微纳电子材料与工艺习题配套电子版

1、澦电子料澧 &61;/39 注：本资料页数指页数1. 真空镀膜（Solution P10 中文P102）2. Si 中的扩散（Solution P21 中文P106）3. SiO2 中的扩散（Solution P22 中文P21）4. Si、SiO2 的晶格结构（Solution P31 中文P106）BAGeaEaaDCFigure 1Q19-2 The (100) plane of the diamond crystal structure.The number of atoms per nm2, n, is therefore:1004 1nm2109 nm/m22100a2.543 1

2、099nm/m= 6.78 atoms/nm2 or 6.78 n1001018atoms/m2The (110) plane is shown below in Fig. 1Q19-3. There are 4 atoms at the corners and shared with neighboring planes (hence each contributing a quarter), 2 atoms on upper and lower sides shared with upper and lowlanes (hence each atom contributing 1/2) a

3、nd 2 atoms wholly withhe plane.Ba2ABA(110)(110)aDCCDFigure 1Q19-3 The (110) plane of the diamond crystal structure.The number of atoms per nm2, n, is therefore:1104 1 2 1 2 4 2 1naa 2 109 nm/m2 1104 1 2 1 2 4 2 1n0.543 109 m0.543 109 m 2 109 nm/m2 110= 9.59 atoms/nm2 or 9.59 n1101018atoms/m2This is

4、the most crowded plane with the most number of atoms per unit area.The (111) plane is shown below in Fig. 1Q19-4:DaCABaE(100)AAa32a2a23030C6060DDC BBa2 2a2 2Figure 1Q19-4 The (111) plane of the diamond crystal structureThe number of atoms per nm2, n, is therefore:1113 60 3 1 360 2 1n 2 3 109 nm/m2 1

5、11 12 2 a a2 23 60 3 1 360 2 1n 3 109 nm/m2 111 1 2 20.543 10m0.543 10m99 2 2 2 c= 7.83 atoms/nm2 or 7.83 n1111018atoms/m2Given:Molar mass of SiO2: Mat = 28.09 10 kg/mol + 2 16 10 kg/mol = 60.09 10 kg/mol-3-3-3Density of SiO2: = 2.27 10 kg m3-3Let n be the number of SiO2 molecules per unit volume, t

6、hen: n MatNA6.022 1023 mol12.27 103 kg m3 N n A60.09 10kg/mol3Matn = 2.27 1028 molecules per m3Or, converting to molecules per nm3:2.27 1028 molecules/m3n 22.7 molecules per nm3109 nm/m3Oxide has less dense packing so is a more open structure. For every 1 micron of oxide formedon the crystal surface

7、, only about 0.5 micron of Si crystal is consumed.5.多层互联的 RC 时间常数（英文P188 中文P163）6.金属金属间接触的隧道效应导电(英文P239，中文P205)7.电子不确定性（英文P224 中文P193）8. 原子尺度的电子（英文版 P229 中文版 P197）9. 掺杂变化产生的内建电势（英文P434 中文P381）10.二极管（英文P455 中文P397）11. 稳态光电导率（英文P427 中文P375）12. pnp 晶体管（英文P524 中文P456）pnp 型硅双极晶体管特性如下：发射极受主掺杂浓度 Na=21018cm

8、-3，基极施主浓度 Nd=11016cm-3.基极空穴迁移率 400cm2V-1S-1，发射极电子迁移率 200cm2V-1S-1当晶体管正常工作时，基极中性区宽度为 2 微米，器件有效截面积 0.02mm2.基极空穴发射极的注入效率为 100%，计算共基极电流传输因子和电流增益。如若发射极电流是 1mA，求 VEB。为 400ns，假设13. 证明发射极注入效率 emitter injection efficiency（英文P525 中文P457）14. 相对介电常数和极化率（Solution P181 中文P583）15. 相对介电常数，键强，带隙和折射率（Solution P182 中文

9、P583）Solutionamond, Si, and Ge, the polarization mechanism is electronic (bond). There are two factorst increase the polarization.is the number of electrons available for displacement and the ease withwhich the field can displace the electrons. The number of electronshe core ss increases fromdiamond

10、 to Ge. Secondly, and most importantly, the bond strength per atom decreases from diamond to Ge, making it easier for valence electronshe bonds to be displaced.bFor diamond, atomic concentration N is:3.52 103 kg/m3 6.022 1023 mol112 103 kg/molDN N = 1.766 1029 m-3AMatThe polarizability can then be f

11、ound from the Clausius-Mossotti equation:r 1 N 23ero3 8 854 10F/m5.8 13 1 orN 2e29 m3r e = 9.256 10-41m2FThe polarizability for Si and Ge can be found similarly, and are summarizedable 7Q2-2:Table 7Q2-2Polarizability values for diamond, Si and Ge6.0010-4010-405.004.0010-403.0010-402.0010-4010-401.00

12、002004006008001000Youngs modulus (GPa)= 5.31110-40 - (5.32710-43)YCorrelation coefficient = 0.9969Figure 7Q2-1 Plot of polarizability per atom versus Youngs modulus.Polarizability per atom (F m2)N (m-3)e (F m )2Diamond1.766 1029 m-39.256 10-41 F m2Si4.995 1028 m-34.170 10-40 F m2Ge4.412 1028 m-35.01

13、7 10-40 F m2As the polarization mechanismhese crystals is due to electronic bond polarization, thedisplacement of electronshe covalent bonds depends on the flexibility or elasticity of these bonds and hence also depends on the elastic modulus.c10-406.005.0010-404.0010-403.0010-4010-402.001.0010-4000

14、123456Bandgap, Eg (eV)= 5.32510-40 - (8.043510-41)EgCorrelation coefficient = 0.9873Figure 7Q2-2 Plot of polarizability versus bandgnergy.Thereseems to be a linear relationship betn polarizability and bandgnergy.To facilie this proof, we can plot a graph of refractive index, n, versus relative permi

15、ttivity, r.d5Log-log plot:er law241020Relative permittivity,rFigure 7Q2-3 Logarithmic plot of refractive index versus relative permittivity.s. The best fit line is n = ArxThe log-log plot exhibits a straight line through the three po(Correlation coefficient is 0.9987) where x = 0.513 1/2 and A = exp

16、(0.02070) 1. Thus n = (r).The refractive index n is an optical propertyt represents the speed of a light wave, or an electromagnetic wave, through the material (v = c/n). The light wave is a high frequency electromagnetic). n and polarizability (or ) will bewave where the frequency is of the order of 1014 1015 Hz (opticalrrelated if the polarization can follow the field oscillations at this frequency (optical). This will be the case inelectronic polarization because electrons are light and raly respond to the fast os