《超大规模集成电路设计》考试习题(含答案)完整版分析
集成电路设计岗位招聘笔试题与参考答案(某大型集团公司)
招聘集成电路设计岗位笔试题与参考答案(某大型集团公司)(答案在后面)一、单项选择题（本大题有10小题，每小题2分，共20分）1、在集成电路设计中，以下哪种类型的设计通常负责处理数字逻辑功能？A、模拟集成电路B、数字集成电路C、混合信号集成电路D、射频集成电路2、以下哪种技术用于在集成电路设计中实现晶体管间的连接？A、光刻技术B、蚀刻技术C、键合技术D、离子注入技术3、在CMOS工艺中，P型MOSFET的阈值电压通常会随着温度的升高而：A. 增加B. 减少C. 不变D. 先增加后减少4、下列哪一项不是减少互连延迟的有效方法？A. 使用更细的金属线B. 使用更高介电常数的绝缘材料C. 减少金属层之间的距离D. 使用铜代替铝作为互连线材料5、集成电路设计中，以下哪种工艺主要用于制造CMOS（互补金属氧化物半导体）逻辑电路？A. 双极型工艺B. 金属氧化物半导体工艺C. 双极型/金属氧化物半导体混合工艺D. 双极型/CMOS混合工艺6、在集成电路设计中，以下哪个参数通常用来描述晶体管的开关速度？A. 饱和电压B. 输入阻抗C. 开关时间D. 集成度7、在集成电路设计中，用于描述电路逻辑功能的硬件描述语言不包括以下哪一种？A. VerilogB. VHDLC. C++D. SystemVerilog8、下列选项中，哪一个不是ASIC（专用集成电路）设计流程中的一个阶段？A. 逻辑综合B. 布局布线C. 系统集成D. 物理验证9、以下哪种工艺技术通常用于制造高性能的集成电路？A. 混合信号工艺B. CMOS工艺C. GaN（氮化镓）工艺D. BiCMOS工艺二、多项选择题（本大题有10小题，每小题4分，共40分）1、在CMOS工艺中，关于阱（well）的概念，下列说法正确的有：A. NMOS晶体管通常位于P型阱中B. PMOS晶体管通常位于N型阱中C. N阱用于隔离不同区域的晶体管，防止电流泄露D. P阱可以与N阱共存于同一层硅片上而不会相互影响2、关于集成电路版图设计中的DRC（Design Rule Check）规则，下列哪些陈述是正确的？A. DRC规则是为了确保电路性能优化B. DRC规则定义了最小特征尺寸、最小间距等制造限制C. 违反DRC规则可能会导致制造缺陷，如短路或开路D. DRC规则在所有半导体制造工艺中都是相同的3、关于集成电路设计，以下哪些是典型的电路设计类型？（）A、模拟电路设计B、数字电路设计C、混合信号电路设计D、射频电路设计E、光电子电路设计4、在集成电路设计中，以下哪些因素会影响电路的功耗？（）A、晶体管的工作状态B、电源电压C、电路的复杂度D、芯片的温度E、外部负载5、在集成电路设计过程中，下列哪些技术用于提高电路的性能？A. 使用更先进的制程技术B. 优化电路布局减少信号延迟C. 增加电源电压以提升速度D. 减少电路层数降低制造成本E. 应用低功耗设计方法6、下列哪些是实现CMOS逻辑门时需要考虑的关键因素？A. 输入电平的阈值B. 输出驱动能力C. 功率消耗D. 静态电流消耗E. 电路的工作频率7、以下哪些技术或方法属于集成电路设计中的模拟设计领域？（）A. 信号处理算法B. 逻辑门电路设计C. 模拟电路仿真D. 功耗分析E. 版图设计8、在集成电路设计中，以下哪些步骤是进行版图设计的必要阶段？（）A. 电路原理图设计B. 布局规划C. 逻辑分割D. 布局布线E. 版图检查9、在CMOS工艺中，影响MOSFET阈值电压的因素有哪些？A. 氧化层厚度B. 衬底掺杂浓度C. 栅极材料类型D. 源漏区掺杂浓度E. 温度F. 器件尺寸三、判断题（本大题有10小题，每小题2分，共20分）1、集成电路设计岗位的工程师需要具备扎实的数学基础和电子工程知识。
5大规模数字集成电路习题解答
自我检测题1．在存储器结构中，什么是“字”什么是“字长”，如何表示存储器的容量解：采用同一个地址存放的一组二进制数，称为字。
字的位数称为字长。
习惯上用总的位数来表示存储器的容量，一个具有n字、每字m位的存储器，其容量一般可表示为n ×m位。
2．试述RAM和ROM的区别。
解：RAM称为随机存储器，在工作中既允许随时从指定单元内读出信息，也可以随时将信息写入指定单元，最大的优点是读写方便。
但是掉电后数据丢失。
ROM在正常工作状态下只能从中读取数据，不能快速、随时地修改或重新写入数据，内部信息通常在制造过程或使用前写入，3．试述SRAM和DRAM的区别。
解：SRAM通常采用锁存器构成存储单元，利用锁存器的双稳态结构，数据一旦被写入就能够稳定地保持下去。
动态存储器则是以电容为存储单元，利用对电容器的充放电来存储信息，例如电容器含有电荷表示状态1，无电荷表示状态0。
根据DRAM的机理，电容内部的电荷需要维持在一定的水平才能保证内部信息的正确性。
因此，DRAM在使用时需要定时地进行信息刷新，不允许由于电容漏电导致数据信息逐渐减弱或消失。
4．与SRAM相比，闪烁存储器有何主要优点解：容量大，掉电后数据不会丢失。
5．用ROM实现两个4位二进制数相乘，试问：该ROM需要有多少根地址线多少根数据线其存储容量为多少解：8根地址线，8根数据线。
其容量为256×8。
6．简答以下问题：（1）CPLD和FPGA有什么不同FPGA可以达到比 CPLD更高的集成度，同时也具有更复杂的布线结构和逻辑实现。
FPGA 更适合于触发器丰富的结构，而 CPLD更适合于触发器有限而积项丰富的结构。
在编程上 FPGA比 CPLD具有更大的灵活性；CPLD功耗要比 FPGA大；且集成度越高越明显；CPLD比 FPGA有较高的速度和较大的时间可预测性，产品可以给出引脚到引脚的最大延迟时间。
CPLD的编程工艺采用 E2 CPLD的编程工艺，无需外部存储器芯片，使用简单，保密性好。
集成电路设计岗位招聘笔试题与参考答案(某大型集团公司)
招聘集成电路设计岗位笔试题与参考答案(某大型集团公司)(答案在后面)一、单项选择题（本大题有10小题，每小题2分，共20分）1.题目：集成电路设计中，以下哪个因素对电路性能影响最大？A. 晶圆尺寸B. 芯片制造工艺C. 电路架构设计D. 印刷电路板布局2.题目：在CMOS工艺中，以下哪种器件用于实现电流放大功能？A. PMOSB. NMOSC. 二极管D. 反相器3.在集成电路设计中，以下哪个因素对电路性能的影响最大？A. 电源电压B. 地线宽度C. 电路布局D. 输入输出信号4.集成电路的晶体管数量与其性能的关系，以下哪个说法是正确的？A. 晶体管数量越多，性能越好B. 晶体管数量越多，性能越差C. 晶体管数量适中，性能最佳D. 晶体管数量与性能无关5.在集成电路设计中，哪个过程是为了验证设计的正确性？A. 电路设计B. 制程模拟C. 芯片测试D. 设计验证6.在集成电路设计中，以下哪个因素是影响功耗的主要因素？A. 芯片尺寸大小B. 工作频率高低C. 电路设计复杂度D. 环境温度高低7、在集成电路设计中，以下哪种技术常用于降低功耗？（）A. 增加时钟频率B. 优化布线结构C. 增加晶体管数量D. 优化算法设计以减少运算次数8、关于数字集成电路的下列说法中，哪项是不正确的？（）A. 数字集成电路通过逻辑门电路实现数字信号的传输和处理。
B. 数字集成电路只能处理二进制信号。
C. 数字集成电路的设计和制造都需要先进的工艺和严格的标准。
D. 数字集成电路不适用于大规模生产，因为生产成本较高。
9.在集成电路设计中，以下哪个工具常用于逻辑综合？A. CAD工具B. 仿真器C. 物理验证工具D. 编程语言编译器 10. 集成电路设计中的时钟树结构有何作用？A. 减少布线资源B. 优化布线路径C. 提高电路性能D. 增加电路功耗二、多项选择题（本大题有10小题，每小题4分，共40分）1.关于集成电路设计流程，下列说法正确的是：A. 集成电路设计首要步骤是电路原理图设计。
硅超大规模集成电路工艺技术—理论、实践与模型_课后习题答案
1.2. Assuming dopant atoms are uniformly distributed in a silicon crystal, how far apart arethese atoms when the doping concentration is a). 1015 cm -3, b). 1018 cm -3, c). 5x1020 cm -3.Answer:The average distance between the dopant atoms would just be one over the cube root of the dopant concentration:x =N A -1/3a)x =1x1015cm-3()-1/3=1x10-5cm =0.1μm =100nmb)x =1x1018cm-3()-1/3=1x10-6cm =0.01μm =10nmc)x =5x1020cm-3()-1/3=1.3x10-7cm =0.0013μm =1.3nm1.3. Consider a piece of pure silicon 100 µm long with a cross-sectional area of 1 µm2. Howmuch current would flow through this “resistor” at room temperature in response to an applied voltage of 1 volt?Answer: If the silicon is pure, then the carrier concentration will be simply n i . At room temperature, n i≈ 1.45 x 1010 cm -3. Under an applied field, the current will be due to drift and hence,I =I n +I p =qAn i μn +μp ()ε=1.6x10-19coul ()10-8cm 2()1.45x1010carrierscm -3()2000cm 2volt -1sec -1()1volt 10-2cm ⎛ ⎝ ⎫⎭=4.64x10-12amps or 4.64pA1.10. A state-of-the-art NMOS transistor might have a drain junction area of 0.5 x 0.5 µm.Calculate the junction capacitance associated with this junction at an applied reverse bias of 2 volts. Assume the drain region is very heavily doped and the substrate doping is 1 x 1016 cm -3. Answer:The capacitance of the junction is given by Eqn. 1.25.C A =εS x d =q εS 2N A ND N A +N D ⎛ ⎝ ⎫ ⎭ ⎪ 1φi ±V ()⎡ ⎣ ⎢⎤ ⎦ ⎥The junction built-in voltage is given by Eqn. 1.24. N D is not specified except that it is very large, so we take it to be 1020 cm -3 (roughly solid solubility). The exact choice for N D doesn't make much difference in the answer.φi =kT q ln N D N An i 2⎛ ⎝ ⎫ ⎭⎪ =.0259volt s ()ln 1020cm -3()1016cm -3()1.45x1010cm-3()2⎛ ⎝ ⎫⎭⎪ ⎪ =0.934 volt sSince N D >> N A in this structure, the capacitance expression simplifies toC A ≅εSW=qεS2N A()1φi±V()⎡⎣⎢⎤⎦⎥=1.6x10-19coul()11.7()1016cm-3()8.86x10-14Fcm-1()2()2.934volt s()⎡⎣⎢⎤⎦⎥ =1.68x10-8Fcm-2Given the area of the junction (0.25 x 10-8 cm2, the junction capacitance is thus 4.2 x 10-17 Farads.3.2. A boron-doped crystal pulled by the Czochralski technique is required to have aresistivity of 10 Ω cm when half the crystal is grown. Assuming that a 100 gm pure silicon charge is used, how much 0.01 Ω cm boron doped silicon must be added to the melt? For this crystal, plot resistivity as a function of the fraction of the melt solidified. Assume k 0 = 0.8 and the hole mobility µp = 550 cm 2 volt -1 sec -1.Answer:Using the mobility value given, and ρ=1q μN A we have:10 Ω cm ⇒ N A = 1.14 x 1015 cm -3 and 0.01 Ω cm ⇒ N A = 1.14 x 1018 cm -3From Eqn. 3.38, C S =C O k O 1-f ()k O -1and we want C S = 1.14 x 1015 cm -3 when f =0.5. Thus, solving for C 0 the initial doping concentration in the melt, we have:C 0=1.14x10150.81-0.5()0.2=1.24x1015cm -3But C 0=I 0V 0=# of impurities unit vol of melt =(Doping)(Vol. of 0.01 Ωcm)Vol 100 gm Si∴ Wgt added of 0.01 Ω cm S i = C 0Doping ⎛ ⎝ ⎫⎭ 100gm ()=0.109gmThe resistivity as a function of distance is plotted below and is given byρx ()=1q μN A x ()=1-f ()1-k 0q μC 0k 0=11.5Ωcm 1-f ()0.2R e s i s t i v i t y0.20.40.60.81Fraction Solidified - f3.3. A Czochralski crystal is pulled from a melt containing 1015 cm -3 boron and 2x1014 cm -3phosphorus. Initially the crystal will be P type but as it is pulled, more and more phosphorus will build up in the liquid because of segregation. At some point the crystal will become N type. Assuming k O = 0.32 for phosphorus and 0.8 for boron, calculate the distance along the pulled crystal at which the transition from P to N type takes place.Answer:We can calculate the point at which the crystal becomes N type from Eqn. 3.38 as follows:C S Phos ()=C 0k 01-f ()k 0-1=2x1014()0.32()1-f ()-0.68C S Boron ()=C 0k 01-f ()k 0-1=1015()0.8()1-f ()-0.2At the point where the cross-over occurs to N type, these two concentrations will be equal. Solving for f, we findf ≅0.995Thus only the last 0.5% of the crystal is N type.3.6. Suppose your company was in the business of producing silicon wafers for thesemiconductor industry by the CZ growth process. Suppose you had to produce the maximum number of wafers per boule that met a fairly tight resistivity specification. a). Would you prefer to grow N type or P type crystals? Why?b). What dopant would you use in growing N-type crystals? What dopant would you use in growing P type crystals? ExplainAnswer:a). Boron has the segregation coefficient closest to unity of all the dopants. Thus it produces the most uniform doping along the length of a CZ crystal. Thus P type would be the natural choice.b). For P type, the obvious (and only real choice) is boron as explained in part a). For N type crystals Fig. 3-18 shows that either P or As would be a reasonable choice since their segregation coefficients are quite close and are better than Sb. Table 3-2 indicates that P might be slightly preferred over As because its k O value is slightly closer to 1.4.1. An IC manufacturing plant produces 1000 wafers per week. Assume that each wafer contains 100 die, each of which can be sold for $50 if it works. The yield on these chips is currently running at 50%. If the yield can be increased, the incremental income is almost pure profit because all 100 chips on each wafer are manufactured whether they work or not. How much would the yield have to be increased to produce an annual profit increase of $10,000,000?Answer:At 1000 wafers per week, the plant produces 52,000 wafers per year. If each wafer has 50 good die each of which sells for $50, the plant gross income is simplyIncome = (52,000)(50)($50) = $130,000,000 per year.To increase this income by $10,000,000 requires that the yield increase by10130≅7.7%4.3. As MOS devices are scaled to smaller dimensions, gate oxides must bereduced in thickness.a. As the gate oxide thickness decreases, do MOS devices become more or lesssensitive to sodium contamination? Explain.b. As the gate oxide thickness decreases, what must be done to the substrate doping (oralternatively the channel V TH implant, to maintain the same V TH? Explain.Answer:a). From the text, Na+ contamination causes threshold voltage instabilities in MOS devices.Also from Eqn. 4.1, the threshold voltage is given byV TH=V FB+2φf+2εS qN A2φf()C OX+qQ MC OXAs the gate oxide thickness decreases, C OX increases, so the same amount of mobile charge Q M will have less effect on V TH as oxides get thinner. Therefore MOS devices are less sensitive to sodium contamination.b). Using the same expression for V TH as in part a), we observe that as the oxide thickness decreases, (C OX increases), to maintain the same V TH, N A will have to increase. N A willactually have to increase by the square root of the oxide thickness decrease to keep V TH constant.4.4. A new cleaning procedure has been proposed which is based on H2O saturated with O2as an oxidant. This has been suggested as a replacement for the H2O2 oxidizing solution used in the RCA clean. Suppose a Si wafer, contaminated with trace amounts of Au, Fe and Cu is cleaned in the new H2O/O2solution. Will this clean the wafer effectively?Why or why not? Explain.Answer:As described in the text, cleaning metal ions off of silicon wafers involves the following chemistry:M↔M z++ze-The cleaning solution must be chosen so that the reaction is driven to the right because this puts the metal ions in solution where they can be rinsed off. Since driving the reaction to the right corresponds to oxidation, we need an oxidizing solution to clean the wafer.H2O/O2 is certainly an oxidizing solution. But whether it cleans effectively or not depends on the standard oxidation potential of the various possible reactions. From Table 4-3 in the text, we have:The stronger reactions (dominating) are at the bottom.Thus the H2O/O2 reaction will clean Fe and Cu, but it will not clean Au off the wafer.4.5. Explain why it is important that the generation lifetime measurement illustrated inFigure 4-19 is done in the dark.Answer:The measurement depends on measuring carriers generated thermally in the silicon substrate (or at the surface). If light is shining on the sample, then absorbed photons can also generate the required carriers. As a result, the extracted generation lifetime with the light on would really be measuring the intensity of the incident light and not a basic property of the silicon material.5.1. Calculate and plot versus exposure wavelength the theoretical resolution and depth offocus for a projection exposure system with a NA of 0.6 (about the best that can be done today). Assume k 1 = 0.6 and k 2 = 0.5 (both typical values). Consider wavelengths between 100 nm and 1000 nm (DUV and visible light). ). Indicate the common exposure wavelengths being used or considered today on your plot (g-line, i-line, KrF and ArF). Will an ArF source be adequate for the 0.13 µm and 0.1 µm technology generations according to these simple calculations?Answer:The relevant equations are simply∴R =k 1λNA =0.6λ0.6 and DOF=±k 2λNA ()2=±0.5λ0.6()2These equations are plotted below. Note that the ArF (193 nm) will not reach 0.13 µm or 0.1 µm resolution according to these simple calculations. In fact, with more sophisticated techniques such as phase shift masks, off axis illumination etc., ArF is expected to reach 0.13 µm and perhaps the 0.1 µm generations.R e s o l u t i o n , D O F 祄20040060080010001200Exposure W avelength nm5.3. An X-ray exposure system uses photons with an energy of 1 keV. If the separation between the mask and wafer is 20 µm, estimate the diffraction limited resolution that is achievable by this system.Answer:The equivalent wavelength of 1 keV x-rays is given byE =h ν=hc λ∴ λ=hc E=4.14x10-15eVsec ()3x1010cmsec -1()103eV=1.24x10-7cm =1.24 nmX-ray systems operate in the proximity printing mode, so that the theoretical resolution is given by Eqn. 5.12:Resolut ion =λg =1.24x10-3μm ()20μm ()=0.15μm5.8. As described in this chapter, there are no clear choices for lithography systems beyondoptical projection tools based on 193-nm ArF eximer lasers. One possibility is an optical projection system using a 157-nm F 2 excimer laser.a. Assuming a numerical aperture of 0.8 and k 1 = 0.75, what is the expected resolution of such a system using a first order estimate of resolution?b. Actual projections for such systems suggest that they might be capable of resolving features suitable for the 2009 0.07 µm generation. Suggest three approaches to actually achieving this resolution with these systems.Answera). The simple formula for resolution isR =k 1λNA =0.750.157μm0.8=0.147μmb). The calculated resolution in part a is a factor of two larger than required for the 0.07 µm generation. Therefore some “tricks” will have to be used to act ually achieve such resolution. There are a number of possibilities:1. Use of phase-shift masks. This technique, discussed in this chapter, has the potential for significant resolution improvements. It works by designing a more sophisticated mask. Simple masks are digital - black or white. Phase shifting adds a second material to the mask features, usually at the edges which shifts the optical phase and sharpens up the aerial image. Sophisticated computer programs are required to design such masks.2. Use of optical proximity correction in the mask design. This is another approach to designing a better mask and as discussed in class, can also improve resolution significantly. The approach involves adding extra features to the mask, usually at corners where features are sharp, to compensate for the high frequency information lost to diffraction effects.3. Off-axis illumination. This allows the optical system to capture some of the higher order diffracted light and hence can improve resolution.5.9. Current optical projection lithography tools produce diffraction limited aerial images. A typical aerial image produced by such a system is shown in the simulation below where a square and rectangular mask regions produce the image shown. (The mask features are the black outlines, the calculated aerial image is the grayscale inside the black rectangles.) The major feature of the aerial image is its rounded corners compared to the sharp square corners of the desired pattern. Explain physically why these features look the way they do, using diffraction theory and the physical properties of modern projection optical lithography tools.Answer:Modern optical projection lithography systems are limited in the resolution they can achieve by diffraction effects. The finite size of the focusing lens means that the high order diffraction components are “lost” and are therefore not available to help in printing a replica of the mask image. But the high frequency spatial components are exactly the components that contain information about “sharp” features, i.e. corners etc. Thus the projected aerial image loses this information and corners become rounded. The only ways to improve the image are by using shorter wavelength light, or a higher NA lens.5.10. Future optical lithography systems will likely use shorter exposure wavelengths toachieve higher resolution and they will also likely use planarization techniques to provide “flat” substrates on which to expose the resist layers. Explain why “flat”substrates will be more important in the future than they have been in the past. Answer:As the wavelength of the exposure system decreases, the depth of focus of the exposure system also decreases. Thus it will be necessary to make sure that the resist in which the image is to be exposed, is flat and does not require much depth of focus. Planarization techniques will be required to accomplish this. This could mean CMP to planarize the substrate before the resist is applied, or it could mean using a spun on resist which planarizes the substrate and which is then covered with a thin, uniform imaging resist layer.6.4. Construct a HF CV plot for a P-type silicon sample, analogous to Fig. 6-9. Explain your plot based on the behavior of holes and electrons in the semiconductor in a similar manner to the discussion in the text for Fig. 6.9. Answer:C ODV GGQ GQ IQ DThe C-V plot looks basically the same as the N substrate example in the text, that we discussed in class, except that the horizontal axis is flipped. For negative applied gate voltages, the majority carrier holes in the substrate are attracted to the surface. This is the accumulation region a) above. We measure just C OX for the capacitance since there is no depletion in the substrate. For + V G, the holes are driven away from the surface creating first a depletion region as in b) and finally an inversion layer of electrons as in c). Themeasured capacitance drops as we move into depletion and finally reaches a minimum value after an inversion layer forms.The C-V curves shown are high frequency curves. As discussed in the text, the capacitance remains at its minimum value for + V G values greater than V TH because the inversion layer electrons cannot be created or destroyed as fast as the signal is changing. Hence the small AC signal must “wiggle” the bottom of the depletion region to balance ∆V G.6.6. In a small MOS device, there may be a statistical variation in V T due to differences inQ F from one device to another. In a 0.13 µm technology minimum device (gate oxide area = 0.1µm x 0.1µm) with a 2.5nm gate oxide, what would the difference in threshold voltage be for devices with 0 or 1 fixed charge in the gate oxide?Answer:The oxide capacitance isC ox=εAd=3.9⨯8.854⨯10-14()0.1⨯10-4()0.1⨯10-4()2.5⨯10-7=1.38⨯10-16The change in threshold voltage is given by∆V T=qQ FC ox=1.6⨯10-19()1()1.38⨯10-16=1.1mVThis shows that a single electron trap in a gate oxide will have a negligible effect on thethreshold voltage at this technology generation.6.12 A silicon wafer is covered by an SiO2 film 0.3 μm thick.a. What is the time required to increase the thickness by 0.5 μm by oxidation in H2Oat 1200˚C?b. Repeat for oxidation in dry O2at 1200˚C.Answer:We will perform the calculation for <111> silicon wafers. For <100> wafers, the linear rate constant should be divided by 1.68.a. At 1200˚C, in H2OB=3.86⨯102exp-0.78 kT⎛ ⎝ ⎫⎭ =0.829μm2/hrB A =1.63⨯108exp-2.05kT⎛⎝⎫⎭ =15.86μm/hrA=0.052μmThe initial oxide, if grown at 1200˚C would have taken this long to growτ=x i2+Ax iB=0.3()2+0.052()0.3()0.829=0.127hrThe time required to grow 0.8μm at 1200˚C isτ=x i 2+Ax i B =0.8()2+0.052()0.8()0.829=0.822hrThus, the time required to add 0.5μm to an existing 0.3μm film is0.822-0.127=0.695hr or 41.7 minutes.b. At 1200˚C, in dry oxygenB =7.72⨯102exp - 1.23k(1200+273)⎛ ⎝ ⎫ ⎭ =0.048μm 2/hrB A =6.23⨯106exp -2.0kT ⎛ ⎝ ⎫ ⎭ =0.899μm /hrA =0.053μmThe initial oxide would have taken 2.206 hours to grow in dry oxygen, it would require 14.217 hours to grow 0.8μm , thus would require an additional 12 hours to add 0.5μm to an existing 0.3μm film.6.13. Suppose an oxidation process is used in which (100) wafers are oxidized in O 2 for threehrs. at 1100˚C, followed by two hrs. in H 2O at 900˚C, followed by two hrs in O 2 at 1200˚C. Use Figs. 6-19 and 6-20 in the text to estimate the resulting final oxide thickness. Explain how you use these figures to calculate the results of a multi-step oxidation like this.Answer:We can use these figures to estimate the oxide thickness as follows. First, we use Fig. 6-19 for the first dry oxidation cycleA three hour oxidation at 1100˚C produces an oxide thickness of about 0.21 µm. We nextuse Fig. 6-20 for the wet oxidation as shown below. The oxidation is 2 hrs in H 2O at 90 ˚C. We start by finding the point on the 900˚C curve that corresponds to 0.21 µm since this is the starting oxide thickness. This is point A. We then move along the 900˚C curve by two hours to point B. This corresponds to a thickness of about 0.4 µm which is the thickness at the end of the wet oxidation.We now go back to Fig. 6-19 for the final dry O 2 cycle. This process is 2 hrs at 1200˚C. Westart by finding the point on the 1200˚C curve that corresponds to a starting oxide thickness of 0.4 µm. This is point A below. We then increment the time by 2 hrs along the 1200˚C curve, to arrive at a final oxide thickness of about 0.5 µm.6.18. Silicon on Insulator or SOI is a new substrate material that is being considered forfuture integrated circuits. The structure, shown below, consists of a thin single crystal silicon layer on an insulating (SiO 2) substrate. The silicon below the SiO 2 provides mechanical support for the structure. One of the reasons this type of material is being considered, is because junctions can be diffused completely through the thin silicon layer to the underlying SiO 2. This reduces junction capacitances and produces faster circuits. Isolation is also easy to achieve in this material, because the thin Si layer can be completely oxidized, resulting in devices completely surrounded by SiO 2. A LOCOS process is used to locally oxidize through the silicon as shown on the right below. Assuming the LOCOS oxidation is done in H 2O at 1000˚C, how long will it take to oxidize through the 0.3 µm silicon layer? Calculate a numerical answer using the Deal Grove model.Answer:To oxidize completely through a 0.3 µm silicon layer, we will need to grow (2.2)(0.3 µm) = 0.66 µm of SiO 2. At 1000˚C in H 2O, the Deal Grove rate constants are given by (Table 6-2):B =3.86x102exp -0.78eV kT ⎛ ⎝ ⎫ ⎭ =0.316μm 2hr-1B A =1.63x1081.68exp -2.05eV kT ⎛ ⎝ ⎫ ⎭ =0.747μmhr -1∴t =0.66()20.316+0.660.747≅2.25 hours6.23. As part of an IC process flow, a CVD SiO 2 layer 1.0 µm thick is deposited on a <100>silicon substrate. This structure is then oxidized at 900˚C for 60 minutes in an H 2O ambient. What is the final SiO 2 thickness after this oxidation? Calculate an answer, do not use the oxidation charts in the text .Answer:At 900˚C in H 2O, the oxidation rate constants are given by:B =3.86x102exp -0.788.62x10-5()1173()⎛ ⎝ ⎫ ⎭ ⎪ μm 2 hr -1=0.17 μm 2 hr -1B A =1.63x1081.68exp - 2.058.62x10-5()1173()⎛ ⎝ ⎫ ⎭ ⎪ μm hr -1=0.152 μm hr -1The initial oxide on the wafer is 1.0 µm thick. This corresponds to a τ ofτ=1()2+1()0.170.152⎛ ⎝⎫⎭ 0.17=12.46 hoursThus the final oxide thickness is given byx o =0.172()0.152()1+13.461.11()24()0.17()-1⎧ ⎨ ⎪ ⎩ ⎪ ⎫⎬ ⎪ ⎭ ⎪=1.064 μmThus not much additional oxide grows.Chapter 7 Problems7.1. A resistor for an analog integrated circuit is made using a layer of deposited polysilicon0.5 µm thick, as shown below.Polysilicon (a) (a) The doping the polysilicon is 1⨯10 cm -3. The carrier mobilityμ=100cm 2V -1sec -1is low because of scattering at grain boundaries. If the resistor has L=100µm, W=10µm, what is its resistance in Ohms?(b) (b) A thermal oxidation is performed on the polysilicon for 2 hours at 900˚C inH 2O . Assuming B/A for polysilicon is 2/3 that of <111> silicon, what is thepolysilicon thickness that remains.(c) (c) Assuming that all of the dopant remains in the polysilicon (i.e. does notsegregate to oxide), what is the new value of the resistor in (a). Assume the mobility does not change.Answer:(a)ρ=1nq μ=11⨯1016()1.6⨯10-19()100()=6.25ΩcmρS =ρx j = 6.250.5⨯10-4=125k ΩR =10010ρS =1.25M Ω(10squares)(b) The linear rate coefficient at 900˚C isB A ⎛ ⎝ ⎫ ⎭ poly =23 1.63⨯108exp -2.05kT ⎛ ⎝ ⎫ ⎭ ⎛ ⎝ ⎫ ⎭=0.170μm hr-1The parabolic rate constant for poly is unchanged:B pol y =3.86⨯102exp -0.78kT ⎛ ⎝ ⎫ ⎭ =0.172μm 2hr-1A poly =1.01μmThe oxide thickness isx o =A 21+tA 2/4B -1⎧ ⎨ ⎩ ⎫ ⎬⎭x o =1.0121+21.01()2/40.172()-1⎧ ⎨ ⎩ ⎫ ⎬ ⎭ =0.27μmThis oxide consumes a silicon thickness of 0.45*0.27=0.12 µm, leaving a remaining polysilicon thickness of 0.5-0.12=0.38 µm and contains all the dopant with a concentration of1⨯1016()0.50.38=1.31⨯1016cm -3(c) Since the concentration has gone up and the thickness has gone down by the same factor, the polysilicon restivity and hence the resistance of the line remains the same.7.4. Suppose we perform a solid solubility limited predeposition from a doped glass sourcewhich introduces a total of Q impurities / cm 2.(a) (a) If this predeposition was performed for a total of t minutes, how long would ittake (total time) to predeposit a total of 3Q impurities / cm 2 into a wafer if the predeposition temperature remained constant.(b) (b) Derive a simple expression for the Dt ()drive -in which would be required todrive the initial predeposition of Q impurities / cm 2 sufficiently deep so that the final surface concentration is equal to 1% of the solid solubility concentration. Thiscan be expressed in terms ofDt ()predep and the solid solubility concentrationC S .Answer:(a)Q =2C SπDt ⇒Q ∝t∴3Q ⇒9t(b)C 0,t ()drive -in =QπDt =0.01C SQ =2C SπDt ()predep∴2πDt ()predep Dt ()drive -in=0.01∴Dt ()drive -in =200π⎛ ⎝ ⎫⎭ 2Dt ()predep7.7. A boron diffusion is performed in silicon such that the maximum boron concentration is1 x 1018 cm -3. For what range of diffusion temperatures will electric field effects and concentration dependent diffusion coefficients be important?Answer:Electric field effects and concentration dependent diffusion are both important when the doping concentration exceeds the intrinsic electron (or hole) concentration. The intrinsic orbackground electron concentration is n i which increases with higher temperature. This provides a background sea of electrons or holes in the lattice at a given temperature. If the doping exceeds this concentration, then these extrinsic effects are important.When the temperature is below the temperature where n i =1⨯1018/cm 3, these effects will become dominant since they often depend on n /n i (where n =N A or n =N D to a first approximation).n i =3.9⨯1016T 32exp -0.605kT⎛ ⎝ ⎫ ⎭By trial and error, n i =1⨯1018/cm 3at T=720C.Therefore, extrinsic effects become important below 720˚C.7.15. A silicon wafer is uniformly doped with boron (2 x 1015 cm -3) and phosphorus (1 x 1015cm -3) so that it is net P type. This wafer is then thermally oxidized to grow about 1 µm of SiO 2. The oxide is then stripped and a measurement is made to determine the doping type of the wafer surface. Surprisingly it is found to be N type. Explain why the surface was converted from P to N type. Hint: Consider the segregation behavior of dopants when silicon is oxidized.Answer:The boron segregates preferentially into the growing oxide, thus depleting the surface concentration in the silicon. The phosphorus on the other hand preferentially segregates (piles-up) on the silicon side of the interface. Both of these effects act in the same direction and tend to make the surface of the silicon more N-type.It is for this reason that a P-type “channel stop” implant is almost always needed under a locally oxidized lightly doped P-type region, to prevent depletion of the P-type dopant in the substrate and in the worst case to prevent an N-type channel from forming.7.20. Fig. 7.38 shows that a wet oxidation produces a significantly higher C I /C I *than doesa dry O 2 oxidation. Explain quantitatively why this should be the case. Answer:BecauseC I ∝dx dta faster oxidation rate produces a higher interstitial supersaturation. Thus, wet oxidationproduces a higher C I /C I *than dry oxidation, for the same time at the same temperature.Chapter 8 Problems8.1. Arsenic is implanted into a lightly doped p-type Si substrate at an energy of 75keV . Thedose is 1⨯1014/cm 2. The Si substrate is tilted 7˚ with respect to the ion beam to make it appear amorphous. The implanted region is assumed to be rapidly annealed so that complete electrical activation is achieved. What is the peak electron concentration produced?Answer:From Fig. 8-3, the range and standard deviation for 75 keV arsenic areR P =0.05μm ∆R P =0.02μmThe peak concentration isC P =Q2π∆R P=1⨯10142π0.02⨯10-4()=2⨯1019cm -3Assuming all the dose is active, then the peak electron concentration is equal to the peak dopant concentration.8.4. How thick does a mask have to be to reduce the peak doping of an implantby a factor of 10,000 at the mask/substrate boundary. Provide an equation in terms of the Range and the Standard Deviation of the implant profile.Answer:We want to reduce the peak doping N P *in the mask at range R P *by 10,000 at the mask/substrate boundary. We will use the equation which describes the profile of an implant in a mask layerN *(d)=N P *exp -d -R P*()22∆R P *2⎡ ⎣ ⎢ ⎢ ⎤ ⎦⎥ ⎥WhenN *(d)N P *=10-4we haved =R P *+4.3∆R P *8.6. The equations below provide a reasonable analytical description for some of thediffusion processes indicated schematically in the diagrams on the following page. Put the equation number (a-f) on each figure that is the best match. Equations may be。
集成电路设计岗位招聘笔试题与参考答案(某大型国企)
招聘集成电路设计岗位笔试题与参考答案(某大型国企)(答案在后面)一、单项选择题（本大题有10小题，每小题2分，共20分）1、集成电路设计中，关于CMOS反相器的描述，以下哪项是正确的？A. CMOS反相器具有低功耗特性，但速度较慢。
B. CMOS反相器具有高速度特性，但功耗较高。
C. CMOS反相器具有低功耗特性，且速度较快。
D. CMOS反相器具有高速度特性，但功耗较低。
2、在集成电路设计中，以下哪个因素对电路的性能有重要影响？A. 晶圆尺寸B. 制造工艺C. 电路规模D. 所有上述因素3、在集成电路设计中，以下哪个因素对电路性能的影响最大？A. 电源电压B. 地线宽度C. 电阻值D. 电容值4、在CMOS工艺中，以下哪种器件主要用于实现电流放大功能？B. NMOSC. 二极管D. 反相器5、（关于集成电路设计基础）以下关于集成电路设计的描述中，哪项是正确的？A. 集成电路设计完全依赖于自动化工具，无需人工干预。
B. 集成电路设计过程中，版图设计是第一步。
C. 集成电路设计主要关注电路的功能实现，而不考虑其物理实现。
D. 在集成电路设计中，功耗和性能同样重要，需要平衡考虑。
6、（关于数字集成电路设计）在数字集成电路设计中，关于时序分析，以下说法错误的是？A. 时序分析是确保电路在规定的时钟周期内正确工作的关键步骤。
B. 时序分析只关注组合逻辑部分，不涉及时序逻辑部分。
C. 时序分析包括建立时序和保持时序的分析。
D. 时序分析是确保芯片性能的重要手段之一。
7、在集成电路设计中，以下哪个因素对电路性能的影响最大？A. 电源电压B. 地址线宽度C. 数据总线宽度D. 输入输出接口8、在CMOS工艺中，以下哪个器件用于实现电流隔离？A. 晶体管C. 互斥开关D. 绝缘层9、下列哪个选项是集成电路设计中常用的EDA工具软件？A. AutoCADB. SolidWorksC. Altium DesignerD. MATLAB 10、在集成电路设计中，关于CMOS工艺的特点描述正确的是？A. CMOS工艺只能用于数字电路的设计B. CMOS工艺功耗大，不适合低功耗应用C. CMOS工艺可以同时实现数字与模拟电路的设计D. CMOS工艺不兼容其他集成工艺类型二、多项选择题（本大题有10小题，每小题4分，共40分）1、关于集成电路设计的基础知识中，下列哪些说法是正确的？（）选项：A. 集成电路设计主要涉及到模拟电路、数字电路和混合信号电路设计。
(完整版)集成电路设计复习题及解答
集成电路设计复习题绪论1．画出集成电路设计与制造的主要流程框架。
2．集成电路分类情况如何？集成电路设计1．层次化、结构化设计概念，集成电路设计域和设计层次2．什么是集成电路设计？集成电路设计流程。
（三个设计步骤：系统功能设计逻辑和电路设计版图设计）3．模拟电路和数字电路设计各自的特点和流程4．版图验证和检查包括哪些内容？如何实现？5．版图设计规则的概念，主要内容以及表示方法。
为什么需要指定版图设计规则？6．集成电路设计方法分类？（全定制、半定制、PLD）7．标准单元/门阵列的概念，优点/缺点，设计流程8．PLD设计方法的特点，FPGA/CPLD的概念9．试述门阵列和标准单元设计方法的概念和它们之间的异同点。
10．标准单元库中的单元的主要描述形式有哪些？分别在IC设计的什么阶段应用？11．集成电路的可测性设计是指什么？Soc设计复习题1.什么是SoC？2.SoC设计的发展趋势及面临的挑战？3.SoC设计的特点?4.SoC设计与传统的ASIC设计最大的不同是什么？5.什么是软硬件协同设计？6.常用的可测性设计方法有哪些？7. IP的基本概念和IP分类8.什么是可综合RTL代码?9.么是同步电路，什么是异步电路，各有什么特点？10.逻辑综合的概念。
11.什么是触发器的建立时间（Setup Time），试画图进行说明。
12.什么是触发器的保持时间（Hold Time），试画图进行说明。
13. 什么是验证，什么是测试，两者有何区别？14.试画图简要说明扫描测试原理。
绪论1、 画出集成电路设计与制造的主要流程框架。
2、集成电路分类情况如何？集成电路设计1． 层次化、结构化设计概念，集成电路设计域和设计层次分层分级设计和模块化设计．将一个复杂的集成电路系统的设计问题分解为复杂性较低的设计级别，⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎧⎪⎪⎪⎪⎩⎪⎪⎪⎪⎨⎧⎩⎨⎧⎩⎨⎧⎪⎪⎪⎪⎩⎪⎪⎪⎪⎨⎧⎪⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎪⎨⎧⎩⎨⎧⎪⎪⎪⎪⎩⎪⎪⎪⎪⎨⎧⎩⎨⎧⎪⎩⎪⎨⎧按应用领域分类数字模拟混合电路非线性电路线性电路模拟电路时序逻辑电路组合逻辑电路数字电路按功能分类GSI ULSI VLSI LSI MSI SSI 按规模分类薄膜混合集成电路厚膜混合集成电路混合集成电路B iCMOS B iMOS 型B iMOS CMOS NMOS PMOS 型MOS双极型单片集成电路按结构分类集成电路这个级别可以再分解到复杂性更低的设计级别；这样的分解一直继续到使最终的设计级别的复杂性足够低，也就是说，能相当容易地由这一级设计出的单元逐级组织起复杂的系统。
超大规模集成电路期终复习题
VLSI总复习题一、选择题：1. 下列哪一步不属于光刻过程（ C ）A 涂胶 B显影 C 离子注入 D刻蚀2. 下列哪一个不属于MOS反相器的类型（ C ）A.(E/D)B.(E/R) C (D/E) D.CMOS3. 上题2中，（ D ）属于无比电路。
4. 当Vgs<0, Vds>Vt时，NMOS晶体管处在（A ）状态A 截止 B饱和 C 线性 D.击穿5. 在工艺尺寸相同，NMOS管与PMOS管的驱动能力之比约为（ B）A．2/5 B.5/2 C.1/5 D.5/16. MOS管的阈值（开启）电压不受下列哪个参数影响（D ）A.衬底掺杂浓度B.栅极绝缘材料C.栅极导电材料D.源漏掺杂浓度7. 衬偏效应对阈值电压的影响（A ）A. 增加B.减小C.不变8. 对于短沟道而言，沟道长度调制效应对Ids的影响（ A ）A. 增加B.减小C.不变9. 以下不是防止寄生晶体管的措施有( C )A.场区掺杂B.增加衬底偏压C.减小栅氧化层厚度D.增厚场区SIO210. 大电容负载通常采用逐级放大的方法驱动，如下图所示，应采用逐级放大，每级放大的等比例因子为（ B ）。
A.2B.3C.4D.511. 下列哪一项不属于引起CMOS静态功耗的电流（C ）A.栅极漏电流B.源漏极反偏漏电流C.场开启漏电流D.亚阈漏电流12.如下图，设P管和N管的驱动能力均为β，若输入A,B为（0,0）时，此二输入或非门导电因子为（A ），若输入A,B为（1,1）时，导电因子为（ B ）A.β/2B.2βC. βD.2/3β13. CMOS传输高电平，当 时NMOS为（C ），PMOS为（A ）。
A.线性状态B. 饱和状态C.截止状态D.击穿状态14.可编逻辑阵列PLA（ B ）A.与固定，或可编程B.与，或均可编程C.与可编程，或固定15.下列哪一项不属于RS触发器的功能（C ）A.置0B.置1C.翻转D.保持16.若存储器存储容量为4K×4（位）等于（C ）字节。
(整理)集成电路设计习题答案1-5章
CH11．按规模划分，集成电路的发展已经经历了哪几代？它的发展遵循了一条业界著名的定律，请说出是什么定律？晶体管-分立元件-SSI-MSI-LSI-VLSI-ULSI-GSI-SOC。
MOORE定律2．什么是无生产线集成电路设计？列出无生产线集成电路设计的特点和环境。
拥有设计人才和技术，但不拥有生产线。
特点：电路设计，工艺制造，封装分立运行。
环境：IC产业生产能力剩余，人们需要更多的功能芯片设计3．多项目晶圆（MPW）技术的特点是什么？对发展集成电路设计有什么意义？MPW：把几到几十种工艺上兼容的芯片拼装到一个宏芯片上，然后以步行的方式排列到一到多个晶圆上。
意义：降低成本。
4．集成电路设计需要哪四个方面的知识？系统，电路，工具，工艺方面的知识CH21．为什么硅材料在集成电路技术中起着举足轻重的作用 ?原材料来源丰富，技术成熟，硅基产品价格低廉2．GaAs和InP材料各有哪些特点? P10,11 3．怎样的条件下金属与半导体形成欧姆接触？怎样的条件下金属与半导体形成肖特基接触？接触区半导体重掺杂可实现欧姆接触，金属与掺杂半导体接触形成肖特基接触4．说出多晶硅在CMOS工艺中的作用。
P13 5．列出你知道的异质半导体材料系统。
GaAs/AlGaAs, InP/ InGaAs, Si/SiGe, 6．SOI材料是怎样形成的，有什么特点?SOI绝缘体上硅，可以通过氧隔离或者晶片粘结技术完成。
特点：电极与衬底之间寄生电容大大减少，器件速度更快，功率更低7. 肖特基接触和欧姆型接触各有什么特点？肖特基接触：阻挡层具有类似PN结的伏安特性。
欧姆型接触：载流子可以容易地利用量子遂穿效应相应自由传输。
8. 简述双极型晶体管和MOS晶体管的工作原理。
P19,21CH31．写出晶体外延的意义，列出三种外延生长方法，并比较各自的优缺点。
意义：用同质材料形成具有不同掺杂种类及浓度而具有不同性能的晶体层。
外延方法:液态生长，气相外延生长，金属有机物气相外延生长2．写出掩膜在IC制造过程中的作用，比较整版掩膜和单片掩膜的区别，列举三种掩膜的制造方法。
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1．集成电路的发展过程经历了哪些发展阶段？划分集成电路的标准是什么？集成电路的发展过程:•小规模集成电路(Small Scale IC，SSI)•中规模集成电路(Medium Scale IC，MSI)•大规模集成电路(Large Scale IC，LSI)•超大规模集成电路(Very Large Scale IC，VLSI)•特大规模集成电路(Ultra Large Scale IC，ULSI)•巨大规模集成电路(Gigantic Scale IC，GSI）划分集成电路规模的标准2．超大规模集成电路有哪些优点？1. 降低生产成本VLSI减少了体积和重量等,可靠性成万倍提高,功耗成万倍减少.2.提高工作速度VLSI内部连线很短,缩短了延迟时间.加工的技术越来越精细.电路工作速度的提高,主要是依靠减少尺寸获得.3. 降低功耗芯片内部电路尺寸小,连线短,分布电容小,驱动电路所需的功率下降.4. 简化逻辑电路芯片内部电路受干扰小,电路可简化.5.优越的可靠性采用VLSI后，元件数目和外部的接触点都大为减少，可靠性得到很大提高。
6.体积小重量轻7.缩短电子产品的设计和组装周期一片VLSI组件可以代替大量的元器件,组装工作极大的节省,生产线被压缩,加快了生产速度.3．简述双阱CMOS工艺制作CMOS反相器的工艺流程过程。
1、形成N阱2、形成P阱3、推阱4、形成场隔离区5、形成多晶硅栅6、形成硅化物7、形成N管源漏区8、形成P管源漏区9、形成接触孔10、形成第一层金属11、形成第一层金属12、形成穿通接触孔13、形成第二层金属14、合金15、形成钝化层16、测试、封装，完成集成电路的制造工艺4．在VLSI设计中，对互连线的要求和可能的互连线材料是什么？互连线的要求低电阻值：产生的电压降最小；信号传输延时最小（RC时间常数最小化）与器件之间的接触电阻低长期可靠工作可能的互连线材料金属（低电阻率），多晶硅（中等电阻率），高掺杂区的硅(注入或扩散)（中等电阻率）5．在进行版图设计时为什么要制定版图设计规则？—片集成电路上有成千上万个晶体管和电阻等元件以及大量的连线。
描述这些基本单元的版图，是一系列几何图形有规则的排列。
为了保证由这些基本单元及其相互连线构成的版图能够在工艺线上生产出来，必须制定版图设计规则。
在芯片尺寸尽可能小的前提下，使得即使存在工艺偏差也可以正确的制造出IC，尽可能地提高电路制备的成品率。
6．版图验证和检查主要包括哪些方面？DRC(Design Rule Check)：几何设计规则检查；对IC的版图做几何空间检查，保证能在特定的工艺条件下实现所设计的电路，并保证一定的成品率；ERC(Electrical Rule Check)：电学规则检查；检查电源(power)/地(ground)的短路，浮空的器件和浮空的连线等指定的电气特性；LVS(Loyout versus Schematic)：网表一致性检查；将版图提出的网表和原理图的网表进行比较，检查电路连接关系是否正确，MOS晶体管的长/宽尺寸是否匹配，电阻/电容值是否正确等；LPE(Layout Parameter Extraction)：版图寄生参数提取；从版图中提取晶体管的尺寸、结点的寄生电容、连线的寄生电阻等参数，并产生SPICE 格式的网表，用于后仿真验证；POSTSIM：后仿真，检查版图寄生参数对设计的影响；提取实际版图参数、电阻、电容，生成带寄生量的器件级网表，进行开关级逻辑模拟或电路模拟，以验证设计出的电路功能的正确性和时序性能等，并产生测试向量。
7．版图设计规则是根据什么制定出来的？为什么说它是集成电路的性能和集成度与成品率之间的折衷？从图形如何精确地光刻到芯片上出发，可以确定一些对几何图形的最小尺寸限制规则，这些规则被称为设计规则。
设计规则是电路性能和成品率之间的折中，设计规则保守则成品率高，但电路面积大、性能差一些；设计规则激进，则电路性能好、面积小，但成品率低。
8．简述λ设计规则与微米设计规则各自的优缺点？以λ为单位：把大多数尺寸（width，space等等）约定为λ的倍数λ与工艺线所具有的工艺分辨率有关，是线宽偏离理想特征尺寸的上限以及掩膜版之间的最大套准偏差，一般等于栅长度的一半。
优点：版图设计独立于工艺和实际尺寸，改变λ值就可以得到不同的设计规则；缺点：容易造成芯片面积浪费和工艺难度增加；以微米为单位：现代IC设计普遍采用的方法，每个尺寸之间没有必然的比例关系，提高每一尺寸的合理程度；简化度不高。
9．标准单元法与门阵列法比较有何优点和缺点？标准单元法与门阵列法比较有明显的优点：(1) 芯片面积的利用率比门阵列法要高。
芯片中没有无用的单元，也没有无用的晶体管。
(2) 可以保证100％的连线布通率。
(3) 单元可以根据设计要求临时加以特殊设计并加入库内，因而可以得到较佳的电路性能。
(4) 可以与全定制设计法相结合功能块。
在芯片内放入经编译得到的宏单元或人工设计的功能块。
标准单元法也存在缺点和问题；(1) 原始投资大。
单元库的开发需要投入大量的人力物力；当工艺变化时，单元的修改工作需要付出相当大的代价。
因而如何建立一个在比较长的时间内能适应技术发展的单元库是一个突出问题。
(2) 成本较高。
由于掩膜版带要全部定制，芯片的加工也要经过全过程，因而成本较高。
因此只有芯片产量达到某一定额(几万至十几万)，其成本才可接受。
10．随着工艺进入深亚微米，IC器件的物理实现出现了哪些方面的变化？随着工艺进入深亚微米，IC器件的物理实现出现了以下3个方面的变化：(1) 逻辑单元的几何尺寸和逻辑单元之间的距离随着特征尺寸的减小而减小，从而使总延时减小。
(2) 由于特征尺寸的减小，导线电阻增加。
为了抵消导线横向尺寸的减小，导线侧向尺寸即厚度被适度增加，以使导线电阻的增加不至于过大，从而导致纵向分布电容和边缘分布电容的增加，这两种分布电容都具有导致导线间耦合的性质。
(3) 连线延时(主要是侧向分布电容和边缘分布电容引入的延时)在总延时中占据了主导地位，而输入延时也由于工作频率的提高而变得不容忽视。
11．FPGA与CPLD有何相似之处和不同之处？FPGA是CPLD的一个发展最快的分支,复杂的可编程逻辑器件CPLD是由PLD或GAL发展而来. CPLD延伸出两大分支,即可擦除可编程的逻辑器件EPLD和现场可编程门阵列器件FPGA.中的数据丢失。
因此，需在FPGA外加EPROM，将配置数据写入其中，系统每次上电自动将数据引入SRAM 中。
CPLD器件一般采用EEPROM存储技术，可重复编程，并且系统掉电后，EEPROM中的数据不会丢失，适于数据的保密。
3.FPGA器件含有丰富的触发器资源，易于实现时序逻辑，如果要求实现较复杂的组合电路则需要几个CLB结合起来实现。
CPLD的与或阵列结构，使其适于实现大规模的组合功能，但触发器资源相对较少。
4.FPGA为细粒度结构，CPLD为粗粒度结构。
5.FPGA为非连续式布线，CPLD为连续式布线。
12．可测试性设计的对象是什么？为什么要从事VLSI的可测试性设计？电路的可控制性和可观察性。
Pin数目有限，大量芯片内部的信息无法访问。
在尽可能少地增加附加引线脚和附加电路，并使芯片性能损失最小的情况下，满足电路可控制性和可观察性的要求。
13． 为什么说MOS 管的工作速度与2L 成反比？提高MOS 管的工作速度方法有哪些？（见18题答案） 14．某CMOS 电路负载电容近似等于g C e 8，g C 为标准反相器栅电容。
已知标准反相器的平均延迟时间ns t av 2=。
试求：（1）用标准反相器直接驱动负载电容的延迟时间。
（2）用逐级放大反相器直接驱动负载电容的最小延迟时间。
15．什么是可测性设计？可测性设计包括哪些技术？可测试性包括哪些重要方面？ 可测性设计: 在尽可能少地增加附加引线脚和附加电路，并使芯片性能损失最小的情况下，满足电路可控制性和可观察性的要求。
可测性设计技术：主要包括分块测试技术、扫描测试技术、特征量分析分块测试技术、边界扫描分块测试技术等。
可测试性的三个重要方面：故障模型的提取：将电路失效抽象为故障模型。
测试生成：产生验证电路的一组测试矢量。
测试设计：考虑测试效率问题，加入适当的附加逻辑或电路以提高芯片的测试效率。
16．目前VLSI 系统设计普遍采用的方法是什么？它的基本思想什么？试列举几种设计方法。
可编程逻辑器件设计方法（PLD 方法）。
用户通过生产商提供的通用器件自行进行现场编程和制造，或者通过对与或矩阵进行掩膜编程，得到所需的专用集成电路。
PLA 、PAL 和GALFPGA 和CPLD17．半定制设计方法可分为哪几种方法？它们各自的特点和不足之处是什么？半定制的设计方法分为门阵列（GA ：Gate Array ）法和门海（GS ：Sea of Gates ）法两种。
门阵列方法的设计特点：设计周期短，设计成本低，适合设计适当规模、中等性能、要为沟道长度。
成正比，与为常数，当不变时，L L V L n dsn 22,τμμτ=求设计时间短、数量相对较少的电路。
不足：设计灵活性较低；门利用率低；芯片面积浪费。
门海方法的设计特点：门利用率高，集成密度大，布线灵活，保证布线布通率。
不足：仍有布线通道，增加通道是单元高度的整数倍，布线通道下的晶体管不可用。
18．试分析提高MOS 管工作速度方法。
1. 提高IC 加工精度 减小沟道长度.2. 加强MOS 管的驱动电压（Vgs-Vt ），可以减小管子的内阻,加快工作速度。
3. 由于 ,所以NMOS 管的工作速度比PMOS 管快得多。
可以用NMOS 工艺代替PMOS 工艺。
19．画出C M O S 反相器电压传输特性曲线图，并写出相应的电流方程。
电流方程如下：设 Vtp=-Vtn()()()22200022i tn n tn i o n n tn i V V V V V i tn V V V i tn i tn o V V I V V V V V ββ⎧⎪--------------------≤<⎪⎪=-------------≤<⎨⎪⎪⎡⎤----+≤⎪⎢⎥⎣⎦⎩----tn 截止+V 饱和 线性()()()222022dd tp i dd p o tp i dd tp p p i o tp V V V V V V V V V i tp dd V V V i tp dd i tp o V V V I V V V V V V ββ⎧⎪-----------------------+<<⎪⎪⎪=-----------+<≤+⎨⎪⎪⎡⎤----<+⎪⎢⎥⎪⎣⎦⎩------ 截止 饱和 线性2.5n p μμ≈。