PROPERTIES OF MATERIALS IN ELECTRICAL ENGINEERING MIME262 IN-TERM EXAM #1

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1 IN-TERM EXAM #1, Oct 17 th, 2007 DURATION: 45 mins Version τ CLOSED BOOK McGill ID Name All students must have one of the following types of calculators: CASIO fx-115, CASIO fx-991, CASIO fx-570ms SHARP EL-520, or the SHARP EL-546. NON-REGULATION CALCULATORS WILL BE REMOVED AND NO REPLACEMENT CALCULATOR WILL BE PROVIDED. Leave your calculator cover in your book-bag. You may take only your calculator, a ruler, a pen and your student ID to your desk. NO equation sheets. Rulers are subject to inspection. McGill University Dept. of Mining, Metals and Materials Engineering PROPERTIES OF MATERIALS IN ELECTRICAL ENGINEERING MIME262 IN-TERM EXAM #1 Examiner: Prof. R.R. Chromik Date: Wednesday, October 17 th, 2007 Time: 45 mins Instructions 1. Read the instructions carefully. 2. Put your name and ID# on this page. Put your NAME on pages Read each question completely before working on each part, (the parts may be interrelated). 4. Write your solutions in the space provided below the relevant question. If necessary, you may continue work on the opposite side of the paper for which the question is printed. 5. Any work you wish to be ignored should be clearly indicated as such. 6. Draw any diagrams as large as possible in the space provided. 7. Label all diagrams, axes, curves etc. 8. For concept type questions, the space provided roughly indicates the level of detail expected. 9. Phase diagrams, a periodic table, graphs of erf(z) & erfc(z) and other useful information are attached at the end of this examination book (pages i through iii). Official USE ONLY P1 P2 P3 P4 P5 Bonus

2 1. True/False Section (1 pt each) - Please clearly indicate your answer by circling either true or false. If marks are made on both choices or neither choice, no points will be awarded. i. Metal atoms can only form metallic bonds. When bonding with non-metals, metals will use ionic bonds. ii. A phase must be made up of at least two components. A pure element is a phase. iii. Vacancy diffusion is the diffusion mechanism for most pure metals. In a pure material, self-diffusion requires vacancies be present. iv. Hexagonal close-packed (HCP) and face-centered cubic (FCC) structures have the same atomic packing fraction (APF). Both are closest-packed. v. Transition metals and oxygen always form fully ionic bonds. There is a percent ionic character that describes how bonds will range from ionic to covalent. vi. A crystalline plane with a high surface energy would have a correspondingly high planar density. High surface energy = fewer bonds than the bulk = lower planar density. vii. Plane A shown below would have Miller Indices of (120). viii. Isomorphous phase diagrams only occur for two components with the same crystal structure. Isomorphous means same crystal structure. ix. Metallic bonds are directional.

3 x. Covalent bonds are directional. xi. Ionic bonds are epitomized by sharing of electrons. Ionic bonds are created by the donation of an electron from one atom to another. xii. Defects and dislocations exist because they lower a material s energy. Features that differ from a perfect crystal will add energy. xiii. The number of vacancies in a metal increases linearly with temperature. The dependence is exponential (see equation sheet). xiv. The wire demo showed the important peritectoid transformation in the Fe-C phase diagram. It was a eutectoid transformation. xv. The Bragg condition states that, for constructive interference, the path difference between two diffracting waves must equal a half-integer number of wavelengths. Constructive interference would be an integer number. xvi. A eutectic microstructure will only occur for an alloy with the eutectic composition. Eutectic microstructures will appear for both hypo- and hypereutectic compositions. xvii. Polycrystalline materials exhibit extreme anisotropy in their physical properties. Averaging over many grains of different orientations leads to more isotropy. xviii. A single crystal only has low angle grain boundaries. A single crystal has NO grain boundaries. xix. A solid solution is a new phase with a crystal structure different from the pure components. Solid solutions have the same crystal structure of at least one of the pure components. xx. Fick s First Law is sufficient to evaluate engineering processes like doping of Si for microelectronics and carburization of steel. These processes require Fick s Second Law.

4 xxi. The picture below, a representation of a planar arrangement of spheres, is not close packed. The platform at Lionel-Groulx is close packed. xxii. By cooling a solid solution, a precipitation strengthened conductor will be made. If the composition is less than the solubility limit, it will remain a solid solution. xxiii. Closest-packed crystal structures are so close packed that they have no interstitial sites large enough for an impurity atom to reside. Small atoms, such as C and O can fit into interstitial sites of metals, many of which are closestpacked (FCC or HCP). xxiv. Ionic solids, such as NaCl, have no geometrical constraints on how the atoms are arranged on the lattice. The ions must be arranged to conserve charge (plus-minus-plus-minus, etc.). xxv. Solid solution and precipitation strengthened conductors have no practical applications because they are less conductive than pure metals. Many examples of applications for strengthened conductors were mentioned in lecture.

5 2. Ag-Ti Phase Diagram (30 pts) Use the attached Ag-Ti phase diagram to complete this problem. Assume that the dashed lines are true, experimentally confirmed phase boundaries (e.g. solid lines). a) Identify the three invariant points labeled on the diagram. Point Type of invariant point i Eutectoid ii Peritectoid iii Peritectic b) You are a prisoner in the lab of Dr. Octopus and are watching him formulate the electronics for his super-strong appendages. While the super-strong part is mostly Ti, he is using Ag with a bit of Ti for a precipitation strengthened conductor. i. You escape Doc-Oc s lab and rush to tell Spiderman about your findings about his nemesis. Doc-Oc is using an alloy of 96.5at% Ag and processes the alloy at 600 C. Complete a lever rule calculation for the composition of this alloy. Name of Phase 1 at% Ag in Phase 1 mol fraction in Phase 1 Name of Phase 2 at% Ag in Phase 2 mol fraction in Phase 2 TiAg Ag Phase I: ( )/(98-50) = 0.03 Phase II: ( )/98-50) = 0.97 ii. Spiderman, amazingly enough, asks you about the Hume-Rothery rules for this phase diagram. Considering the phase diagram and the table below, discuss how the Hume- Rothery rules do or do not explain solubility and compound formation in this system. You do not need to consider the valence. Crystal Structure Atomic Radius (nm) Electronegativity Ag Face-Centered Cubic Ti Hexagonal Close-Packed ) Compare crystal structure: different => complete solubility not possible This observation agrees with phase diagram. 2) Radii difference: ( )/0.16 * 100 = 12.5% < 15 Therefore, some solubility is likely This makes sense, from the phase diagram, Ag can dissolve in Ti and Ti can dissolve in Ag up to some amount. 3) Electronegativity: = 0.4 A difference of 0.4 is borderline and may or may not encourage compound formation. There are two compounds on the phase diagram (Ti 2 Ag and TiAg), which is reasonable considering this electronegativity difference.

6 3. Diffusion of P in Si (25 pts) A device engineer at AMD is trying to optimize their doping process at a new temperature. Working with a pure Si wafer, she exposes it to a constant surface dose of 8x10 19 atoms/m 3. For a two minute exposure, a P concentration of 3.22x10 19 atoms/m 3 is measured at a depth of 250 nm. a) From this information, calculate the diffusion coefficient for P in Si. C s = 8x10 19 atoms/m 3, C o = zero, C x = 3.22x10 19 atoms/m 3 x = 250 nm = 2.5x10-7 m, t = 2 min = 120 second ( x, t) C C s C C o o = 1 erf 2 x Dt C x /C s = 3.22x10 19 atoms/m 3 /8x10 19 atoms/m 3 = = erfc (z) Therefore, from the graph, z = 0.6 x / 2(Dt) 0.5 = 0.6 Rearranging, D = (2.5x10-7 m / 0.6) 2 /(4*120 sec) = 3.6x10-16 m 2 /s = 3.6x10-12 cm 2 /s b) The company s diffusion database gives a pre-factor of 1.41x10-5 cm 2 /s and an activation energy of 1.8 ev/atom. What new operating temperature has the engineer chosen? D = D o exp (-Q/k b T) Rearranging, ln D = ln D o Q/k b T, and therefore, T = Q*(ln D o ln D)/k b T = (1.8 ev/atom / 8.62x10-5 ev/atom K) * (ln 1.41x10-5 cm 2 /s ln 3.6x10-12 cm 2 /s) = 1376 K

7 4. Bonding Energy (10 pts) A plot of bonding energy (E) versus interatomic spacing (r) can be used to explain the bulk phenomenon of thermal expansion. Discuss how this works, referring to the graph below. Repulsive energy E R Interatomic separation r Net energy E N Attractive energy E A Example Answer: At the initial temperature, the atom sits in the potential well and has an average atomic spacing marked by the vertical dashed line. As temperature increases, the energy increases (marked by the horizontal line). We can consider that there is now a range of equilibrium atomic spacings bounded by the left and right-hand sides of the well. The purple curve marks this average. Due to the shape of the potential energy curve, the average atomic spacing will increase as energy, or temperature, is increased. This explains, through a consideration of the potential energy between atoms, the bulk phenomenon of thermal expansion. 5. Conductivity (10 pts) You are given two Cu wires. One was just drawn from a die and has a high density of dislocations. The other is Cu-0.5at%Ag and was annealed to remove almost all of the dislocations. Discuss the conductivity of both wires. Can you say which has better conductivity? Example Answer: For both cases we consider Matthieson s rule, where both defects (and dislocations) and impurities will increase resistivity (and decrease conductivity). The wire with a high density of dislocations will have a reduced conductivity from the presence of these defects. The wire with the 0.5wt% impurities will also have a reduced conductivity from the presence of the impurities. Without measuring the electrical properties directly, we cannot say which would have a better conductivity.

8 BONUS (4 pts) Describe two engineering design requirements for microelectronics packaging. See lecture notes and Ch. 22 of Callister. DO NOT WRITE BELOW THIS LINE ON THIS PAGE

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11 ( x, t) C C s C C o o = 1 erf 2 x Dt erfc(z) = 1 erf(z) erf (-z) = - erf (z) Other possibly useful constants, equations and relations: The gas constant, R = J/mol/K. Boltzmann s constant, k b = 8.62x10-5 ev/k = 1.38x10-23 J/K. Avogadro s number is 6.022x10 23 e = and π = D = D o exp (-Q / RT) D = D o exp (-Q / k b T) N v = N exp (-Q v / RT) N v = N exp (-Q v / k b T) E A = A/r and E R = B/r n %ionic character = {1-exp[-(0.25)(X A X B ) 2 ]} x 100