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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)整理为word格式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
4、)，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 the pa
5、rtial 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
6、.6610整理为word格式-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 (c
7、m)01020304045CS (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 silicon整理为word格式M / M
8、0 = 0.5The density of Si is 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 give
9、n as(b) The average 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 is整理为word格式The maximum weight that can be supported by the se
10、ed equals the product of the 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
11、than that of the melt. Therefore, 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-e
12、nd.7. The reason is that the 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
13、1.0solidified 0.05 0.06 0.08 0.12 0.23 9. The segregation coefficient of Ga in Si is 8 10-3From Eq. 18整理为word格式We 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
14、varies from 2.51013 to 41013 cm-3. Therefore 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 整理为word格式Therefore, the fraction of liquid remained f can be obtained as following.13. From the Fig.11, 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 i
