半导体物理与器件4
半导体物理与器件(尼曼第四版)答案
半导体物理与器件(尼曼第四版)答案第一章：半导体材料与晶体1.1 半导体材料的基本特性半导体材料是一种介于导体和绝缘体之间的材料。
它的基本特性包括：1.带隙：半导体材料的价带与导带之间存在一个禁带或带隙，是电子在能量上所能占据的禁止区域。
2.拉伸系统：半导体材料的结构是由原子或分子构成的晶格结构，其中的原子或分子以确定的方式排列。
3.载流子：在半导体中，存在两种载流子，即自由电子和空穴。
自由电子是在导带上的，在外加电场存在的情况下能够自由移动的电子。
空穴是在价带上的，当一个价带上的电子从该位置离开时，会留下一个类似电子的空位，空穴可以看作电子离开后的痕迹。
4.掺杂：为了改变半导体材料的导电性能，通常会对其进行掺杂。
掺杂是将少量元素添加到半导体材料中，以改变载流子浓度和导电性质。
1.2 半导体材料的结构与晶体缺陷半导体材料的结构包括晶体结构和非晶态结构。
晶体结构是指材料具有有序的周期性排列的结构，而非晶态结构是指无序排列的结构。
晶体结构的特点包括：1.晶体结构的基本单位是晶胞，晶胞在三维空间中重复排列。
2.晶格常数是晶胞边长的倍数，用于描述晶格的大小。
3.晶体结构可分为离子晶体、共价晶体和金属晶体等不同类型。
晶体结构中可能存在各种晶体缺陷，包括：1.点缺陷：晶体中原子位置的缺陷，主要包括实际缺陷和自间隙缺陷两种类型。
2.线缺陷：晶体中存在的晶面上或晶内的线状缺陷，主要包括位错和脆性断裂两种类型。
3.面缺陷：晶体中存在的晶面上的缺陷，主要包括晶面位错和穿孔两种类型。
1.3 半导体制备与加工半导体制备与加工是指将半导体材料制备成具有特定电性能的器件的过程。
它包括晶体生长、掺杂、薄膜制备和微电子加工等步骤。
晶体生长是将半导体材料从溶液或气相中生长出来的过程。
常用的晶体生长方法包括液相外延法、分子束外延法和气相外延法等。
掺杂是为了改变半导体材料的导电性能，通常会对其进行掺杂。
常用的掺杂方法包括扩散法、离子注入和分子束外延法等。
半导体物理与器件第四版课后习题标准答案
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-第四章 平衡半导体
16
4.1 半导体中的载流子
4.1.3 本征载流子浓 度
P81例4.3
ni随温度的升高而明显增大。
• 与温度关系很大： • 温升150度时，浓度增大４个数量级。
17
4.1 半导体中的载流子
4.1.4 本征费米能级位置
由电中性条件：n0=p0
禁带中央
本征费米能级精确位于禁带中央；
本征费米能级会稍高于禁带中央； 本征费米能级会稍低于禁带中央；
平征半导体(Intrinsic Semiconductor)
本征激发：共价键上的电子激发成为准自由电子，也就是 价带电子获得能量跃迁到导带的过程。
本征激发的特点：成对的产生导带电子和价带空穴。
14
4.1 半导体中的载流子
4.1.3 本征载流子浓度
说明： 本征半导体中电子的浓度=空穴的浓度即n0=p0 （电中性条件）记为ni=pi
3、施主杂质原子增加导带电子，但并不产生价带空穴，因此，这样的半导体称为 n型半导体。
22
4.2掺杂原子与能级 施主杂质
■ 电子脱离施主杂质的束缚成为导电电子的过程称为施主电 离，所需要的能量
ΔED=Ec-Ed 称为施主杂质电离能。ΔED的大小与半导体材料和杂质种类
有关，但远小于Si和Ge的禁带宽度。 ■ 施主杂质未电离时是中性的，称为束缚态或中性态，电离后
4.4施主和受主的统计学分布 4.4.2完全电离和束缚态
与室温条件相反，当T=0K时，杂质原子没有电离： 1、对n型半导体，每个施主原子都包含一个电子，nd=Nd
费米能级高于施主能级
2、对p型半导体，杂质原子不包含外来电子，na=Na,费米能级低于受主能级
束缚态：
没有电子从施主能态热激发到导带 中，
半导体物理与器件(尼曼第四版)答案之第一部分-半导体属性
半导体物理与器件(尼曼第四版)答案之第一部分-半导体属性
1. 导电性：
半导体材料是指在电声信号强度及温度变化范围内，具有显著能量带隙、静电屏蔽能力和较强导电性的半导体物质。
其导电性取决于半导体物质的原子结构和物理性质。
值得注意的是，半导体材料具有非常高的电阻率，其电阻率取决于半导体材料中存在的空穴和电子的数量及相应的电子移动速率。
在常温下，半导体物质的电阻率可以达到106到1012欧姆之间的数字，而在低温和高温下，电阻率几乎可以忽略不计。
2. 光电效应：
半导体物质具有光电效应，即半导体物质可以在受到光照时发生微小变化。
由于半导体物质具有光电效应，因此，当光照在半导体物质上时，可以产生电压，从而使半导体物质的电阻率发生变化，产生静电效应。
这种光电效应可以被用于光电器件的研制中，例如太阳能电池，光敏电阻等等，具有十分广阔的应用范围。
3. 热敏性：
半导体物质具有高的热敏性，当温度发生变化时，半导体物质的性质也会发生变化。
当温度提高时，半导体物质开始呈现出热电效应，其电阻率会随着温度提高而减小，而当温度降低时，会出现负热效应，其电阻会随着温度降低而增加。
因此，半导体物质的热敏性可以被利用于研制热敏电阻、热敏电容等等的器件中。
《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。
最新半导体物理与器件-课件-教学PPT-作者-裴素华-第4章-MOS场效应晶体管精品课件
2.表面(biǎomiàn)势与表面(biǎomiàn)耗尽区
下图给出了P型半导体MOS结构在栅极电压UG>>0情况下更 为(ɡènɡ wéi)详细的能带图。
第五页，共74页。
在下面(xià mian)的讨论中，定义与费米能级相对应的费
米势为
F
(Ei
EF )体内 q
因此(yīncǐ)，对于P型半 导体，
MOS 电容(diànróng)等效示意图 第十页，共74页。
在平带条件(tiáojiàn)下对应的总电容称为MOS 结构的平带 电容CFB
C FB
t OX
OX 0
1 2
OX S
LD
右图表示(biǎoshì)了 P型半导体MOS结构 的理想C-U曲线
线
第十一页，共74页。
MOS电容-电压(diànyā)曲
UDS较小时(xiǎoshí)，导电沟道随UGS的变化
a) UGS< UT 没有沟道 b) UGS> UT 出现沟道 c) UGS>>UT 沟道增厚
第二十四页，共74页。
2. 饱和(bǎohé)工作区
此时的电流-电压特性(tèxìng)对应与特性(tèxìng)图中UGS=5V曲线的 AB段。
导电沟道(ɡōu dào)随UDS的变化
ns
ni
exp
q(
s
T
F
ps
ni
exp
q(
F T
s
第七页，共74页。
通过以上讨论，以下各区间的表面电势可以区分为： Ψs<0空穴积累（能带向上弯曲）； Ψs=0平带情况； ΨF>Ψs>0空穴耗尽（能带向下弯曲）； ΨF=Ψs 表面上正好(zhènghǎo)是本征的ns=ps=ni ΨF<Ψs 反型情况（反型层中电子积累，能带向下弯曲）。
半导体器件物理课件四
02 半导体器件的基本概念
半导体的定义和特性
半导体：导电性 能介于导体和绝 缘体之间的材料
半导体的特性： 具有可调节的导 电性，可以通过 掺杂、光照、温 度等外部因素改
变其导电性能
半导体的分类： 分为N型半导体 和P型半导体， N型半导体中的 电子是主要的载 流子，P型半导 体中的空穴是主
要的载流子
军事装备：如雷达、导弹、 电子战等
集成电路的应用
计算机： CPU、内 存、存储 设备等
通信设备： 手机、基 站、路由 器等
家电：电 视、冰箱、 洗衣机等
汽车电子： 发动机控 制、安全 系统、导 航系统等
医疗设备： 心电图仪、 CT扫描仪、 超声波诊 断仪等
航空航天： 卫星、火 箭、飞机 等
太阳能电池的应用
半导体材料的选择和处理
半导体材料的选择：根据器件性能和成本要求选择合适的半导体材料
半导体材料的处理：对半导体材料进行清洗、抛光、腐蚀等处理，以获得所需的半导体 表面
半导体材料的掺杂：通过掺杂工艺将杂质引入半导体材料中，以改变其电学性质
半导体材料的热处理：对半导体材料进行热处理，以改善其电学性质和机械性能
半导体光电器件：如光电二极管、光电三极管等，用于光电转换、光电检测等应用
半导体器件的应用领域
汽车电子：如汽车导航、汽 车音响等
通信设备：如基站、路由器 等
电子设备：如手机、电脑、 电视等
医疗设备：如医疗仪器、医 疗电子设备等 航空航天：如卫星、火箭等
军事领域：如雷达、导弹等
03 半导体器件的基本原理
半导体器件物理课件 四
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半导体物理与器件第四版课后习题答案(供参考).doc
Chapter 44.1n i 2E gN c N expkTT 3E gexpN cO N O300kTwhere N cO and N Oare the values at 300 K.(a) SiliconT (K) kT (eV) n i (cm 3) 200 0.01727 7.68 104 400 0.03453 2.38 1210 6000.05189.74 1014(c) GaAs(b) GermaniumT (K)n i (cm 3 ) n i (cm 3 ) 200 2.16 10101.38 4008.60 1014 3.28 109 6003.82 10165.72 1012_______________________________________ 4.2Plot_______________________________________4.3(a) n i 2 N c NexpE gkT31121919T5 2.8 1.04 101010300exp1.120.0259 T 300T 32.5 10 232.912 10 38300exp1.12 3000.0259 TBy trial and error, T 367.5 K(b)n i25 10 1222.5 10 2532.912 10 38T exp 1.12 300300 0.0259 TBy trial and error,T 417.5 K _______________________________________4.4At T200 K, kT0.02592003000. 017267eVAt T400 K, kT0.02594003000. 034533eVn i 2400 7.70 101023.025 10 17n i 2 2001.40 10 2 23400expE g3000.0345333200Egexp300 0.017267E gE g8 exp0.0345330.0172673.025 10178 exp E g 57 .9139 28.9578orE g 28.9561ln 3.025 1017 38.17148 or E g 1.318 eVNow7.70 1010N co N o340023001.318 exp0.03453321N co N o 2.370 175.929 10 2.658 10so N co N o 9.41 10 37 cm 6_______________________________________4.5exp 1.10n i kT 0.20Bexpn i A 0.90 kTexp kTFor T 200 K, kT 0.017267 eVFor T 300 K, kT 0.0259 eVFor T 400 K, kT 0.034533 eV(a) For T 200K,n i B exp 0.20 9.325 10 6n i A 0.017267(b) For T 300K,n i Bexp 0.204.43 10 4n i A 0.0259 (c) For T 400K,n i Bexp 0.203.05 10 3n i A 0.034533_______________________________________ 4.6(a) g c f FE E FE E c expkTThen g c f F x expxkTTo find the maximum value:d g c f F 1 x1 / 2 exp xdx 2 kT1 x1 /2 exp x 0kT kTwhich yields1/ 21 x kT2x1/ 2 x 2kTThe maximum value occurs atEkTE c2(b)g 1 f FE F EE E expkTE EE E expkTexpE F EkTLet E E xThen g 1 f F x expxkTTo find the maximum valued g 1 f F d xdx dxx expkTSame as part (a). Maximum occurs atxkT2E E c exp E E ckTorkTE E2E c EF expkTLet E E c x _______________________________________ 4.7E1 E c exp E1 E cn E1 kTn E2E2 E c exp E2 E c kTwhereE1 E c 4kT and E 2 E c kT 2Thenn E1 4kTexp E1 E2n E2 kT kT22 2 exp 4 12 exp 3.522orn E10.0854n E 2_______________________________________ 4.8Plot_______________________________________4.9Plot_______________________________________ 4.10E Fi E midgap 3kT ln m*pm n* 4Silicon: m*p 0.56 m o , m n* 1.08m oE Fi E midgap 0.0128 eVGermanium: m*p 0. 37m o ,*0.55m om nE Fi E midgap 0 .0077 eVGallium Arsenide: m*p 0.48m o ,m n* 0.067m oE Fi E midgap 0 .0382 eV_______________________________________ 4.11E Fi E midgap 1 kT ln N2 N c1kT ln 1.04 1019 0.4952 kT2 2.8 1019T (K) kT (eV) ( E Fi E midgap )(eV) 200 0.01727 0.0086 400 0.03453 0.0171 600 0.0518 0.0257_______________________________________4.12(a) E Fi E midgapm*p3 kT ln4 m n*3 0.0259 ln0.704 1.2110.63 meV(b) E Fi E midgap 3 0.0259 ln0.754 0.08043.47 meV_______________________________________4.13Let g c E K constantThenn o g c E f F E dEE cK1dEE E FEc 1 expkTK expE E FdEkTE cLetE E cso that dE kT dkTWe can writeE EF E c E F E E cso thatE E Fexp E c E FexpexpkTkTThe integral can then be written asn o K kT exp E c E Fexp d kTwhich becomesn o K kTE c EF expkT_______________________________________4.14Let g c E C1E E c for E E cThenn o g c E f F E dEE cC1 E E cdEE c 1exp E EF kTC1 EE E FdE E C expE ckTLetE E cdE kT dso thatkTWe can writeE EF E E c E c E FThenE c E Fn o C1 expkTE E cE E cdE expE ckT orn oE c EF C1 expkTkT exp kT d 0We find thatexp d exp 1 1So2 E c E Fn o C1 kT expkT_______________________________________4.15r1 m oWe have rm*a oFor germanium, r 16 , m* 0.55m oThenr1 16 1 a o 29 0.530.55oror1 15.4 AThe ionization energy can be written asm*2E o 13.6 eVm o s0.552 13.6 E 0.029 eV16_______________________________________ 4.16We have r1 m orm*a oFor gallium arsenide, r 13.1 , *m0.067 m o1or1 13.1 104 A0.530.067The ionization energy ism*20.067E o 13.6 13.6m o s 13.1 2orE0.0053 eV_______________________________________4.17Nc(a) E c E F kT ln2.8 10190.0259 ln 157 100.2148 eV(b) E F E E g E c E F1.12 0.2148 0.90518eV(c) p o NE F E expkT1.04 19 0.9051810 exp0.02596.90 103cm 3(d) Holesn o(e) E F E Fi kT lnn i710 150.0259 ln1.5 10100.338 eV_______________________________________4.18N(a) E F E kT lnp o190.0259 ln 1.0410210160.162 eV(b) E c E F E g E F E1.12 0.162 0.958 eV(c) n o 2.8 19 0.95810 exp0.02592.41 103cm3p o(d) E Fi E F kT lnn i2 10 160.0259 ln 101.5 100.365eV_______________________________________4.19Nc(a) E c E F kT ln0.0259 ln 2.810192 1050.8436 eVE F E E g E c E F1.12 0.8436E F E 0.2764 eV(b) p o 1.04 1019 exp 0.276370.02592.414 1014cm3(c)p-type_______________________________________4.20(a) kT3750.032375 eV0.02593003 / 2n o 4.7 10 17 375 exp 0.28300 0.0323751.15 1014cm3E F E E g E c E F 1.42 0.281.14 eV375 3 / 2 1.14 p o 7 18 exp10300 0.0323754.99 103cm 3(b) E c E F 0.0259 ln 4.7 10171.15 10 140.2154 eVE F E E g E c E F 1.42 0.21541.2046 eVp o 7 10 18 exp 1.20460.02594.42 10 2cm 3_______________________________________ 4.21(a) kT 0.0259 3750.032375 eV 300375 3 / 2 0.28n o 2.8 19 exp10300 0.0323756.86 1015cm 3E F E E g E c E F 1.12 0.280.840 eV375 3 / 20.840p o 1.04 1019 exp300 0.0323757.84 107cm 3(b) E c E F kT ln N cn o0.0259 ln2.8 10196.862 10 150.2153 eVE F E 1.12 0.2153 0.9047 eVp o 1.04 10 19 exp 0.9046680.02597.04 103 cm 3_______________________________________4.22(a) p-typeE g(b) E F E1.124 0.28 eV4p o N exp E F EkT1.04 10 19 exp 0.280.02592.10 1014cm 3E c EF E g E F E1.12 0.28 0.84 eVn o N c exp E c E FkT2.8 1019exp0.840.02592.30 105cm 3_______________________________________4.23(a) n o n iE F E FiexpkT1.5 1010 exp 0.220.02597.3313cm310p oE Fi E Fn i expkT1.5 1010 exp 0.220.02593.07 106cm 3(b) n o n iE F E FiexpkT1.8 10 6 exp 0.220.02598.80 109cm 3p o n i expE Fi E FkT1.8 106 exp 0.220.02593.68 102cm 3_______________________________________4.24(a) E F ENkT lnp o0.0259 ln1.04 10 195 10 150.1979 eV(b) E c E F E g E F E1.12 0.19788 0.92212 eV(c) n o 2.8 1019 exp 0.922120.02599.66 103cm 3(d) Holesp o(e) E Fi E F kT lnn i510 150.0259 ln1.5 10100.3294 eV _______________________________________4.25kT 0.0259 4000.034533 eV 3003 / 2N 1.04 10 19400300 1.601 1019cm 33 / 2N c 2.8 1019400300 4.3109 1019cm 30.2642 eV _______________________________________4.26(a) p o 7 1018 exp 0.250.02594.50 1014cm 3E c EF 1.42 0.25 1.17 eVn o 4.7 10 17 exp 1.170.02591.13 10 2cm 3(b)kT 0.034533eV3 / 2N 7 10184003001.078 1019cm 33 / 217 400N c 4.7 103007.236 1017cm3expn i 2 4.3109 10 19 1.601 10191.12NE F E kT lnp o19 0.0345335.67022410n i 2.381 1012 cm 3(a) E F ENkT lnp o0.034533 ln 1.601 10195 1015 0.2787 eV(b) E c E F 1.12 0.27873 0.84127 eV(c) n o 4.3109 10 19 exp 0.841270.0345331.134 109cm3(d) Holes(e) E Fi E F kT ln p on i510150.034533 ln2.381 10120.034533 ln1.078104.50 10 140.3482 eVE c EF 1.42 0.3482 1.072 eVn o 7.236 1017 exp 1 .071770. 0345332.40 104cm 3_____________________________________4.27(a) p o 1.04 1019 exp 0.250.02596.68 1014cm 3E c EF 1.12 0.25 0.870 eVn o 2.8 10 19 exp 0.8700.0259n o7.2310 4 cm 3(b)kT0.034533 eV3 / 2N 1.04 10194003001.601 1019cm 33 / 2N c 2.8 1019 4003004.311 1019cm 3NE F E kT lnp o1.60110 190.034533ln6.6810140.3482 eVE c EF 1.12 0.34820.7718 eVn o 4.311 1019 exp 0.771750.0345338.49 109cm 3_______________________________________4.282(a) n o N c F1 / 2 FFor E F E c kT 2 ,E F E c kT 2 FkT 0.5kTThen F1/ 2 F 1.0n o 2 2.8 1019 1.03.16 1019cm 3(b) n o 2 N c F1 / 2 F24.7 1017 1.05.30 1017cm 3_______________________________________ 4.29p o 2 N F1/2 F5 1019 2 1.04 1019 F1/2 FSo F1/ 2 F 4.26We find F 3.0E E FkTE EF 3.0 0.0259 0.0777 eV_______________________________________4.30E F E c 4kT(a) F 4kT kTThen F1 / 2 F 6.02N c F1 / 2n o F2 2.8 1019 6.01.90 10 20 cm 3(b) n o 2 4.7 1017 6.03.18 1018cm 3_______________________________________ 4.31For the electron concentrationn E g c E f F EThe Boltzmann approximation applies, so4 * 3 / 22m nE E cn Eh3E E FexpkTor4 2m n* 3 / 2 E c E Fexpn E h3kTE E c E E ckT expkTkTDefinexEE ckTThenn E n x K x exp xTo find maximumn E n x , setdn x 0 K 1 x 1 / 2 exp xdx 2x 1 / 21 expxorKx 1 / 2 expx1 x2which yieldsx1 E E cE E c12kTkT2For the hole concentrationp Eg E 1f F EUsing the Boltzmann approximation4 2m p * 3 / 2p EEEh 3E F EexpkT or3 / 242m *p E F Ep Eh 3expkTE E E EkTexpkTkTDefinexE EkTThenp xK x exp xTo find maximum value ofp Ep x ,setdp xUsing the results from0 dxabove,we find the maximum at1E E kT2_______________________________________4.32 (a) Silicon:We haven oN c expE cE FkTWe can writeE c E FE c E d E d E FForE c E d 0.045 eV andE dE F3kT eVwe can writen o2.8 1019 exp 0.04530.02592.8 1019exp 4.737or10 17 cm3n o2.45 We also havep oN expE F EkTAgain, we can writeE FEE FE aE aEForE FE a3kTandE aE0.045eVThenp o1.04 1019 exp 3 0.0450.02591.04 1019 exp4.737orp o9.12 10 16 cm 3(b) GaAs: assume E c E d0.0058eVThenn o4.7 1017 exp0.0058 30.025917exp 3.2244.7 10orn o1.87 1016 cm3Assume E a E 0.0345 eVThenp o71018 exp0.0345 30.02597 1018 exp 4.332orp o9.20 1016 cm 3_______________________________________ 4.33Plot_______________________________________4.34 10 151015 cm 3(a)p o415 31.5 10 10 2n o7.5 10 4 cm33 10153(b) n oN d316cm1010 2p o1.5 107.5 10 3cm 33 1016 (c)n op on i 1.5 10 10cm33(d) n i 22.8 10 19 1.041019 375300 exp1.12 3000.0259 375n i7.334 1011 cm3p o N a4 10 15 cm 37.334 10 11 2n o1.34 10 8 cm34 10 153(e) n i 22.8 10 19 1.04 10 19 4503001.12 300exp0.0259 450133n i1.722 10 cm14142n o1.722 10 1310102221.029 1014 cm 31.722 1013 2p o2.88 1012 cm 31.029 1014_______________________________________(a) p oN aN d4 101510153 1015 cm 3n i 2 1.8 10 6 2n o1.08 10 3cm 3p o3 1015(b) n oN d 3 10 16 cm 3p o1.8 10 6 2 1.08 10 4 cm33 10163(c) n o p on i1.8 10 6cm375 3(d) n i 24.7 1017 7.0 10 18300 exp1.42 3000.0259 375n i 7.580 10 8 cm 3p o N a4 1015 cm 38 2n o7.580 10 1.44 10 2 cm 34 10 153 (e) 2 4.7 10 17 7.0 18450 n i 10 300 exp1.42 3000.0259 450n i 3.853 1010 cm3n oN d10 14 cm 33.853 1010 2p o1.48 10 7 cm 310 14_______________________________________4.3610 13 cm 3(a) Ge: n i2.42(i) n oN dN dn i 22 22 10152 210152.4 13 22210or2 1015 cm 3n oN d4.35n i 2 2.4 1013 2p o2 1015n o2.88 1011 cm 3(ii) p o N a N d 10167 10153 1015 cm 32n i22.4 10 13n op o310 151.92 1011cm3(b) GaAs: n i 1.8 10 6cm3(i) n o N d2 1015 cm62p o1.8 10 1.62 10 3cm32 10 15(ii) p oN aN d3 10 15 cm 362n o1.8 101.08 10 3cm 33 1015 (c) The result implies that there is only one 33minority carrier in a volume of 10 cm ._______________________________________4.37(a) For the donor leveln d 1N d1 1exp EdE F2kT11 1 exp 0.2020.0259orn d8.85 10 4N d (b) We havef F E1E E F1expkTNowE E FE E cE c E ForE EF kT 0.245Thenf F E10.2451 exp 1 0.0259orf F E 2.87 10 5_______________________________________4.38N aN d(a) p-type(b) Silicon:10131013p oN aN d 2.5 1 or1013 cm 3p o1.5Thenn i 21.5 10 10 210 7cm 3n o1.5p o 1.5 1013 Germanium:N aN d N a N d 2p o2n i 221.5131.5 10 1322.4 101310222or3.26 10 13 cm 3p oThen2n i 22.4 10 13n o1.76 10 13p o3.264 1013cm 3Gallium Arsenide:p oN a N d1.5 10 13 cm 3and2n i 21.8 10 6n o0.216 cm 3p o1.5 1013_______________________________________4.39 (a) N d N an-type(b) n oN d N a 2 10151.2 10158 1014 cm 3n i 21.5 101022.81 10 5cm 3p o8 14n o10(c)p o N aN a N d4 1015N a 1.2 10 152 1015N a 4.8 10 15 cm31.5 10 102n o5.625 10 4cm 3 4 1015_______________________________________4.40n i21.5 101021. 153n o2 10 5 125 10cmp on o p on-type_______________________________________4.413n i 21.04 10196.0 10 18 250300 exp0.660.0259250 3001.8936 102412n i 1.376cm310 n on i 2 n i 2n o 21n i 2p o4n o 4n o1n i2Son o 6.88 1011 cm 3 ,Then p o2.75 1012cm3N a N a 2p on i 222N a22.752 10122N a21.8936 10 24227.5735 10 242.752 10 12 N aN a2N a 21.8936 10 242so that N a 2.064 1012cm 3_______________________________________4.42Plot_______________________________________4.43Plot_______________________________________4.44Plot_______________________________________ 4.45N d N aN dN a 2n o2n i 2214141.1 1014 2 10 1.2 102 2 10141.2 1014 2n i 221.1 10144 10 1324 10132n i 24.9 10 271.6 10 27n i2so n i5.74 10 13 cm 3p on i 23.3 10 273 133n o 1.1 10 1410 cm_______________________________________4.46(a)N a N d p-typeMajority carriers are holesp o N a N d16163 101.5 101.5 1016 cm 3Minority carriers are electrons210 10 2n on i 1.5 1.5 10 4 cm 3p o 1.5 1016(b) Boron atoms must be addedp o N a N aN d5 1016N a 3 10161.5 1016So N a3.5 10 16 cm 31.5 10 102n o4.5 10 3cm 35 10 16_______________________________________4.47p on i (a)n-type(b) p on i 2 n on i 2n op o1.5 10 1021016 cm3n o4 1.125 2 10electrons are majoritycarriersp o2 10 4cm3holes are minority carriers(c) n oN d N a1.125 101615N d 7 10so N d1.825 1016 cm3_______________________________________4.48E Fi E FkT lnp on iFor GermaniumT (K)kT (eV)n i (cm 3)200 0.01727 2.16 1010400 0.03453 8.60 1410 6000.05183.82 1016N aN a 2p o n i 2and22N a10 15 cm 3T (K)p o (cm3)E Fi EF (eV)200 1.0 1015 0.1855 4001.49 1015 0.01898 6003.87 10160.000674_______________________________________4.49(a) E c E FkT lnN cN d0.0259 ln 2.8 1019N dFor 1014cm 3 , E cE F 0.3249eV15 cm 3 ,E cE F0.2652eV1016cm 3, E c E F 0.2056eV 101017 cm 3 , E c E F0.1459eV(b) E F E FikT lnN dn i0.0259 lnN d1.51010For 1014cm 3 , E FE Fi 0.2280 eV15cm 3, E F E Fi 0.2877 eV10 1016 cm 3 , E F E Fi 0.3473 eV 1017 cm 3 ,E F E Fi0.4070 eV_______________________________________ 4.50N d N d 2(a) n on i 222n o1.05N d1.05 10 15 cm 31.05 10150.5 10 1520.5 10152n i2son i 25.25 10 28Now3n i 22.8 1019 1.04 1019T300exp1.120.0259 T 30035.25 10 28 2.912 10 38 T300exp 12972.973TBy trial and error, T 536.5K(b) At T 300 K,E c EF kT ln N cn oE c EF 0.0259 ln 2.8 1019 1015T 536.5 K, 0.2652 eVAt536.5kT0.02590.046318 eV3003 / 2N c 2.8 1019 536.53006.696 1019cm 3E c E FN c kT lnn oE c E F6.696 10 19 0.046318 ln10151.050.5124 eVthen E c E F 0.2472 eV(c)Closer to the intrinsic energy level._______________________________________4.51p oE Fi EF kT lnn iAt T 200K, kT 0.017267 eVT 400 K, kT 0.034533 eVT 600 K, kT 0.0518 eV At T 200K,22.8 10191019 200n i 1.04300exp1.120.017267n i 7.638 10 4 cm 3At T 400 K,3n i 2 2.8 1019 1.04 10 19 4003001.12exp0.034533n i 2.381 1012 cm 3At T 600 K,322.8 1019 19 600n i 1.04 10300exp 1.120.0518n i 9.740 1014 cm 3At T 200 K and T 400 K,p o N a 3 1015 cm 3At T 600 K,N a N a2p o n i22 23 15 3 10 15 2 9.740 10 1410 22 23.288 1015cm3Then, T 200K, E Fi E F 0.4212eVT 400K,E Fi EF 0.2465 eVT600K,E Fi EF 0.0630 eV_______________________________________4.52(a)N a N aE Fi EF kT ln 0.0259 ln6n i 1.8 10For N a10 14 cm 3 ,E FiE F0.4619 eVN a 10 15 cm 3,E FiE F0.5215 eV163,N a 10 cmE FiE F0.5811 eVN a 10 17cm 3,E FiE F 0.6408 eV(b)E FEN7.0 1018kT ln0.0259 lnN aN aFor N a10 14 cm 3 ,E F E0.2889 eVN a 10 15 cm 3 ,E FE0.2293 eV163,N a 10 cmE F E0.1697 eVN a 10 17 cm3,E F E 0.1100 eV_______________________________________ 4.53(a) E Fi3 m *p E midgapkT ln4m n *3 0.0259 ln 104 orE Fi E midgap 0.0447 eV(b) Impurity atoms to be added soE midgap EF 0.45 eV(i) p-type, so add acceptor atoms(ii)E Fi EF 0.0447 0.45 0.4947 eVThenp oE FiE Fn i expkT10 5exp 0.49470.0259 or10 13 cm3p o N a1.97_______________________________________4.54n oN d N aN c expE c E FkTsoN d 5 10 15 2.8 10 19 exp0.2150.025951015 6.95 1015orcm 3N d 1.2 1016_______________________________________4.55(a) Silicon(i) E cE F N ckT lnN d0.0259 ln 2.8 10 190.2188 eV6 1015(ii) E cE F0.2188 0.0259 0.1929 eVN dN c expE c E FkT2.8 10 19 exp0.19290.0259N d1.631 1016 cm3N d 6 1015N d1.031 10 16 cm 3Additional donor atoms(b) GaAs(i) E c E F0.0259 ln4.7101710150.15936eV(ii) E cE F0.15936 0.0259 0.13346 eVN d4.7 1017 exp0.133460.02592.718 1015 cm 3N d 1015N d1.718 10 15 cm3Additionaldonor atoms_______________________________________ 4.56(a) E Fi E FN kT lnN a0.0259 ln 1.04 10190.1620 eV2 1016(b) E F E Fi kT ln N c N d0.0259 ln 2.8 1019 0.1876 eV2 10 16(c) For part (a);p o 2 1016 cm 3n i2 1.5 1010 2n op o 2 10161.125 104cm3For part (b):3n o 2 1016 cmn i 2 1.5 1010 2p on o 2 10 161.125 104cm3_______________________________________ 4.57n oE F E Fin i expkT1.8 10 6 exp 0.550.02593.0 1015cm 3Add additional acceptor impuritiesn o N d N a3 10 15 7 10 15 N aN a 4 10 15 cm 3_______________________________________(a) E Fi E F kT lnpon i0.02593 10 150.3161 eVln10 101.5(b) E F E Fin okT lnn i0.02593 10160.3758 eVln10 101.5(c) E F E Fi(d) E Fi E Fp okT lnn i0.0259 375 ln 4 1015300 7.334 10 110.2786 eV(e) E F E Fi kT lnnon i140.0259 450 ln 1.029 10300 1.722 10 130.06945eV_______________________________________4.59(a) E F ENkT lnp o0.0259 ln7.0 10180.2009 eV3 1015(b) E F E 0.0259 l n7.0 10 181.08 10 41.360 eV(c) E F E 0.0259 l n 7.0 10181.8 10 60.7508 eV4.58(d) E F E 0.0259 375300ln 7.0 10 18 375 300 3 / 24 10 150.2526 eV(e) E F E 0.0259 450 300ln 7.0 10 18 450 300 3/ 21.48 10 71.068 eV_______________________________________4.60n-typeE F E Fi kT ln n o n i0.02591.125 10 16ln100.3504 eV1.5 10______________________________________ 4.61N a N a 2 p o 22 2 n i5.08 1015 5 101525 10 15 2n i225.08 10 15 2.5 10 15 22.5 1015 2n i26.6564 10 30 6.25 10 30 n i2n i 2 4.064 10 29n i2 N c N expE gkTkT 0.02593500.030217 eV3003502N c 1.2 10 19 1.633 1019 cm 33003502N 1.8 1019 2.45 10 19 cm 3300Now4.064 10 29 1.633 1019 2.45 1019E gexp0.030217SoE g 0.030217 ln 1.633 10 19 2.45 10 194.064 10 29E g 0.6257 eV_______________________________________4.62(a) Replace Ga atoms Silicon acts as adonorN d0.05 7 1015 3.5 10 14 cm 3Replace As atoms Silicon acts asanacceptorN a 0.95 7 1015 6.65 10 15 cm 3(b) N a N d p-type(c) p o N a N d 6.65 1015 3.5 10146.3 1015cm 3n i 2 1. 810 6 2n o 5.14 10 4 cm 3 p o 6 .3 1015(d) E Fi E F kT ln p o n i0.0259 ln 6.3 10 150.5692 eV1.8 10 6_______________________________________。
半导体物理与器件第四版答案
半导体物理与器件第四版答案半导体物理与器件第四版答案【篇一：半导体物理第五章习题答案】>1. 一个n型半导体样品的额外空穴密度为1013cm-3，已知空穴寿命为100?s，计算空穴的复合率。
解：复合率为单位时间单位体积内因复合而消失的电子-空穴对数，因此1013u1017cm?3?s ?6100?102. 用强光照射n型样品，假定光被均匀吸收，产生额外载流子，产生率为gp，空穴寿命为?，请①写出光照开始阶段额外载流子密度随时间变化所满足的方程；②求出光照下达到稳定状态时的额外载流子密度。
解：⑴光照下，额外载流子密度?n=?p，其值在光照的开始阶段随时间的变化决定于产生和复合两种过程，因此，额外载流子密度随时间变化所满足的方程由产生率gp和复合率u的代数和构成，即 d(?p)?p gp? dt?d(?p)0，于是由上式得⑵稳定时额外载流子密度不再随时间变化，即dtp?p?p0?gp?3. 有一块n型硅样品，额外载流子寿命是1?s，无光照时的电阻率是10??cm。
今用光照射该样品，光被半导体均匀吸收，电子-空穴对的产生率是1022/cm3?s，试计算光照下样品的电阻率，并求电导中少数载流子的贡献占多大比例？解：光照被均匀吸收后产生的稳定额外载流子密度p??n?gp??1022?10?6?1016 cm－3取?n?1350cm2/(v?s)，?p?500cm/(v?s)，则额外载流子对电导率的贡献2pq(?n??p)?1016?1.6?10?19?(1350?500)?2.96 s/cm无光照时?0?10.1s/cm，因而光照下的电导率02.96?0.1?3.06s/cm相应的电阻率 ??110.33??cm 3.06少数载流子对电导的贡献为：?p?pq?p??pq?p?gp?q?p代入数据：?p?(p0??p)q?p??pq?p?1016?1.6?10?19?500?0.8s/cm∴p?00.80.26?26﹪ 3.06即光电导中少数载流子的贡献为26﹪4.一块半导体样品的额外载流子寿命? =10?s，今用光照在其中产生非平衡载流子，问光照突然停止后的20?s时刻其额外载流子密度衰减到原来的百分之几？解：已知光照停止后额外载流子密度的衰减规律为p(t)??p0e?因此光照停止后任意时刻额外载流子密度与光照停止时的初始密度之比即为t??p(t)e? ?p0t当t?20?s?2?10?5s时20??p(20)e10?e?2?0.135?13.5﹪ ?p05. 光照在掺杂浓度为1016cm-3的n型硅中产生的额外载流子密度为?n=?p= 1016cm-3。
半导体物理与器件 第四章
第四章平衡半导体4.0本章概要在上一章中，我们讨论了一般晶体，运用量子力学的概念对其进行了研究，确定了单晶晶格中电子的一些重要特性。
在这一章中，我们将运用这些概念来专门研究半导体材料。
我们将利用导带与价带中的量子态密度函数以及费米-狄拉克分布函数确定导带与价带中电子与空穴的浓度。
另外，我们将在半导体材料中引入费米能级的概念。
注意，本章中所涉及的半导体均处于平衡状态。
所谓平衡状态或者热平衡状态，是指没有外界影响（如电压、电场、磁场或者温度梯度等）作用于半导体上的状态。
在这种状态下，材料的所有特性均与时间无关。
本章目标：（1）推导半导体中热平衡电子浓度和空穴浓度关于费米能级的表达式。
（2）讨论通过在半导体中添加特定杂质原子来改变半导体材料性质的过程。
（3）推导半导体材料中热平衡电子浓度和空穴浓度关于添加到半导体中的掺杂原子浓度的表达式。
（4）求出费米能级的位置，其为添加到半导体中的掺杂原子浓度的函数。
简单说来，本章讨论的重点是：在不掺杂和掺杂的情况下，分别求平衡半导体中电子和空穴的浓度值，以及费米能级位置。
4.1半导体中的载流子我们知道：电流从本质上来说是电荷移动的速率。
在半导体中有两种载流子——电子和空穴——有能力产生电流。
载流子的定义：在物理学中，载流子指可以自由移动的带有电荷的物质微粒，如电子和离子。
如半导体中的自由电子与空穴，导体中的自由电子，电解液中的正、负离子，放电气体中的离子等。
既然半导体中的电流很大程度上取决于导带中电子与价带中空穴的数量，那么我们关心的半导体的一个重要参数就是这些载流子的密度。
联想我们之前学习的知识，我们不难知道电子和空穴的密度与态密度函数、费米-狄拉克分布函数都有关。
在接下来的章节中，我们会从更严谨的数学推导出发，导出电子与空穴的热平衡浓度，定性地讨论这些关系。
4.1.1电子与空穴的热平衡分布导带中电子关于能量的分布，我们可以从允带量子态密度函数乘以量子态被电子占据的概率函数（分布函数）得出。
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4.5.2平衡电子和空穴浓度
4.5.2平衡电子和空穴浓度
• 正负电荷密度相等即电中性条件：
n0+N-a=p0+Nd+ n0+(Na-pa)=p0+(Nd-nd) 假设完全电离： nd 和pa均为零， 则 n0+Na=p0+Nd
4.和温度的变化
4.1.3本征费米能级位置
4.2掺杂原子与能级
掺杂半导体称为非本征半导体 施主donor杂质
施主donor杂质原子：杂质原子向导带提供了电子 ；
N型半导体：由于施主杂质原子增加导带电子并不产生价带空穴；
由于N型半导体中电子浓度大于空穴浓度，电子称为多数载流子， 空穴称为少数载流子。
受主acceptor杂质
4.1半导体中的载流子
4.1.1电子和空穴的平衡分布
导带电子分布
价带空穴分布
4.1.2 n0方程和P0方程
导带电子
定义导带有效状态密度
价带空穴
定义价带有效状态密度
4.1.3 本征载流子浓度
EFi本征费米能级 ni本征载流子浓度 ni2=n0p0
4.1.3本征费米能级位置
本征半导体 n0=p0
r1约为硅晶格常数四倍
4.2.2 电离能
4.3非本征Extrinsic半导体
4.3.1 电子和空穴的平衡状态分布
4.3.1 电子和空穴的平衡状态分布
4.3.2 n0和p0的乘积
4.3.4 简并与非简并半导体
非简并Nondegenerate半导体：杂质原子在n型半导体中引入分立 的、无相互作用的施主能级，而在p型半导体中引入分立的、无相 互作用的受主能级；
Pd空穴占据受主能级的密度， Ed受主能级的能量，g简并因子
4.4.2 完全电离和束缚态（Complete Ionization and Freeze-out）
导带电子：
4.4.2 完全电离和束缚态（Complete Ionization and Freeze-out）
完全电离
4.4.2 完全电离和束缚态（Complete Ionization and Freeze-out）
第四章 平衡半导体
平衡半导体（The Semiconductor in Equilibrium) 是指半导体处于平衡状态或热平衡状态，是指 没有外界影响（如电压、电场、磁场或温度梯 度等）作用于半导体上的状态。 本征半导体(Intrinsic Semiconductor)是指没有杂 质原子和晶体结构缺陷的纯净半导体。 非本征半导体(Extrinsic Semiconductor)是指进行 了定量施主(Donor)或受主(Acceptor)掺杂，从而 使电子浓度或空穴浓度偏离本征载流子浓度产 生多数载流子(Majority Carrier)电子（n-type） 或多数载流子空穴（p-type）的半导体。
n型简并半导体：导带中的电子浓度超过了状态密度Nc时，费米能 级位于导带内部；并不产生价带空穴；
p型简并半导体：价带中的空穴浓度超过了状态密度Nv时，费米能 级位于价带内部；
4.4 施主和受主的统计分布
4.4.1概率分布函数
电子占据施主能级的概率函数：
nd电子占据施主能级的密度， Ed施主能级的能量
4.6.2 费米能级的应用
作业
• • • • P106 P107 P108 P109 4.4 4.21 4.33 4.51, 4.56
4.5 电中性状态(Charge Neutrality)
4.5.1补偿半导体(Compensated Semiconductors)
补偿半导体：是指在同一区域内同时含有施主和受主 杂质原子的半导体。 Nd>Na，n型补偿半导体； Na > Nd ，p型补偿半导体； Na = Nd ，完全补偿半导体。
受主acceptor杂质原子：杂质原子从价带获得电子 ；
P型半导体：由于受主杂质原子在价带产生空穴，但不在导带产生 电子；
由于P型半导体中空穴浓度大于电子浓度，空穴称为多数载流子， 电子称为少数载流子。
4.2.2 电离能
静电力与离心力平衡 角动量量子化
波尔半径 M*电导有效质量 见刘恩科第四版P95
在室温下，施主能级基本上处于完全电离状态 ；受主原子 也基本上处于完全电离状态。
4.4.2 完全电离和束缚态（Complete Ionization and Freeze-out）
在T=0K下，没有电子从施主能级热激发到导带，这种现象称为束缚 态 ；没有电子从价带跃迁到受主能级，这种现象也称为束缚态。