=bi V V Then 35.069.134.1-=-=-=PO bi P V V V V We then obtain ()()GS GS DS V V sat V +=--=35.034.169.1 For 0=GS V , ()35.0=sat V DS V For 175.021-==P GS V V V, ()175.0=sat V DS V (c)()⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--=PO GS bi P D V V V I sat I 3111 ⎥⎥⎦⎤⎪⎪⎭⎫ ⎝⎛--⨯POGS bi V V V 321where ()L W a eN I s d n P ∈=6321μ ()()()[]()()14216191085.81.136102106.18000--⨯⨯⨯=()()434410101035.01030---⨯⨯⨯⨯or 515.11=P I mA Then()()⎢⎣⎡⎪⎪⎭⎫⎝⎛--=69.134.131515.11GS D V sat I ⎥⎥⎦⎤⎪⎪⎭⎫ ⎝⎛--⨯69.134.1321GS V (mA) For 0=GS V , ()0506.01=sat I D mAandFor 175.0-=GS V V,()0124.01=sat I D mA_______________________________________13.14 ⎪⎪⎭⎫ ⎝⎛--=PO GS biPO P mS V V V V I g 131 We have03.11=P I mA, 93.1=PO V V, 874.0=bi V VThe maximum transconductance occurs when0=GS V . Then()()⎪⎪⎭⎫ ⎝⎛-=93.1874.0193.103.13m ax mS g or ()524.0max =mS g mS For μ400=W m, we have ()410400524.0m ax -⨯=mS gor()1.13max =mS g mS/cm = 1.31 mS/mm _______________________________________13.15The maximum transconductance occurs for 0=GS V , so we have (a) ()⎪⎪⎭⎫ ⎝⎛-=PO bi POP mS V V V I g 13m ax 1which can be written as ()⎪⎪⎭⎫ ⎝⎛-=PO bi O mS V V G g 1m ax 1 We found 69.21=O G mS, 34.1=bi V V, 69.1=PO V V Then ()()⎪⎪⎭⎫ ⎝⎛-=69.134.1169.2m ax mS g or ()295.0max =mS g mS This is for a channel length of μ10=L m.(b) If the channel length is reduced toμ2=L m, then ()()47.12102947.0max =⎪⎭⎫ ⎝⎛=mS g mS_______________________________________ 13.16n-channel MESFET - GaAs(a) sdPO N ea V ∈=22()()()()()141624191085.81.132105.1105.0106.1---⨯⨯⨯⨯= or 59.2=PO V VNow n Bn bi V φφ-=where ()⎪⎪⎭⎫ ⎝⎛⨯⨯=⎪⎪⎭⎫ ⎝⎛=1617105.1107.4ln 0259.0ln d c t n N N V φ or 0892.0=nφVso that811.00892.090.0=-=bi V V Then59.2811.0-=-=PO bi T V V V or78.1-=T V V(b )If 0<T V for an n-channel device, the device is a depletion mode MESFET._______________________________________13.17 n-channel MESFET - GaAs (a) We want 10.0+=T V VThen PO n Bn PO bi T V V V V --=-=φφsos dd c t T N ea N N V V ∈-⎪⎪⎭⎫ ⎝⎛-==2ln 89.010.02which can be written as ()⎪⎪⎭⎫ ⎝⎛⨯d N 17107.4ln 0259.0 ()()()()10.089.01085.81.1321035.0106.1142419-=⨯⨯⨯+---dNor()⎪⎪⎭⎫ ⎝⎛⨯d N17107.4ln 0259.0 ()79.010453.817=⨯+-d NBy trial and error,15101.8⨯=d N cm 3- (b) At 400=T K()()54.13004003004002/3=⎪⎭⎫⎝⎛=c c N NThen()()()54.1107.440017⨯=c N171024.7⨯=cm 3- Also()03453.03004000259.0=⎪⎭⎫⎝⎛=t VThen()⎪⎪⎭⎫ ⎝⎛⨯⨯-=1517101.81024.7ln 03453.089.0T V ()()1517101.810453.8⨯⨯-- which becomes 050.0+=T V V_______________________________________13.18(a) 2/12⎥⎦⎤⎢⎣⎡∈=d pO s eN V a()()()()()2/1161914102106.15.11085.81.132⎥⎦⎤⎢⎣⎡⨯⨯⨯=- 51030.3-⨯=cm μ330.0=m(b) pO bi T V V V -= We find ()⎪⎪⎭⎫⎝⎛⨯⨯=⎪⎪⎭⎫ ⎝⎛=1617102107.4ln 0259.0ln d c t n N N V φ 0818.0=V 788.00818.087.0=-=-=n Bn bi V φφV Then 712.05.1788.0-=-=T V V(c) ()2/122⎥⎦⎤⎢⎣⎡-+∈=d GS DS bi s eN V V V h ()()()()()2/1161914102106.11085.81.132⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯-+⨯=--GS DS bi V V V (i) ()()[]2/11024.00788.010246.7-+⨯=-h510677.1-⨯=cm μ1677.0=m μ1623.01677.0330.02=-=-h a m (ii)()()[]2/11024.00.1788.010246.7-+⨯=-h510171.3-⨯=cm μ3171.0=m μ0129.03171.0330.02=-=-h a m (iii)()()[]2/11024.00.4788.010246.7-+⨯=-h51064.5-⨯=cm μ564.0=m022=-⇒>h a a h_______________________________________13.19(a) sdpO N ea V ∈=22()()()()()141524191085.81.1321051050.0106.1---⨯⨯⨯⨯=8626.0=V We find()⎪⎪⎭⎫⎝⎛⨯⨯=1517105107.4ln 0259.0n φ1177.0=V1177.087.0-=-=n Bn bi V φφ 7523.0=V8626.07523.0-=-=pO bi T V V V 1103.0-=V(b) ()0713.0103107.4ln 0259.01617=⎪⎪⎭⎫⎝⎛⨯⨯=n φV0713.087.0-=-=n Bn bi V φφ7987.0=VpO bi T V V V -=or ()1103.07987.0--=-=T bi pO V V V 909.0=VThen 2/12⎥⎦⎤⎢⎣⎡∈=d pO s eN V a ()()()()()2/1161914103106.1909.01085.81.132⎥⎦⎤⎢⎣⎡⨯⨯⨯=-- 510095.2-⨯=cm μ2095.0=m_______________________________________13.20PO n Bn PO bi T V V V V --=-=φφ We want 5.0=T V V, so PO n V --=φ85.05.0 Now()⎪⎪⎭⎫ ⎝⎛⨯=d n N17107.4ln 0259.0φ andsdPO N ea V ∈=22()()()()1424191085.81.1321025.0106.1---⨯⨯⨯=dNor()d PO N V 171031.4-⨯= Then()⎪⎪⎭⎫⎝⎛⨯-=d N17107.4ln 0259.085.05.0 ()d N 171031.4-⨯-By trial and error151045.5⨯=d N cm 3-_______________________________________13.21n-channel MESFET - silicon(a) For a gold contact, 82.0=Bn φV. We find()206.010108.2ln 0259.01619=⎪⎪⎭⎫⎝⎛⨯=n φVand 614.0206.082.0=-=-=n Bn bi V φφV With 0=DS V and 35.0=GS V V, we find410075.0-⨯=-h a ()2/12⎥⎦⎤⎢⎣⎡-∈-=d GS bi s eN V V aso that 410075.0-⨯=a()()()()()2/116191410106.135.0614.01085.87.112⎥⎦⎤⎢⎣⎡⨯-⨯+-- or 41026.0-⨯=a cm μ26.0=m Nows dPO bi T N ea V V V ∈-=-=2614.02 or ()()()()()141624191085.87.112101026.0106.1614.0---⨯⨯⨯-=T VWe obtain092.0=T V V (b)()()GS bi PO DS V V V sat V --=()()GS bi T bi V V V V ---= or()092.035.0-=-=T GS DS V V sat V which yields()258.0=sat V DS V_______________________________________13.22(a) ()⎪⎪⎭⎫⎝⎛⨯⨯=1617102107.4ln 0259.0n φ0818.0=V(i) 0818.090.0-=-=n Bn bi V φφ 818.0=V(ii) sdpO N ea V ∈=22()()()()()141624191085.81.1321021065.0106.1---⨯⨯⨯⨯=83.5=V(iii) 83.5818.0-=-=pO bi T V V V 012.5-=V(b) ()()GS bi pO DS V V V sat V --=(i) ()()()0.1818.083.5---=sat V DS01.4=V (ii) ()()()0.2818.083.5---=sat V DS 01.3=V(iii) ()()()0.3818.083.5---=sat V DS01.2=V_______________________________________13.23 (a) aLWk s n n 2∈=μ ()()()()()()44414105.11025.021*******.81.136500----⨯⨯⨯⨯=310206.1-⨯=A/V 2206.1=mA/V 2(b) ()()21T GS n D V V k sat I -=(i) ()()()2115.025.0206.1-=sat I D 01206.0=mA μ06.12= A(ii) ()()()2115.045.0206.1-=sat I D1085.0=mA (c) ()T GS DS V V sat V -=(i) ()10.015.025.0=-=sat V DS V(ii) ()30.015.045.0=-=sat V DS V _______________________________________13.24(a) ()[]2T GS n GSGS D ms V V k V V I g -∂∂=∂∂=()T GS n V V k -=2 ()15.045.0225.1-=n k 083.2=⇒n k mA/V 2 aLWk s n n 2∈=μ()()()()()44143105.11025.021085.81.13650010083.2----⨯⨯⨯=⨯W310073.2-⨯=⇒W cm μ73.20=m (b) ()()21T GS n D V V k sat I -= (i) ()()()2115.025.0083.2-=sat I D 02083.0=mA μ83.20= A(ii) ()()()2115.045.0083.2-=sat I D1875.0=mA_______________________________________13.25 Plot_______________________________________13.26Plot _______________________________________13.27()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯=2101618105.110310ln 0259.0bi V8424.0=Vsd pO N ea V ∈=22 ()()()()()141624191085.87.1121031050.0106.1---⨯⨯⨯⨯=795.5=V (a) ()()GS bi pO DS V V V sat V --= 953.48424.0795.5=-=V()()2/12⎥⎦⎤⎢⎣⎡-∈=∆d DS DS s eN sat V V L()()()()()2/1161914103106.1953.4101085.87.112⎥⎦⎤⎢⎣⎡⨯⨯-⨯=-- 510666.4-⨯=∆L cm Now()90.0211=∆-='LL L L ()()10.0210666.410.025-⨯=∆=L L410333.2-⨯=L cm μ333.2=m (b) ()()GS bi pO DS V V V sat V --= ()38424.0795.5+-=953.1=V()()()()()2/1161914103106.1953.1101085.87.112⎥⎦⎤⎢⎣⎡⨯⨯-⨯=∆--L510892.5-⨯=cm Then()()10.0210892.510.025-⨯=∆=L L410946.2-⨯=cm μ946.2=m_______________________________________13.28 We have that ()⎪⎪⎭⎫ ⎝⎛∆-='L L LI I D D 2111 Assuming that we are in the saturation region,then ()sat I I D D11'='and ()sat I I D D 11=. We can write()()L L sat I sat I D D ∆⋅-⋅='211111 If L L <<∆, then()()⎥⎦⎤⎢⎣⎡∆⋅+≅'L L sat I sat I D D 21111 We have that()()2/12⎥⎦⎤⎢⎣⎡-∈=∆d DS DS s eN sat V V L()2/112⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-∈=DS DS d DS s V sat V eN V which can be written as ()2/112⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-∈=∆DS DS DS d s DS V sat V V eN V L If we write()()()DS D DV sat I sat I λ+='111 then by comparing equations, we have()2/11221⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-∈=DS DS DSd sV sat V V eN L λ The parameter λis not independent of DS V . Define()sat V V x DS DS≡and consider the function⎪⎭⎫⎝⎛-=x x f 111So that λ is nearly a constant._______________________________________ 13.29 (a) Saturation occurs when 4101⨯=E V/cm.As a first approximation, let L V DS =EThen ()()21021044=⨯=⋅E =-L V DS V (b) We have that ()2/122⎥⎦⎤⎢⎣⎡-+∈==d GS DS bi s sat eN V V V h h and()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101618105.1104105ln 0259.0bi V or8915.0=bi V V For 0=GS V , we obtain()()()()()2/1161914104106.128915.01085.87.112⎥⎦⎤⎢⎣⎡⨯⨯+⨯=--sat h or410306.0-⨯=sat h cm μ306.0=m(c) We then find ()()W h a eN sat I sat sat d D -=υ1 ()()()7161910104106.1⨯⨯=-()()()44103010306.050.0--⨯-⨯or()72.31=sat I D mA (d) For 0=GS V , we have ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-=PO bi PO biP D VV V V I sat I 3213111 Now()LW a eN I s d n P ∈=6321μ()()()[]()()14216191085.87.116104106.11000--⨯⨯⨯=()()()4344102105.01030---⨯⨯⨯⨯or36.121=P I mAAlso sdPO N ea V ∈=22 ()()()()()141624191085.87.112104105.0106.1---⨯⨯⨯⨯= or 726.7=PO V V Then ()()⎢⎣⎡⎪⎭⎫ ⎝⎛-=726.78915.03136.121sat I D⎥⎥⎦⎤⎪⎪⎭⎫ ⎝⎛-⨯726.78915.0321 or()05.91=sat I D mA_______________________________________13.30(a) If μ1=L m, then saturation will occur when()()11011044=⨯=⋅E =-L V DS V We find()2/122⎥⎦⎤⎢⎣⎡-+∈==d GS DS bi s sat eN V V V h hWe have 8915.0=bi V V and for 0=GS V , we obtain()()()()()2/1161914104106.118915.01085.87.112⎥⎦⎤⎢⎣⎡⨯⨯+⨯=--sathor410247.0-⨯=sat h cm μ247.0=m Then()()W h a eN sat I sat sat d D -=υ1()()()7161910104106.1⨯⨯=-()()()44103010247.050.0--⨯-⨯or()86.41=sat I D mAIf velocity saturation did not occur, then from the previous problem, we would have()()1.181205.91=⎪⎭⎫⎝⎛=sat I D mA(b) If velocity saturation occurs, then the relation ()()L sat I D 11∝does not apply._______________________________________13.31 (a)()()731041058000⨯=⨯=E =n μυcm/s Then 1274105104102--⨯=⨯⨯==υL t d s or 5=d t ps(b) Assume 710==sat υυcm/s Then117410210102--⨯=⨯==satd Lt υs or20=d t ps_______________________________________13.32 (a)()()7410101000==E =n μυcm/s Then117410210102--⨯=⨯==υL t d s or20=d t ps(b) For 710==sat υυcm/s 117410210102--⨯=⨯==satd Lt υs or20=d t ps_______________________________________13.33The reverse-bias current is dominated by the generation current. We have PO bi P V V V -= We find()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯=2101618105.1103105ln 0259.0bi Vor884.0=bi V VAlso sd PON ea V ∈=22 ()()()()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯⨯=---141624191085.87.112103103.0106.1or 086.2=PO V V Then 20.1086.2884.0-=-=P V VLet 20.1-=GS V VNow()2/12⎥⎦⎤⎢⎣⎡-+∈=d GS DS bi s n eN V V V x ()()()⎢⎣⎡⨯⨯=--1914106.11085.87.112 ()()()2/11610320.1884.0⎥⎦⎤⨯--+⨯DS Vor()()[]2/110084.210314.4DS n V x +⨯=- (a) For 0=DS V , μ30.0=n x m (b) For 1=DS V V, μ365.0=n x m (c) For 5=DS V V, μ553.0=n x mThe depletion region volume at the drain is()()()()()W a x W L a Vol n 22+⎪⎭⎫⎝⎛=()()44410302104.2103.0---⨯⎪⎪⎭⎫⎝⎛⨯⨯= ()()()441030106.0--⨯⨯+n xor()8121018108.10--⨯+⨯=n x Vol(a) For 0=DS V , 111062.1-⨯=Vol cm 3 (b) For 1=DS V V, 1110737.1-⨯=Vol cm 3 (c) For 5=DS V V, 1110075.2-⨯=Vol cm 3The generation current at the drain isVol n e I O i DG ⋅⎪⎪⎭⎫⎝⎛=τ2 ()()Vol ⋅⎥⎦⎤⎢⎣⎡⨯⨯⨯=--810191052105.1106.1 or()Vol I DG ⋅⨯=-2104.2(a) For 0=DS V , 39.0=DG I pA(b) For 1=DS V V, 42.0=DG I pA (c) For 5=DS V V, 50.0=DG I pA_______________________________________13.34 (a) The ideal transconductance for 0=GS V is ⎪⎪⎭⎫ ⎝⎛-=PO bi O mS V V G g 11 where L W a N e G d n O μ=1 ()()()41619105.11074500106.1--⨯⨯⨯= ()()44103.0105--⨯⨯⨯ or04.51=O G mS We findsdPO N ea V ∈=22()()()()()141624191085.81.132107103.0106.1---⨯⨯⨯⨯=or347.4=PO V V We have()049.0107107.4ln 0259.01617=⎪⎪⎭⎫⎝⎛⨯⨯=n φVso that841.0049.089.0=-=-=n Bn bi V φφV Then()⎪⎪⎭⎫ ⎝⎛-=347.4841.0104.5mS g or82.2=mS g mS(b) With a source resistancesm m m s m m m r g g g r g g g +='⇒+='111For()sm m r g g 823.21180.0+=='we obtain Ω=6.88s r (c)A L A L r s σρ==()An e L n μ=so ()()()()16191074500106.156.88⨯⨯=-L()()44105103.0--⨯⨯⨯ or41067.0-⨯=L cm μ67.0=m _______________________________________13.35Considering the capacitance charging time,we haveG m T C g f π2= where a W L C s G ∈= ()()()()44414103.0105.11051085.81.13----⨯⨯⨯⨯=or 15109.2-⨯=G C F We must use mg ', so we obtain ()()()153109.2280.01082.2--⨯⨯=πT f 1110238.1⨯= Hz We can also write TC C T f f πττπ2121=⇒= so()121110285.110238.121-⨯=⨯=πτC s The channel transit time is 1174105.110105.1--⨯=⨯=t t s The total time constant is121110285.1105.1--⨯+⨯=τ 1110629.1-⨯= s Taking into account the channel transit time and the capacitance charging time, we find()1110629.12121-⨯==πτπT f or 91077.9⨯=T f Hz 77.9=GHz _______________________________________13.36(a) For constant mobility 222L a N e f s d n T ∈=πμ ()()()()()()()2414241619102.11085.81.1321030.010********.1----⨯⨯⨯⨯⨯=π111012.4⨯=T f Hz 412=GHz (b) For saturation velocity model ()47102.12102-⨯==ππυL f sat T101033.1⨯=T f Hz 3.13=GHz_______________________________________13.37 222L a N e f s d n T ∈=πμ()()()()()()2142416191085.87.1121040.010********.1L ---⨯⨯⨯⨯=π 2975.786L f T = (a) ()24103975.786-⨯=T f 91074.8⨯=Hz 74.8=GHz (b) ()24105.1975.786-⨯=T f 101050.3⨯=Hz 0.35=GHz_______________________________________ 13.38 222L a N e f s a p T ∈=πμ or2/122⎥⎥⎦⎤⎢⎢⎣⎡∈=T s a p f a N e L πμ ()()()()()()2/1142416191085.87.1121040.010*******.1⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯⨯=---T f π T f L 18.18=(a) 910518.18⨯=L 41057.2-⨯=cm μ57.2=m (b) 9101218.18⨯=L41066.1-⨯=cm μ66.1=m_______________________________________13.39 (a)2P cB off V eE V -∆-=φwhereNdd P d eN V ∈=222()()()()()142818191085.82.12210350103106.1---⨯⨯⨯⨯=or72.22=P V V Then72.224.089.0--=off V or07.2-=off V V (b)()()off g NS V V d d e n -∆+∈=For 0=g V , we have()()()()()()07.21080350106.11085.82.1281914---+⨯⨯=Sn or121025.3⨯=S n cm 2-_______________________________________13.40(a) We have()()()sO off g N D V V V d d Wsat I υ⋅--∆+∈=We find()()d d W sat I V W g sN D g mS ∆+∈=⎥⎦⎤⎢⎣⎡∂∂=⎪⎪⎭⎫ ⎝⎛υ()()()()()871410803501021085.82.12--+⨯⨯=or02.5=⎪⎪⎭⎫⎝⎛W g mS S/cm 502=mS/mm(b) At 0=g V , we obtain ()()()s O off N D V V d d Wsat I υ⋅--∆+∈= ()()()()()()7812102107.210803501085.82.12⨯-+⨯=-- or()37.5=Wsat I D A/cm 537=mA/mm _______________________________________13.412P cB off V eE V -∆-=φWe want 3.0-=off V V, so 222.085.030.0P V --=- or93.02=P V V We haveNdd P d eN V ∈=222ordP N d eN V d 222∈=()()()()()181914102106.193.01085.82.122⨯⨯⨯=-- We then obtain61051.2-⨯=d d cm oA 251=_______________________________________。

半导体物理与器件第四版课后习题答案4

4.44
Plot
_______________________________________
4.45
so cm
cm
_______________________________________
4.46
(a) p-type
Majority carriers are holes
cm
Minority carriers are electrons
(c)
so cm
_______________________________________
4.48
For Germanium
(K)
(eV)
(cm )
and
cm
(K)
(cm )
(eV)
_______________________________________
4.49
(a)
For cm , eV
4.29
So
We find
eV
_______________________________________
4.30
(a)
Then
cm
(b)
cm
_______________________________________
4.31
For the electron concentration
The Boltzmann approximation applies, so
4.22
(a)p-type
(b) eV
cm
eV
cm
_______________________________________
4.23
(a)

《半导体物理与器件》第四版答案第十章

Chapter 1010.1(a) p-type; inversion (b) p-type; depletion (c) p-type; accumulation (d) n-type; inversion_______________________________________ 10.2(a) (i) ⎪⎪⎭⎫⎝⎛=i a t fp n N V ln φ ()⎪⎪⎭⎫ ⎝⎛⨯⨯=1015105.1107ln 0259.0 3381.0=V 2/14⎥⎦⎤⎢⎣⎡∈=a fp s dTeN x φ()()()()()2/1151914107106.13381.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--51054.3-⨯=cm or μ354.0=dT x m(ii) ()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1103ln 0259.0fp φ3758.0=V ()()()()()2/1161914103106.13758.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dTx51080.1-⨯=cmor μ180.0=dT x m(b) ()03022.03003500259.0=⎪⎭⎫⎝⎛=kT V⎪⎪⎭⎫⎝⎛-=kT E N N n g c i exp 2υ ()()319193003501004.1108.2⎪⎭⎫⎝⎛⨯⨯=⎪⎭⎫⎝⎛-⨯03022.012.1exp221071.3⨯=so 111093.1⨯=i n cm 3-(i)()⎪⎪⎭⎫⎝⎛⨯⨯=11151093.1107ln 03022.0fp φ3173.0=V()()()()()2/1151914107106.13173.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x51043.3-⨯=cm or μ343.0=dT x m(ii) ()⎪⎪⎭⎫⎝⎛⨯⨯=11161093.1103ln 03022.0fp φ3613.0=V()()()()()2/1161914103106.13613.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x51077.1-⨯=cm or μ177.0=dT x m_______________________________________ 10.3(a) ()2/14m ax ⎥⎦⎤⎢⎣⎡∈=='d fn s d dT d SDeN eN x eN Q φ()()[]2/14fns d eN φ∈=1st approximation: Let 30.0=fn φV Then()281025.1-⨯()()()()()()[]30.01085.87.114106.11419--⨯⨯=dN 141086.7⨯=⇒d N cm 3-2nd approximation:()2814.0105.11086.7ln 0259.01014=⎪⎪⎭⎫⎝⎛⨯⨯=fn φV Then ()281025.1-⨯()()()()()()[]2814.01085.87.114106.11419--⨯⨯=d N 141038.8⨯=⇒d N cm 3-(b) ()2831.0105.11038.8ln 0259.01014=⎪⎪⎭⎫⎝⎛⨯⨯=fn φV()566.02831.022===fn s φφV _______________________________________10.4 p-type silicon (a) Aluminum gate ⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛++'-'=fp g m ms e E φχφφ2 We have ⎪⎪⎭⎫ ⎝⎛=i a t fp n N V ln φ ()334.0105.1106ln 0259.01015=⎪⎪⎭⎫ ⎝⎛⨯⨯=V Then()[]334.056.025.320.3++-=ms φ or 944.0-=ms φV (b) +n polysilicon gate ⎪⎪⎭⎫⎝⎛+-=fp g ms e E φφ2()334.056.0+-= or 894.0-=ms φV (c) +p polysilicon gate ()334.056.02-=⎪⎪⎭⎫⎝⎛-=fp g ms 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-_______________________________________10.7 From Problem 10.5, 9932.0-=ms φV ox ssms FB C Q V '-=φ (a) ()()814102001085.89.3--⨯⨯=∈=ox ox ox t C 710726.1-⨯=F/cm 2()()7191010726.1106.11059932.0--⨯⨯⨯--=FB V 040.1-=V (b) ()()81410801085.89.3--⨯⨯=ox C 710314.4-⨯=F/cm 2 ()()7191010314.4106.11059932.0--⨯⨯⨯--=FB V012.1-=V _______________________________________10.8 (a) 42.0-≅ms φV 42.0-==ms FB V φV(b) ()()781410726.1102001085.89.3---⨯=⨯⨯=ox C F/cm 2 (i)()()7191010726.1106.1104--⨯⨯⨯-='-=∆ox ss FB C Q V 0371.0-=V (ii)()()7191110726.1106.110--⨯⨯-=∆FB V 0927.0-=V(c) 42.0-==ms FB V φV ()()781410876.2101201085.89.3---⨯=⨯⨯=ox C F/cm 2 (i)()()7191010876.2106.1104--⨯⨯⨯-=∆FB V 0223.0-=V (ii)()()7191110876.2106.110--⨯⨯-=∆FB V0556.0-=V _______________________________________10.9 ⎪⎪⎭⎫ ⎝⎛++'-'=fp g mms e E φχφφ2 where()365.0105.1102ln 0259.01016=⎪⎪⎭⎫ ⎝⎛⨯⨯=fp φV Then ()365.056.025.320.3++-=ms φor975.0-=ms φVNowox ss ms FB C Q V '-=φor ()ox FB ms ss C V Q -='φ We have()()814104501085.89.3--⨯⨯=∈=ox ox ox t C or 81067.7-⨯=ox C F/cm 2 So now ()[]()81067.71975.0-⨯⋅---='ssQ 91092.1-⨯=C/cm 2or10102.1⨯='e Q ss cm 2- _______________________________________10.10 ()3653.0105.1102ln 0259.01016=⎪⎪⎭⎫ ⎝⎛⨯⨯=fp φV ()()()()()2/1161914102106.13653.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x510174.2-⨯=cm()dT a SDx eN Q ='m ax ()()()5161910174.2102106.1--⨯⨯⨯=810958.6-⨯=C/cm 2()()781410301.2101501085.89.3---⨯=⨯⨯=ox C F/cm 2()fp ms ox ss 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 φV 136.012.19843.0-=-=TN V V(b) p + poly gate on p-type: 28.0+≅ms φV 26.128.09843.0+=+=TN V V (c) Al gate on p-type: 95.0-≅ms φV0343.095.09843.0+=-=TN V V_______________________________________10.11 ()3161.0105.1103ln 0259.01015=⎪⎪⎭⎫ ⎝⎛⨯⨯=fn φV ()()()()()2/1151914103106.13161.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x 510223.5-⨯=cm ()dT d SDx eN Q ='m ax ()()()5151910223.5103106.1--⨯⨯⨯= 810507.2-⨯=C/cm 2 ()()781410301.2101501085.89.3---⨯=⨯⨯=ox C F/cm 2 ()fn ms ox ss SDTP C Q Q V φφ2m ax -+⎥⎥⎦⎤⎢⎢⎣⎡'+'-= ()()⎥⎦⎤⎢⎣⎡⨯⨯⨯+⨯-=---71019810301.2107106.110507.2 ()3161.02-+ms φ ms TP V φ+-=7898.0(a) n + poly gate on n-type: 41.0-≅ms φV 20.141.07898.0-=--=TP V V(b) p + poly gate on n-type: 0.1+≅ms φV 210.00.17898.0+=+-=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=⎪⎪⎭⎫ ⎝⎛⨯⨯=fp φV The 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()()()()4151910413.0105106.1m ax --⨯⨯⨯='SDQ or()810304.3m ax -⨯='SDQ C/cm 2 We also find()()814104001085.89.3--⨯⨯=∈=ox ox ox t C or810629.8-⨯=ox C F/cm 2 Then90.0659.010629.810304.388-+⨯⨯=--T Vor142.0+=T V V_______________________________________10.13()()814102201085.89.3--⨯⨯=∈=ox ox ox t C 710569.1-⨯=F/cm 2()()1019104106.1⨯⨯='-ssQ 9104.6-⨯=C/cm 2By trial and error, let 16104⨯=a N cm 3-.Now ()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1104ln 0259.0fp φ3832.0=V()()()()()2/1161914104106.13832.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x 510575.1-⨯=cm ()m ax SDQ ' ()()()5161910575.1104106.1--⨯⨯⨯= 710008.1-⨯=C/cm 294.0-≅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 ox t C 7109175.1-⨯=F/cm 3- ()()1019104106.1⨯⨯='-ssQ 9104.6-⨯=C/cm 2By trial and error, let 16105⨯=d N cm 3- Now()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1105ln 0259.0fn φ3890.0=V()()()()()2/1161914105106.13890.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x510419.1-⨯=cm()m ax SDQ ' ()()()5161910419.1105106.1--⨯⨯⨯= 710135.1-⨯=C/cm 3-10.1+≅ms φV Then()()fn ms ox ss SDTP C Q Q V φφ2max -+'+'-= ()797109175.1104.610135.1---⨯⨯+⨯-= ()3890.0210.1-+Then 303.0-=TP V V, which is within thespecified value. _______________________________________ 10.15 We have 710569.1-⨯=ox C F/cm 2 9104.6-⨯='ssQ C/cm 2 By trial and error, let 14105⨯=d N cm 3-Now()⎪⎪⎭⎫⎝⎛⨯⨯=1014105.1105ln 0259.0fn φ 2697.0=V()()()()()2/1141914105106.12697.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x 410182.1-⨯=cm ()m ax SDQ ' ()()()4141910182.1105106.1--⨯⨯⨯= 910456.9-⨯=C/cm 233.0-≅ms φVThen ()()fn ms oxss SDTP C Q Q V φφ2max -+'+'-= ⎪⎪⎭⎫⎝⎛⨯⨯+⨯-=---79910569.1104.610456.9 ()2697.0233.0--970.0=V Then 970.0-=TP V V 975.0-≅ V which meets the specification._______________________________________ 10.16(a) 03.1-≅ms φV()()814101801085.89.3--⨯⨯=ox C 7109175.1-⨯=F/cm 2Now oxss ms FB C Q V '-=φ()()71019109175.1106106.103.1--⨯⨯⨯--= 08.1-=FB V V(b) ()⎪⎪⎭⎫ ⎝⎛⨯=1015105.110ln 0259.0fp φ 2877.0=V ()()()()()2/115191410106.12877.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dT x 510630.8-⨯=cm()m ax 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, so2/14⎥⎦⎤⎢⎣⎡∈==d fn s C dT eN t x φ where()288.0105.110ln 0259.01015=⎪⎪⎭⎫ ⎝⎛⨯=fn φVThen()()()()()2/115191410106.1288.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dT x or 410863.0-⨯==C dT t x cm μ863.0=m (b)()()fn ms ox ox ss SD T t Q Q V φφ2max -+⎪⎪⎭⎫⎝⎛∈'+'-= For an +n polysilicon gate, ()288.056.02--=⎪⎪⎭⎫ ⎝⎛--=fn g ms e E φφ or272.0-=ms φV Now ()()()()4151910863.010106.1m ax --⨯⨯='SD Q or ()81038.1m ax -⨯='SDQ C/cm 2 We have()()91019106.110106.1--⨯=⨯='ssQ C/cm 2 We now find ()()()()81498105001085.89.3106.11038.1----⨯⨯⨯+⨯-=T V ()288.02272.0--or 07.1-=T V V _______________________________________ 10.18 (b) ⎪⎪⎭⎫⎝⎛++'-'=fp g m ms e E φχφφ2 where 20.0-='-'χφm V and()3473.0105.110ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯=fp φV Then()3473.056.020.0+--=ms φ or 107.1-=ms φV (c) For 0='ss Q ()fp ms ox ox SDTN t Q V φφ2max ++⎪⎪⎭⎫⎝⎛∈'= We find()()()()()2/116191410106.13473.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dT xor 41030.0-⨯=dT x cm μ30.0=m Now()()()()416191030.010106.1m ax --⨯⨯='SDQ or ()810797.4m ax -⨯='SDQ C/cm 2Then()()()()14881085.89.31030010797.4---⨯⨯⨯=T V()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 ox t C 710876.2-⨯=F/cm 2a st s ox ox oxFB eNV t C ∈⎪⎪⎭⎫ ⎝⎛∈∈+∈=' ()()()()()()()16191481410106.11085.87.110259.07.119.3101201085.89.3----⨯⨯⎪⎭⎫ ⎝⎛+⨯⨯= 710346.1-⨯='FB C F/cm 2 dTs ox ox oxx t C ⋅⎪⎪⎭⎫ ⎝⎛∈∈+∈='minNow ()3473.0105.110ln 0259.01016=⎪⎪⎭⎫ ⎝⎛⨯=fp φV ()()()()()2/116191410106.13473.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯=--dTx51000.3-⨯=cmThen ()()()5814min 1000.37.119.3101201085.89.3---⨯⎪⎭⎫ ⎝⎛+⨯⨯='C 810083.3-⨯=F/cm 2 C '(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 ='m ax ()()()516191000.310106.1--⨯⨯=81080.4-⨯=C/cm 2 ()3473.0210.110876.21080.478+-⨯⨯=--TN V 2385.0-=TN V V_______________________________________10.24(a) 1=f Hz (low freq), ()()814101201085.89.3--⨯⨯=∈=ox ox oxt C 710876.2-⨯=F/cm 2a st s ox ox oxFB eNV t C ∈⎪⎪⎭⎫ ⎝⎛∈∈+∈='()()()()()()()141914814105106.11085.87.110259.07.119.3101201085.89.3⨯⨯⨯⎪⎭⎫ ⎝⎛+⨯⨯=---- 810726.4-⨯='FBC F/cm 2 dTs ox ox oxx t C ⋅⎪⎪⎭⎫⎝⎛∈∈+∈='minNow()2697.0105.1105ln 0259.01014=⎪⎪⎭⎫⎝⎛⨯⨯=fn φV()()()()()2/1141914105106.12697.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x 410182.1-⨯=cmThen()()()4814min 10182.17.119.3101201085.89.3---⨯⎪⎭⎫ ⎝⎛+⨯⨯='C 910504.8-⨯=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 ='m ax ()()()4141910182.1105106.1--⨯⨯⨯= 910456.9-⨯=C/cm 2Then()2697.0295.010876.210456.979-+⨯⨯-=--TP V378.0+=TP V V_______________________________________10.25The amount of fixed oxide charge at x is ()x x ∆ρ C/cm 2By lever action, the effect of this oxide charge on the flatband voltage is()x x t x C V ox ox FB ∆⎪⎪⎭⎫⎝⎛-=∆ρ1 If we add the effect at each point, we must integrate so that ()dx t x x C V oxt oxoxFB⎰-=∆01ρ _______________________________________10.26 (a) We have ρx Q t SS ()='∆ Then∆V C x x t dx FB ox ox ox t=-()z 10ρ ≈-'F H G I K J F H I K-z 1C t t Q t dx ox ox oxox oxSSt t t ∆∆b g =-'--=-'F H I K 1C Q t t t t Q C ox SS ox ox SSox ∆∆a for ∆V Q t FB SS ox ox=-'∈F H G I K J =-⨯⨯⨯⨯---()16108102001039885101910814...b g b g b gb gor∆V FB =-00742.V(b) We have ρx Q t SS ox()='=⨯⨯⨯--16108102001019108.b g b g =⨯=-64103.ρONow ∆V C x x t dx C t xdx FB oxox oxOox oxoxt t =-=-()zz10ρρor ∆V t FB O oxox=-∈ρ22=-⨯⨯⨯---()6410200102398851038214...bg b g b gor∆V FB =-00371.V (c) ρρx x t O ox()F H G I KJ =We find12216108102001019108t Q ox O SS O ρρ='⇒=⨯⨯⨯--.b gb g or ρO =⨯-128102. Now ∆V C t x x t dx FB ox ox O ox t ox =-⋅⋅F H G I KJ z110ρ =-⋅z122C t x dx ox O oxox t ρaf which becomes ∆V t t x t FB ox oxO oxox O oxox t =-∈⋅⋅=-∈F H G I KJ 1332302ρρaf Then∆V FB =-⨯⨯⨯---()12810200103398851028214...b g b 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 0259.0or695.0-=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 snd x eN C ∈-=1 so()n s dx x eN +∈-=E for 0≤≤-x x n In the oxide, 0=ρ, so=E ⇒=E 0dxd constant. From the boundary conditions, in the oxidesn d x eN ∈-=E In the p-region,2C x eN eN dx d sa sa s+∈=E ⇒∈+=∈-=Eρ 0=E at ()p ox x t x +=, then ()[]x x teN p oxsa-+∈-=EAt ox t x =, snd sp a x eN x eN ∈-=∈-=E So 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 eN x V n s d n '+⎪⎪⎭⎫ ⎝⎛⋅+∈=22Arbitrarily, 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 the voltagedrop across the n-region. Because ofsymmetry, p n V V =. Then for zero bias, wehavebi ox n V V V =+2 which can be written as bi sox n d s n d V t x eN x eN =∈+∈2or 02=∈-+ds bi 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 by G 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 cmNow 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 have731210110210200---⨯=⨯⨯=ox C F/cm 2Also ()()7141011085.89.3--⨯⨯=∈=⇒∈=ox ox ox ox ox ox C t t C or 61045.3-⨯=ox t cm 5.34=nm o A 345= (c)oxssms FB C Q V '-=φ or 71050.080.0-'--=-ssQwhich yields8103-⨯='ssQ C/cm 21110875.1⨯=cm 2- (d) ⎪⎪⎭⎫ ⎝⎛∈⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛∈∈+∈='d s s ox ox ox FB eN e kT t C()()[][6141045.31085.89.3--⨯÷⨯= ()()()()()⎥⎥⎦⎤⨯⨯⨯⎪⎭⎫ ⎝⎛+--161914102106.11085.87.110259.07.119.3 which yields81082.7-⨯='FBC F/cm 2 or156=FB C pF_______________________________________10.31 (a) Point 1: Inversion 2: Threshold3: Depletion4: Flat-band5: Accumulation_______________________________________10.32 We 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 n V V C Q ()()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡+-'+'+fp ms ox ss SD C Q Q φφ2m ax For a p-type substrate, ()max SDQ ' is a negative value, so we can write()⎩⎨⎧--='DS GS ox n V V C Q()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡++'-'-fp ms ox ss SD C Q Q φφ2m ax Using the definition of threshold voltage 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 the drain terminal._______________________________________10.33(a) ()[]222DS DS T GS n D V V V V L W k I --⋅'= ()()()()[]22.02.04.08.028218.0--⎪⎭⎫ ⎝⎛= 0864.0=mA (b) ()22T GS n D V V LW 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 LW k 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-⎪⎭⎫ ⎝⎛=D I06.3=mA(c) ()[]222DSDS T GS n D V V V V L W 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 W k 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---⨯=⨯⨯=ox C F/cm 2 ()()()()2.122010138.342527-⨯==L W C K ox n n μ310111.1-⨯=A/V 2=1.111 mA/V 2(a) 0=GS V , 0=D I 6.0=GS V V, ()15.0=sat V DS V, ()()()245.06.0111.1-=sat I D 025.0=mA2.1=GS V V, ()75.0=sat V DS V, ()()()245.02.1111.1-=sat I D 625.0=mA8.1=GS V V, ()35.1=sat V DS V,()()()245.08.1111.1-=sat I D 025.2=mA4.2=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 ox t C 710138.3-⨯=F/cm 2L 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 D 060.0=mA2.1=SG V V, ()85.0=sat V SD V()()()235.02.1961.0-=sat I D 694.0=mA 8.1=SG V V, ()45.1=sat V SD V()()()235.08.1961.0-=sat I D02.2=mA4.2=SG V V, ()05.2=sat V SD V()()()235.04.2961.0-=sat I D04.4=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 I154.0=mA8.1=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 Problem 10.37,111.1=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 always biased 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 V10.43From 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 g For 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) GS D m V I g ∂∂=()()[]{}22DS DS 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 I125.3=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 C or81012.8-⨯=ox C F/cm 2We are given 10=L W . From the graph, for 3=GS V V, we have ()033.0≅sat I D , then ()2.032033.0-⋅=LC W oxn μ or310139.02-⨯=LC W oxn μor()()3810139.01012.81021--⨯=⨯n μwhich 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 DS n T GS n D 22=-= so()244102n K =⨯- which yields μ5.12=n K A/V 2 (c) ()2.18.02=-=-=T GS DS V V sat V Vso ()sat V V DS DS > ()()()258.021025.1-⨯=-sat I Dor ()μ18=sat I D A(d)()sat V V DS DS <()[]22DS DS T GS n D V V V V K I --= ()()()()[]25118.0321025.1--⨯=-orμ5.42=D I A_______________________________________10.47(a) ()()814101801085.89.3--⨯⨯=ox C 7109175.1-⨯=F/cm 2(i)()()7109175.1450-⨯=='ox n nC k μ 510629.8-⨯=A/V 2 or μ29.86='nk A/V 2 (ii)()()22T GS nD V V L W k sat I -⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛'= ()24.02208629.08.0-⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=L W24.7=⇒L W(b) (i) ()()7109175.1210-⨯=='ox p p C k μ 510027.4-⨯=A/V 2or 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_______________________________________ 10.48 From Problem 10.37, 111.1=n K mA/V 2(a) ()()[]{}22DS DS T GS n GS mL V V V V K V g --∂∂= ()()()()1.02111.12==DS n V K so 222.0=mL g mA/V (b) (){}2T GS n GS ms V V K V g -∂∂=()()()45.05.1111.122-=-=T GS n V V K so 33.2=ms g mA/V _______________________________________10.49From Problem 10.38, 961.0=p K mA/V 2(a) ()()[]{}22SD SD T SG p SGmL V V V V K Vg -+∂∂= ()()()()1.02961.02==SD p V K or 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 K or 21.2=ms g mA/V_______________________________________10.50 (a) oxa s C N e ∈=2γNow ()()814101501085.89.3--⨯⨯=oxC 710301.2-⨯=F/cm 2 Then()()()()716141910301.21051085.87.11106.12---⨯⨯⨯⨯=γ 5594.0=γV 2/1 (b) ()3890.0105.1105ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=fpφV (i)()()()()()2/1161914105106.13890.01085.87.114⎥⎦⎤⎢⎣⎡⨯⨯⨯=--dT x510419.1-⨯=cm()m ax SDQ ' ()()()5161910419.1105106.1--⨯⨯⨯=710135.1-⨯=C/cm 2 ()fp FB oxSDTO V C Q V φ2max ++'= ()3890.025.010301.210135.177+-⨯⨯=-- 7713.0=VL WC K ox n n 2μ=()()()()2.12810301.24507-⨯=410452.3-⨯=A/V 2 or 3452.0=n K mA/V 2 For 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=⎪⎪⎭⎫⎝⎛⨯=fp φ V[]fpSBfpT 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--⨯⨯=ox C710726.1-⨯=F/cm 2oxds C N e ∈=2γ ()()()()715141910726.11051085.87.11106.12---⨯⨯⨯⨯= 2358.0=γV 2/1 (b) ()3294.0105.1105ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=fnφV []fn BS fnT 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()()()()4151910863.010106.1m ax --⨯⨯='SDQ or()81038.1m ax -⨯='SDQ C/cm 2 Also()()814104001085.89.3--⨯⨯=∈=ox ox ox t C or81063.8-⨯=ox C F/cm 2 We find ()()91019108105106.1--⨯=⨯⨯='ss Q C/cm 2 Then ()fp ms oxss SD T C Q Q V φφ2m ax ++'-'=()288.020.11063.81081038.1898+-⎪⎪⎭⎫ ⎝⎛⨯⨯-⨯=--- or 357.0-=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[]fp SB 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 oxoxn V V t L W -∈=μ ()()()()()65.0510*******.89.340010814-⨯⨯=--or26.1=m g mS Nowsm m m s m m m r g g g r g g g +=='⇒+='118.01which yields⎪⎭⎫⎝⎛-=⎪⎭⎫ ⎝⎛-=18.0126.1118.011m s g r or 198.0=s r k Ω (b) For 3=GS V V, 683.0=m g mS Then ()()602.0198.0683.01683.0=+='m g mS or 88.0683.0602.0=='m m g g which is a 12% reduction._______________________________________10.56 (a) The ideal cutoff frequency for no overlap capacitance is,()222L V V C g f T GS n gs m T πμπ-==()()()24102275.04400-⨯-=π or 17.5=T f GHz (b) Now ()M gsT m T C C g f +=π2 where ()L m gdT M R g C C +=1 We find()()4410201075.0--⨯⨯=ox gdT C C()()814105001085.89.3--⨯⨯= ()()4410201075.0--⨯⨯⨯ or1410035.1-⨯=gdT C F Also ()T GS oxn m V V LC W g -=μ()()()()()()84144105001021085.89.34001020----⨯⨯⨯⨯= ()75.04-⨯or3108974.0-⨯=m g SThen ()1410035.1-⨯=M C ()()[]331010108974.01⨯⨯+⨯- or 1310032.1-⨯=M C F Now()()W L C C ox gsT 41075.0-⨯+= ()()814105001085.89.3--⨯⨯= ()()44410201075.0102---⨯⨯+⨯⨯ or1410797.3-⨯=gsT C F We now find ()M gsT mTC C g f +=π2 ()1314310032.110797.32108974.0---⨯+⨯⨯=π or 01.1=T f GHz _______________________________________10.57 (a) For the ideal case()4610221042-⨯⨯==ππυL f ds Tor 18.3=T f GHz(b) With overlap capacitance (using the values from Problem 10.56), ()MgdT mT C C g f +=π2 We findds ox m W C g υ= ()()()()86144105001041085.89.31020---⨯⨯⨯⨯= or3105522.0-⨯=m g S We have()L m gdT M R g C C +=1 ()1410035.1-⨯=()()[]331010105522.01⨯⨯+⨯- or 1410750.6-⨯=M C F。

半导体物理与器件第四版课后答案第七章

17
or
x p 0.0213 10 4 cm 0.0213 m
We have
max
19
eN d x n s
15 4

1.6 10 5 10 0.426 10 11.7 8.85 10
14
or
max 3.29 10 4 V/cm

7.4 (a) n-side
Nd E F E Fi kT ln n i

5 1015 1017
1 17 10 5 1015

1/ 2
5 1015 0.0259 ln 1.5 1010 or E F E Fi 0.3294 eV p-side Na E Fi E F kT ln n i 10 0.0259 ln 1.5 1010 or E Fi E F 0.4070 eV (b) Vbi 0.3294 0.4070 or Vbi 0.7364 V (c)
(a) N d 1014 cm 3 , N a 1017 cm 3 ' Si: Vbi 0.635 V Ge: Vbi 0.253 V GaAs: Vbi 1.10 V
; 1017 (c) Silicon (400 K),
n i 2.38 1012 cm 3
Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ For N a N d 1014 cm 3 ; Na Nd Vbi Vt ln n2 Vbi 0.2582 V i

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Chapter 3If a o were to increase, thebandgap energywould decrease and the materialwould beginto behave less like asemiconductor and morelike a metal. If a o were todecrease, thebandgap energy would increase andthematerial would begin to behavemore like aninsulator._______________________________________Schrodinger's wave equation is:2 2 x, tV x x, t2m x2j x, t tAssume the solution is of the form:x, t u x exp j kx E tRegion I: V x 0 . Substituting theassumed solution into the wave equation, weobtain:2E2m xjku x exp j kx tu xexp j kx E txj jE u x exp j kx E t which becomes22 u E2mjk x exp j kx t2 jk u xexp j kx E tx2 u xexp jEtx2kxEu x exp j kx E tThis equation may be written ask 2 u xu x 2 u x 2mE2 jkx2 2 u x 0xSetting u x u1 x for region I,the equationbecomes:d 2 u1 x 2 jk du1 x k 2 2 u1 x 0dx2 dxwhere2 2mE2In Region II, V x V O. Assume thesameform of the solution:x, t u x exp j kx E tSubstituting into Schrodinger'swaveequation, we find:22 u Ejk x exp j kx t2m2 jku xexp j kx E tx22xexp ju kx E txV O u x exp j kx E tEu x exp j kxEtThis equation can be written as:k 2u x 2 jk u x 2u xx x22mV Ou x2mEu x 02 2Setting u x u 2 x for region II, thisequation becomesd 2u 2 x du 2 x dx 2 2 jkdxk 22 2mV Ou 2 x 02where again22mE2We have22 jk du 1 xd u 1 x k 22u 1 x 0dx 2dxAssume the solution is of the form:u 1 xA exp jk xB exp jk xThe first derivative isdu 1 x jk A exp jk xdxjk B expjk xand the second derivative becomesd 2 u 1 x jk2k xdx 2A exp j2 B expj k jk xSubstituting these equations into thedifferential equation, we findk 2 A exp jk x2B exp jk xk2 jk j k A exp jk x jk B exp jk xk 22A exp jk xB exp jk xCombining terms, we obtain22 k k22kkk22A exp jk x22 k k 22kkk 22B exp jk xWe find that0 0For the differential equation in u 2 xand theproposed solution, the procedure is exactlythe same as above._______________________________________We have the solutionsu 1 xA exp j k xB exp jk xfor0 x a andu 2 x C exp jk xD exp jk xfor b x 0 .The first boundary condition isu 1 0 u 2 0which yieldsA B C DThe second boundary condition isdu 1 du 2dxx 0dxx 0which yieldsk Ak Bk Ck D 0The third boundary condition isu 1 a u 2 bwhich yieldsA exp jk aB exp jk a C exp jkbD exp jkband can be written asA exp jk aB exp jk aC exp jk bD exp jk bThe fourth boundary condition is du 1 du 2dxx adxxbwhich yieldsjk A exp jk ajk B exp j k ajk C exp jkbjk D exp j kband can be written ask A exp jk ak B exp jk ak C exp jk bk D exp j k b 0_______________________________________(b) (i) First point:aSecond point: By trialand error,a 1.729(ii) First point:a2Second point: By trialand error,a 2.617_______________________________________(b) (i) First point:aSecond point: By trialand error,a 1.515(ii) First point:a 2Second point: By trialand error,a 2.375_______________________________________Psin a cos a coskaaLet kay ,a xThenP sin xcos x cos yxConsiderd of this function.dyd P x1cos xsin ydy sin xWe findP1 x2sin x dx x 1 cos x dxdydy sin xdxsin ydyThendx P 21sin x cos xsin xsin ydy xxFory ka n ,n 0, 1, 2, ... sin y0 So that, in general,dx 0d a ddyd kadkAnd2mE2Sod 11/ 2dE2mE2mdk2 22dkThis implies thatd 0 dE forkn dk dka_______________________________________(a)1a2m o E 12a2 221.054 10 34 2E 12 9.11 10 31 4.2 101022m o a 23.4114 10 19JFrom Problem2 a 1.7292m o E 2a1.72921.72921.054 10 34 2E 22 9.11 10 314.2 10 10 21.0198 1018JE E 2E 11.0198 10 18 3.4114 10 196.7868 10 19JorE6.7868 10 194.24 eV1.610 193 a 2(b)2m o E 3 22a2 2 1.05410 34 2221.054 10 34 2E 310 31 4.2 10 10 2E 310314. 2 10 10 22 9.112 9.111.3646 10 18Jka1.3646 10 18JFrom Problem ,At. From Problem ,4a 2.6172a 1.7292m o E 42.617 2m o E 2a1.7292a22.6172 1.054103421.72921.054 10 34 2 E 4E 210 314.2 10 10 22 9.11 10 314.2 10 10 22 9.112.3364 10 18J1.0198 10 18JEE 4 E 3E E 3E 22.3364 10 18 1.3646 10 18 1.3646 10 18 1.0198 10 189.718 10 19J3.4474 10 19Jor9.718 10 196.07 eVorE 3.4474 10 19 2.15 eVE 1.6 10 191.6 1019_______________________________________ _______________________________________(a) At ka, 1a2m o E 1 a221.054 10 34 2E 12 9.11 10314.2 10 1023.4114 10 19JAt ka 0 , By trial anderror,oa 0.8590.85921.054 10 34 2E o2 9.11 10 31 4.210 1022.5172 1019JE E 1 E o3.4114 10 19 2.5172 10 198.942 10 20J8.942 10 20orE0.559 eV1.6 10 19(b) At ka2 ,3 a22m o E 3a22(a)1a2m o E 1a221.054 10 34 2E 110 31 4.2 10 10 22 9.113.4114 10 19JFrom Problem , 2 a 1.5152m o E 2a 1.51521.515 21.054 10342E 22 9.1110314.2 1010 27.830 10 19JE E 2E 17.830 10 19 3.4114 10 194.4186 10 19J or E4.4186 10 192.76 eV1.610 19(b)3a 22m o E3a 222 2 1.054 10 34 2E32 9.11 10 31 4.2 10 1021.3646 10 18JFrom Problem , 4 a 2.3752m o E 4a 2.37522.375 2 1.054 10 34 2E42 9.11 10 31 4.2 10 1021.9242 1018E E4 E3J1.9242 10 18 1.3646 10 185.597 10 19Jor E 5.597 10 193.50 eV 1.6 10 19_____________________________________(a) At ka , 1 a2m o E1a22 1.054 10 34 2E131 10 22 9.11 10 4.2 10ka 0 3.4114 10 19JAt , By trial and error,or1.6084 10 191.005 eVE10 191.6(b) At ka 2 , 3 a 22m o E 3a 222210 342E 31.05410 31 4.2 10 1022 9.111.3646 10 18JAt ka , From Problem ,2 a 1.5152m o E 22 a 1.5151.515 2 1.054 10 34 2E 210 34 4.2 10 1022 9.117.830 10 19JE E3 E21.3646187.8301910 105.816 10 19Jor5.816 10 193.635 eVE1.6 10 19_______________________________________For T 100K,4.73 10 4 2E g1001.170100636o a 0.7272m o E oa 0.72720.727 2 1.054 10 34 2E o31 10 22 9.11 10 4.2 101.8030 10 19J E g 1.164 eVT 200 K,T 300 K,T 400 K,T 500 K,T 600 K,E g 1.147 eVE g 1.125 eVE g 1.097 eVE g 1.066 eVE g 1.032 eVE E1E o3.4114 10 19 1.8030 10 191.6084 10 19J _______________________________________The effective mass is given by11m *d 2 E2dk 2We haved 2 E curve A d 2 E curveBdk2 dk 2so that m* curve A m* curve B_______________________________________The effective mass for a hole isgiven by1 d2 E 1m*p2dk 2 We have thatd 2 Ecurve A d 2 Edk 2 dk 2curve Bso that m*p curve A m*p curve B_______________________________________Points A,B: dE0 velocity in -x dkdirectionPoints C,D: dE0 velocity in +x dkdirectionPoints A,D: d 2 E0 dk 2negative effective massPoints B,C: d 2 E0 dk 2positive effective mass_______________________________________For A:E C i k 2At k 0.08 10 10m1, E 0.05eV OrE 0.05 1.6 10 19 8 10 21 JSo 8 10 21 C1 0.08 10102C1 1.25 10 382 1.054 10 342Now m2C1 2 1.25 10 384.44 10 31kgor m4.4437 10 31m o9.11 10 31m 0.488 m oFor B: E C i k 2At k 0.08 10 10m1, E 0.5 eV OrE 0.5 1.6 10 19 8 10 20 JSo 8 10 20 C1 0.08 10102C1 1.25 10 3721.054 10 342 Now m2C1 2 1.25 10 374.44 10 32kgor m4.4437 10 32m o9.11 10 31m 0.0488 m o_______________________________________For A: E E C 2 k 20.025 1.6 10 19 C 2 0.08 1010 2C 2 6.25 10 3921.054 10 342m2C 2 2 6.25 10 398.8873 10 31kg8.887331or m10m o9.11 10 31m 0.976 m oFor B: E E C 2 k 20.3 1. 610 19 C 2 0.08 10 102C 2 7.5 10 3821.054 10 34 2m2 7.5 10 382C 27.406 10 32kgor7.406 10 32m om10 319.11m 0.0813 m o_______________________________________(a) (i) Eh10 19 orE 1.42 1.6h6.625 10 343.429 1014Hz(ii)hcc3 1010E3.429 10148.75 10 5cm 875nm1019(b) (i)E 1.12 1.6h6.625 10342.705 1014 Hz(ii)c3 10102.705 14101.109 10 41109 nm cm_______________________________________m *22E 1_______________________________________(a) m dn42/ 3m t21 / 3m l42/321/ 30.082m o 1.64m om dn 0.56m o (b)3 2 1 21m cnm tm l0.082 m o 1.64m o24.39 0.6098 m om om cn 0.12m o_______________________________________3 / 23/22/3(a) m dp m hhm lh(c) Curve A: Effective mass is a constantCurve B: Effective mass ispositivearoundk0 ,and is negativearoundk.2 _______________________________________mdp(b) m cpmcp0.45m o 3 / 23 / 2 2 / 30.082 m o0.30187 2 / 3m o0.023480.473 m o3 / 23 / 2m hh m lh1/ 2 1/ 2m hhm lh3 / 23 / 20.450.082 m o1 / 21 / 20.45 0.0820.34m oE E OE 1 cos k k OThendE E 1 sin k k OdkE 1 sink k Oandd 2E E 1 2cos k k O dk 2Then1 1 d 2EE 12 m * 2dk 2 k k o2or_______________________________________For the 3-dimensional infinite potential well,V x 0 when 0 x a , 0 y a ,and0 z a . In this region, thewave equationis:22 2x, y, z x, y, z x, y, z x 2y 2z 22mEx, y, z 02Use separation of variables technique, so letx, y, z X x Y y Z zSubstituting into the waveequation, we have2X2Y2ZYZ2 XZ2 XY2x y zwhere22mEk 2We can then write2mE2XYZ 0k2mEDividing byXYZ , we obtain12X1 2Y12Z2mEX x2Yy2Z z22LetTaking the differential, we obtain1 2Xk x22Xk x 2X 0X x 2x2The solution is of the form:X xAsin k x x B cos k x xSincex, y, z0 atx 0 , thenX 0 0so that B 0 .Also, x, y, z 0 at x a , so11 1dk2mdE2 ESubstituting these expressionsinto the densityof states function, we haveg T E dEa 32mE 132Noting that1mdE2Em dE2EthatX a 0 . Thenk x a n x wheren x 1, 2, 3, ...Similarly, we have221 Y k y 2and1 Zk z2Y y 2Z z 2From the boundary conditions, we findk y a n yandk z a n zwheren y 1, 2, 3, ... and n z 1, 2, 3, ...From the wave equation, we can write2222mE0 k x k yk z2The energy can be written as22E E n x n y n zn x 2 n y 2 n z 2a2m_______________________________________The total number of quantum states in the3-dimensional potential well is given(in k-space) byk 2dka 3g T k dk3h2this density of states functioncan besimplified and written asg T E dE4 a 3 3 / 2E dE32mhDividing by a 3 will yield thedensity of states so that4 3 / 2g E 2m Eh 3_______________________________________For a one-dimensional infinitepotential well,2m n En 2 2k 22a2Distance between quantum statesk n 1 k nn 1naaaNow2 dk g T k dkaNow1 k2m n E1 1 2m ndEdkE 2Then(ii) AtT 400 K,400 kT 0.02593002a 1 2m n g T E dEdE2EDivide by the "volume"a , so0.034533 eV0.034533 1.610191 2m n g EESog E12 0.067 9.11 10 311.054 1034Eg E1.055 1018m 3 J 1E_______________________________________(a) Silicon,m n 1.08m o4 2m n 3 / 2g c EE E ch 34 2m n3 / 2 E c 2kTg c E E c dEh3E c4 2m n3 / 22E c 2kTE E c 3 / 2h33E c4 2m n3 / 223 / 2h 32kT34 2 1.08 9.11 10 31 3 / 223 / 26.625 10 34 33 2kT7.953 10552kT3 / 2(i) AtT 300 K, kT 0.0259 eV0.0259 1.6 10194.144 10 21JTheng c7.953 10552 4.144 10213 / 26.0 10 25m 3org c6.0 10 19 cm 35.5253 10 21JTheng c7.953 10 552 5.5253 10 213 / 29.239 1025 m 3org c 9.24 1019 cm 3(b) GaAs, m n 0.067 m o4 2 0.067 9.11 10313 / 222kT 3 / 2g c6.625 1034 331.2288 10 542kT 3 / 2(i) AtT 300 K, kT4.144 10 21Jg c1.2288 10542 4.144 10 213 / 29.272 10 233mor g c9.27 1017 cm3(ii) AtT 400 K, kT 5.5253 10 21 Jg c1.2288 10542 5.5253 10 213 / 21.427 1024 m 3g c1.43 1018 cm 3_______________________________________(a) Silicon,m p 0.56m o4 2m p 3 / 2g EEEh343 / 2Eg 2m pEEdEh 3E3 kT4 2m p3 / 22EE E3 / 2h 33E3 kT4 2m p3 / 223 / 2h333kT31 3 / 24 2 0.56 9.11 1023 / 2 33kT6.625 10 3432.969 10553 / 23kT(i)At T 300 K, kT 4.144 10 21Jg 2.969 10 55 3 4.144 10 21 3 / 24.116 1025m3or g 4.12 1019 cm 3(ii)At T 400 K, kT 5.5253 10 21Jg 2.969 10 55 3 5.5253 10 21 3 / 26.337 1025m3or g 6.34 1019 cm 3(b) GaAs, m p 0.48 m o4 2 0.48 9.11 10 31 3 / 22 3kT3 / 2g 36.625 10 34 32.3564 10 55 3kT 3 / 2(i)At T 300 K, kT 4.144 10 21Jg 2.3564 10 55 3 4.144 10 21 3 / 23.266 1025m3or g 3.27 1019 cm 3(ii)At T 400 K, kT 5.5253 10 21Jg 2.3564 10 55 3 5.5253 10 21 3 / 25.029 10 25m3or g 5.03 1019 cm 3_______________________________________4 3 / 2(a) g c E 2m nE E c h34 2 1.08 9.11 10 31 3 / 2E E c6.625 10 34 31.1929 10 56 E E cFor E E c; g c 0E E c 0.1eV;g c 1.509 10 46 m 3 J 1E E c 0.2 eV;2.134 1046m3J 1E E c 0.3 eV;2.614 1046m3J 1E E c 0.4 eV;3.018 1046m3J 14 2m p3 / 2(b) g E Eh 34 2 0.56 9.11 10 313 / 2E E6.625 10 3434.4541 10 55 E EFor E E ; g 0E E 0.1 e V;g 5.634 10 45 m 3 J 1E E 0.2 eV;7.968 10 45m3J 1E E 0.3 eV;9.758 10 45 m 3 J 1E E 0.4 eV;1.127 1046m3J 1_______________________________________g c 3 / 2 3 / 2m n 1.08(a) 3 / 2 2.68g m p 0.56g c 3 / 2 3 / 2m n 0.067(b) 3 / 2 0.0521g m p 0.48_______________________________________Plot_______________________________________(a) W i g i ! 10!N i ! 7! 10 7!N i ! g i10 9 8 7! 10 9 81207! 3! 3 2 1(b) (i) W i12! 12 11 10!10! 12 10 ! 10!2166(ii)12!121110 98!W i8!12 8!8!4321495_______________________________________f E1E E F1 exp kT(a) E E F kT ,1f E1 exp 1f E0.2691 (b)E E F5kT , f E1exp 5f E 6.69 10 31 (c)E E F10kT , f E1exp 102.08 10 6E c 2kT ; fF exp0.30 2 0.02590.02591.26 10 6(b) 1 f F 11E E F1 expkTexpE F EkTE E ; 1 fF exp 0.25 6.43 10 50.0259f E 4.54 10 5 E kT ; _______________________________________1 f E 11E E F1 expkTor210.25 0.0259 2f F exp0.02593.90 10 5E kT ; 1 fF exp0.25 0.02590.02591 f E1E F E1 expkT(a) E E kT , 1 f E 0.2692.36 10 E3kT ;25F(b) E F E3 5kT , 1 f E 6.69 10(c)E F E 10kT , 1 f E 4.54 105_______________________________________(a) f F exp E E F kTE E c; fF exp 0.30 9.32 10 60.0259E c kT ; fF exp 0.30 0.0259 20.025925.66 10 6E c kT ; fF exp 0.30 0.02590.02593.43 10 6E c 3kT; 2f F exp 0.30 3 0.0259 20.02590.25 3 0.0259 21 f F exp0.02591.43 10 5E 2kT ;1 f F exp0.25 2 0.02590.02598.70 10 6_______________________________________f FE EF E c kT E Fexp expkTkTand1 f F expE F EkTE F E kTexpkTSoE c kT E FexpkTE FE kTexpkTThen E c kT E F E F EkTOrE c EE FEmidgap2_______________________________________2 22E nn2ma 2 Forn 6 , Filled state1.05410 34222 E6610 31 1210 1022 9.111.5044 10 18J1.504418orE 6109.40 eV1.6 10 19Forn 7 , Empty state1.054 10 34 2722E 710 311210 1022 9.112.048 10 18JorE2.048 10 1812.8 eV 71.6 10 19Therefore9.40 E F 12.8 eV_______________________________________(a) For a 3-D infinite potential well2mE22222n x n y n zaFor 5 electrons, the 5thelectronoccupies the quantum state n x 2, n y2, n z 1 ; so22n x 2n y 2n z2E 52ma1.054 10 34 222 2 22 122 9.11 10 31 1210 10 23.761 10 19Jor3.761 10 19 2.35 eVE 510 191.6n x1, n y 2, n z 2 . This quantumstate is at the same energy, soE F 2.35 eV(b) For 13 electrons, the 13thelectron occupies the quantum staten x3, n y 2, n z 3 ; so1.054 10 34 223 2 2 2 32E132 9.11 10 31 12 1010 29.194 10 19JorE 13 9.194 10195.746 eV1.6 10 19The 14 th electron would occupy thequantum staten x 2, n y 3, n z3 .This state is at the same energy, soE F 5.746 eV_______________________________________The probability of a state atE 1 E FEbeing occupied isf 1 11E 1EE 1 E F1 exp1 expkTkTThe probability of a state atE 2 EF Ebeing empty is11 f2 E 2 11 exp E2 E FkTE1exp1kTE E1exp1 expkTkTor1 f2 E 21E1expkT(a)so f 1E 1 1 f 2 E 2energyE 1 , we wantFor the next quantum state, whichis empty, the quantum state is1 1E1E F E1E Fexp 1 expkT kT0.0111exp E1 E FkTThis expression can be written as1E1 E Fexp kT1 0.01E1 E FexpkT or1E1 E F 0.01 expkTThenE1 E F kT ln 100orE1 E F 4.6kT(b)At E E F 4.6kT ,f E11 1expE1 EF 1 exp 4.6 1kTwhich yieldsf E1 0.00990 0.01_______________________________________(a)f F exp E E Fexp5.80 5.50 kT 0.02599.32 10 6(b) kT 0.0259 7000.060433 eV 300f F exp 0.30 6.98 10 30.060433(c) 1 f F exp E F E kT0.02 exp0.25kTor0.25 150exp0.02kT0.25ln 50kTor0.250.063906TkT 0.0259ln 50 300which yields T 740 K_______________________________________(a)10.00304f E7.15 7.01 exp0.0259or %(b)At T 1000 K, kT 0.08633 eV Thenf E 1 0.1496exp 7.151 7.00.08633or %(c) f E 1 0.9976.8517.0exp0.0259or %(d)At E E F, f E1for all2temperatures_______________________________________(a) For E E1f E1expE1 E FE1 E F kT1 expkTThenf E1 exp 0.30 9.32 10 60.0259For E E2,E F E2 1.12 0.30 0.82 eVThen1 f E 110.821 exp0.0259or1 f E 11 exp0.82 0.0259exp0.82 1.78 10 140.0259(b) For E F E 2 0.4 eV,E 1 E F0.72 eV At E E 1 ,f Eexp E 1 E F 0.72kTexp0.0259orf E8.45 10 13At E E 2 ,1 f EE F E 2 expkT0.4exp0.0259or1 f E 1.96 107_______________________________________(a) At EE 1f EE 1 E F0.30 expexp0.0259kTorf E9.32 10 6At E E 2 , E F E 21.42 0.3 1.12 eVSo1 f EE FE 2expf EexpE 1 E F1.02kTexp0.0259orf E187.88 10 At E E 2 ,1 f EexpE F E 2kTexp0.40.0259or 1 f E1.96 10 7_______________________________________EE F 1f E1 expkTsodf EE EF 21 1 expdEkT1 exp E E FkTkT or1 exp E E Fdf EkTkTdE21 expE E FkT(a) At T 0 K, ForE E FexpdfdEE E FexpdfdEAtEE FdfdE(b) At T300 K, kT 0.0259 eVkTexp1.12For EE F ,dfdE0.0259or1 f E1.66 1910(b) For E F E 2 0.4 ,E 1 E F1.02 eV At EE 1 ,ForE EF ,dfAtE EF ,dE11df 0.025919.65 (eV)dE21 1(c) At T 500 K, kT 0.04317 eVFor E E F, df0 dEFor E E F, df0 dEAt E E F,1df 0.04317 115.79 (eV)dE2 1 1_______________________________________(b)Using the results of Problem ,the answers to part (b) are exactlythe same as those given in part (a)._______________________________________(a) f F expE E FkT10 8 exp 0.60kTor 0.60 ln 10 8kTkT0.600.032572 eVln 1080.032572 0.0259T(a) At E E midgap ,f E1 1E EF E g1 exp 1 expkT 2kTSi: E g 1.12 eV,300 so T 377K(b)10 6 exp0.60kT0.60 ln 10 6kTf E 11.12 kT 0.606 0.0434291exp2 0.0259 ln 10Torf E 4.07 10 100.043429 0.0259300or T 503 K_______________________________________Ge: E g0.66 eV1f E0.661 exp2 0.0259 (a) At T 200 K,kT2000.0259 0.017267 eVorf E 2.93 10 6300f F 0.05 1GaAs: E g 1.42 eV1f E1.421exp2 0.02591 exp E E FkTE EF 11 19 expkT 0.05orf E 1.24 10 12E EF kT ln 19 0.017267 ln 190.05084eV By symmetry, for f F 0.95,E EF 0.05084eVThen E 2 0.05084 0.1017 eV(b)T 400K, kT 0.034533 eVFor f F 0.05 , from part (a),E EF kT ln 19 0.034533 ln 190.10168eVThen E 2 0.10168 0.2034 eV_______________________________________。