施敏 半导体器件物理英文版 第一章习题

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(施敏)半导体器件物理(详尽版)

(施敏)半导体器件物理(详尽版)
十分纯净 不含任何杂质 晶格中的原子严格 按周期排列的
实际应用中的
半导体材料
原子并不是静止在具有严格周期性 的晶格的格点位置上,而是在其平 衡位置附近振动
并不是纯净的,而是含有若干杂质, 即在半导体晶格中存在着与组成半 导体的元素不同的其他化学元素的 原子
晶格结构并不是完整无缺的,而存 在着各种形式的缺陷
在晶体中,不但外层价电 子的轨道有交叠,内层电 子的轨道也可能有交叠, 它们都会形成共有化运动;
半导体中的电子是在周期性排列 且固定不动的大量原子核的势场 和其他大量电子的平均势场中运动。 这个平均势场也是周期性变化的, 且周期与晶格周期相同。
但内层电子的轨道交叠较 少,共有化程度弱些,外 层电子轨道交叠较多,共 有化程度强些。
思考
• 既然半导体电子和空穴都能导电,而导 体只有电子导电,为什么半导体的导电 能力比导体差?
江西科技师范大学
半导体器件物理
●导带底EC 导带电子的最低能量
●价带顶EV 价带电子的最高能量
●禁带宽度 Eg
Eg=Ec-Ev
●本征激发 由于温度,价键上的电子 激发成为准自由电子,亦 即价带电子激发成为导带 电子的过程 。
江西科技师范大学
半导体器件物理
如图,晶面ACC’A’在 坐标轴上的
截距为1,1,∞,
其倒数为1,1,0,
此平面用密勒指数表示 为(110),
此晶面的晶向(晶列指 数)即为[110];
晶面ABB’A’用密勒指 数表示为( 100 );
晶面D’AC用密勒指数 表示为( 111 )。
江西科技师范大学
半导体器件物理
图1-7 一定温度下半导体的能带示意图 江西科技师范大学
半导体器件物理

施敏-半导体器件物理英文版-第一章习题

施敏-半导体器件物理英文版-第一章习题

施敏-半导体器件物理英文版-第一章习题施敏-半导体器件物理英文版-第一章习题施敏半导体器件物理英文版第一章习题1. (a )求用完全相同的硬球填满金刚石晶格常规单位元胞的最大体积分数。

(b )求硅中(111)平面内在300K 温度下的每平方厘米的原子数。

2. 计算四面体的键角,即,四个键的任意一对键对之间的夹角。

(提示:绘出四个等长度的向量作为键。

四个向量和必须等于多少?沿这些向量之一的方向取这些向量的合成。

)3. 对于面心立方,常规的晶胞体积是a 3,求具有三个基矢:(0,0,0→a/2,0,a/2),(0,0,0→a/2,a/2,0),和(0,0,0→0,a/2,a/2)的fcc 元胞的体积。

4. (a )推导金刚石晶格的键长d 以晶格常数a 的表达式。

(b )在硅晶体中,如果与某平面沿三个笛卡尔坐标的截距是10.86A ,16.29A ,和21.72A ,求该平面的密勒指数。

5. 指出(a )倒晶格的每一个矢量与正晶格的一组平面正交,以及(b )倒晶格的单位晶胞的体积反比于正晶格单位晶胞的体积。

6. 指出具有晶格常数a 的体心立方(bcc )的倒晶格是具有立方晶格边为4π/a 的面心立方(fcc )晶格。

[提示:用bcc 矢量组的对称性:)(2x z y a a -+=,)(2y x z a b -+=,)(2z y x a c -+= 这里a 是常规元胞的晶格常数,而x ,y ,z 是fcc 笛卡尔坐标的单位矢量:)(2z y a a ρρρ+=,)(2x z a b ρρρ+=,)(2y x a c ρρρ+=。

] 7. 靠近导带最小值处的能量可表达为.2*2*2*22++=z z y y xx m k m k m k E η 在Si 中沿[100]有6个雪茄形状的极小值。

如果能量椭球轴的比例为5:1是常数,求纵向有效质量m*l 与横向有效质量m*t 的比值。

8. 在半导体的导带中,有一个较低的能谷在布里渊区的中心,和6个较高的能谷在沿[100] 布里渊区的边界,如果对于较低能谷的有效质量是0.1m0而对于较高能谷的有效质量是1.0m0,求较高能谷对较低能谷态密度的比值。

半导体物理章习题全英文版

半导体物理章习题全英文版

212k mE =)(1222k V E mE -= )(2232k V E mE -= Chapter 11.22 Calculate the density of valence electrons in silicon.1.23 The structure of GaAs is the zincblende lattice. The lattice constant is 5.65 ︒A . Calculate thedensity of valence electrons in GaAs.1.24 (a) If 17105⨯ phosphorus atoms per 3cm are add to silicon as a substitutional impurity, determine the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.(b)Repeat part (a) for 15102⨯ boron atoms per 3cm added to silicon.1.25 (a)Assume that 16102⨯-3cm of boron atoms are distributed homogeneously throughout single crystal silicon. What is the fraction by weight of boron in the crystal? (b)If phosphorus atoms, at a concentration of 1810-3cm , are added to the material in part (a), determine the fraction by weight of phosphorus.1.26 If 16102⨯-3cm boron atoms are added to silicon as a substitutional impurity and are distributed uniformly throughout the semiconductor, determine the distance between boron atoms in terms of the silicon lattice constant.(Assume the boron atoms are distributed in a rectangular or cubic array.)1.27 Repeat Problem 1.26 for 15104⨯-3cm phosphorus atoms being added to silicon.Chapter 22.32Consider a proton in a one-dimensional infinite potential well shown in Figure2.6.(a)Derive the expression for the allowed energy states of the proton.(b)Calculate the energy state for (i)a=4︒A ,and (ii)a=0.5cm.2.33 For the step potential function shown in Figure P2.33, assume that 0V E > and that particles are incident from the +x direction traveling in the -x direction.(a) Write the wave solutions for each region.(b) Derive expressions for the transmission and reflection coefficients.2.38 An electron with energy E is incident on a rectangular potential barrier as shown in Figure2.9. The potential barrier is of width a and height E V ≥0.(a)Write the form of the wave function in each of the three regions. (b)For this geometry ,determine what coefficient in the wave function solutions is zero.(c)Derive the expression for the transmission coefficient for the electron(tunneling probability ).(d)Sketch the wave function for the electron in each region.2.39 A potential function is shown in Figure P2.39 with incident particles coming from -∞ with a total energy 2V E >.The constants k are defined asAssume a special case for which πn a 2k 2=,n=1,2,3... Derive the expression, in terms of the constants ,,,21k k and 3k , for the transmission coefficient .The transmission coefficient is defined as the ratio of the flux of particles in region Ⅲ to the incident flux in region Ⅰ.2.40 Consider the one-dimensional potential function shown in FigureP2.40.Assume the total energy of an electron is 0V E <.(a )Write the wave solutions that apply in each region.(b)Write the set of equations that result from applying the boundary conditions.(c)Show explicitly why not , the energy levels of the electron are quantized.Chapter 33.25 Derive the density of status function for a one-dimension electron gas in GaAs(m n *=0.067m 0).Note that the kinetic energy may be written as E=(±p)2/2m n *,which means that there are two momentum states for each energy level.3.26 (a)Determine the total number (#/cm 3) of energy states in silicon between C E and at (i)T=300K and (ii)T=400K.(b) Repeat part (a) for GaAs.3.27 (a)Determine the total number (#/cm 3) of energy states in silicon between v E and kT E v 3- at (i) T=300K and (ii)T=400K.(b) Repeat part (a) for GaAs.3.28 (a)Plot the density of states in the conduction band of silicon over the range . (b) Repeat part (a) for the density of states in the valence band over the range v v E E eV E <<-4.0.3.29 (a)For silicon,find the ratio of the density of states in the conduction band at E=Ec+KT to the density of states in the valence band at E=Ev-KT. (b)Repeate part (a) for GaAs.Chapter 44.49 Consider silicon at T =300 K with donor concentrations of N d =1014,1015,1016,and1017,cm -3. Assume N a =0.(a )Calculate the position of the Fermi energy level with respect to the conduction band for these donor concentrations.(b )Determine the position of the Fermi energy level with respect to the intrinsic Fermi energy level for the donor concentrations given in part (a ).4.52 Consider GaAs at T=300K with N d =0. (a)Plot the position of the Fermi energy level with respect to the intrinsic Fermi energy level as a function of the acceptor impurity concentration over the range of 1014≤N a ≤1017 cm -3. (b)Plot the position of the Fermi energy level with respect to the valence-band energy over the same acceptor impurity concentration as given in part(a).4.53 For a particular semiconductor , eV E g 50.1= ,**10n p m m =, T=300K,and 35101-⨯=cm n i .(a) Determine the position of the intrinsic Fermi energy level with respect to the center of the bandgap.(b) Impurity atoms are added so that the Fermi energy level is 0.45eV below the center of the bandgap. (i) Are acceptor or donor atoms added? (ii) What's the concentration of impurity atoms added?4.55 (a) Silicon at T=300K is doped with donor impurity atoms at a concentration of 315106-⨯=cm N d . (i) Determine E c - E F . (ii) Calculate the concentration of additional donor impurity atoms that must be added to move the Fermi energy level a distance KT closer to the conduction band edge. (b) Repeat Part (a) for GaAs if the original donor impurity concentration is 315101-⨯=cm N d .4.56 (a) Determine the position of the Fermi energy level with respect to the intrinsic Fermi level in silicon at T=300K that is doped with boron atoms at a concentration of 316102-⨯=cm N d .(b) Repeat Part (a) if the silicon is doped with phosphorus atoms at a concentration of N d =2x1016cm -3 . (c) Calculate n 0 and p 0 in parts (a) and (b).Chapter 55.8 (a) A silicon semiconductor resistor is in the shape of a rectangular bar with a cross-sectional area of 8.5×10-4cm 2, a length of 0.075 cm, and is doped with a concentration of 2×1016cm -3, boron atoms. Let T=300K. A bias of 2 volts is applied across the length of the silicon device. Calculate the current in the resistor. (b) Repeat part (a) if the length is increased by a factor of three. (c) Determine the average drift velocity of holes in the parts (a) and (b).5.18 An n-type silicon resistor has a leng th L=150μm, which W=7.5μm, and thickness T=1μm. A voltage of 2 V is applied across the length of the resistor. The donor impurity concentration varies linearly through the thickness of the resistor with N d =2×1016cm -3at the top surface and N d =2×1015cm -3at the bottom surface. Assume an average carrier mobility of n μ=750cm 2/V -s . (a)s V cm N d n⋅⨯+=-/)1051(135032/116μ)()(2min p n pn i μμμμσσ+=What is the electric field in the resistor? (b) Determine the average conductivity of the silicon. (c) Calculate the current in the resistor. (d) Determine the current density near the top surface and the current density near the bottom surface.5.22 A semiconductor material has electron and hole mobilitiesμn andμp , respectively. When the conductivity is considered as a function of the hole concentration p 0, (a) show that the minimum value of conductivity, σmin , can be written aswhere i σis the intrinsic conductivity, and (b) show that the corresponding hole concentration is2/10)/(p n n p μμ=.5.23 Consider three samples of silicon at T=300K. The n-type sample is doped with arsenic atoms to a concentration of 316105-⨯=cm N d , The p-type sample is doped with boron atoms to a concentration of 316102-⨯=cm N a .The compensated sample is doped with both the donors and acceptors described in the n-type and p-type sample. (a) Find the equilibrium electron and hole concentrations in each sample. (b) determine the majority carrier mobility in each sample. (c) calculate the conductivity of each sample. (d) and determine the electric field required in each sample to induce a drift current density of J=120A/cm 2 .5.28 (a) Assume that the electron mobility in an n-typesemiconductor is given by Where N d is the donor concentration in cm -3. Assuming complete ionization, plot the conductivity as a function of N d over the range 318151010-≤≤cm N d . (b) Compare the results of part (a) tothat if the mobility were assumed to be a constant equal to 1350 s V cm ⋅-/3. (c) If an electricfield of E=10V/cm is applied to the semiconductor. Plot the electron drift current density of parts (a) and (b).5.36 The total current in a semiconductor is constant and equal to J=-10A/cm -3. The total current is composed of a hole drift current. Assume that the hole concentration is a constant and equal to 1016 cm -3 and assume that electron concentration is given by 3/15102)(--⨯=cm e x n L x where L=15m μ. The electron diffusion coefficient is D n =27 cm 2/s and the hole mobility is s V cm p ⋅=/4202μ . Calculate (a) the electron diffusion current density for x>0, (b) the hole drift current density for, and (c) the required electric field for x>0 .Chapter 66.3 An n-type silicon sample contains a donor concentration of 31610-=cm N d . The minority carrier hole lifetime is found to be . (a) What is the lifetime of the majority carrier electrons? (b) Determine the thermal-equilibrium generation rate for electrons and holes in this material. (c) Determine the thermal-equilibrium recombination rate for electrons and holes in this material.6.5 Derive Equation (6.27) from Equations (6.18) and (6.20).6.6 Consider a one-dimensional hole flux as shown in Figure 6.4. If the generation rate of holes in this differential volume is 132010--⋅=s cm g p and the recombination rate is 1319102--⋅⨯s cm ,what must be the gradient in the particle current density to maintain a steady-state hole concentration?6.7 Repeat Problem 6.6 if the generation rate becomes zero.6.17 (a)Consider a silicon sample at T=300K doped with 1016cm -3 donor atoms. Let τp0=5x10-7s.Alight source turns on at t=0 producing excess carriers with a uniform generation rate of g'=5x1020cm -3s -1.At t=5x10-7s,the light source turns off.(i)Derive the expression(s) for the excess carrier concentration as a function of time over the range 0≤t≤∞.(ii) What is the value of the excess concentration when the light source turns off. (b) Repeat Part (a) for the case when the light source turns off at t=2x10-6s. (c) Sketch the excess minority carrier concentrations versus time for parts (a) and (b).6.18 A semiconductor is uniformly doped with 1710-3cm acceptor atoms and has the following properties :s cm D n /272=,s cm D p /122=,s n 170105-⨯=τ,and s p 17010-=τ.An external source has been turned on for 0<t producing a uniform concentration of excess carriers at a generation rate of 132110'--=s cm g .The source turns off at time t=0 and back on at time s t 6102-⨯=(a)Derive the expressions for the excess carrier concentration as a function of time for ∞≤≤t 0.(b)Determine the value of excess carrier concentration at (i)t=0 (ii)s t 6102-⨯=,and (iii)∞=t (c)Plot the excess carrier concentration as a function of time.Chapter 77.4 An abrupt silicon pn junction at zero bias dapant concentrations of N a =1017cm -3 and N d =5×1015cm -3. T=300K. ﹙a ﹚Calculate the Fermi level on each side of the junction with respect to the intrinsic Fermi level. ﹙b ﹚ Sketch the equilibrium energy band diagram for the junction and determine V bi from the diagram and the results of part ﹙a ﹚. ﹙c ﹚Calculate V bi using Equation ﹙7.10﹚, and compare the results to part ﹙b ﹚. ﹙d ﹚Determine xn, xp, and the peak electric field for this junction.7.6 A Silicon pn junction in thermal equilibrium at T=300K is doped such that E F -E Fi =0.365eV in the n region and E Fi -E F =0.330eV in the p region ﹙a ﹚Sketch the energy-band diagram for the pn junction. ﹙b ﹚Find the impurity doping concentration in each region. ﹙c ﹚Determine V bi .7.18 An ideal one-sided silicon p +n junction at T=300K is uniformly doped on both sides of the metallurgical junction. It is found that the doping relation is N a =80N d and the build-in potential barrier is V bi =0.740V . A reverse-biased voltage of VR=10V is applied. Determine ﹙a ﹚N a , N d ;﹙b ﹚x p , x n ;﹙c ﹚|E max |; and ﹙d ﹚C’j .7.19 A silicon n +p junction is biased at V R =5V. ﹙a ﹚Determine the change in built-in potential barrier if the doping concentration in the p increases by a factor of 3.﹙b ﹚Determine the ratio of junction capacitance when the acceptor doping is 3N a compared to that when the acceptor doping is N a .﹙c ﹚Why does the junction capacitance increase when the doping concentration increase ?7.20 ﹙a ﹚The peak electric field in a reverse-biased silicon pn junction is |E max |=3×105 V/cm. The doping concentrations are N d =4×1015cm -3 and N a =4×1017cm -3 .Find the magnitude of the reverse-biased voltage.﹙b ﹚Repeat part ﹙a ﹚ for N d =4×1016cm -3 and N a =4×1017cm -3﹙c ﹚Repeat part ﹙a ﹚N d =N a =4×1017cm -3Chapter 88.2 A silicon pn junction has impurity doping concentrations of N d =2×1015cm -3 and N a =8×1015cm -3. Determine the minority carrier concentrations at the edges of the space charge region for ﹙a ﹚V a =0.45 V ﹙b ﹚V a =0.55 V , and ﹙c ﹚V a =-0.55V .8.3 The doping concentrations in a G a A s pn junction are N d =1016cm -3 and N a =4×1016cm -3.Find the minority carrier concentrations at the edges of the space charge region for ﹙a ﹚V a =0.90 V ﹙b ﹚V a =1.10 V, and﹙c﹚V a =-0.95V.8.5 Consider a G a A s pn junction with doping concentrations N a=5×1016cm-3 and N d=1016cm-3. The junction cross-sectional area is A=10-3cm2and the applied forward-bias voltage is Va=1.10V. Calculate the﹙a﹚minority electron diffusion current at the space charge region, ﹙b﹚minority hole diffusion current at the edge of the space charge region. and﹙c﹚total current in the pn junction diode.8.6An n+p silicon dione at T=300K has the following parameters N d=1016cm-3, D n=25cm2/s,D p=10cm2/s , N d=1016cm-3, Τn0=Τp0=1 μs, and A=10-4cm2. Determine the diode current for ﹙a﹚a forward-bias voltage of 0.5V and ﹙b﹚a reverse-biased voltage of 0.5V.8.7 An ideal germanium pn junction diode has the following parameters: N a=4×1015cm-3, N a=2×1017cm-3, D p=48cm2/s, D n=90cm2/s, Τp0=Τn0=2×10-6s, and A=10-4cm2 . Determine the diode current for ﹙a﹚a forward-bias voltage of 0.25V and ﹙b﹚a reverse-biased voltage of 0.25V.Chapter 99.7 A Schottky diode with n-type GaAs at T = 300 K yields the 1/C’2 versus V R plot shown in Figure P9.7,where C’ is the capacitance per cm2. Determine (a)V bi (b)N d , (c) υn ,and (d)υBO.9.8 Consider a W-n-type silicon Schottky barrier at T = 300 K with N d=5×1015cm-3Use the data in Figure 9.5 to determine the barrier height. (a) determine V bi, χn, and |E max| for(i) V R = 1 V and (ii) V R=5 V . (b) Using the values of |E max| from part (a),determine the Schottky barrier lowering parameters ΔФ and χm .9.14 A Schottky diode at T= 300 K is formed with Pt on n-type silicon with a doping concentration of N d=5×1015cm-3 . the barrier height is found to beυBn = 0.89V . Determine (a) υn ,(b) V bi , (c) J sT ,and (d) V a such at J n = 5 A/cm2 . (Neglect the barrier lowering effect)9.15(a) Consider a Schottky diode at T= 300 K that is formed with tungsten on n-type silicon .Use Figure 9.5 to determine the barrier height. Assume a doping concentration of N d = 1016cm-3and assume across-sectional area A=10-4 cm2. Determine the forward-bias voltage required to induce a current of (i) 10μA ,(ii) 100μA ,and(iii) 1mA .(b) Repeat part (a) for a temperature of T = 350 K (Neglect the barrier lowering effect)9.16 An Au-n-GaAs Schottky diode at T = 300 K has a doping concentration of N d = 1016cm-3 .(a) Using Figure 9.5, to determine the barrier height. (b) Calculate the reverse-biased saturation current J sT . (c) Determine the forward-bias voltage required to induce a current density of J n = 10A/cm2. (d )what is the change in forward-bias voltage necessary to double the current density? (Neglect the Schottky barrier lowering)Chapter 1010.10Conder a MOS device with a p-type silicon substrate with N a=2×1016cm-3.The oxide thickness t ox=15nm=150Å and the equvialent oxide charge is Q,ss=7×1010cm-2.Calculate the threshold voltage for(a)an n+ polysilicon gate.(b)a p+ polysilicon gate,and(c)an aluminum gate. 10.11 Repeat Problem 10.10 for an n-type silicon substrate with a doping of N d=3×1015 cm-310.12 A 400-Å oxideis grown on p-type silicon with N a=5×1015cm-3.The flat-band voltage is -0.9V.Calculate the surface potential at the threshold inversion point as well as the threshold voltage asssuming negligible oxide charge. Also find the maximum space charge width for this device.10.13 A MOS device with an aluminum gate is fabricated on a p-type silicon substrate. The oxide thickness t ox=22nm=220Å and the trapped oxide charge is Q,ss=4×1010cm-2.The measure threshold voltage is V T=+0.45V. Determine the p-type doping concentration.10.14 Consider a MOS device with the following paremeter :p+ polysilicon gate, n-type silicon substrate, t ox=18nm=180Å,and Q,ss=4×1010cm-2. Determine the silicon doping concentration such that the threshold voltage is in the range -0.35≦V TP≦-0.25V.。

施敏-课后习题答案

施敏-课后习题答案

硅在300K时的晶格常数为5.43Å,
所以硅中最相邻原子距离=
35.432.35Å 4
-
(b)计算硅中(100),(110),(111)三平面 上每平方厘米的原子数。
-
❖ (1) 从(100)面上看,每个单胞侧面上有 1 41 2 个原子 4
❖ 所以,每平方厘米的原子数= a22(5.4 321 08)26.7 81104
(EE)cN(E)F(E)dE Ec
而导带单位体积总的电子数为
N(E)F(E)dE Ec
-
导带中电子平均动能:
(EEc)N(E)F(E)dE Ec N(E)F(E)dE Ec
=3/2kT
-
14. 一半导体的本征温度为当本征载流子浓度等 于杂质浓度时的温度。找出掺杂1015 磷原子/立 方厘米的硅样品的本征温度。
-
6. (a)计算砷化镓的密度(砷化镓的晶格常数为 5.65 Å,且砷及镓的原子量分别为69.72及 74.92克/摩尔)。
❖ 砷化镓为闪锌矿晶体结构
其中,每个单胞中有
18164 82
个As原子,和4个Ga原子
所以,每立方厘米体积中的As和Ga原子数均为
a 4 3(5.6 5 41 0 8)32.2120c 2 m 3
1
1
qp p 1 .6 1 1 0 95 110 54
-
密度 = 每立方厘米中的原子数× 原子量/阿伏伽德罗常数
2.2120 2(6.7 9 27.9 4)2 g/c3 m 6.0 2 1203
2.214.644g/cm3 60.2
5.29g/cm3
-
(b)一砷化镓样品掺杂锡。假如锡替代了晶格中镓 的位置,那么锡是施主还是受主? 为什么? 此 半 导体是n型还是p型?

半导体器件物理习题与参考文献

半导体器件物理习题与参考文献

半导体器件物理习题与参考文献第一章习题1–1.设晶格常数为a的一维晶体,导带极小值附近能量为Ec(k):?2k2?2(k?k1)2 Ec(k)??3mm?2k23?2k2?价带极大值附近的能量为:Ev(k)?式中m为电子能量,6mmk1??a?,试求:,a?禁带宽度;导带底电子的有效质量;价带顶空穴的有效质量。

1–2.在一维情况下:利用周期性边界条件证明:表示独立状态的k值数目等于晶体的原胞数;?2k2设电子能量为E?,并考虑到电子的自旋可以有两种不同的取向,试*2mn*2mn证明在单位长度的晶体中单位能量间隔的状态数为N(E)?E?12。

h1–3.设硅晶体电子中电子的纵向有效质量为mL,横向有效质量为mt 如果外加电场沿[100]方向,试分别写出在[100]和[001]方向能谷中电子的加速度;如果外加电场沿[110]方向,试求出[100]方向能谷中电子的加速度和电场之间的夹角。

?2k21–4.设导带底在布里渊中心,导带底Ec附近的电子能量可以表示为E(k)?Ec? *2mn*式中mn是电子的有效质量。

试在二维和三维两种情况下,分别求出导带附近的状态密度。

1–5.一块硅片掺磷10原子/cm。

求室温下的载流子浓度和费米能级。

1–6.若n 型半导体中Nd?ax,式中a为常数;Nd?N0e?ax推导出其中的电场。

1–7.一块硅样品的Nd?1015cm?3,?p?1?s,GL?5?10cms,计算它的电导率和准费米能级。

求产生10个空穴/cm的GL值,它的电导率和费米能级为若干?1–8.一半导体Na?1016cm?3,?n?10?s,ni?1010cm?3,以及GL?10cms,计算300K 时的准费米能级。

1–9.一块半无限的n 型硅片受到产生率为GL的均匀光照,写出此条件下的空穴连续方程。

若在x?0处表面复合速度为S,解新的连续方程证明稳定态的空穴分布可用下式表示18?3?115?315319?3?1pn(x)?pn0??pGL(1? ?pSe?x/LpLp?S?p) 1–10.于在一般的半导体中电子和空穴的迁移率不同的,所以在电子和空穴数目恰好相等的本征半导体中不显示最高的电阻率。

半导体器件物理及工艺办法施敏答案

半导体器件物理及工艺办法施敏答案

半导体器件物理及工艺办法+施敏++答案半导体器件物理及工艺是半导体科学与工程领域的重要分支,涉及半导体器件的基本原理、结构和制造工艺等方面。

本文将介绍施敏的《半导体器件物理及工艺》一书,并给出相应的答案。

一、半导体器件物理及工艺概述半导体器件物理及工艺是研究半导体器件的基本原理、结构和制造工艺的学科。

半导体器件具有高灵敏度、高可靠性、高速度等优点,在电子、通信、自动化等领域得到广泛应用。

半导体器件物理及工艺的主要研究对象包括半导体材料、半导体器件的原理和结构、制造工艺等。

二、施敏《半导体器件物理及工艺》简介施敏的《半导体器件物理及工艺》是一本经典的教材,系统地介绍了半导体器件的基本原理、结构和制造工艺。

全书分为十章,包括半导体材料、半导体器件的基本原理、PN结二极管、双极晶体管、金属氧化物半导体场效应晶体管、光电器件、半导体集成电路等。

三、施敏《半导体器件物理及工艺》答案1.什么是半导体?请列举出三种常见的半导体材料。

答:半导体是指导电性能介于导体和绝缘体之间的材料。

常见的半导体材料包括硅、锗、砷化镓等。

2.简述PN结的形成及其基本性质。

答:PN结是由P型半导体和N型半导体相互接触形成的势垒区。

PN结的基本性质包括单向导电性、电容效应和光电效应等。

3.解释双极晶体管的工作原理。

答:双极晶体管是由P型半导体和N型半导体组成的三明治结构,通过控制基极电流来控制集电极和发射极之间的电流,实现放大作用。

4.什么是金属氧化物半导体场效应晶体管?请简述其工作原理。

答:金属氧化物半导体场效应晶体管是一种电压控制型器件,通过改变栅极电压来控制源极和漏极之间的电流,实现放大作用。

其工作原理是基于MOS结构的电容效应和隧道效应。

5.光电器件的基本原理是什么?请举例说明其应用。

答:光电器件的基本原理是光电效应,即光照射在物质表面上时,物质会吸收光能并释放电子,产生电流。

光电器件的应用包括太阳能电池、光电传感器等。

6.请简述半导体的基本制备工艺流程。

半导体器件物理施敏答案

半导体器件物理施敏答案

半导体器件物理施敏答案【篇一:施敏院士北京交通大学讲学】t>——《半导体器件物理》施敏 s.m.sze,男,美国籍,1936年出生。

台湾交通大学电子工程学系毫微米元件实验室教授,美国工程院院士,台湾中研院院士,中国工程院外籍院士,三次获诺贝尔奖提名。

学历:美国史坦福大学电机系博士(1963),美国华盛顿大学电机系硕士(1960),台湾大学电机系学士(1957)。

经历:美国贝尔实验室研究(1963-1989),交通大学电子工程系教授(1990-),交通大学电子与资讯研究中心主任(1990-1996),国科会国家毫微米元件实验室主任(1998-),中山学术奖(1969),ieee j.j.ebers奖(1993),美国国家工程院院士(1995), 中国工程院外籍院士 (1998)。

现崩溃电压与能隙的关系,建立了微电子元件最高电场的指标等。

施敏院士在微电子科学技术方面的著作举世闻名,对半导体元件的发展和人才培养方面作出了重要贡献。

他的三本专著已在我国翻译出版,其中《physics of semiconductor devices》已翻译成六国文字,发行量逾百万册;他的著作广泛用作教科书与参考书。

由于他在微电子器件及在人才培养方面的杰出成就,1991年他得到了ieee 电子器件的最高荣誉奖(ebers奖),称他在电子元件领域做出了基础性及前瞻性贡献。

施敏院士多次来国内讲学,参加我国微电子器件研讨会;他对台湾微电子产业的发展,曾提出过有份量的建议。

主要论著:1. physics of semiconductor devices, 812 pages, wiley interscience, new york, 1969.2. physics of semiconductor devices, 2nd ed., 868 pages, wiley interscience, new york,1981.3. semiconductor devices: physics and technology, 523 pages, wiley, new york, 1985.4. semiconductor devices: physics and technology, 2nd ed., 564 pages, wiley, new york,2002.5. fundamentals of semiconductor fabrication, with g. may,305 pages, wiley, new york,20036. semiconductor devices: pioneering papers, 1003 pages, world scientific, singapore,1991.7. semiconductor sensors, 550 pages, wiley interscience, new york, 1994.8. ulsi technology, with c.y. chang,726 pages, mcgraw hill, new york, 1996.9. modern semiconductor device physics, 555 pages, wiley interscience, new york, 1998. 10. ulsi devices, with c.y. chang, 729 pages, wiley interscience, new york, 2000.课程内容及参考书:施敏教授此次来北京交通大学讲学的主要内容为《physics ofsemiconductor device》中的一、四、六章内容,具体内容如下:chapter 1: physics and properties of semiconductors1.1 introduction 1.2 crystal structure1.3 energy bands and energy gap1.4 carrier concentration at thermal equilibrium 1.5 carrier-transport phenomena1.6 phonon, optical, and thermal properties 1.7 heterojunctions and nanostructures 1.8 basic equations and exampleschapter 4: metal-insulator-semiconductor capacitors4.1 introduction4.2 ideal mis capacitor 4.3 silicon mos capacitorchapter 6: mosfets6.1 introduction6.2 basic device characteristics6.3 nonuniform doping and buried-channel device 6.4 device scaling and short-channel effects 6.5 mosfet structures 6.6 circuit applications6.7 nonvolatile memory devices 6.8 single-electron transistor iedm,iscc, symp. vlsi tech.等学术会议和期刊上的关于器件方面的最新文章教材:? s.m.sze, kwok k.ng《physics of semiconductordevice》,third edition参考书:? 半导体器件物理(第3版)(国外名校最新教材精选)(physics of semiconductordevices) 作者:(美国)(s.m.sze)施敏 (美国)(kwok k.ng)伍国珏译者:耿莉张瑞智施敏老师半导体器件物理课程时间安排半导体器件物理课程为期三周,每周六学时,上课时间和安排见课程表:北京交通大学联系人:李修函手机:138******** 邮件:lixiuhan@案2013~2014学年第一学期院系名称:电子信息工程学院课程名称:微电子器件基础教学时数: 48授课班级: 111092a,111092b主讲教师:徐荣辉三江学院教案编写规范教案是教师在钻研教材、了解学生、设计教学法等前期工作的基础上,经过周密策划而编制的关于课程教学活动的具体实施方案。

半导体器件物理与工艺英文版(施敏著)苏州大学出版社课后答案

半导体器件物理与工艺英文版(施敏著)苏州大学出版社课后答案

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半导体器件物理课后习题(施敏)

半导体器件物理课后习题(施敏)

因为热平衡时,样品内部没有载流子的净流动,所以有
J n漂移 J n扩散 J n 0
根据欧姆定律的微分形式
J n漂移 E ( x)
(a) q
E
J n扩散 ( x)


Dn N 0 exp( ax)
a q kT n N 0 exp( ax) q a kT n N 0 exp( ax) a kT n N D q n N D a kT q
ni (9.65 109 ) 2 4 3 n 1 . 86 3.57cm 19 15 qp p 1.6 10 5 10 350
(c) 51015硼原子/cm3、1017砷原子/cm3及1017镓 原子/cm3

答:因为镓为III族元素,最外层有3个电子;锡为IV族元 素,最外层有4个电子,所以锡替换镓后作为施主提供电 子,此时电子为多子,所以该半导体为n型。
12. 求出在300K时一非简并n型半导体导带中电
子的动能。
解:在能量为dE范围内单位体积的电子数 N(E)F(E)dE, 而导带中每个电子的动能为E-Ec 所以导带中单位体积电子总动能为
(b)
注,可用题十中的公式:

kT dN D ( x) 1 E(x) q N ( x) dx D
a kT 6 E ( x) 110 0.026 260V / cm q
12. 一个厚度为L的n型硅晶薄片被不均匀地掺杂了施主磷, 其中浓度分布给定为ND(x) = No + (NL - No) (x/L)。当样品在 热平衡状态下且不计迁移率及扩散系数随位置的变化,前后 表面间电势能差异的公式为何?对一个固定的扩散系数及迁 移率,在距前表面x的平面上的平衡电场为何?

施敏-课后习题答案

施敏-课后习题答案


1 1 (3) 从(111)面上看,每个面上有 3 3 2 个原子 6 2
所以,每平方厘米的原子数=

4 7.831014 3 ( 2a ) 2 3 (5.43108 ) 2 4
2
2.
假如我们将金刚石晶格中的原子投影到底部,原 子的高度并以晶格常数为单位表示,如下图所示。 找出图中三原子(X, Y, Z)的高度。
第三章
载流子输运现象
4. 对于以下每一个杂质浓度,求在300 K时硅 晶样品的电子及空穴浓度、迁移率及电阻率:(a) 51015硼原子/cm3

(a)300K时,杂质几乎完全电离:
p N A 5 1015 cm3
ni (9.65109 ) 2 n 1.86104 cm3 15 p 5 10
2

1 1 2.78cm qp p 1.6 1019 5 1015 450

注意:双对数坐标! 注意:如何查图?NT?
(b) 21016硼原子/cm3及1.51016砷原子/cm3
p N A N D 2 1016 1.5 1016 5 1015 cm3
20. 对一掺杂1016 cm-3磷施主原子,且施主能级ED= 0.045 eV的n型硅样品而言,找出在77K时中性施主 浓度对电离施主浓度的比例;此时费米能级低于导 带底部0.0459eV(电离施主的表示式可见问题19)。
题19公式:
n = N D [1 - F (E D ) ] =
1 e E F -E D / kT

(1) 低温情况(77K) 由于低温时,热能不 足以电离施主杂质,大部 分电子仍留在施主能级, 从而使费米能级很接近施 主能级,并且在施主能级 之上。(此时,本征载流 子浓度远小于施主浓度)

(完整版)半导体器件物理试题库.docx

(完整版)半导体器件物理试题库.docx

西安邮电大学微电子学系商世广半导体器件试题库常用单位:在室温( T = 300K )时,硅本征载流子的浓度为n i = 1.510×10/cm3电荷的电量 q= 1.6 ×10-19Cn2/V sp2/V s μ=1350 cmμ=500 cmε0×10-12F/m=8.854一、半导体物理基础部分(一)名词解释题杂质补偿:半导体内同时含有施主杂质和受主杂质时,施主和受主在导电性能上有互相抵消的作用,通常称为杂质的补偿作用。

非平衡载流子:半导体处于非平衡态时,附加的产生率使载流子浓度超过热平衡载流子浓度,额外产生的这部分载流子就是非平衡载流子。

迁移率:载流子在单位外电场作用下运动能力的强弱标志,即单位电场下的漂移速度。

晶向:晶面:(二)填空题1.根据半导体材料内部原子排列的有序程度,可将固体材料分为、多晶和三种。

2.根据杂质原子在半导体晶格中所处位置,可分为杂质和杂质两种。

3.点缺陷主要分为、和反肖特基缺陷。

4.线缺陷,也称位错,包括、两种。

5.根据能带理论,当半导体获得电子时,能带向弯曲,获得空穴时,能带向弯曲。

6.能向半导体基体提供电子的杂质称为杂质;能向半导体基体提供空穴的杂质称为杂质。

7.对于 N 型半导体,根据导带低E C和 E F的相对位置,半导体可分为、弱简并和三种。

8.载流子产生定向运动形成电流的两大动力是、。

9.在 Si-SiO 2系统中,存在、固定电荷、和辐射电离缺陷 4 种基本形式的电荷或能态。

10.对于N 型半导体,当掺杂浓度提高时,费米能级分别向移动;对于P 型半导体,当温度升高时,费米能级向移动。

(三)简答题1.什么是有效质量,引入有效质量的意义何在?有效质量与惯性质量的区别是什么?2.说明元素半导体Si 、 Ge中主要掺杂杂质及其作用?3.说明费米分布函数和玻耳兹曼分布函数的实用范围?4.什么是杂质的补偿,补偿的意义是什么?(四)问答题1.说明为什么不同的半导体材料制成的半导体器件或集成电路其最高工作温度各不相同?要获得在较高温度下能够正常工作的半导体器件的主要途径是什么?(五)计算题1.金刚石结构晶胞的晶格常数为a,计算晶面( 100)、( 110)的面间距和原子面密度。

半导体器件物理施敏

半导体器件物理施敏
• 是快速器件和光电器件的关键构成要素。 一种由p型和 n型半导体接触形成的结,是大局部半导体器件的关键根底构造;
由两种不同材料的半导体接触形成的结; 一种由p型和 n型半导体接触形成的结,是大局部半导体器件的关键根底构造; 结合三个p-n结就可以形成p-n-p-n构造,叫做可控硅器件。 〔三〕异质结(heterojunction) 金属-氧化物界面和氧化物-半导体界面结合的构造; 可以用来做整流接触,具有单向导电性; 结合三个p-n结就可以形成p-n-p-n构造,叫做可控硅器件。 加上另一个p型半导体就可以形成一个p-n-p双极型晶体管; 〔二〕p-n结(junction) 〔一〕金属半导体接触 〔三〕异质结(heterojunction)
半导体器件物理施敏
〔一〕金属半导体接触
• 可以用来做整流接触,具有单向导 电性;
• 也可以用来做欧姆接触,电流双向 通过。
〔二〕p-n结(junction)
• 一种由p型和 n型半导体接触形成的结,是 〔一〕金属半导体接触
〔二〕p-n结(junction) 用MOS构造当作栅极,再用两个p-n结分别当作漏极和源极,就可以制作出金氧半场效应晶体管(MOSFET);
也可以用来做欧姆接触,电流双向通过。 〔三〕异质结(heterojunction) 〔二〕p-n结(junction) 结合三个p-n结就可以形成p-n-p-n构造,叫做可控硅器件。 加上另一个p型半导体就可以形成一个p-n-p双极型晶体管; 金属-氧化物界面和氧化物-半导体界面结合的构造; 由两种不同材料的半导体接触形成的结;
是快速器件和光电器件的关键构成要素。 一种由p型和 n型半导体接触形成的结,是大局部半导体器件的关键根底构造; 结合三个p-n结就可以形成p-n-p-n构造,叫做可控硅器件。 金属-氧化物界面和氧化物-半导体界面结合的构造;

半导体物理章习题全英文版

半导体物理章习题全英文版

212k mE =)(1222k V E mE -= )(2232k V E mE -= Chapter 11.22 Calculate the density of valence electrons in silicon.1.23 The structure of GaAs is the zincblende lattice. The lattice constant is 5.65 ︒A . Calculate thedensity of valence electrons in GaAs.1.24 (a) If 17105⨯ phosphorus atoms per 3cm are add to silicon as a substitutional impurity, determine the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.(b)Repeat part (a) for 15102⨯ boron atoms per 3cm added to silicon.1.25 (a)Assume that 16102⨯-3cm of boron atoms are distributed homogeneously throughout single crystal silicon. What is the fraction by weight of boron in the crystal? (b)If phosphorus atoms, at a concentration of 1810-3cm , are added to the material in part (a), determine the fraction by weight of phosphorus.1.26 If 16102⨯-3cm boron atoms are added to silicon as a substitutional impurity and are distributed uniformly throughout the semiconductor, determine the distance between boron atoms in terms of the silicon lattice constant.(Assume the boron atoms are distributed in a rectangular or cubic array.)1.27 Repeat Problem 1.26 for 15104⨯-3cm phosphorus atoms being added to silicon.Chapter 22.32Consider a proton in a one-dimensional infinite potential well shown in Figure2.6.(a)Derive the expression for the allowed energy states of the proton.(b)Calculate the energy state for (i)a=4︒A ,and (ii)a=0.5cm.2.33 For the step potential function shown in Figure P2.33, assume that 0V E > and that particles are incident from the +x direction traveling in the -x direction.(a) Write the wave solutions for each region.(b) Derive expressions for the transmission and reflection coefficients.2.38 An electron with energy E is incident on a rectangular potential barrier as shown in Figure2.9. The potential barrier is of width a and height E V ≥0.(a)Write the form of the wave function in each of the three regions. (b)For this geometry ,determine what coefficient in the wave function solutions is zero.(c)Derive the expression for the transmission coefficient for the electron(tunneling probability ).(d)Sketch the wave function for the electron in each region.2.39 A potential function is shown in Figure P2.39 with incident particles coming from -∞ with a total energy 2V E >.The constants k are defined asAssume a special case for which πn a 2k 2=,n=1,2,3... Derive the expression, in terms of the constants ,,,21k k and 3k , for the transmission coefficient .The transmission coefficient is defined as the ratio of the flux of particles in region Ⅲ to the incident flux in region Ⅰ.2.40 Consider the one-dimensional potential function shown in FigureP2.40.Assume the total energy of an electron is 0V E <.(a )Write the wave solutions that apply in each region.(b)Write the set of equations that result from applying the boundary conditions.(c)Show explicitly why not , the energy levels of the electron are quantized.Chapter 33.25 Derive the density of status function for a one-dimension electron gas in GaAs(m n *=0.067m 0).Note that the kinetic energy may be written as E=(±p)2/2m n *,which means that there are two momentum states for each energy level.3.26 (a)Determine the total number (#/cm 3) of energy states in silicon between C E and at (i)T=300K and (ii)T=400K.(b) Repeat part (a) for GaAs.3.27 (a)Determine the total number (#/cm 3) of energy states in silicon between v E and kT E v 3- at (i) T=300K and (ii)T=400K.(b) Repeat part (a) for GaAs.3.28 (a)Plot the density of states in the conduction band of silicon over the range . (b) Repeat part (a) for the density of states in the valence band over the range v v E E eV E <<-4.0.3.29 (a)For silicon,find the ratio of the density of states in the conduction band at E=Ec+KT to the density of states in the valence band at E=Ev-KT. (b)Repeate part (a) for GaAs.Chapter 44.49 Consider silicon at T =300 K with donor concentrations of N d =1014,1015,1016,and1017,cm -3. Assume N a =0.(a )Calculate the position of the Fermi energy level with respect to the conduction band for these donor concentrations.(b )Determine the position of the Fermi energy level with respect to the intrinsic Fermi energy level for the donor concentrations given in part (a ).4.52 Consider GaAs at T=300K with N d =0. (a)Plot the position of the Fermi energy level with respect to the intrinsic Fermi energy level as a function of the acceptor impurity concentration over the range of 1014≤N a ≤1017 cm -3. (b)Plot the position of the Fermi energy level with respect to the valence-band energy over the same acceptor impurity concentration as given in part(a).4.53 For a particular semiconductor , eV E g 50.1= ,**10n p m m =, T=300K,and 35101-⨯=cm n i .(a) Determine the position of the intrinsic Fermi energy level with respect to the center of the bandgap.(b) Impurity atoms are added so that the Fermi energy level is 0.45eV below the center of the bandgap. (i) Are acceptor or donor atoms added? (ii) What's the concentration of impurity atoms added?4.55 (a) Silicon at T=300K is doped with donor impurity atoms at a concentration of 315106-⨯=cm N d . (i) Determine E c - E F . (ii) Calculate the concentration of additional donor impurity atoms that must be added to move the Fermi energy level a distance KT closer to the conduction band edge. (b) Repeat Part (a) for GaAs if the original donor impurity concentration is 315101-⨯=cm N d .4.56 (a) Determine the position of the Fermi energy level with respect to the intrinsic Fermi level in silicon at T=300K that is doped with boron atoms at a concentration of 316102-⨯=cm N d .(b) Repeat Part (a) if the silicon is doped with phosphorus atoms at a concentration of N d =2x1016cm -3 . (c) Calculate n 0 and p 0 in parts (a) and (b).Chapter 55.8 (a) A silicon semiconductor resistor is in the shape of a rectangular bar with a cross-sectional area of 8.5×10-4cm 2, a length of 0.075 cm, and is doped with a concentration of 2×1016cm -3, boron atoms. Let T=300K. A bias of 2 volts is applied across the length of the silicon device. Calculate the current in the resistor. (b) Repeat part (a) if the length is increased by a factor of three. (c) Determine the average drift velocity of holes in the parts (a) and (b).5.18 An n-type silicon resistor has a leng th L=150μm, which W=7.5μm, and thickness T=1μm. A voltage of 2 V is applied across the length of the resistor. The donor impurity concentration varies linearly through the thickness of the resistor with N d =2×1016cm -3at the top surface and N d =2×1015cm -3at the bottom surface. Assume an average carrier mobility of n μ=750cm 2/V -s . (a)s V cm N d n⋅⨯+=-/)1051(135032/116μ)()(2min p n pn i μμμμσσ+=What is the electric field in the resistor? (b) Determine the average conductivity of the silicon. (c) Calculate the current in the resistor. (d) Determine the current density near the top surface and the current density near the bottom surface.5.22 A semiconductor material has electron and hole mobilitiesμn andμp , respectively. When the conductivity is considered as a function of the hole concentration p 0, (a) show that the minimum value of conductivity, σmin , can be written aswhere i σis the intrinsic conductivity, and (b) show that the corresponding hole concentration is2/10)/(p n n p μμ=.5.23 Consider three samples of silicon at T=300K. The n-type sample is doped with arsenic atoms to a concentration of 316105-⨯=cm N d , The p-type sample is doped with boron atoms to a concentration of 316102-⨯=cm N a .The compensated sample is doped with both the donors and acceptors described in the n-type and p-type sample. (a) Find the equilibrium electron and hole concentrations in each sample. (b) determine the majority carrier mobility in each sample. (c) calculate the conductivity of each sample. (d) and determine the electric field required in each sample to induce a drift current density of J=120A/cm 2 .5.28 (a) Assume that the electron mobility in an n-typesemiconductor is given by Where N d is the donor concentration in cm -3. Assuming complete ionization, plot the conductivity as a function of N d over the range 318151010-≤≤cm N d . (b) Compare the results of part (a) tothat if the mobility were assumed to be a constant equal to 1350 s V cm ⋅-/3. (c) If an electricfield of E=10V/cm is applied to the semiconductor. Plot the electron drift current density of parts (a) and (b).5.36 The total current in a semiconductor is constant and equal to J=-10A/cm -3. The total current is composed of a hole drift current. Assume that the hole concentration is a constant and equal to 1016 cm -3 and assume that electron concentration is given by 3/15102)(--⨯=cm e x n L x where L=15m μ. The electron diffusion coefficient is D n =27 cm 2/s and the hole mobility is s V cm p ⋅=/4202μ . Calculate (a) the electron diffusion current density for x>0, (b) the hole drift current density for, and (c) the required electric field for x>0 .Chapter 66.3 An n-type silicon sample contains a donor concentration of 31610-=cm N d . The minority carrier hole lifetime is found to be . (a) What is the lifetime of the majority carrier electrons? (b) Determine the thermal-equilibrium generation rate for electrons and holes in this material. (c) Determine the thermal-equilibrium recombination rate for electrons and holes in this material.6.5 Derive Equation (6.27) from Equations (6.18) and (6.20).6.6 Consider a one-dimensional hole flux as shown in Figure 6.4. If the generation rate of holes in this differential volume is 132010--⋅=s cm g p and the recombination rate is 1319102--⋅⨯s cm ,what must be the gradient in the particle current density to maintain a steady-state hole concentration?6.7 Repeat Problem 6.6 if the generation rate becomes zero.6.17 (a)Consider a silicon sample at T=300K doped with 1016cm -3 donor atoms. Let τp0=5x10-7s.Alight source turns on at t=0 producing excess carriers with a uniform generation rate of g'=5x1020cm -3s -1.At t=5x10-7s,the light source turns off.(i)Derive the expression(s) for the excess carrier concentration as a function of time over the range 0≤t≤∞.(ii) What is the value of the excess concentration when the light source turns off. (b) Repeat Part (a) for the case when the light source turns off at t=2x10-6s. (c) Sketch the excess minority carrier concentrations versus time for parts (a) and (b).6.18 A semiconductor is uniformly doped with 1710-3cm acceptor atoms and has the following properties :s cm D n /272=,s cm D p /122=,s n 170105-⨯=τ,and s p 17010-=τ.An external source has been turned on for 0<t producing a uniform concentration of excess carriers at a generation rate of 132110'--=s cm g .The source turns off at time t=0 and back on at time s t 6102-⨯=(a)Derive the expressions for the excess carrier concentration as a function of time for ∞≤≤t 0.(b)Determine the value of excess carrier concentration at (i)t=0 (ii)s t 6102-⨯=,and (iii)∞=t (c)Plot the excess carrier concentration as a function of time.Chapter 77.4 An abrupt silicon pn junction at zero bias dapant concentrations of N a =1017cm -3 and N d =5×1015cm -3. T=300K. ﹙a ﹚Calculate the Fermi level on each side of the junction with respect to the intrinsic Fermi level. ﹙b ﹚ Sketch the equilibrium energy band diagram for the junction and determine V bi from the diagram and the results of part ﹙a ﹚. ﹙c ﹚Calculate V bi using Equation ﹙7.10﹚, and compare the results to part ﹙b ﹚. ﹙d ﹚Determine xn, xp, and the peak electric field for this junction.7.6 A Silicon pn junction in thermal equilibrium at T=300K is doped such that E F -E Fi =0.365eV in the n region and E Fi -E F =0.330eV in the p region ﹙a ﹚Sketch the energy-band diagram for the pn junction. ﹙b ﹚Find the impurity doping concentration in each region. ﹙c ﹚Determine V bi .7.18 An ideal one-sided silicon p +n junction at T=300K is uniformly doped on both sides of the metallurgical junction. It is found that the doping relation is N a =80N d and the build-in potential barrier is V bi =0.740V . A reverse-biased voltage of VR=10V is applied. Determine ﹙a ﹚N a , N d ;﹙b ﹚x p , x n ;﹙c ﹚|E max |; and ﹙d ﹚C’j .7.19 A silicon n +p junction is biased at V R =5V. ﹙a ﹚Determine the change in built-in potential barrier if the doping concentration in the p increases by a factor of 3.﹙b ﹚Determine the ratio of junction capacitance when the acceptor doping is 3N a compared to that when the acceptor doping is N a .﹙c ﹚Why does the junction capacitance increase when the doping concentration increase ?7.20 ﹙a ﹚The peak electric field in a reverse-biased silicon pn junction is |E max |=3×105 V/cm. The doping concentrations are N d =4×1015cm -3 and N a =4×1017cm -3 .Find the magnitude of the reverse-biased voltage.﹙b ﹚Repeat part ﹙a ﹚ for N d =4×1016cm -3 and N a =4×1017cm -3﹙c ﹚Repeat part ﹙a ﹚N d =N a =4×1017cm -3Chapter 88.2 A silicon pn junction has impurity doping concentrations of N d =2×1015cm -3 and N a =8×1015cm -3. Determine the minority carrier concentrations at the edges of the space charge region for ﹙a ﹚V a =0.45 V ﹙b ﹚V a =0.55 V , and ﹙c ﹚V a =-0.55V .8.3 The doping concentrations in a G a A s pn junction are N d =1016cm -3 and N a =4×1016cm -3.Find the minority carrier concentrations at the edges of the space charge region for ﹙a ﹚V a =0.90 V ﹙b ﹚V a =1.10 V, and﹙c﹚V a =-0.95V.8.5 Consider a G a A s pn junction with doping concentrations N a=5×1016cm-3 and N d=1016cm-3. The junction cross-sectional area is A=10-3cm2and the applied forward-bias voltage is Va=1.10V. Calculate the﹙a﹚minority electron diffusion current at the space charge region, ﹙b﹚minority hole diffusion current at the edge of the space charge region. and﹙c﹚total current in the pn junction diode.8.6An n+p silicon dione at T=300K has the following parameters N d=1016cm-3, D n=25cm2/s,D p=10cm2/s , N d=1016cm-3, Τn0=Τp0=1 μs, and A=10-4cm2. Determine the diode current for ﹙a﹚a forward-bias voltage of 0.5V and ﹙b﹚a reverse-biased voltage of 0.5V.8.7 An ideal germanium pn junction diode has the following parameters: N a=4×1015cm-3, N a=2×1017cm-3, D p=48cm2/s, D n=90cm2/s, Τp0=Τn0=2×10-6s, and A=10-4cm2 . Determine the diode current for ﹙a﹚a forward-bias voltage of 0.25V and ﹙b﹚a reverse-biased voltage of 0.25V.Chapter 99.7 A Schottky diode with n-type GaAs at T = 300 K yields the 1/C’2 versus V R plot shown in Figure P9.7,where C’ is the capacitance per cm2. Determine (a)V bi (b)N d , (c) υn ,and (d)υBO.9.8 Consider a W-n-type silicon Schottky barrier at T = 300 K with N d=5×1015cm-3Use the data in Figure 9.5 to determine the barrier height. (a) determine V bi, χn, and |E max| for(i) V R = 1 V and (ii) V R=5 V . (b) Using the values of |E max| from part (a),determine the Schottky barrier lowering parameters ΔФ and χm .9.14 A Schottky diode at T= 300 K is formed with Pt on n-type silicon with a doping concentration of N d=5×1015cm-3 . the barrier height is found to beυBn = 0.89V . Determine (a) υn ,(b) V bi , (c) J sT ,and (d) V a such at J n = 5 A/cm2 . (Neglect the barrier lowering effect)9.15(a) Consider a Schottky diode at T= 300 K that is formed with tungsten on n-type silicon .Use Figure 9.5 to determine the barrier height. Assume a doping concentration of N d = 1016cm-3and assume across-sectional area A=10-4 cm2. Determine the forward-bias voltage required to induce a current of (i) 10μA ,(ii) 100μA ,and(iii) 1mA .(b) Repeat part (a) for a temperature of T = 350 K (Neglect the barrier lowering effect)9.16 An Au-n-GaAs Schottky diode at T = 300 K has a doping concentration of N d = 1016cm-3 .(a) Using Figure 9.5, to determine the barrier height. (b) Calculate the reverse-biased saturation current J sT . (c) Determine the forward-bias voltage required to induce a current density of J n = 10A/cm2. (d )what is the change in forward-bias voltage necessary to double the current density? (Neglect the Schottky barrier lowering)Chapter 1010.10Conder a MOS device with a p-type silicon substrate with N a=2×1016cm-3.The oxide thickness t ox=15nm=150Å and the equvialent oxide charge is Q,ss=7×1010cm-2.Calculate the threshold voltage for(a)an n+ polysilicon gate.(b)a p+ polysilicon gate,and(c)an aluminum gate. 10.11 Repeat Problem 10.10 for an n-type silicon substrate with a doping of N d=3×1015 cm-310.12 A 400-Å oxideis grown on p-type silicon with N a=5×1015cm-3.The flat-band voltage is -0.9V.Calculate the surface potential at the threshold inversion point as well as the threshold voltage asssuming negligible oxide charge. Also find the maximum space charge width for this device.10.13 A MOS device with an aluminum gate is fabricated on a p-type silicon substrate. The oxide thickness t ox=22nm=220Å and the trapped oxide charge is Q,ss=4×1010cm-2.The measure threshold voltage is V T=+0.45V. Determine the p-type doping concentration.10.14 Consider a MOS device with the following paremeter :p+ polysilicon gate, n-type silicon substrate, t ox=18nm=180Å,and Q,ss=4×1010cm-2. Determine the silicon doping concentration such that the threshold voltage is in the range -0.35≦V TP≦-0.25V.。

半导体器件物理习题

半导体器件物理习题

半导体器件物理习题第一章1 设晶体的某晶面与三个直角坐标轴的截距分别为2a,3a,4a,其中a 为晶格常数,求该晶面的密勒指数。

2 试推导价带中的有效态密度公式232]2[2h kT m N p V π=。

提示:价带中的一个状态被空穴占据的几率为1-F (E ),其中F (E )为导带中电子占据能量E 的几率函数。

3 室温300K 下,硅的价带有效态密度为1.04×1019cm -3,砷化镓的为7×1018cm -3,求相应的空穴有效质量,并与自由电子的质量相比较。

4 计算在液氮温度下77K 、室温300K 及100℃下硅中Ei 的位置,设m p =0.5m 0,m n =0.3m 0。

并说明Ei 位于禁带中央的假设是否合理。

5 求300K 时下列两种情况下硅的电子和空穴浓度及费米能级:(a) 掺1×1016原子/cm 3的硼,(b) 掺3×1016原子/cm 3的硼及2.9×1016原子/cm 3的砷。

6 假定满足杂质完全电离的条件,求出在掺磷浓度分别为1015 、1017、1019原子/cm 3时,硅在室温下的费米能级。

根据计算结果得到的费米能级验证这三种情况下杂质完全电离的假设是否成立。

7 计算300K 时,迁移率为1000cm 2/Vs 的电子平均自由时间和平均自由程,设m n =0.26m 0。

8 在均匀n 型半导体样品的某一点注入少数载流子(空穴),样品的两端加50V/cm 的电场,电场使少数载流子在100μs 中运动1cm ,求少子的漂移速度和扩散系数。

9 求本征硅及本征砷化镓在300K 时的电阻率。

10 一不知掺杂浓度的样品硅,用霍耳测量得到下述的数据:W=0.05cm ,A=1.6×10-3cm 2,I=2.5mA ,磁场为30nT(1T=104Wb/cm 2)。

若测得霍耳电压为10mV ,求该半导体样品的霍耳系数、导电类型、多数载流子浓度、电阻率和迁移率。

半导体器件物理习题

半导体器件物理习题
| Q (max) | exdT N d 4 s eN d fn
' SD
Nd fn Vt ln ni s 2 fn
11.3、4、6、8
Eg ms ( fp ) 2q
' Al '
Eg ms ' ( ' fp ) 2q Eg ms ' Eg ( ' fp ) 2q
• MOSFET击穿特性(栅调制击穿、沟道雪崩倍增、寄生
npn-BJT击穿、源漏穿通 )
• 等比例缩小(Scaling down)规则:CE律
复习
• MOSFET短沟热载流子效应*? • MOSFET噪声特性*? • 薄膜场效应晶体管TFT*?
P型衬底MOS电容器CV特性
11.3、4、6、8
Z 1 2 nCox [(VGS VT )VDS ( sat ) VDS ( sat )] L 2 1
12.11、12.12、12.33、12.34
12.33、
VTN (| Q (max) | Q )(
' SD ' ss
Eg ms ( fp ) 2q
ox
tox
12.34、
gm
I D VGS
VDS

W nCox (VGS VT ) L
12.1、12.9
12.1
计算同样VGS值下的芯片总功耗 P=VDD*ID*106
12.1、12.9
饱和区(a)
I D (VGS VT )
Z nCox 2L
1/ 3
( b)
Cox (VGS VT ) Z ID 0 1/ 6 VGS 2 L EC ox

半导体器件物理课后习题(施敏)

半导体器件物理课后习题(施敏)

20. 对一掺杂 16 cm-3磷施主原子,且施主能级 D= 对一掺杂10 磷施主原子,且施主能级E 0.045 eV的n型硅样品而言,找出在 型硅样品而言, 的 型硅样品而言 找出在77K时中性施主 时中性施主 浓度对电离施主浓度的比例; 浓度对电离施主浓度的比例;此时费米能级低于导 带底部0.0459eV(电离施主的表示式可见问题 )。 带底部 (电离施主的表示式可见问题19)。
半导体器件物理 习题讲解
第二章
热平衡时的能带和载流子浓度
1. (a)硅中两最邻近原子的距离是多少? 硅中两最邻近原子的距离是多少? 硅中两最邻近原子的距离是多少
解答: (a) 硅的晶体结构是金刚石 晶格结构,这种结构也 属于面心立方晶体家族, 而且可被视为两个相互 套构的面心立方副晶格, 此两个副晶格偏移的距 离为立方体体对角线的 1/4(a /4的长度) 3
解:根据题意有
ni =
N c N v exp(-E g /2kT), N D = 1015 cm−3
3 2
本征温度时,Ni=ND
将NV ≡2(2πmpkT/h2)3/2和N C ≡ 12( 2πm n kT / h 2 )
1 2
代入上式并化简,得
− Eg 3 2πkT 3 2 ni = 24× (mpmn ) × ( 2 ) × exp( ) h 2kT
因为霍耳电压为正的,所以该样品为p型半导体(空穴导电) 多子浓度:
霍耳系数:
IBZW 2.5×10−3 × 30×10−4 × 0.05 p= = ≈ 1.46×1017 cm−3 qVH A 1.6×10−19 ×10×10−3 ×1.6×10−3
1 1 RH = = ≈ 42.8cm 3 / C qp 1.6 × 10−19 × 1.46 × 1017

(施敏)半导体器件物理(详尽版)

(施敏)半导体器件物理(详尽版)

江西科技师范大学
半导体器件物理
光照与半导体
光照对半导体材料的导电能力也有很大的影响。
例如,硫化镉(CdS)薄膜的暗电阻为几十兆欧, 然而受光照后,电阻降为几十千欧,阻值在受光照以 后改变了几百倍。
光敏电阻 成为自动化控制中的一个重要元件。
江西科技师范大学
半导体器件物理
其他因素与半导体
除温度、杂质、光照外,电场、磁场及其他

江西科技师范大学
半导体器件物理 例1-2
硅(Si)在300K时的晶格常数为5.43Å。请计算出每立方厘米体 积中硅原子数及常温下的硅原子密度。(硅的摩尔质量为 28.09g/mol)

江西科技师范大学
半导体器件物理
晶体的各向异性
沿晶格的不同方向,原子排列的周期 性和疏密程度不尽相同,由此导致晶体在 不同方向的物理特性也不同 。
图1-7 一定温度下半导体的能带示意图 江西科技师范大学
半导体器件物理
注意三个“准”
• 准连续 • 准粒子 • 准自由
江西科技师范大学
半导体器件物理
练习
• 整理空带、满带、半满带、价带、导带、 禁带、导带底、价带顶、禁带宽度的概 念。
• 简述空穴的概念。
江西科技师范大学
半导体器件物理 1.4 半导体中的杂质和缺陷 理想的半导体晶体
● 空带
没有电子填充的一系列准 连续的能量状态 空带也不导电
图1-5 金刚石结构价电子能带图(绝对零度) 江西科技师范大学
半导体器件物理
●导带
有电子能够参与导电的能带, 但半导体材料价电子形成的高 能级能带通常称为导带。
●价带
由价电子形成的能带,但半导体 材料价电子形成的低能级能带通 常称为价带。
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施敏 半导体器件物理英文版 第一章习题1. (a )求用完全相同的硬球填满金刚石晶格常规单位元胞的最大体积分数。

(b )求硅中(111)平面内在300K 温度下的每平方厘米的原子数。

2. 计算四面体的键角,即,四个键的任意一对键对之间的夹角。

(提示:绘出四个等长度的向量作为键。

四个向量和必须等于多少?沿这些向量之一的方向取这些向量的合成。

)3. 对于面心立方,常规的晶胞体积是a 3,求具有三个基矢:(0,0,0→a/2,0,a/2),(0,0,0→a/2,a/2,0),和(0,0,0→0,a/2,a/2)的fcc 元胞的体积。

4. (a )推导金刚石晶格的键长d 以晶格常数a 的表达式。

(b )在硅晶体中,如果与某平面沿三个笛卡尔坐标的截距是10.86A ,16.29A ,和21.72A ,求该平面的密勒指数。

5. 指出(a )倒晶格的每一个矢量与正晶格的一组平面正交,以及(b )倒晶格的单位晶胞的体积反比于正晶格单位晶胞的体积。

6. 指出具有晶格常数a 的体心立方(bcc )的倒晶格是具有立方晶格边为4π/a的面心立方(fcc )晶格。

[提示:用bcc 矢量组的对称性:)(2x z y a a -+=,)(2y x z a b -+=,)(2z y x a c -+= 这里a 是常规元胞的晶格常数,而x ,y ,z 是fcc 笛卡尔坐标的单位矢量:)(2z y a a +=,)(2x z a b +=,)(2y x a c +=。

] 7. 靠近导带最小值处的能量可表达为.2*2*2*22⎪⎪⎭⎫ ⎝⎛++=z z y y xx m k m k m k E 在Si 中沿[100]有6个雪茄形状的极小值。

如果能量椭球轴的比例为5:1是常数,求纵向有效质量m*l 与横向有效质量m*t 的比值。

8. 在半导体的导带中,有一个较低的能谷在布里渊区的中心,和6个较高的能谷在沿[100] 布里渊区的边界,如果对于较低能谷的有效质量是0.1m0而对于较高能谷的有效质量是1.0m0,求较高能谷对较低能谷态密度的比值。

9. 推导由式(14)给出的导带中的态密度表达式。

(提示:驻波波长λ与半导体的长度L 相关,按L/λ=nx ,这里nx 是某整数。

按德布罗依假设波长可表达为λ = h/px ,考虑三维边长L 的立方体]10. 计算n-型非简并半导带内电子的平均动能。

态密度由式(14)给出。

11. 说明.e x p 211-+⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛+=kT E E N N D F D D [提示:占据概率是 1exp 1)(-⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛-+=kT E E gh E F F 这里h 是物理上占据能级E 的电子数,而g 是能被能级接受的电子数,也称之为施主杂质能级的基态简并(g=2)。

]12. 如果某硅样品掺杂有1016/cm 3的磷,求77K 温度时的离化施主密度。

假设磷施主杂质的电离能和电子的有效质量与温度无关。

(提示:首先选择用以计算费米能级的+D N 值,然后求出相应的+D N ,如果不一致,则选择另一个+D N 值,重复该过程直到获得一致的+D N 值。

)13. 用图解法确定杂质浓度为1015/cm 3掺硼硅样品在300K 温度时的费米能级(注意ni=9.65×109cm -3)。

14. 费米-狄拉克分布函数是]/)exp[(11)(kT E E E F F -+=。

F(E)对于能量的微分是F ’(E)。

求F ’(E)的宽度,也即是⎥⎦⎤⎢⎣⎡-)'21()'(2max max F at E atF E 这里max 'F 是)('E F 的最大值。

15. 求硅样品在300K 时费米能级相对于导带底的位置(Ec-Ef ),其掺杂有2×1010cm -3完全离化的施主。

16. 硅中的金在带隙中有两个能级:EC-EF=0.54 eV ,ED-EV=0.29 eV ,假设第三个能级ED-EV=0.35 eV 是不活跃的。

(a )在高浓度掺杂硼原子的硅中金能级电荷状态将会是怎样的?(b )金对电子和空穴浓度的效应是什么?17. 由图13,估计和确定何种杂质原子被用来掺杂硅样品?18. 对于n-型掺杂有2.86×1016cm -3的磷原子的硅样品,求在300K (EC-ED=0.045eV )中性施主原子对离化施主的比例。

19. (a )假设硅中迁移率比μn/μp ≡b 是与杂质浓度无关的常数,借助于300K 时的本征电阻率ρi 求最大电阻率ρm 。

如果b=3且本征硅的空穴迁移率是450cm 2/V-s ,计算ρi 和ρm 。

(b )求GaAs 在300K 时具有5×1015锌原子/cm 3,1017硫原子/cm 3,和1017碳原子/cm 3时的电子和空穴的浓度、迁移率和电阻率。

20. 伽马函数定义为10()exp().n n x x dx ∞-Γ=-⎰(a )求Г(1/2),以及(b )证明Г(n)=(n-1)Г(n-1).21. 考虑T=300K 的补偿型n-型硅,具有电导率σ=16 S/cm ,且受主掺杂浓度为1017cm -3。

确定其施主浓度和电子的迁移率。

(补偿半导体是在相同的区域内既含有施主又含有受主杂质原子的)。

22. 求在300K 掺杂有1.0×1014cm -3的磷原子、8.5×1012cm -3的砷原子和1.2×1013cm -3的硼原子硅样品的电阻率。

假设杂质完全离化且迁移率是 μn=1500 cm 2/V-s,μp=500cm 2,与杂质浓度无关。

23. 电阻率为1.0 Ω-cm 的半导体且霍尔系数是-1250 cm 2/库伦。

假设现在仅有一种载流子且平均自由时间正比于载流子能量也即τ∝E ,求载流子密度和迁移率。

24. 推导如式(92)给出的非直接复合的复合率。

(提示:参考图25b ,电子由复合中心的俘获率正比于Re ∝nNt (1-F ),其中n 是导带中电子的密度,Nt 是复合中心的密度,F 是费米分布,且Nt/(1-F)是没有占据对电子俘获有效的复合中心的密度。

)25.由式92给出的复合速率,在低注入条件下,U 可以表示为(pn-pn0)/τr,这里τr 是复合寿命,如果σn=σp=σ0,nn0=1015cm -3,且τr0≡(νth σ0N t )-1,求复合寿命τr 为2τr0时的(Et-Ei )值。

26. 对于电子与空穴具有相同俘获截面的单能级复合,求在载流子完全耗尽的条件下,每单位体积每个产生率下的俘获中心数。

假设俘获中心的位置在带隙的中间,σ=2ⅹ10-16cm 2,及νth =107 cm/s.27. 在半导体某个区域,载流子完全耗尽(即,n<<ni ,p<<ni ),电子-空穴对由中心的子和空穴而产生的。

推导发生在这些发射过程的平均时间,(假设σn=σp=σ);并求出对于σ=2×10-16cm2,νth=107cm/s,且Et=Ei(T=300K)的平均时间。

28. 对于单一能级的复合过程,求发生在Si半导体某区域中各个复合过程的平均时间。

Si样品参数:n=p=1013cm-3,σn=σp=2×10-16cm2,νth=107cm/s,Nt=1016cm-3,以及(Et-Ei)=5kT。

29. (a)推导式(123)。

(提示:假设原子为直线链且原子间的相互作用仅由最邻近的原子产生。

偶数原子质量为m1且奇数原子的质量为m2);(b)对于Si晶体有m1=m2以及√(αf/m1)=7.63×1012Hz,求在布里渊区边界处的光学波声子的能量。

力学常数是αf。

30.假设Ga0.5In0.5As在500℃与InP衬底晶格匹配。

当样品冷却到27℃时,求它们层间的晶格失配程度。

31. 求异质结Al0.4Ga0.6As/GaAs导带的中断对Al0.4Ga0.6As带隙之比。

32.在Haynes-Shockley的实验中,少数载流子浓度在t1=25us和t2=100us时最大幅度相差10倍因子。

求少数载流子的寿命。

33.根据在Haynes-Shockley实验中的描述载流子漂移和扩散的表达式,求载流子在t=1s时的脉冲半宽度。

假设扩散系数是10cm2/s。

34.过剩载流子被注入在具有长度W=0.05mm的n-型硅薄片的表面(x=0)且在反向表面Pn(W)=pn0提取空穴.如果载流子的寿命是50us,求由扩散到达反面的注入电流份额。

35.某n-型GaAs样品ND=5×1015cm-3被照明,均匀吸收的光产生1017电子-空穴对/cm3-s。

寿命τp是10-7s,Lp=1.93×10-3cm,表面复合速率Sp是105cm/s.求在表面单位时间单位面积复合的空穴数。

36.某n-型半导体具有过剩空穴10cm,在体材料中少子寿命是10-6s,在表面的少子寿命的10-7s,假设所加的电场为零,且令Dp=10cm2/s。

确定稳态过剩载流子浓度作为以半导体表面(x=0)距离的函数。

(本章习题完)21.。

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