微电子实用工艺习题参考解答.docx

1、微电子实用工艺习题参考解答CRYSTAL GROWTH AND EXPITAXY1画出一50cm长的单晶硅锭距离籽晶10cm、20cm、30cm、40cm、45cm时砷的掺杂分布。单晶硅锭从融体中拉出时，初始的掺杂浓度为1017cm-32硅的晶格常数为5.43假设为一硬球模型： (a)计算硅原子的半径。 (b)确定硅原子的浓度为多少(单位为cm-3)? (c)利用阿伏伽德罗(Avogadro)常数求出硅的密度。3假设有一l0kg的纯硅融体，当硼掺杂的单晶硅锭生长到一半时，希望得到0.01 cm的电阻率，如此需要加总量是多少的硼去掺杂?4一直径200mm、厚1mm的硅晶片，含有5.41mg的硼均

2、匀分布在替代位置上，求： (a)硼的浓度为多少? (b)硼原子间的平均距离。5用于柴可拉斯基法的籽晶，通常先拉成一小直径(5.5mm)的狭窄颈以作为无位错生长的开始。如果硅的临界屈服强度为2106g/cm2，试计算此籽晶可以支撑的200mm直径单晶硅锭的最大长度。6在利用柴可拉斯基法所生长的晶体中掺入硼原子，为何在尾端的硼原子浓度会比籽晶端的浓度高?7为何晶片中心的杂质浓度会比晶片周围的大?8对柴可拉斯基技术，在k0=0.05时，画出Cs/C0值的曲线。9利用悬浮区熔工艺来提纯一含有镓且浓度为51016cm-3的单晶硅锭。一次悬浮区熔通过，熔融带长度为2cm，如此在离多远处镓的浓度会低于510

3、15cm-3？10从式，假设ke=0.3，求在x/L=1和2时，Cs/C0的值。11如果用如右图所示的硅材料制造p+-n突变结二极管，试求用传统的方法掺杂和用中子辐照硅的击穿电压改变的百分比。12由图10.10，假如Cm=20，在Tb时，还剩下多少比例的液体?13用图10.11解释为何砷化镓液体总会变成含镓比拟多?14空隙ns的平衡浓度为Nexp-Es/(kT)，N为半导体原子的浓度，而Es为形成能量。计算硅在27、900和1 200的ns (假设Es=2.3eV)15假设弗兰克尔缺陷的形成能量(Ef)为1.1eV，计算在27、900时的缺陷密度弗兰克尔缺陷的平衡密度是，其中N为硅原子的浓度(

4、cm-3)，N为可用的间隙位置浓度(cm-3)，可表示为N=11027cm-316在直径为300mm的晶片上，可以放多少面积为400mm2的芯片？解释你对芯片形状和在周围有多少闲置面积的假设17求在300K时，空气分子的平均速率(空气相对分子质量为29)图 10.10. Phase diagram for the gallium- 图 10.11. Partial pressure of gallium and arsenic arsenic system. over gallium arsenide as a function of temperature. Also shown is th

5、e partial pressure of silicon.18淀积腔中蒸发源和晶片的距离为15cm，估算当此距离为蒸发源分子的平均自由程的10时系统的气压为多少?19求在严密堆积下(即每个原子和其他六个邻近原子相接)，形成单原子层所需的每单位面积原子数Ns假设原子直径d为4.6820假设一喷射炉几何尺寸为A=5cm2与L=12cm (a)计算在970下装满砷化镓的喷射炉中，镓的到达速率和MBE的生长速率； (b)利用同样形状大小且工作在700，用锡做的喷射炉来生长，试计算锡在如前述砷化镓生长速率下的掺杂浓度(假设锡会完全进入前述速率生长的砷化镓中，锡的摩尔质量为118.69；在700时，锡的

6、压强为2.6610-6Pa)21求铟原子的最大比例，即生长在砷化镓衬底上而且并无任何错配的位错的GaxIn1-xAs薄膜的x值，假定薄膜的厚度是10nm22薄膜晶格的错配f定义为，f=a0(s)-a0(f)/a0(f)a0/a0。a0(s)和a0(f)分别为衬底和薄膜在未形变时的晶格常数，求出InAs-GaAs和Ge-Si系统的f值Solution1.C0 = 1017 cm-3k0(As in Si) = 0.3CS= k0C0(1 - M/M0)k0-1= 0.31017(1- x)-0.7 = 31016/(1 - l/50)0.7x00.20.40.60.80.9l (cm)01020

7、304045CS (cm-3)310163.510164.2810165.6810161.0710171.510172. (a) The radius of a silicon atom can be expressed as (b) The numbers of Si atom in its diamond structure are 8.So the density of silicon atoms is(c) The density of Si is = 2.33 g / cm3. 3. k0 = 0.8 for boron in siliconM / M0 = 0.5The densi

8、ty of Si is 2.33 g / cm3.The acceptor concentration for = 0.01 cm is 91018 cm-3.The doping concentration CS is given byTherefore The amount of boron required for a 10 kg charge is boron atomsSo that.4. (a) The molecular weight of boron is 10.81. The boron concentration can be given as(b) The average

9、 occupied volume of everyone boron atoms in the wafer is We assume the volume is a sphere, so the radius of the sphere ( r ) is the average distance between two boron atoms. Then.5. The cross-sectional area of the seed isThe maximum weight that can be supported by the seed equals the product of the

10、critical yield strength and the seeds cross-sectional area:The corresponding weight of a 200-mm-diameter ingot with length l is6. The segregation coefficient of boron in silicon is 0.72. It is smaller than unity, so the solubility of B in Si under solid phase is smaller than that of the melt. Theref

11、ore, the excess B atoms will be thrown-off into the melt, then the concentration of B in the melt will be increased. The tail-end of the crystal is the last to solidify. Therefore, the concentration of B in the tail-end of grown crystal will be higher than that of seed-end.7. The reason is that the

12、solubility in the melt is proportional to the temperature, and the temperature is higher in the center part than at the perimeter. Therefore, the solubility is higher in the center part, causing a higher impurity concentration there.8. We haveFractional 0 0.2 0.4 0.6 0.8 1.0solidified 0.05 0.06 0.08

13、 0.12 0.23 9. The segregation coefficient of Ga in Si is 8 10-3From Eq. 18We have10. We have from Eq.18So the ratio = = at x/L = 2.11. For the conventionally-doped silicon, the resistivity varies from 120 -cm to 155 -cm. The corresponding doping concentration varies from 2.51013 to 41013 cm-3. There

14、fore the range of breakdown voltages of p+ - n junctions is given byFor the neutron irradiated silicon, = 148 1.5 -cm. The doping concentration is 31013 (1%). The range of breakdown voltage is.12. We have Therefore, the fraction of liquid remained f can be obtained as following.13. From the Fig.11,

15、we find the vapor pressure of As is much higher than that of the Ga. Therefore, the As content will be lost when the temperature is increased. Thus the position of liquid GaAs always bees gallium rich.14. = = = .15. = =at 27oC = 300 K =2.141014 at 900oC = 1173 K.16. 37 4 = 148 chipsIn terms of litho

16、-stepper considerations, there are 500 m space tolerance between the mask boundary of two dice. We divide the wafer into four symmetrical parts for convenient dicing, and discard the perimeter parts of the wafer. Usually the quality of the perimeter parts is the worst due to the edge effects.17. Whe

17、re M: Molecular mass k: Boltzmann constant = 1.3810-23 J/k T: The absolute temperature: Speed of molecular So that.18. .19. For close-packing arrange, there are 3 pie shaped sections in the equilateral triangle. Each section corresponds to 1/6 of an atom. Therefore = =.20. (a) The pressure at 970C (

18、=1243K) is 2.910-1 Pa for Ga and 13 Pa for As2. The arrival rate is given by the product of the impringement rate and A/L2 : Arrival rate = 2.641020 = 2.641020 = 2.91015 Ga molecules/cm2 s The growth rate is determined by the Ga arrival rate and is given by (2.91015)2.8/(61014) = 13.5 /s = 810 /min

19、.(b)The pressure at 700C for tin is 2.6610-6 Pa. The molecular weight is 118.69. Therefore the arrival rate is If Sn atoms are fully incorporated and active in the Ga sublattice of GaAs, we have an electron concentration of21. The x value is about 0.25, which is obtained from Fig. 26.22. The lattice

20、 constants for InAs, GaAs, Si and Ge are 6.05, 5.65,5.43, and 5.65 , respectively (Appendix F). Therefore, the f value for InAs-GaAs system is And for Ge-Si system isTHERMAL OXIDATION AND FILM DEPOSITION1一p型掺杂、方向为的硅晶片，其电阻率为10cm，置于湿法氧化的系统中，其生长厚度为0.45m，温度为1 050试决定氧化的时间2习题1中第一次氧化后，在氧化膜上定义一个区域生长栅极氧化膜，其生

21、长条件为1000，20 min试计算栅极氧化膜的厚度与场氧化膜的总厚度3试推导方程式(11)当时间较长时，可化简为x2=Bt；时间较短时可化简为x=4试计算在方向为的硅晶片上，温度980与latm下进展干法氧化的扩散系数D5(a)在等离子体式淀积氮化硅的系统中，有20的氢气，且硅与氮的比值为1.2，试计算淀积SiNxHy，中的x与y (b)假设淀积薄膜的电阻率随51028exp(-33.3)而改变(当208)，其中为与氮的比值试计算(a)中薄膜的电阻率6SiO2、Si3N4与Ta2O5的介电常数约为3.9、7.6与25试计算以Ta2O5与SiO2：Si3N4：SiO2作为介质的电容的比值其中介

22、质厚度均相等，且SiO2：Si3N4：SiO2的比例亦为1：1：17续习题6，假如选择介电常数为500的BST来取代Ta2O5。试计算欲维持相等的电容值，面积所减少的比例假设两薄膜厚度相等8续习题6，试以SiO2的厚度来计算Ta2O5的等效厚度假设两者有一样的电容值。9在硅烷与氧气的环境下，淀积未掺杂的氧化膜当温度为425时，淀积速率为15nm/min在多少温度时，淀积速率可提高一倍?10磷硅玻璃回流的工艺需高与1000在ULSI中，当器件的尺寸缩小时，必须降低工艺温度试建议一些方法，可在温度小于900的情形下，淀积外表平坦的二氧化硅绝缘层来作为金属层间介质11为何在淀积多晶硅时，通常以硅烷为

23、气体源，而不以硅氯化物为气体源?12解释为何一般淀积多晶硅薄膜的温度普遍较低，大约在600650之间。13一电子束蒸发系统淀积铝以完成MOS电容的制作假如电容的平带电压因电子束辐射而变动0.5V，试计算有多少固定氧化电荷(氧化膜厚度为50nm)?试问如何将这些电荷去除?14一金属线长20m，宽0.25m，薄层电阻值为5/ 请计算此线的电阻值15计算TiSi2与CoSi2的厚度，其中Ti与Co的初始厚度为30nm16比拟TiSi2与CoSi2在自对准金属硅化物应用方面的优、缺点17一介质置于两平行金属线间其长度L=lcm，宽度W=0.28m，厚度T=0.3m两金属间距s为0.36m. (a)计算

24、RC时间延迟。假设金属材料为铝，其电阻率为2.67cm，介质为氧化膜，其介电常数为3.9 (b)计算RC时间延迟。假设金属材料为铜，其电阻率为1.7cm，介质为有机聚合物，其介电常数为2.8 (c)比拟(a)、(b)中结果，我们可以减少多少RC时间延迟?18重复计算习题17(a)与(b)假设电容的边缘因子(fringing factor)为3，边缘因子是由于电场线分布超出金属线的长度与宽度的区域19为防止电迁移的问题，最大铝导线的电流密度不得超过5105 A/cm2假设导线长为2mm,宽为1m，最小厚度为1m，此外有20的线在台阶上，该处厚度为0.5m试计算此线的电阻值假设电阻率为310-6c

25、m并计算铝线两端可承受的最大电压20在布局金属线时假如要使用铜，必须克制以下几点困难：铜通过二氧化硅层而扩散；铜与二氧化硅层的附着性；铜的腐蚀性有一种解决的方法是使用具有包覆性、附着性的薄膜来保护铜导线考虑一被包覆的铜导线，其横截面积为0.5m0.5m与一样尺寸大小的TiN/Al/TiN导线相比(其中上层TiN厚度为40 nm，下层为60 nm)，其最大包覆层的厚度为多少?(假设被包覆的铜线与TiN/A1/TiN线的电阻相等)1. From Eq. 11 (with =0)x2+Ax = Bt From Figs. 6 and 7, we obtain B/A =1.5 m /hr, B=0.

26、47 m 2/hr, therefore A= 0.31 m. The time required to grow 0.45m oxide is .2. After a window is opened in the oxide for a second oxidation, the rate constants are B = 0.01 m 2/hr, A= 0.116 m (B/A = 6 10-2 m /hr). If the initial oxide thickness is 20 nm = 0.02 m for dry oxidation, the value ofcan be o

27、btained as followed: (0.02)2 + 0.166(0.02) = 0.01 (0 +)or = 0.372 hr.For an oxidation time of 20 min (=1/3 hr), the oxide thickness in the window area is x2+ 0.166x = 0.01(0.333+0.372) = 0.007or x = 0.0350 m = 35 nm (gate oxide).For the field oxide with an original thickness 0.45 m, the effectiveis

28、given by =x2+ 0.166x = 0.01(0.333+27.72) = 0.28053or x = 0.4530 m (an increase of 0.003m only for the field oxide).3. x2 + Ax = B when t , t , then, x2 = Bt similarly, when t , t , then, x = 4. At 980(=1253K) and 1 atm, B = 8.510-3 m 2/hr, B/A = 410-2 m /hr (from Figs. 6 and 7). Since A 2D/k , B/A =

29、 kC0/C1, C0 = 5.21016 molecules/cm3 and C1 = 2.21022 cm-3 , the diffusion coefficient is given by 5.(a) For SiNxHyx = 0.83 atomic % y = 0.46The empirical formula is SiN0.83H0.46. (b) = 5 1028e-33.31.2 = 2 1011 -cmAs the Si/N ratio increases, the resistivity decreases exponentially. 6. Set Ta2O5 thic

30、kness = 3t, 1 = 25 SiO2 thickness = t, 2 = 3.9 Si3N4 thickness = t, 3 = 7.6, area = A then.7. Set BST thickness = 3t, 1 = 500, area = A1 SiO2 thickness = t, 2 = 3.9, area = A2 Si3N4 thickness = t, 3 = 7.6, area = A2 then8. Let Ta2O5 thickness = 3t, 1 = 25 SiO2 thickness = t, 2 = 3.9 Si3N4 thickness = t, 3 = 7.6 area = A then9. The deposition rate can be ex