微电子工艺习题参考解答

1、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-3）2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，则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的硼均匀分布在替代位置上，求：

2、 (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，则在离多远处镓的浓度会低于51015cm-3？10从式，

3、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，若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较多?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(cm-3)，N为可用的间隙

4、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 the partial pre

5、ssure 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求在紧密堆积下(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及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，锡的压强为2.6610-6Pa

6、)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)01020304045CS (c

7、m-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 density of Si is

8、 2.33 g / cm3.The acceptor concentration for r = 0.01 Wcm 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 occupie

9、d 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 critica

10、l 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. Therefore, th

11、e 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 solubil

12、ity 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 0.12 0.

13、23 9. The segregation coefficient of Ga in Si is 8 10-3From Eq. 18We have10. We have from Eq.18 So the ratio = = at x/L = 2.11. For the conventionally-doped silicon, the resistivity varies from 120 W-cm to 155 W-cm. The corresponding doping concentration varies from 2.51013 to 41013 cm-3. Therefore

14、the range of breakdown voltages of p+ - n junctions is given byFor the neutron irradiated silicon, r = 148 1.5 W-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, we

15、 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 composition of liquid GaAs always becomes gallium rich.14. = = = .15. = =at 27oC = 300 K =2.141014 at 900oC = 1173 K.16. 37 4 = 148 chipsIn terms of l

16、itho-stepper considerations, there are 500 mm 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

17、. Where M: Molecular mass k: Boltzmann constant = 1.3810-23 J/k T: The absolute temperature n: Speed of molecular So that.d18. .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

18、 970C (=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/pL2 : 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 = 8

19、10 /min .(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. Th

20、e lattice 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

21、长栅极氧化膜，其生长条件为1000，20 min试计算栅极氧化膜的厚度及场氧化膜的总厚度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及latm下进行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及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及Ta2O5的介电常数约为3.9、7.6及25试计算以Ta2O5与SiO2：Si3N4：SiO2作为介质

22、的电容的比值其中介质厚度均相等，且SiO2：Si3N4：SiO2的比例亦为1：1：17续习题6，若选择介电常数为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相同的电容值。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，淀积表面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

23、时，通常以硅烷为气体源，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电容的制作若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比较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.

24、 (a)计算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)比较(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)及(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为避免电迁移的问题，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

25、310-6cm并计算铝线两端可承受的最大电压20在布局金属线时若要使用铜，必须克服以下几点困难：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与相同尺寸大小的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 = BtFrom Figs. 6 and 7, we obtain B/A =1.5 m /hr,

26、B=0.47 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

27、be obtained 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 effective

28、is 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 = Btsimilarly,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 SiNxHy x = 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, e1 = 25 SiO2 thickness = t, e2 = 3.9 Si3N4 thickness = t, e3 = 7.6, area = A then .7. Set BST thickness = 3t, e1 = 500, area = A1 SiO2 thickness = t, e2 = 3.9, area = A2 Si3N4 thickness = t, e3 = 7.6, area = A2 then 8. Let Ta2O5 thickness = 3t, e1 = 25 SiO2 thickness = t, e2 = 3.9 Si3N4

31、thickness = t, e3 = 7.6 area = A then9. The deposition rate can be expressed as r = r0 exp (-Ea/kT) where Ea = 0.6 eV for silane-oxygen reaction. Therefore for T1 = 698 K ln 2 = T2 =1030 K= 757 .10. We can use energy-enhanced CVD methods such as using a focused energy source or UV lamp. Another meth

32、od is to use boron doped P-glass which will reflow at temperatures less than 900. 11. Moderately low temperatures are usually used for polysilicon deposition, and silane decomposition occurs at lower temperatures than that for chloride reactions. In addition, silane is used for better coverage over

33、amorphous materials such SiO2.12. There are two reasons. One is to minimize the thermal budget of the wafer, reducing dopant diffusion and material degradation. In addition, fewer gas phase reactions occur at lower temperatures, resulting in smoother and better adhering films. Another reason is that

34、 the polysilicon will have small grains. The finer grains are easier to mask and etch to give smooth and uniform edges. However, for temperatures less than 575 C the deposition rate is too low.有两个原因。一是减少硅片的热预算，降低掺杂剂扩散和材料的降解。此外，少气相反应在较低的温度下发生，导致更顺畅，更好的粘合膜。另一个原因是，多晶硅将有小颗粒。细颗粒容易掩模蚀刻给光滑和均匀的边缘。然而，温度低于575

35、C沉积速率太低。13. The flat-band voltage shift is = 0.5 V . Number of fixed oxide charge is To remove these charges, a 450 heat treatment in hydrogen for about 30 minutes is required. 14.20/0.25 = 80 sqs.Therefore, the resistance of the metal line is 550 = 400 W .15. For TiSi2 30 2.37 = 71.1nm For CoSi2 30

36、 3.56 = 106.8nm.16. For TiSi2: Advantage:low resistivityIt can reduce native-oxide layersTiSi2 on the gate electrode is more resistant to high-field-induced hot-electron degradation. Disadvantage: bridging effect occurs.Larger Si consumption during formation of TiSi2Less thermal stabilityFor CoSi2:A

37、dvantage:low resistivityHigh temperature stabilityNo bridging effectA selective chemical etch exitsLow shear forcesDisadvantage:not a good candidate for polycides17. (a) (b) (c) We can decrease the RC delay by 55%. Ratio = = 0.45. 18. (a) RC = 3.2 103 8.7 10-13 = 2.8 ns.(b) 19. (a) The aluminum runn

38、er can be considered as two segments connected in series: 20% (or 0.4 mm) of the length is half thickness (0.5 m) and the remaining 1.6 mm is full thickness (1m). The total resistance is = 72 . The limiting current I is given by the maximum allowed current density times cross-sectional area of the t

39、hinner conductor sections: I = 5105 A/cm2 (10-40.510-4) = 2.510-3 A = 2.5 mA. The voltage drop across the whole conductor is then = 0.18V.Cu0.5 mm0.5 mm20.40 nm60 nm Al = h: height , W : width , t : thickness, assume that the resistivities of the cladding layer and TiN are much larger than When Then

40、 t = 0.073 mm = 73 nm .LITHOGRAPHY AND ETHING1对等级为100的洁净室，试依粒子大小计算每单位立方米中尘埃粒子总数 (a)0.5m到1m； (b)1m到2m； (c)比2m大 2试计算一有9道掩模版工艺的最后成品率其中有4道平均致命缺陷密度为0.1/cm2，4道为0.25cm2。，1道为1.0/cm2芯片面积为50 mm2 3一个光学光刻系统，其曝光功率为0.3mW/cm2。正性光刻胶要求的曝光能量为140mJ/cm2。，负性光刻胶为9mJ/cm2。假设忽略装载与卸载晶片的时间，试比较正性光刻胶与负性光刻胶的产率 4(a)对波长为193nm的ArF-

41、准分子激光光学光刻系统，其DNA=0.65，k1=0.60，k2=0.50此光刻机理论的分辨率与聚焦深度为多少? (b)实际上我们可以如何修正DNA、k1与k2参数来改善分辨率? (c)相移掩模版(PSM)技术改变哪一个参数可改善分辨率? 5右图为光刻系统的反应曲线(response curves)： (a)使用较大值的光刻胶有何优缺点? (b)传统的光刻胶为何不能用于248nm或193rim光刻系统? 6(a)解释在电子束光刻中为何可变形状电子束比高斯电子束拥有较高的产率? (h)电子束光刻图案如何对准?为何X射线光刻的图案对准如此困难? (c)X射线光刻比电子束光刻的潜在优点有哪些? 7(

42、a)为何光学光刻系统的工作模式由邻近影印法进化到投影，最后进化到5：1的步进重复投影法? (b)X射线光刻系统是否可能使用重复扫描系统?并说明原因 8如果掩蔽层与衬底不能被某一腐蚀剂腐蚀，试画出下列几种情形薄膜厚度为hf的各向异性腐蚀图案的侧边轮廓； (a)刚好完全腐蚀； (b)100过度腐蚀； (c)200过度腐蚀 9一个晶向硅晶片，利用KOH溶液腐蚀一个利用二氧化硅当掩蔽层的1.5ml.5m窗，垂直于晶面的腐蚀速率为0.6m/min而：晶面的腐蚀速率比为100：16：1画出20s、40s与60s的腐蚀轮廓10续上题，一个晶向硅晶片利用薄的SiO2当掩蔽层，在KOH溶液中藕蚀画出硅的腐蚀轮廓

43、11一个直径150mm晶向硅晶片厚度为625m晶片上有1 000m1 000m的IC这些IC是利用各向异性腐蚀的方式来隔开试用两种方法来完成此工艺，并计算使用这两种工艺方法损失的面积所占的比例 12粒子碰撞平均移动的距离称为平均自由程，51 0-3/p(cm)，其中P为压强，单位为Torr一般常用的等离子体，其反应腔压强范围为1Pa150Pa其相关的气体浓度(cm-3)与平均自由程是多少? 13氟原子(F)刻蚀硅的刻蚀速率为： 刻蚀速率(nm/min)=2.8610-13nFT1/2exp(-Ea/RT)其中nF为氟原子的浓度(cm-3)，T为绝对温度(K)，Ea与R分别为激活能(10.416

44、kJ/mol)与气体常数(8.345JK)如果nF为3l015cm-3，试计算室温时硅的刻蚀速率1 4续上题，利用氟原子一样可以刻蚀SiO2，刻蚀速率可表示为 刻蚀速率(nm/min)=0.61410-13nFT1/2exp(-Ea/RT)其中nF为31015cm-3，Ea为15.12kJ/mol计算室温时SiO2的刻蚀速率及SiO2对Si的刻蚀选择比1 5可以用多重步骤的刻蚀工艺来刻蚀薄栅极氧化层上的多晶硅栅极如何设计一个刻蚀工艺使之满足：没有做掩蔽效应(micrornasking)、各向异性刻蚀、对薄的栅极氧化层有适合的选择比?16刻蚀400 nm多晶硅而不会移去1 nm厚的底部栅氧化层，

45、试找出所需的刻蚀选择比?假设多晶硅的刻蚀工艺有10的刻蚀速率均匀度171 um厚的A1薄膜淀积在平坦的场氧化层区域上并且利用光刻胶来定义图案接着金属层利用Helicon刻蚀机，混合BCI3/Cl2气体。在温度为70 C来刻蚀A1与光刻胶的刻蚀选择比维持在3假设有30的过度刻蚀，试问为确保顶部的金属不被侵蚀所需光刻胶的最薄厚度为多少?18在ECR等离子体中，一个静磁场B驱使电子沿着磁场随一个角频率做圆周运动e=qB/me，其中q为电荷、me为电子质量如果微波的频率为2.45GHz，试问所需的磁场太小为多少?19传统的反应离子刻蚀与高密度等离子体(ECR，ICP等)相比，最大的区别是什么?20叙述

46、如何消除Al金属线在氯化物等离子体刻蚀后所造成的腐蚀1. With reference to Fig. 2 for class 100 clean room we have a total of 3500 particles/m3 with particle sizes 0.5 m = 735 particles/m2 with particle sizes 1.0 m= 157 particles/m2 with particle sizes 2.0 mTherefore, (a) 3500-735 = 2765 particles/m3 between 0.5 and 1 m(b) 735

47、-157 = 578 particles/m3 between 1 and 2 m(c) 157 particles/m3 above 2 m.2.A = 50 mm2 = 0.5 cm2 . 3. The available exposure energy in an hour is 0.3 mW2/cm2 3600 s =1080 mJ/cm2For positive resist, the throughput isFor negative resist, the throughput is .4. (a) The resolution of a projection system is

48、 given by m= 0.228 m(b) We can increase NA to improve the resolution. We can adopt resolution enhancement techniques (RET) such as optical proximity correction (OPC) and phase-shifting Masks (PSM). We can also develop new resists that provide lower k1 and higher k2 for better resolution and depth of

49、 focus.(c) PSM technique changes k1 to improve resolution.5. (a) Using resists with high value can result in a more vertical profile but throughput decreases.(b) Conventional resists can not be used in deep UV lithography process because these resists have high absorption and require high dose to be

50、 exposed in deep UV. This raises the concern of damage to stepper lens, lower exposure speed and reduced throughput.6. (a)A shaped beam system enables the size and shape of the beam to be varied, thereby minimizing the number of flashes required for exposing a given area to be patterned. Therefore,

51、a shaped beam can save time and increase throughput compared to a Gaussian beam. (b) We can make alignment marks on wafers using e-beam and etch the exposed marks. We can then use them to do alignment with e-beam radiation and obtain the signal from these marks for wafer alignment.X-ray lithography

52、is a proximity printing lithography. Its accuracy requirement is very high, therefore alignment is difficult.(c)X-ray lithography using synchrotron radiation has a high exposure flux so X-ray has better throughput than e-beam.7. (a) To avoid the mask damage problem associated with shadow printing, projection printing exposure tools have been developed to project an image from the mask. With a 1:1 projection printing system is much more difficult to produce defect-free masks than it is with a 5:1 reduction step-and-r