半导体物理答案第三章
半导体物理与器件(尼曼第四版)答案
半导体物理与器件(尼曼第四版)答案第一章：半导体材料与晶体1.1 半导体材料的基本特性半导体材料是一种介于导体和绝缘体之间的材料。
它的基本特性包括：1.带隙：半导体材料的价带与导带之间存在一个禁带或带隙，是电子在能量上所能占据的禁止区域。
2.拉伸系统：半导体材料的结构是由原子或分子构成的晶格结构，其中的原子或分子以确定的方式排列。
3.载流子：在半导体中，存在两种载流子，即自由电子和空穴。
自由电子是在导带上的，在外加电场存在的情况下能够自由移动的电子。
空穴是在价带上的，当一个价带上的电子从该位置离开时，会留下一个类似电子的空位，空穴可以看作电子离开后的痕迹。
4.掺杂：为了改变半导体材料的导电性能，通常会对其进行掺杂。
掺杂是将少量元素添加到半导体材料中，以改变载流子浓度和导电性质。
1.2 半导体材料的结构与晶体缺陷半导体材料的结构包括晶体结构和非晶态结构。
晶体结构是指材料具有有序的周期性排列的结构，而非晶态结构是指无序排列的结构。
晶体结构的特点包括：1.晶体结构的基本单位是晶胞，晶胞在三维空间中重复排列。
2.晶格常数是晶胞边长的倍数，用于描述晶格的大小。
3.晶体结构可分为离子晶体、共价晶体和金属晶体等不同类型。
晶体结构中可能存在各种晶体缺陷，包括：1.点缺陷：晶体中原子位置的缺陷，主要包括实际缺陷和自间隙缺陷两种类型。
2.线缺陷：晶体中存在的晶面上或晶内的线状缺陷，主要包括位错和脆性断裂两种类型。
3.面缺陷：晶体中存在的晶面上的缺陷，主要包括晶面位错和穿孔两种类型。
1.3 半导体制备与加工半导体制备与加工是指将半导体材料制备成具有特定电性能的器件的过程。
它包括晶体生长、掺杂、薄膜制备和微电子加工等步骤。
晶体生长是将半导体材料从溶液或气相中生长出来的过程。
常用的晶体生长方法包括液相外延法、分子束外延法和气相外延法等。
掺杂是为了改变半导体材料的导电性能，通常会对其进行掺杂。
常用的掺杂方法包括扩散法、离子注入和分子束外延法等。
半导体物理习题答案(1-3章)
第1章 半导体中的电子状态1. 设晶格常数为a 的一维晶格，导带极小值附近能量()c E k 和价带极大值附近能量()v E k 分别为2222100()()3c h k k h k E k m m -=+，22221003()6v h k h k E k m m =-0m 为电子惯性质量，112k a =， 0.314a =nm 。
试求：1) 禁带宽度；2) 导带底电子有效质量； 3) 价带顶电子有效质量；4) 价带顶电子跃迁到导带底时准动量的变化。
解：1) 禁带宽度g E ，根据22100()2()202c dE k h k k h k dk m m -=+=，可求出对应导带能量极小值min E 的k 值：m i n 134k k =， 由题目中()c E k 式可得：min 12min 3104()4c k k k h E E k k m ====； 根据20()60v dE k h k dk m =-=，可以看出，对应价带能量极大值max E 的k 值为：k max = 0；可得max 221max 00()6v k k h k E E k m ====，所以2221min max 2001248g h k h E E E m m a=-== 2) 导带底电子有效质量m n由于2222200022833c d E h h h dk m m m =+=，所以202238nc m h md E dk== 3) 价带顶电子有效质量vn m由于22206v d E h dk m =-，所以20226v nv m h m d E dk ==- 4) 准动量的改变量min max 133()48hh k h k k hk a∆=-==2. 晶格常数为0.25 nm 的一维晶格，当外加102V/m 、107V/m 的电场时，试分别计算电子自能带底运动到能带顶所需的时间。
解：设电场强度为E ，电子受到的力f 为dkf hqE dt==（E 取绝对值），可得h dt dk qE =， 所以12012ta h h t dt dk qE qE a===⎰⎰，代入数据得： 34619106.62108.310()1.6102(2.510)t s E E----⨯⨯==⨯⨯⨯⨯⨯ 当E = 102V/m 时，88.310t s -=⨯；当E = 107V/m 时，138.310t s -=⨯。
半导体物理与器件 尼曼 第四版第三章课后答案
Chapter 33.1If o a were to increase, the bandgap energy would decrease and the material would begin to behave less like a semiconductor and more like a metal. If o a were to decrease, the bandgap energy would increase and thematerial would begin to behave more like an insulator._______________________________________ 3.2Schrodinger's wave equation is:()()()t x x V xt x m ,,2222ψ⋅+∂ψ∂- ()tt x j ∂ψ∂=, Assume the solution is of the form:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Region I: ()0=x V . Substituting theassumed solution into the wave equation, we obtain:()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧∂∂-t E kx j x jku x m exp 22 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⋅⎪⎭⎫ ⎝⎛-=t E kx j x u jE j exp which becomes()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+=t E kx j x Eu exp This equation may be written as()()()()0222222=+∂∂+∂∂+-x u mE x x u x x u jk x u kSetting ()()x u x u 1= for region I, the equation becomes:()()()()021221212=--+x u k dx x du jk dxx u d α where222mE=α Q.E.D.In Region II, ()O V x V =. Assume the same form of the solution:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Substituting into Schrodinger's wave equation, we find:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+t E kx j x u V O exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=t E kx j x Eu exp This equation can be written as:()()()2222x x u x x u jk x u k ∂∂+∂∂+- ()()02222=+-x u mEx u mV OSetting ()()x u x u 2= for region II, this equation becomes()()dx x du jk dxx u d 22222+ ()022222=⎪⎪⎭⎫ ⎝⎛+--x u mV k O α where again222mE=α Q.E.D._______________________________________3.3We have ()()()()021221212=--+x u k dx x du jk dx x 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 +-+-ααexp and the second derivative becomes()()[]()[]xk j A k j dx x u d --=ααexp 2212 ()[]()[]x k j B k j +-++ααexp 2Substituting these equations into thedifferential equation, we find ()()[]x k j A k ---ααexp 2()()[]x k j B k +-+-ααexp 2(){()[]x k j A k j jk --+ααexp 2()()[]}x k j B k j +-+-ααexp ()()[]{x k j A k ---ααexp 22 ()[]}0exp =+-+x k j B α Combining terms, we obtain()()()[]222222αααα----+--k k k k k ()[]x k j A -⨯αexp()()()[]222222αααα--++++-+k k k k k ()[]0exp =+-⨯x k j B α We find that00= Q.E.D. For the differential equation in ()x u 2 and the proposed solution, the procedure is exactly the same as above._______________________________________ 3.4We have the solutions ()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp for a x <<0 and()()[]x k j C x u -=βexp 2()[]x k j D +-+βexp for 0<<-x b .The first boundary condition is ()()0021u u =which yields 0=--+D C B AThe second boundary condition is 0201===x x dx du dx duwhich 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 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=a Second point: By trial and error,πα515.1=a (ii) First point: πα2=aSecond point: By trial and error, πα375.2=a_______________________________________ 3.7 ka a a a P cos cos sin =+'ααα Let y ka =, x a =αThen y x x x P cos cos sin =+' Consider dy d of this function.()[]{}y x x x P dy d sin cos sin 1-=+⋅'- We find()()()⎭⎬⎫⎩⎨⎧⋅+⋅-'--dy dx x x dy dx x x P cos sin 112y dydxx sin sin -=- Theny x x x x x P dy dx sin sin cos sin 12-=⎭⎬⎫⎩⎨⎧-⎥⎦⎤⎢⎣⎡+-'For πn ka y ==, ...,2,1,0=n 0sin =⇒y So that, in general,()()dk d ka d a d dy dxαα===0 And 22 mE=α Sodk dEm mE dk d ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-22/122221 α This implies thatdk dE dk d ==0α for an k π= _______________________________________ 3.8 (a) πα=a 1 π=⋅a E m o 212 ()()()()2103123422221102.41011.9210054.12---⨯⨯⨯==ππa m E o 19104114.3-⨯=J From Problem 3.5πα729.12=a π729.1222=⋅a E m o ()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE 18100198.1-⨯=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 _______________________________________3.16 For 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 2122Then 221222*11 αE dk E d m o k k =⋅== or 212*αE m = _______________________________________ 3.21(a) ()[]3/123/24l t dn m m m =* ()()[]3/123/264.1082.04o o m m = o dn m m 56.0=*(b) oo l t cn m 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 o m m += []om ⋅+=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 have222222z ZXY y Y XZ x X YZ ∂∂+∂∂+∂∂ 022=⋅+XYZ mEDividing by XYZ , we obtain 021*********=+∂∂⋅+∂∂⋅+∂∂⋅ mEz Z Z y Y Y x X XLet 01222222=+∂∂⇒-=∂∂⋅X k x X k x X X x x 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 nSimilarly, we have 2221y k y Y Y -=∂∂⋅ and 2221z k z Z 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 E mdE 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 that π2h = this density of states function can be simplified and written as ()()dE E m hadE E g T ⋅⋅=2/33324π Dividing by 3a will yield the density of states so that()()E hm E g ⋅=32/324π _______________________________________ 3.25 For a one-dimensional infinite potential well, 222222k a n E m n ==*πDistance between quantum states()()aa n a n k k n n πππ=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=-+11 Now()⎪⎭⎫ ⎝⎛⋅=a dk dk 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 ()()()()()E E g 31341011.9067.0210054.11--⨯⋅⨯=π ()E E g 1810055.1⨯= m 3-J 1- _______________________________________3.26(a) Silicon, o n m m 08.1=*()()c n c E E h m E g -=*32/324π ()dE E E h m g kT E E c n c c c⋅-=⎰+*232/324π()()kT E E c n c c E 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, ()⎪⎭⎫⎝⎛=3004000259.0kT 034533.0=eV ()()19106.1034533.0-⨯=21105253.5-⨯=J Then ()()[]2/32155105253.5210953.7-⨯⨯=c g 2510239.9⨯=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 +-=-+υ Or midgap c F E E E E =+=2υ_______________________________________ 3.3622222ma n E n π= For 6=n , Filled state()()()()()2103122234610121011.92610054.1---⨯⨯⨯=πE 18105044.1-⨯=Jor 40.9106.1105044.119186=⨯⨯=--E eV For 7=n , Empty state()()()()()2103122234710121011.92710054.1---⨯⨯⨯=πE 1810048.2-⨯=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---⨯⨯++⨯=π 1910761.3-⨯=J or 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, so35.2=F E eV(b) For 13 electrons, the 13th electron occupies the quantum state3,2,3===z y x n n n ; so ()()()()()2103122222341310121011.9232310054.1---⨯⨯++⨯=πE1910194.9-⨯=Jor 746.5106.110194.9191913=⨯⨯=--E eV The 14th electron would occupy the quantum state 3,3,2===z y x n n n . This state is atthe same energy, so746.5=F E eV _______________________________________ 3.38The probability of a state at E E E F ∆+=1being occupied is ()⎪⎭⎫ ⎝⎛∆+=⎪⎪⎭⎫ ⎝⎛-+=kT E kT E E E f F exp 11exp 11111 The probability of a state at E E E F∆-=2being empty is()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F 222exp 1111⎪⎭⎫ ⎝⎛∆-+⎪⎭⎫ ⎝⎛∆-=⎪⎭⎫ ⎝⎛∆-+-=kT E kT E kT E exp 1exp exp 111or()⎪⎭⎫ ⎝⎛∆+=-kT E E f exp 11122so ()()22111E f E f -= Q.E.D. _______________________________________3.39 (a) At energy 1E , we want 01.0exp 11exp 11exp 1111=⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-+-⎪⎪⎭⎫ ⎝⎛-kT E E kT E E kT E E F F FThis expression can be written as 01.01exp exp 111=-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫⎝⎛-+kT E E kT E E F F or()⎪⎪⎭⎫ ⎝⎛-=kT E E F 1exp 01.01 Then()100ln 1kT E E F += or kT 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 eV 31098.6060433.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f(c) ()⎥⎦⎤⎢⎣⎡--≅-kT E E f F F exp 1 ⎥⎦⎤⎢⎣⎡-=kT 25.0exp 02.0 or 5002.0125.0exp ==⎥⎦⎤⎢⎣⎡+kT ()50ln 25.0=kT or ()()⎪⎭⎫⎝⎛===3000259.0063906.050ln 25.0T kTwhich yields 740=T K _______________________________________3.41 (a) ()00304.00259.00.715.7exp 11=⎪⎭⎫⎝⎛-+=E f or 0.304% (b) At 1000=T K, 08633.0=kT eVThen ()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 kT E E E f F F 11exp exp 11Then ()611032.90259.030.0exp -⨯=⎪⎭⎫ ⎝⎛-=E f For 2E E =, 82.030.012.12=-=-E E F eVThen ()⎪⎭⎫ ⎝⎛-+-=-0259.082.0exp 1111E for()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛---≅-0259.082.0exp 111E f 141078.10259.082.0exp -⨯=⎪⎭⎫ ⎝⎛-=(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 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=dE df At F E E =,()()79.511104317.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1- _______________________________________3.45(a) At midgap E E =,()⎪⎪⎭⎫⎝⎛+=⎪⎪⎭⎫ ⎝⎛-+=kT E kT E E E f g F 2exp 11exp 11 Si: 12.1=g E eV,()()⎥⎦⎤⎢⎣⎡+=0259.0212.1exp 11E for ()101007.4-⨯=E fGe: 66.0=g E eV()()⎥⎦⎤⎢⎣⎡+=0259.0266.0exp 11E f or ()61093.2-⨯=E fGaAs: 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 ()⎪⎭⎫⎝⎛=3000259.0032572.0T so 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 _______________________________________。
半导体物理导论课后习题答案第1-3章
有效原子数为
3
13（顶角原子）
3
12（面心原子）
5 2
个原子
等边三角形的面积为
S
1 2
a
2
2
a
10 2
a2 4
5
第1章
所以，(111)面的原子面密度为
第1章
3.已知Si的晶格常数或单胞的边长a=5.43089 Å, 求： （1）Si的原子体密度。 （2）(111)面、(110)面以及(100)面的原子面密度，比较哪个晶面的面 密度最大？哪个晶面的面密度最小？
解:(1)每个晶胞中有8个原子，晶胞体积为a3， 每个原子所占的空间体积为a3/8，因此每立方
图3
第2章
2. 图中的A、B 分别为两种半导体材料价带中载流子的E-k关系抛物 线，其中哪个材料的电子有效质量小？为什么？试确定两种电子的 有效质量之比(以自由电子质量为单位) 。
解：由图可知，B材料的能带极值附近的曲率变化较大，故其电子 有效质量小。所以，A、B材料的电子有效质量之比为0.26，B材料 的电子有效质量小。
Eg h
1.12 4.135 1015
2.711014 (s1)
hc Eg
4.1351015 31017 1.12
1107.1(nm)
第3章
1.掺杂半导体与本证半导体之间有何差异？试举例说明掺杂对半导体的 导电性能的影响。
2.试简述杂质在半导体中的几种作用，并分别在能带图上标志出其在半 导体中的跃迁过程 3.试说明浅能级杂质和深能级杂质的物理意义及特点。 4.何谓非平衡载流子？非平衡状态与平衡状态的差异何在
半导体物理 刘恩科 第三章答案
第三章习题讲解7.Ec − E F 解： Q n0 = N c exp(− ) k0T∴ ∴ Q ∴ EF Nc = E c − k 0 T ln = E c − 0 . 017 eV n0E F − E c = − 0 . 017 eV E c − E D = 0 . 01 eV E F − E D = − 0 . 007 eVND n0 = 1 + 2 exp[( E F − ED ) / k0T ] ⇒ N D = 1.7 ×1017 cm −3ND 9. E F = Ec + k 0T ln Nc N D1 E F 1 = Ec + k 0T ln = Ec − 0.206eV Nc EF 2 EF 3 N D2 = Ec + k 0T ln = Ec − 0.087eV Nc N D3 = Ec + k 0T ln = Ec − 0.027eV NcEc − E D = 0.05eVEF1远在ED之下，故此时全电离方法1 方法2(Ec-EF2) /k0T=1.4,故此时不能全电离 EF3在ED之上，故此时全电离+ nD 1 = N D 1 + 2 exp[−( ED − EF ) / k 0T ]1 = 是否大于90％ 1 + 2 exp[−( Ec − ΔED − EF ) / k 0T ]算出电离度分别为1，67％，55％。
所以第二、三种 不能认为全电离。
10. 解：Ge 在300K时的本征载流子浓度ni = 2.4 × 10 cm13−3要以杂质电离为主，其杂质浓度最低应 高于ni一个数量级，即：N D ,min = 1014 cm −3最高浓度为：⎛ ΔE D ⎞ ⎛D N ⎞ N D = ⎜ − C ⎟ exp⎜ − ⎜ k T ⎟= ⎟ ⎝ 2 ⎠ 0 ⎝ ⎠ ⎛ 0.1 × 1.05 × 1019 ⎞ 0.0127 ⎞ 17 −3 ⎜ ⎟ exp⎛ − ⎟ = 3.22 × 10 cm ⎜ ⎜ ⎟ 2 ⎝ 0.026 ⎠ ⎝ ⎠11. 根据未电离杂质占总掺杂比例的定义：ΔE D 2N D ⇒ D− = exp Nc k 0T ⎡⎛ D ΔE D 1 3 = ln T + ln ⎢⎜ − ⎜ k0 T 2 ⎢⎝ N D ⎣∗ 3/ 2 k 0 mn 3⎞ 2π ⎟ ⎟ h ⎠(∗ 3/ 2 k 0 mn 3)(2π 其中)⎤ ⎥ ⎥ ⎦h＝1015ΔE D ＝116 k0将上两个常数代入，得 ： ⎛ D− 116 3 = ln T + ln ⎜ ⎜N T 2 ⎝ D ⎞ ⎟＋15 ln 10 ⎟ ⎠(1) 将N D＝1014 cm −3D−＝1 ％代入116 3 ⎛ 0.01 ⎞ = ln T + ln⎜ 14 ⎟＋15 ln 10 T 2 ⎝ 10 ⎠ 3 ＝ ln T－2.3 2 ⇒ T = 37.1K 将N D＝1017 cm −3 D−＝1 ％代入 116 3 ⎛ 0.01 ⎞ = ln T + ln⎜ 17 ⎟＋15 ln 10 T 2 ⎝ 10 ⎠ 3 ln T－9.2 2 ⇒ 533K(1) 将N D＝1014 cm −3D−＝10％代入116 3 3 ⎛ 0.1 ⎞ = ln T + ln⎜ 14 ⎟＋15 ln 10＝ ln T T 2 2 ⎝ 10 ⎠ ⇒ T = 24.3 将N D＝1017 cm −3 D−＝10％代入 116 3 ⎛ 0.1 ⎞ = ln T + ln⎜ 17 ⎟＋15 ln 10 T 2 ⎝ 10 ⎠ 3 ln T－6.9 2 ⇒ T = 160.5 K50%电离时，不能再用上方法，须用：ND ⎛ ED − EF 1 1 + exp⎜ ⎜ kT 2 0 ⎝⎞ ⎟ ⎟ ⎠ND = ⎛ EF − ED ⎞ 1 + 2 exp⎜ ⎜ kT ⎟ ⎟ 0 ⎝ ⎠⎛ EF − ED ⎞ ⎛ ED − EF ⎞ ⎟ ⎟ = 4 exp⎜ ⇒ exp⎜ ⎜ kT ⎟ ⎜ kT ⎟ 0 0 ⎠ ⎝ ⎠ ⎝ ED − EF EF − ED ⇒ = ln 4 + k 0T k 0T ⇒ E F = E D − k 0T ln 2 (1)⎛ E F − Ec ⎞ ND n0 = = N c exp⎜ ⎟ ⎜ kT ⎟ 2 0 ⎠ ⎝ ⎛ ND ⎞ ⇒ E F = Ec + k 0T ln⎜ ⎟ ⎜ 2N ⎟ c ⎠ ⎝ ( 2)联立（1）、（2）两式可得：⎛ ND ⎞ ⎟ E D − k 0T ln 2 = Ec + k 0T ln⎜ ⎜ 2N ⎟ c ⎠ ⎝ ⎛ N c ⎞ 116 ⎛ Nc ⎞ ⎟⇒ ⎟ ⇒ ΔE D = k 0T ln⎜ = ln⎜ ⎜N ⎟ ⎜N ⎟ T ⎝ D⎠ ⎝ D⎠ 116 = ln 2 × 1015 × T 3 / 2 − ln N D T[()]分别将 N D＝1014 cm −3 ，N D＝1017 cm −3代入116 15 3/ 2 = ln 2 × 10 × T − ln N D T[()]可得其分别对应得温度为16K和55K。
半导体物理学第三章习题和答案
时 Eg=0.76eV。求这两个温度时锗的本征载流子浓度。②77K 时，锗的电子浓度为 1017cm-3 ，假定受主浓度为零，而 Ec-ED=0.01eV，求锗中施主浓度 ED 为多少？
3 k 0Tmn ) 2 2 2
7 （ .1 ）根据N c 2( N v 2( k 0Tm p 2
' ' N（ C 77 K） 3 T N（ T C 300 K） ' NC NC （
77 3 77 3 ） 1.05 1019 （ ） 1.37 1018 / cm 3 300 300
' NV NV （
77 3 77 3 ） 3.9 1018 （ ） 5.08 1017 / cm 3 300 300
5. 利用表 3-2 中的 m*n，m*p 数值，计算硅、锗、砷化镓在室温下的 NC , NV 以及本征载
流子的浓度。
3 2koTmn 2 N 2 ( ) C 2 h 2koTm p 32 5 N v 2( ) h2 Eg 1 2 koT 2 n i ( N c N v ) e Ge : mn 0.56m0 ; m p o.37 m0 ; E g 0.67ev si : mn 1.08m0 ; m p o.59m0 ; E g 1.12ev Ga As : mn 0.068m0 ; m p o.47 m0 ; E g 1.428ev
0.037
nD ND
30%不成立
80%10%不成立 0.023 1 0.026 1 e 2 ' （2） 求出硅中施主在室温下全部电离的上限 2N E D ( D )e D (未电离施主占总电离杂质数的百分比） NC koT 10% 0.1N C 0.026 2 N D 0.05 e , ND e 2.5 1017 / cm 3 N C 0.026 2
半导体物理第三章习题答案
第3章 半导体中载流子的统计分布2. 试证明具有类似于Ge 、Si 能带结构的半导体的导带底附近状态密度公式为式(3-6)。
证明：设导带底能量为C E ，具有类似结构的半导体在导带底附近的电子等能面为旋转椭球面，即2222122()()2C t lk k k h E k E m m +=++与椭球标准方程2221122221k k k a b c ++= 相比较，可知其电子等能面的三个半轴a 、b 、c 分别为1222()[]t C m E E a b h -==1222()[]l C m E E c h-=于是，K 空间能量为E 的等能面所包围的体积即可表示为13222344(8)()33l t C V abc m m E E hππ==-能量为E 和E +d E 的两个等能面之间的体积即为1122232(8)()l t C dV m m E E dE hπ=-设晶体体积为V ，则其K 空间的量子态密度在考虑自旋的情况下为2V ，能量为E 和E +d E 的两个等能面之间的量子态数即为11222322(8)()l t C dZ V m m E E dE hπ=-设导带底的等效状态数为S ，则状态密度112223(8)()4()l tC C S m m dZg E V E E dE hπ==⋅- 令13222(8)(2)l tdn S m m m =，则21233()dn l tm S m m =，代入上式即使式(3-6)得证。
3. 当E －E F =1.5kT 、4kT 、10kT 时，分别用费米分布函数和玻耳兹曼分布函数计算电子占据这些能级的概率，并分析计算结果说明了什么问题。
解：已知费米分布函数1()1FE E kTf E e-=+；玻耳兹曼分布函数F E E kTB f e--=当E －E F =1.5kT 时： 1.51()0.18241f E e ==+， 1.50.223B f e -==； 当E －E F =4kT 时：41()0.017991f E e==+，40.0183B f e -==； 当E －E F =10kT 时：5101() 4.54101f E e-==⨯+，1054.510B f e --==⨯； 计算结果表明，两种统计方法在E －E F ＜2kT 时误差较大，反之误差较小；E －E F 高于kT 的倍数越大，两种统计方法的误差越小。
半导体物理分章答案第三章
（1）杂质离化区
特征：本征激发可以忽略，导带电子主要由电离杂质 提供。
电中性条件n0=p0+nD+可近似为： n0 =nD+
（9）
ECEF
NCe k0T
ND
EDEF
12e k0T
（10）
EC
EV
• 低温弱电离区
特征：nD+<< ND
弱电离
EDEF
e k0T 1
所以电中性条件简化为：
EC EF
n0 N C e k0T
EF EV
p0 N V e k0T
导带底有效状态密度 价带顶有效状态密度
（3） （4）
可以见到：NC T3/2 和 NV T3/2
且，
E CE V
E g
n0p0N CN Ve k0T N CN Vek0T
（5）
§3.3 本征半导体的载流子浓度
Carriers Density of Intrinsic Semiconductors
（3）p型半导体的计算方法
§3.5 一般情况下的载流子统计分布
重点讨论半导体中同时含有施主和受主杂质时，载流 子浓度及费米能级EF。 热平衡状态下的电中性条件：
p0 + nD+ = n0 + pA利用该方程确定EF和n0、 p0 。
对于掺入不同类型的浅能级杂质，可先确定有效掺杂浓度， 近似按照只掺一种类型杂质的方法确定EF和n0、 p0 。
1
ND2 4ni2
ND 2
ni
ni
p0
n0
ND
ND 2
ni
ni
显然这时过渡区接近于本征激发区。
半导体器件物理学参考答案(第三章)
Chapter 3Terminology and knowledge3.1Integrated semiconductor companiesCompanies that design and fabricate integrated circuitsFablessWith no fabrication/processing facilityFoundriesCompanies that specialize in processing wafers to produce silicon devicesWafer fabFabrication facility where wafers are processed to produce silicon devicesIntegrated circuitsSystem of transistors manufactured on silicon wafersCPUCentral processor unit that is an active part of computer containing the datapath and controlDRAMDynamic random access memoryFlat-panel displaysDisplay panels with flat screensMEMSMicro-electro-mechanical-systemDNA chipsSilicon chips used for DNA screeningDry oxidationGrowth of SiO2 using oxygen gasWet oxidationGrowth of SiO2 using water vaporHorizontal furnaceHorizontally oriented oxidation furnaceVertical furnaceVertically oriented oxidation furnacePhotolithography/Optical lithographyProcess in which the resist is optically patterned and selectively removed from designated areas on a waferWafer stepperEquipment used in lithography processPhotoresistUltraviolet-light sensitive materialPhotomaskQuartz photo-plate containing a copy of pattern to be transferred to Si or SiO2 surfaceNegative resistPhotoresist that becomes polymerized and resistant to a developer when exposed to an UV lightPositive resistPhotoresist whose stabilizer breaks down when exposed to an UV light, leading to the preferential removal of exposed regions in a developerStripRemoval of photoresistAsherSystem that removes the resist on a wafer by oxidizing it in oxygen plasma or UV ozone systemLithography fieldSmall area exposed to an UV light during the exposure through a photomask and an optical reduction systemStepperAnother name for lithography equipmentStep-and-repeat actionThe process of exposing different parts of a wafer until the whole wafer has been exposedPhase-shift photomaskPhotomask that produces 180 degree phase difference in neighboring clear features so that their diffraction fringes cancel each otherOptical Proximity Correction (OPC)Printing a slightly different shape on the photomask to correct distortions resulting from an exposure processOverlayAlignment between 2 separate lithography stepsExtreme UV lithography (EUVL)Lithography that uses 13nm wavelength and is expected to result in much higher resolutionSoft-x-ray lithographyOld name of EUVLElectron-beam lithography (EBL)Lithography using a focused stream of electronsElectron projection lithography (EPL)EBL that exposes a complex pattern simultaneously using a mask and a reduction electron lens systemWet etchingRemoval of SiO2 using hydrofluoric acidIsotropicWithout preference in directionDry etching (also known as plasma etching or reactive-ion etching (RIE)) Removal of SiO2 using plasma and reactive ionsAnisotropicWith preference in directionSelectivityThe extent in which an etching process distinguishes between different materials End-point detectorDetector that monitors the emission of characteristic light from the etching products so as to signal when etching should endPlasma process induced damage / Wafer charging damageDamage to devices on wafers due to the use of plasmaAntenna EffectSensitivity of the damage to the size of the conductorIon implantationMethod of doping in which ions of impurity are accelerated and shot into thesemiconductor surfaceGas-source dopingMethod of doping in which a gas reacts with silicon and liberates phosphorus so that phosphorous diffuses into the silicon substrateSolid-source diffusionMethod of doping in which the dopants from the thin film coated on the siliconsurface diffuse into siliconAnnealHeating of wafers for dopant activation and damage removalDopant activationMaking dopants behave as donors and acceptors by heating the wafersImplantation doseTotal number of implanted ions/cm2Depth/ RangeThe location of peak concentration below the surface of siliconStraddleSpread of dopant concentration profile3.2 DiffusionThe movement of molecules from an area of high concentration to an area of low concentrationJunction depthThickness of diffusion layerDiffusivityConstant that describes how quickly a given impurity diffuses in silicon for a given furnace temperaturePredepositionPortion of diffusion process step with the source presentDrive-inPortion of diffusion process step without the sourceFurnace annealingHeating of wafers in a furnace for dopant activation and damage removalRapid thermal annealing (RTA)Annealing process in which a wafer is rapidly heated to high temperature and cooled quickly down to the room temperatureRapid thermal oxidationOxidation process in which a wafer is heated to the designated temperature quickly, oxidized, and then cooled rapidly down to the room temperatureRapid thermal chemical vapor deposition (CVD)Chemical vapor deposition process in which a wafer is heated to the designated temperature quickly, the material is deposited on the wafer, and the wafer is cooled rapidly down to the room temperatureLaser annealingAnnealing process in which a silicon wafer is heated with a pulsed laser Transient enhanced diffusion (TED)Diffusion phenomena in which diffusion rate is increased by crystal damage due to ion implantationInterconnectMetal connection between devices in integrated circuitsInter-metal dielectricsMaterials used for electrical isolation between metal layersCrystallineMolecular structure with nearly perfect periodicityPolycrystallineMolecular structure composed of densely packed crystallites or grains of single-crystalsAmorphousStructure with no atomic or molecular orderingGrain boundaryInterface between crystal grainsThin-film transistors (TFT)Transistors made of amorphous or polycrystalline silicon, widely used in flat-panel displaysSputtering targetThe source material for sputteringReactive sputteringSputtering accompanied by chemical reaction of sputtered ions. For example, Ti sputtered in nitrogen gas forms TiN film on the Si waferPhysical vapor deposition (PVD)Another name for sputteringStep coverage problemThe inability of sputtering to deposit uniform films in small holes or vertical features on waferHigh-temperature oxide (HTO)Very conformal oxide formed by a CVD process at a high temperatureLow-pressure chemical vapor deposition (LPCVD)CVD process at low pressure, which yields good thickness uniformity and low gas consumptionPlasma-enhanced chemical vapor deposition (PECVD)CVD using plasma; low deposition temperaturesIn-situ dopingDoping process in which dopant species are introduced during the CVD deposition of polycrystalline siliconSpin-onProcess in which liquid materials are spun onto the waferEpitaxySpecial type of thin-film deposition technology that produces a crystalline layer on silicon surfaces that is an extension of the underlying semiconductor crystal arrangementMetallizationInterconnection of individual devices by metal linesVia/ PlugElectrical connection between adjacent metal layersElectromigrationMigration of metal along the crystal-grain boundaries in a quasi-random manner, causing voids to occur in metal interconnectsDamascene processProcess used to form copper interconnect linesChemical-mechanical polishing (CMP)Process in which a polishing pad and slurry are used to polish away material and leave a very flat surfaceBack-end processMetallization; last step of IC fabricationFront-end processesOxidation, diffusionPlanarizationProcess to obtain a flat surface to improve lithography and etchingMulti-chip modulesMultiple chips put into one packageSolder bumpsElectrical connection between chip and packageFlip-chip bondingMelting pre-formed solder bumps on IC pads to make all connectionssimultaneouslyBurn-inSubjecting IC packages to higher than normal voltage and temperatures; identify potential failuresQualificationRoutine used to verify the quality of manufacturing and reliabilityOperation life testPart of qualification process; to find out if the devices last over one thousandoperating hours3.3 (a) Lithography fieldA small area having the best optical resolution (The beam intensity is uniformwithin a lithography field.)(b) M isalignmentLayer to layer mismatch (Each new mask level should be aligned to one of theprevious levels)(c) SelectivityThe ratio of etching rate of the film to be etched to that of the substrate filmbelow(d) End-point detectionDetecting the exposed substrate film after the removal of the desired film.(e) Step coverageThe ratio of film thickness deposited on the flat surface to that deposited on thenon-flat surface.(f) ElectromigrationThe movement of atoms in a metal film due to momentum transfer from theelectrons carrying the current.3.4 (a) Wet. Otherwise, it would take too long to grow the thick oxide.(b) Dry because it provides better control of the gate (channel) length and verticalwall profile.(c) Arsenic because- it is a donor ion (group V),- it reduces R p and ∆R p, and- it reduces diffusivity.(d) PECVD(e) Sputtering, dry etching, plasmas containing chlorine.(f)Oxidation3.5 Wet oxidation is faster than dry oxidation because the solid solubility of H 2O steaminto SiO 2 is higher than that of O 2 gas into SiO 2. This creates a very sharpconcentration profile that causes H 2O to diffuse towards the SiO 2-Si interface muchmore effectively than O 2 under the same conditions.3.6 (a) hr hrum um um um um 239.6/0117.02.0*165.02.0*2.02=+=τSolving the given equation for t ox and using quadratic formula,um t B A A T ox 256.02)(42=+++-=τ. (b) Linear approximation: )(τ+≈t AB T ox => 0.655um (156% error). Quadratic approximation: => 0.320um (25% error). )(2τ+≈t B T oxIn this case , neither linear nor quadratic approximations would bevalid.ox T A ≈Deposition3.7 Poly Silicon:SiH 4 → Si(s) + 2H 2Silicon Nitride:3SiH 2Cl 2 + 4NH 3 → Si 3N 4 + 6HCl + 6H 2LTO:SiH 4 + O 2 → SiO 2 + 2HCl + 2H 2HTO:SiH 2Cl 2 + 2H 2O → SiO 2 + 2HCl + 2H 2Diffusion3.8 (a) )(Dt Dt x x Dt x j j j ∆+=∆+⇒=())()(222Dt x Dt Dt x x x x j j j j j ∆+=∆+=∆+∆+())(2Dt x x x j j j ∆=∆+∆Since ,j j x x ∆>>j j j j x Dt x Dt x x 2)()(2∆≈∆⇒∆≈∆. (b) For boron,D (500K) = 10.5⨯Exp[-3.69/(8.614⨯10-5⨯773)]cm2/sec = 9.0⨯10-24 cm 2/sec, and500 years = 1.58⨯1010 sec.Hence,m cm x j μ9131042.11042.1--⨯=⨯=∆.Our assumption (j j x x ∆>>) in part (a) is correct.3.9 (a) Junction depth is distance from surface where the dopant concentration equalsthe substrate concentration.31510-===cm N N N junction d sub (Required junction dopant concentration) From Eq. (3.6.1),31540102--==cm e DtN N Dt x junction j π UsingeV E and cm N cm D e D D a kT E a 7.3,10sec,/5.10,215020/0====--,cm X j 41097.1-⨯=.(b) X j has the form2/1ln 2⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛=Dt k Dt X j . Taking the derivative,j j j X Dt X Dt k Dt k Dt dDt dX 12ln 21ln 12/12/1-=⎪⎪⎭⎪⎪⎬⎫⎪⎪⎩⎪⎪⎨⎧⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛=. 14491039.91098.11102---⨯=⨯-⨯⨯=cm 101.98dDt dX -4j ()()240637370.310996.3360010.5.10cm eDt --⨯==∆ Hence, .01075.3.35≈⨯=∆∂∂=∆-cm Dt Dt X X j jVisualization3.10 (a) (b) 1.0μm resist Silicon substrate 1.0μm oxide 1.0μm oxide Silicon substrate(c) (d) 1.0μm oxide 1.0μm resist Silicon substrateSilicon substrate 1.0μm oxide3.11Contact 2 MaskContact 1 MaskChanging the polarity of the contact 1 mask results in the same cross section as problem 3.10 (d). The same cross-section is obtained by using a negative resist and the reversed mask of contact 1, which is opaque in the inside of contact 1 area.。
半导体物理 课后习题答案
第一章习题1．设晶格常数为a 的一维晶格，导带极小值附近能量E c (k)和价带极大值附近能量E V (k)分别为：E c =0220122021202236)(,)(3m k h m k h k E m k k h m k h V -=-+ 0m 。
试求：为电子惯性质量，nm a ak 314.0,1==π（1）禁带宽度;（2） 导带底电子有效质量; （3）价带顶电子有效质量;（4）价带顶电子跃迁到导带底时准动量的变化 解：（1）eV m k E k E E E k m dk E d k m kdk dE Ec k k m m m dk E d k k m k k m k V C g V V V c 64.012)0()43(0,060064338232430)(2320212102220202020222101202==-==<-===-==>=+===-+ 因此：取极大值处，所以又因为得价带：取极小值处，所以：在又因为：得：由导带：043222*83)2(1m dk E d mk k C nC===sN k k k p k p m dk E d mk k k k V nV/1095.7043)()()4(6)3(25104300222*11-===⨯=-=-=∆=-== 所以：准动量的定义：2. 晶格常数为0.25nm 的一维晶格，当外加102V/m ，107 V/m 的电场时，试分别计算电子自能带底运动到能带顶所需的时间。
解：根据：tkhqE f ∆∆== 得qE k t -∆=∆sat sat 137192821911027.810106.1)0(1027.810106.1)0(----⨯=⨯⨯--=∆⨯=⨯⨯--=∆ππ第三章习题和答案1. 计算能量在E=E c 到2*n 2C L 2m 100E E π+= 之间单位体积中的量子态数。
解322233*28100E 21233*22100E 0021233*231000L 8100)(3222)(22)(1Z VZZ )(Z )(22)(2322C 22C L E m h E E E m V dE E E m V dE E g V d dEE g d E E m V E g c nc C n l m h E C n l m E C n n c n c πππππ=+-=-====-=*++⎰⎰**）（）（单位体积内的量子态数）（2. 试证明实际硅、锗中导带底附近状态密度公式为式（3-6）。
半导体物理导论课后习题答案第1-3章
半导体物理导论课后习题答案第1-3章1.倒格子的实际意义是什么？一种晶体的正格矢和相应的倒格矢是否有一一对应的关系？解答：倒格子的实际意义是由倒格子组成的空间，实际上是状态空间空间，在晶体的X 射线衍射照片上的斑点实际上就是倒格子所对应的点子。
由正格子的基矢(a 1,a 2,a 3)就得到倒格子的矢量(b 1,b 2,b 3)，其中其中Ω是晶格原胞的体积。
由此可以唯一地确定相应的倒格子空间。
显然，倒格子与正格子之间有如下关系：所以一种晶体的正格矢和相应的倒格矢有一一对应的关系。
Ω⨯=Ω⨯=Ω⨯=213132321222a a b a a b a a b πππ，，ij i i πδ2=⋅b a (i,j=1,2,3)2.假设有一立方晶体，画出以下各晶面（1）（100）；（2）（110）；（3）（111）；（4）（100）；（5）（110）；（6）（111）3.已知Si的晶格常数或单胞的边长a=5.43089 Å, 求：（1）Si的原子体密度。
（2）(111)面、(110)面以及(100)面的原子面密度，比较哪个晶面的面密度最大？哪个晶面的面密度最小？解:(1)每个晶胞中有8个原子，晶胞体积为a 3，每个原子所占的空间体积为a 3/8，因此每立方厘米体积中的硅原子数为：原子体密度=8/a 3=8/(5.43×108)3=5×1022（个原子/cm 3）(2)（111）面为一个边长为的等边三角形，有效原子数为等边三角形的面积为个原子（面心原子）（顶角原子）25213313=⨯+⨯4521022212a a a S =⨯⨯=2a所以，(111)面的原子面密度为(110)面为一个边长为 的长方形，有效原子数为长方形的面积为所以，(110)面的原子面密度为22524525aa ==等边三角形面积有效原子数2a a ⨯个原子（体对角线原子）（面心原子）（顶角原子）42212414=+⨯+⨯222a a a S =⨯=222224aa ==等边三角形面积有效原子数(100)面为一个边长为 的正方形，有效原子数为正方形的面积为所以，(100)面的原子面密度为因此，(111)面的原子面密度∶(110)面的原子面密度∶(100)面的原子面密度为 ∶ ∶ = ∶ ∶1说明(111)面的原子面密度最高，(100)面的原子面密度最低。
半导体物理课后习题(保密)
解: 须先求出本征载流子浓度ni，即
代入数据得，ni=1.86 ×1013cm-3 根据电中性条件有 p0+ND+=n0+NA-
ni
q( n p )
i
1 i ( n p )
联立 载流子浓度公式
n0p0=ni2
可求解得 n0=3.89 ×1013cm-3, p0=8.89 ×1012cm-3 所以样品的电导率为：
解: 由图3-7查得T=500k时，Si的本征载流子浓度ni=3.5×1014cm-3 联立方程
p0=ni2/n0
解得， ND=3.5×1014cm-3时，n0≈4.3×1014cm-3, p0=2.8×1014cm-3 —— n0,p0差别不显著，杂质导电特性不很明显 ND=1012cm-3时，n0≈ni=3.5×1014cm-3, p0=3.5×1014cm-3,即n0=p0. —— 进入本征 半导体材料在某一温度下所处的区域与杂质浓度相关 或 杂质浓度不同，材料进入同一区域所需要的温度不一样。
m0为电子惯性质量，k1=1/2a; a=0.314nm。试求： （1）禁带宽度； （2）导带底电子有效质量； （3）价带顶电子有效质量； （4）价带顶电子跃迁到导带底时准动量的变化。
练习2-课后习题2
第一章 半导体中的电子状态
2.晶格常数为0.25nm的一维晶格，当外加102V/m和107V/m 的电 场时，试分别计算电子自能带底运动到能带顶所需的时间。
作业-课后习题14
第三章 半导体中载流子的统计分布
计算含有施主杂质浓度ND=9×1015cm-3与受主杂质浓度为1.1×1016cm-3 的硅在室温时的电子和空穴浓度以及费米能级的位置。
半导体物理学(第7版)第三章习题和答案
第三章习题和答案1. 计算能量在E=E c 到 之间单位体积中的量子态数。
2*n 2C L 2m 100E E π+=解：2. 试证明实际硅、锗中导带底附近状态密度公式为式（3-6）。
322233*28100E 21233*22100E 0021233*231000L 8100)(3222)(22)(1Z VZZ )(Z )(22)(2322C22CL E m h E E E m V dE E E m V dE E g Vd dEE g d E E m V E g cn c C nlm h E C nlm E C nn c n c πππππ=+-=-====-=*++⎰⎰**）（）（单位体积内的量子态数）（)(21)(,)"(2)()(,)(,)()(2~.2'213''''''2'21'21'21'2222222C a a lt tz y x ac c zla z y t ay x t a xz t y x C C e E E m hk V m mm m k g k k k k k m h E k E k m m k k m m k k m m k mlk m k k h E k E K IC E G si -=⎪⎪⎭⎫ ⎝⎛+∙=+++====+++=*****系中的态密度在等能面仍为球形等能面系中在则：令）（关系为）（半导体的、证明：[]3123221232'2123231'2'''')()2(4)()(111100)()(24)(4)()(~ltn c nc l t t z m m sm VE E hm E sg E g si V E E h m m m dE dz E g dkk k g Vk k g d k dE E E =-==∴-⎥⎥⎦⎤⎢⎢⎣⎡+∙∙==∴∙=∇∙=+**πππ）方向有四个，锗在（旋转椭球，个方向，有六个对称的导带底在对于即状态数。
半导体物理答案 第三章
任课老师：吴海霞作业章次：第三章完成日期：2010年4月9日姓名：高磊学号： 20070682班级： 01530701老师点评：6.计算硅在-78℃、27℃、300℃时本征半导体费米能级、假定它在禁带中线处合理吗？()()假设它在中线处合理约为的时，时，时，，其中即本征半导体不带电解：∴⨯-+==⨯-+==⨯-+==∴==++=++=∴====∴-------eV E Si eV E E E K T eV E E E K T eV E E E K T m m m m m m T k E E N N T k E E E h Tk mN hT k m N e N eN p n g vc F vc F vc F p n n pv c c v v c F pv ncTk E E v Tk E E c v F F c 1)(102.22573)(1016.12300)(105.7219559.0,08.1ln 432ln 2222,222230*0***0032/30*32/30*00 ππ10.以施主杂质电离90%作为强电离的标准，求掺砷的n 型锗在300K 时以杂质电离为主的饱和区掺杂的浓度范围⎪⎭⎫⎝⎛⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛∆-≤∴≤⎪⎪⎭⎫ ⎝⎛∆⎪⎪⎭⎫ ⎝⎛=⨯=⨯⨯=≥--026.0exp 21005.1%10exp 2%10%10exp 2104.2104.21010190031413T k E N N T k E NN D cm n N D c D D cD i D 解：强电离时 11.若锗中施主杂质电离能DE ∆=0.01eV ，施主杂质浓度分别为D N =31410-cm 及31710-cm 。
计算1、99%电离；2、90%电离；3、50%电离时温度各为多少？⎪⎩⎪⎨⎧====⎪⎩⎪⎨⎧====⎪⎩⎪⎨⎧====⎪⎪⎩⎪⎪⎨⎧=-==+=∴=∆=⎪⎪⎭⎫ ⎝⎛--∴==-==⎪⎪⎭⎫ ⎝⎛--+=⎪⎪⎩⎪⎪⎨⎧=-====⎪⎪⎩⎪⎪⎨⎧=-==-==∴==∆⎪⎪⎭⎫⎝⎛∆⎪⎪⎭⎫ ⎝⎛=--------++---------）（）（电离时）（）（电离时）（）（电离时）（）（电离时，不在强电离区当）（）（电离时，）（）（电离时，代入已知常量得电离时，处于强电离区、解：当317314317314317314317314000003173143173142/3150010551016%50103201010%90105331037%99109.3ln 23116103ln 23116ln 2exp 22ln 2exp 21%50109.6ln 2311610ln 23116%10%90102.9ln 23116103.2ln 23116%1%9910ln 116,2ln exp 2%90%99cm N KT cm N K T cm N K T cm N K T cm N K T cm N K T cm N T Tcm N T TN N T k E N Tk E E N N n n T k E E N T k E E N n cm N T Tcm N T T D cm N T Tcm N T T D N T D T N N D T k E T k E N N D D D D D D D D D DcDD Fc c DD D F DF D D D D D D D D D c DD c D14.计算含有施主杂质浓度D N =315/109cm ⨯及受主杂质浓度A N =316/101.1cm ⨯的硅在300K 时的电子和空穴浓度及费米能级位置eVN N T k E E cm pn n kT E N N D cm N N N vAv F i D c A D A A 224.0ln /1013.1%1exp 2/102'0352'315'=-=⨯===⎪⎭⎫ ⎝⎛∆=⨯=-=-故杂质完全电离净杂质浓度为20.制造晶体管一般在高杂质浓度的n 型衬底上外延一层n 型外延层，再在外延层中扩散硼磷而成。
半导体物理第三章习题答案
第三篇 题解 半导体中载流子的统计分布刘诺 编3-1、对于某n 型半导体，试证明其费米能级在其本征半导体的费米能级之上。
即E Fn >E Fi 。
证明：设n n 为n 型半导体的电子浓度，n i 为本征半导体的电子浓度。
显然n n > n iin i n F F F c c F c c E E T k E E N T k E E N >⎪⎪⎭⎫⎝⎛--⋅>⎪⎪⎭⎫ ⎝⎛--⋅则即00exp exp即得证。
3-2、试分别定性定量说明：（1） 在一定的温度下，对本征材料而言，材料的禁带宽度越窄，载流子浓度越高；（2） 对一定的材料，当掺杂浓度一定时，温度越高，载流子浓度越高。
解：(1) 在一定的温度下，对本征材料而言，材料的禁带宽度越窄，则跃迁所需的能量越小，所以受激发的载流子浓度随着禁带宽度的变窄而增加。
由公式Tk E v c i g eN N n 02-=也可知道，温度不变而减少本征材料的禁带宽度，上式中的指数项将因此而增加，从而使得载流子浓度因此而增加。
（2）对一定的材料，当掺杂浓度一定时，温度越高，受激发的载流子将因此而增加。
由公式可知，这3-3、若两块Si 样品中的电子浓度分别为2.25×1010cm -3和6.8×1016cm -3，试分⎪⎪⎭⎫⎝⎛--=⎪⎪⎭⎫ ⎝⎛--⋅=Tk E E N p Tk E E N n V F V Fc c 0000exp exp 和别求出其中的空穴的浓度和费米能级的相对位置，并判断样品的导电类型。
假如再在其中都掺入浓度为2.25×1016cm -3的受主杂质，这两块样品的导电类型又将怎样？解：由 200i n p n = 得()()()()⎪⎪⎩⎪⎪⎨⎧⨯≈⨯⨯==⨯=⨯⨯==--3316210022023101021001201103.3108.6105.1100.11025.2105.1cm n n p cm n n p i i可见，型半导体本征半导体n p n p n →>→≈02020101又因为 Tk E E v v F e N p 00--=，则⎪⎪⎩⎪⎪⎨⎧+=⎪⎪⎭⎫ ⎝⎛⨯⨯⋅+=⎪⎪⎭⎫ ⎝⎛⋅+=+≈⎪⎪⎭⎫⎝⎛⨯⨯⋅+=⎪⎪⎭⎫ ⎝⎛⋅+=eV E E p N T k E E eV E E p N T k E E v v n v F v v v v F 331.0103.3101.1ln 026.0ln 234.0100.1101.1ln 026.0ln 319020210190101 假如再在其中都掺入浓度为2.25×1016cm -3的受主杂质，那么将出现杂质补偿，第一种半导体补偿后将变为p 型半导体，第二种半导体补偿后将近似为本征半导体。
半导体物理第三章习题参考答案
NA
解得：
p
NA 2
1
1
4ni2
N
2 A
1
2
1 叶良修，半导体物理学（第二版），上册，129 页。
(1) T 300K 时，硼原子全部电离，此时本征载流子浓度 ni 1.51010cm-3 有：
NA ，
p
NA
1014 cm-3 ， n
ni2 p
2.3106 cm-3 ；
(2) T 400K 时，此时本征载流子浓度23 ni 1.31014 cm-3 NA ，本征激发已不 能忽略，有：
答：当T 300K 时，有：
3
3
NC
2
2 mnkT h2
2
2.509 1019
mn m0
2
cm-3
3
3
NV
2
2
mp h2
kT
2
2.509
1019
mp m0
2
cm-3
ni
NC NV
1
2
exp
Eg 2kT
代入数据得到：
Si GaAs
NC cm-3 2.7581019 4.351017
由波尔兹曼分布近似：
n NA ND nD
n
NC
exp
EC EF kT
以及施主能级上的电子的分布规律：
有：
nD
ND
1 gD
exp
EF ED kT
1
1 gD
exp
EF ED kT
n NA n n ND nD
ND NA n
nD
n
ND nD
1
n
1 gD
exp
《半导体物理学》刘恩科课后答案
代入数据得：
t＝
6.62 ×10-34
＝ 8.3 ×10−6 （s）
2 ×1.6 ×10−19 × 2.5 ×10−10 × E
E
当 E＝102 V/m 时，t＝8.3×10－8（s）；E＝107V/m 时，t＝8.3×10－13（s）。
3. 如果 n 型半导体导带峰值在[110]轴上及相应对称方向上,回旋共振实验结果应 如何? [解] 根据立方对称性,应有下列 12 个方向上的旋转椭球面:
(6.625
×
10
−34
)
2
( 5.7
×
1018
)
2 3
=
2 2 × 3.14 ×1.38 ×10−23 × 300
= 3.39173 ×10−31 Kg
﹟求 77k 时的 Nc 和 Nv：
3
2(2π ⋅ mn*k0T ') 2
N
' c
=
h3
Nc
3
2(2π ⋅ mn*k 0T ) 2
=
(
T' T
)
3 2
d 2 EC dk 2
= 2h2 3m0
+ 2h2 m0
= 8h2 3m0
；∴
mn＝
h2
/
d 2 EC dk 2
=
3 8
m0
③价带顶电子有效质量 m’
d 2 EV dk 2
=
− 6h2 m0
，∴ mn'
=
h2Leabharlann /d 2 EV dk 2
=
−
1 6
m0
④准动量的改变量
h △k＝ h (kmin-kmax)=
半导体物理学(第7版)第三章知识题和答案解析
第三章习题和答案1. 计算能量在E=E c 到2*n 2C L 2m 100E E π+= 之间单位体积中的量子态数。
解：2. 试证明实际硅、锗中导带底附近状态密度公式为式（3-6）。
322233*28100E 21233*22100E 0021233*231000L 8100)(3222)(22)(1Z VZZ )(Z )(22)(2322C22CL E m h E E E m V dE E E m V dE E g Vd dEE g d E E m V E g cn c C nlm h E C nlm E C nn c n c πππππ=+-=-====-=*++⎰⎰**）（）（单位体积内的量子态数）（)(21)(,)"(2)()(,)(,)()(2~.2'213''''''2'21'21'21'2222222C a a lt tz y x ac c zla z y t ay x t a x z t y x C C e E E m k V m m m m k g k k k k k m h E k E k m m k k m m k k m m k mlk m k k h E k E K IC E G si -=⎪⎪⎭⎫ ⎝⎛+•=+++====+++=*****系中的态密度在等能面仍为球形等能面系中在则：令）（关系为）（半导体的、证明：[]3123221232'2123231'2'''')()2(4)()(111100)()(24)(4)()(~ltn c nc l t t z m m sm VE E hm E sg E g si V E E h m m m dE dz E g dkk k g Vk k g d k dE E E =-==∴-⎥⎥⎦⎤⎢⎢⎣⎡+••==∴•=∇•=+**πππ）方向有四个，锗在（旋转椭球，个方向，有六个对称的导带底在对于即状态数。
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任课老师：吴海霞
作业章次：第三章
完成日期：2010年4月9日姓名：高磊
学号： 20070682
班级： 01530701
老师点评：
6.计算硅在-78℃、27℃、300℃时本征半导体费米能级、假定它在禁带中线处合理吗？
()()
假设它在中线处合理
约为的时，时，时，，其中即本征半导体不带电解：∴⨯-+==⨯-+==⨯-+==∴==++=++=∴====∴-----
--eV E Si eV E E E K T eV E E E K T eV E E E K T m m m m m m T k E E N N T k E E E h T
k m
N h
T k m N e
N e
N p n g v
c F v
c F v
c F p n n p
v c c v v c F p
v n
c
T
k E E v T
k E E c v F F c 1)
(102.22
573)
(1016.12300)(105.7219559.0,08.1ln 432ln 2222
,22
2230
*0*
*
*
003
2
/30*3
2
/30*
00 ππ
10.以施主杂质电离90%作为强电离的标准，求掺砷的n 型锗在300K 时以杂质电离为主的饱和区掺杂的浓度范围
⎪⎭⎫
⎝⎛⨯⨯⨯=⎪⎪⎭
⎫ ⎝⎛∆-≤∴≤⎪⎪⎭
⎫ ⎝⎛∆⎪⎪⎭⎫ ⎝⎛=⨯=⨯⨯=≥--026.0exp 21005.1%10exp 2%10%10exp 2104.2104.21010190031413T k E N N T k E N
N D cm n N D c D D c
D i D 解：强电离时 11.若锗中施主杂质电离能D
E ∆=0.01eV ，施主杂质浓度分别为D N =31410-cm 及31710-cm 。
计算1、99%电离；2、90%电离；3、50%电离时温度各为多少？
⎪⎩⎪⎨⎧====⎪⎩⎪⎨⎧====⎪⎩⎪⎨⎧====⎪⎪⎩⎪⎪⎨
⎧=-==+=∴=∆=⎪⎪⎭
⎫ ⎝⎛--
∴=
=-==
⎪⎪⎭
⎫ ⎝
⎛--+=
⎪⎪⎩⎪⎪⎨
⎧=-====⎪⎪⎩⎪⎪⎨
⎧=-==-==∴==∆⎪⎪⎭
⎫
⎝⎛∆⎪⎪⎭⎫ ⎝⎛=--------+
+---------）
（）（电离时）（）（电离时）（）（电离时）
（）（电离时，不在强电离区
当）（）（电离时，）（）（电离时，代入已知常量得电离时，处于强电离区、解：当317314317314317314317314000003173
143173
142/3150010551016%50103201010%90105331037%99109.3ln 23116103ln 2
3
116ln 2exp 22
ln 2
exp 21%50109.6ln 2311610ln 23116%
10%90102.9ln 23116103.2ln 23116%
1%9910ln 116
,2ln exp 2%90%99cm N K
T cm N K T cm N K T cm N K T cm N K T cm N K T cm N T T
cm N T T N N T k E N T
k E E N N n n T k E E N T
k E E N n cm N T T
cm N T T D cm N T T
cm N T T D N T D T N N D T k E T k E N N D D D D D D D D D D
c
D
D
F
c c D
D D F D F
D D
D D D D D D D c D
D c D
14.计算含有施主杂质浓度D N =315/109cm ⨯及受主杂质浓度
A N =316/101.1cm ⨯的硅在
300K 时的电子和空穴浓度及费米能级
位置
eV
N N T k E E cm p
n n kT E N N D cm N N N v
A
v F i D
c A D A A 224.0ln /1013.1%1exp 2/102'03
52'
315'
=-=⨯===⎪⎭
⎫
⎝⎛∆=⨯=-=-故杂质完全电离
净杂质浓度为
20.制造晶体管一般在高杂质浓度的n 型衬底上外延一层n 型外延层，再在外延层中扩散硼磷而成。
设n 型硅单晶衬底是掺锑的，锑的电离能为0.039eV ， 300K 时F E 位于导带底下面0.026eV 处，计算锑的浓度和导带中电子浓度。
3
1902/1003
1802/1/1037.1)exp()exp(212/105.92
cm T k E E F T k E T k E E N N cm T k E E F N n F c D c F c
D F c c ⨯=⎪⎪⎭
⎫ ⎝⎛--⎥⎦⎤⎢⎣⎡
∆-+=⨯=⎪⎪⎭
⎫ ⎝⎛--=ππ锑的浓度为导带中电子浓度。