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.docx

1、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 are these 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

2、of the dopant concentration: a) b) c) 1.3. Consider a piece of pure silicon 100 m long with a cross-sectional area of 1 m2. How much 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 concentrati

3、on will be simply ni. At room temperature, ni 1.45 x 1010 cm-3. Under an applied field, the current will be due to drift and hence, 1.10. A state-of-the-art NMOS transistor might have a drain junction area of 0.5 x 0.5 m. Calculate the junction capacitance associated with this junction at an applied

4、 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.The junction built-in voltage is given by Eqn. 1.24. ND is not specified except that it is very large, so we take it to be 1

5、020 cm-3 (roughly solid solubility). The exact choice for ND doesnt make much difference in the answer.Since ND NA in this structure, the capacitance expression simplifies toGiven 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 crysta

6、l pulled by the Czochralski technique is required to have a resistivity of 10 cm when half the crystal is grown. Assuming that a 100 gm pure silicon charge is used, how much 0.01 cm boron doped silicon must be added to the melt? For this crystal, plot resistivity as a function of the fraction of the

7、 melt solidified. Assume k0 = 0.8 and the hole mobility p = 550 cm2 volt-1 sec-1. Answer:Using the mobility value given, and we have:10 cm NA = 1.14 x 1015 cm-3 and 0.01 cm NA = 1.14 x 1018 cm-3From Eqn. 3.38, and we want CS = 1.14 x 1015 cm-3 when f = 0.5. Thus, solving for C0 the initial doping co

8、ncentration in the melt, we have:The resistivity as a function of distance is plotted below and is given by 3.3. A Czochralski crystal is pulled from a melt containing 1015 cm-3 boron and 2x1014 cm-3 phosphorus. Initially the crystal will be P type but as it is pulled, more and more phosphorus will

9、build up in the liquid because of segregation. At some point the crystal will become N type. Assuming kO = 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 crysta

10、l becomes N type from Eqn. 3.38 as follows:At the point where the cross-over occurs to N type, these two concentrations will be equal. Solving for f, we findThus 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 the semiconducto

11、r 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

12、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 expl

13、ained 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 kO value is slightly closer to 1. 4.1. An IC

14、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 w

15、afer 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 inc

16、ome is simplyIncome = (52,000)(50)($50) = $130,000,000 per year.To increase this income by $10,000,000 requires that the yield increase by4.3. As MOS devices are scaled to smaller dimensions, gate oxides must be reduced in thickness. a. As the gate oxide thickness decreases, do MOS devices become mo

17、re or less sensitive to sodium contamination? Explain. b. As the gate oxide thickness decreases, what must be done to the substrate doping (or alternatively the channel VTH implant, to maintain the same VTH? Explain.Answer:a). From the text, Na+ contamination causes threshold voltage instabilities i

18、n MOS devices. Also from Eqn. 4.1, the threshold voltage is given byAs the gate oxide thickness decreases, COX increases, so the same amount of mobile charge QM will have less effect on VTH as oxides get thinner. Therefore MOS devices are less sensitive to sodium contamination.b). Using the same exp

19、ression for VTH as in part a), we observe that as the oxide thickness decreases, (COX increases), to maintain the same VTH, NA will have to increase. NA will actually have to increase by the square root of the oxide thickness decrease to keep VTH constant. 4.4. A new cleaning procedure has been prop

20、osed which is based on H2O saturated with O2 as 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/O2 solution. Will this clean the wafer effectivel

21、y? Why or why not? Explain.Answer: As described in the text, cleaning metal ions off of silicon wafers involves the following chemistry: 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 d

22、riving 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,

23、we have:Oxidant/ReductantStandard OxidationPotential (volts)Oxidation-Reduction ReactionSiO2/Si0.84/Fe0.17/Cu-0.34/H2O-1.23/Au-1.42The 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 impo

24、rtant that the generation lifetime measurement illustrated in Figure 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 requir

25、ed 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 of focus for a project

26、ion exposure system with a NA of 0.6 (about the best that can be done today). Assume k1 = 0.6 and k2 = 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,

27、i-line, KrF and ArF). Will an ArF source be adequate for the 0.13 m and 0.1 m technology generations according to these simple calculations?Answer:The relevant equations are simplyThese equations are plotted below. Note that the ArF (193 nm) will not reach 0.13 m or 0.1 m resolution according to the

28、se simple calculations. In fact, with more sophisticated techniques such as phase shift masks, off axis illumination etc., ArF is expected to reach 0.13 m and perhaps the 0.1 m generations.5.3. An X-ray exposure system uses photons with an energy of 1 keV. If the separation between the mask and wafe

29、r is 20 m, estimate the diffraction limited resolution that is achievable by this system.Answer:The equivalent wavelength of 1 keV x-rays is given byX-ray systems operate in the proximity printing mode, so that the theoretical resolution is given by Eqn. 5.12:5.8. As described in this chapter, there

30、 are no clear choices for lithography systems beyond optical projection tools based on 193-nm ArF eximer lasers. One possibility is an optical projection system using a 157-nm F2 excimer laser. a. Assuming a numerical aperture of 0.8 and k1 = 0.75, what is the expected resolution of such a system us

31、ing 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 m generation. Suggest three approaches to actually achieving this resolution with these systems.Answer a). The simple formula for resolution is b). The calculated resolution in part a is a factor of two larger than required for the 0.07 m generation. Therefore some “tricks” will have to be used to actually achieve such res