University of Texas Arlington Department of Electrical Engineering. Nanotechnology Microelectromechanical Systems Ph.D. Diagnostic Examination

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1 University of Texas Arlington Department of Electrical Engineering Nanotechnology Microelectromechanical Systems Ph.D. Diagnostic Examination Fall 2012 November 17, 2012 Question # To be filled by the student Check to have this question graded. Check only 2. To be filled by the graders Grade Grade Average TOTAL: GRADE OUT of 100: THIS EXAM PACKET HAS 9 SHEETS INCLUDING THIS SHEET AND THE HELPFUL EQUATIONS / CONSTANTS.

2 1. (50 pts) The E-k relationship characterizing electrons confined to a two-dimensional semiconductor is given by: k k x y E Ec m m 2m 2m Assuming the dimensions of the two-dimensional semiconductor are much larger than the atomic spacing, derive an expression for the density of states per unit area function g2 ( E ) for the electrons. Show all your work. D

3 2. (50 points) Please answer all sections A. If you have two wafers, one un-doped and one with boron doping. Which wafer has a higher rate of oxidation? And why? (10 points) B. If 1 cm 3 of silicon doped with boron at concentration of cm -3 is turned into oxide through thermal oxidation, such that 0.5 cm 3 of silicon remains at the center. What will happen to boron concentration in silicon and SiO 2? Explain. (10 points) C. If we need to grow thermal oxide between two active areas of a device to isolate the active areas from each other and try to avoid further redistribution of the doped area, what technique of oxidation is suitable? What technique is called and how that can be achieved? (10 points) D. What wafer orientation is preferable for the fabrication of MOS devices? What are the reasons for the choice of this orientation? (10 points) E. What approaches are used to introduce water vapor during wet oxidation? Can we use steam to carry out wet oxidation? (10 points)

4 3. (50 pts) Assuming a cubic crystal system, answer the following questions. a) Make a sketch of (16 pts) b) Find and sketch the equivalent directions of <100>. (18 pts) c) Find the angle between 100 and 111. Show all your work. (16 pts)

5 COLOR CHART FOR THERMALLY GROWN SiO 2 FILMS (OBSERVED PERPENDICULARLY UNDER DAYLIGHT FLURORESCENT LIGHTING) OXIDE NITRIDE Film Thickness Order (Microns) (5450 A) Color and Comments A Tan Brown Dark Violet to red violet Royal blue Light blue to metallic blue Metallic to very light yellow-green Light gold or yellow slightly metallic Gold with slight yellow-orange Orange to Melon Red-Violet Blue to violet-blue Blue Blue to blue-green Light green Green to yellow-green Yellow-green Green- yellow Yellow Light orange Carnation pink Violet-red Red-violet Violet Blue Violet Blue Blue-green Green (Broad) Yellow-green Green-yellow Yellow to Yellowish (not yellow but is in the position where yellow is to be expected. At times is appears to be light creamy gray or metallic) Light orange or yellow to pink borderline Carnation pink Violet-red OXIDE NITRIDE Film Thickness Order (Microns) (5450 A) Bluish (Not blue but borderline between violet and blue-green. It appears more like a Mixture between violet -red and blue-green and over-all looks grayish Blue-green to green (quite broad Yellowish Orange (rather broad for orange) Salmon Dull, light red-violet Violet Blue-violet Blue Blue-green Dull yellow-green Yellow to Yellowish Orange Carnation Pink Violet-red Red-violet Violet Blue-violet Green Yellow-green Green Violet Red-violet Violet-red Carnation Pink-Salmon Orange Yellowish Sky blue to green-blue Orange Violet Blue-violet 1.50 Blue Dull Yellow-green

6 Physical Constants (in units frequently used in semiconductor electronics) Electronic charge q x C Speed of light in vacuum c x cm s -1 Permittivity of vacuum ε x F cm -1 Free electron mass m x kg Planck's constant h x J s x ev s Boltzmann's constant k 1.38 x J K x 10-5 ev K -1 Avogadro's number A x molecules (g mole) -1 Thermal voltage V t =kt/q at 80.6 F (300K) mv 68 F (293K) mv Conversion Factors 1 Ǻ = 10-8 cm = 0.1nm 1 mil = 10-3 inch = 25.4 µm 1 ev = x J 1 J = 10 7 erg

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11 Q 1(a) If you have two wafers, one un-doped and one with boron doping. Which wafer has a higher rate of oxidation? And why? (20 points) Ans: Heavy doping changes the oxidation characteristics. Boron doping increases the parabolic rate constant (B) but has little effect on the linear rate constant (B/A). Long oxidation see increased oxidation rate due to Boron. Q 1(b) If 1 cm 3 of silicon doped with boron at concentration of cm -3 is turned into oxide through thermal oxidation, such that 0.5 cm 3 of silicon remains at the center. What will happen to boron concentration in silicon and SiO 2? Explain. (20 points) Ans: The boron will diffuse slowly through the oxide. It will deplete from silicon and remain in oxide. At the end the concentration of boron will be much more in oxide than in silicon. Q 1(c) If we need to grow thermal oxide between two active areas of a device to isolate the active areas from each other and try to avoid further redistribution of the doped area, what technique of oxidation is suitable? What technique is called and how that can be achieved? (20 points) Ans: We would use Shallow Trench Isolation to achieve the isolation between the devices. Process flow in diagram. Full points if explained with diagram. Q 1(d) What wafer orientation is preferable for the fabrication of MOS devices? What are the reasons for the choice of this orientation? (20 points) Ans: The <100> wafer orientation results into smallest number of unsatisfied silicon bonds at the Si- SiO2 interface and this orientation yields the lowest number of interfacial traps. Hence <100> is mostly used for MOS devices. Q 1(e) What approaches are used to introduce water vapor during wet oxidation? Can we use steam to carry out wet oxidation? (20 points) Ans: The water vapor can be introduced by passing oxygen through a bubbler with DI water heated to 95 C. High-purity water can slo be obtained by burning hydrogen and oxygen in the furnace tube. Steam may not be used as it may cause pitting on the silicon.

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