sheet1 - semi we m Samm nms an 1 For[he FeFeJC phase...

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Unformatted text preview: semi... we) m) Samm- nms an 1) For [he FeFeJC phase diagram shown above: (15 pts.) a) Whatisihc sohihilitylrmitofCinFeat NWT? [7) What is the composition uflh: Fe- FEJC alloy for which the mess Reaction eta-ferrite is 0.90 at 726‘C? e) What is die composition of the Fe-Fegc alloy for Wind] the mass fraction Dfpfllfllcmiti wfarile is 0.99 at. 726%?! a) The solubility limit is obtained simply by reading the figure, starting from 0 Ivt‘Va C at 1000'C and finding the maximum C concentration that can he dissolved in the 1 phase. By inspection, this is about 1.8 “1% C. b) Using the Ivar rule as usual: Cf,.c * C], W: = 7 6.70 e c, Ch; — c, : 6.7070322 Cx : 0.69 wta/n C c) Using the lever rule, but remembering that the pro-eutectoid phase precipitates prior to the eutectnid reaction: C a C o 767 C w , (726°C) : w (728‘6‘) : # 030 : ¥ m E I Cs... 4;, 037670.022 0.90 an AD 50 so 10 1500 26m 1400 1300 2400 E E 2 2 E 1200 2200 E i s i 1100 —zooo 1000 man 900 70 Gnlrlutsitiuu (M15 Tl) From the inn phase diagram above: (15 pts ) a) What are the likely chemical formulas of the two intermetallic compounds? 12) Fora 30 wP/h Ti alloy at 1100“C, what is the mass fraction nfTi Within the 7 phase? c) Fora30 w1% Ti alloy at llUU”C,whatisthemass fraction ufdie fiphnse? The chemical formulas can he guessed by reading from the upper x—axis, where composition is given in atom “/u. The vertical line next to the 1' phase corresponds to exactly 50 atom “/- Ti. Therefore the 1- phase corresponds in the compound NiTi. Similarly, the 5 phase corresponds to 67 atom “A Ti, so this corresponds to the compound NiTiz. The mass fraction 0! each phase is read from the end points oi the horizontal tie line. The right side of this tie line intersects with the 7 phase neginn at n . ahout39wt/oTl. 7397:“) w , :Cr—Cs 5 3970 Cris“7 w,:0.23 Using the lever rule again: "’5 Giver the information below for Cu, what 1: the numba of free electrons per Cu Example Problem: amen {lipts) a = 6.0x107(fl-m)'1 p = [5.94 glen? p, o 0030 nflvsee NA: 6.023x1 A : 63.55 gjmnl You can also see Equation (5 2), N : concentration of free electrons: a WP. 3:107 0—1": Ir : 1151:1029 electrons/m3 Nip N = e| = 1.602x10'” c 3 alums/mole A tungsten light hull: filament is 9 mm long and 10 um in diameter. What is the amen! in the filament when opaoting at 1000“C with a line voltage of 110 W Here p9 = -]'l.5x1()'9 thin and a = 2.4sx10'1“ n-mPK. Reading this problem carefully, we are really being asked to determine the resistancehesistiviw of the tungsten filament at 1000°C. lfwe know the filament resistance and line voltage current calculation follows trivially from Ohm‘s law. We know that Pr = Po 4"”- p(1ooo°C) = —17.5x10*n — m +(2.4sx1tr‘"Q — ml°KX1273°K) p(10l]0°('): 2.9sx10'7r2 — m If you instead do this calculation at room temperature, p(20°C) : 5..’>lxlC|'a fi-m. Note that 3 3 a the resistivity is > 57: higher Than at room temperature” New from Ohni‘s law and the N = 6.023x10 atoms More 8.94 girm lflfl cmfm) definition efresisn'vity: (63.55 gimme N = 8.47::10" mom: :1 1.25110“ electrons f m 3 i n — : 1.48 electronsfatom N 8.4-7.1:leimmune:3 The micron-manure below is observed at 716T. What might be me C composition amisr-‘eregc anew This phase diagram is shown in Problem #1. (5 pit) nor mm: c mg 0.10 an. %C 0.76 wt.% C 1.30 wt.% C Eutectalfl o 6.70 wr.% C For the MgO/AlsOJ phase diagram below. what is the primaryphase for a 60 wt% V VA (110V)£(10x10"m)2 1:}:— I: 774 73 :3.22A Pl l2.98x10 n — m i9x10 ml Example Prohlem: Starting from an ambient temperature of 300 K. what temperature increase is necessary to double the conductivity of pure Ge? For the sake of simplicity, assume that temperature }p,mme affects only the carrier concentration. not the electron and hole mobility. If the only effect of tempera-mire is on the carrier concentration. we can write that: E d=dflex — 1 2k? Pluellleclnid a A1201 alloy cooled li'om lhe liquid phase to room temperature? (5 pas.) 9 llnwm rt) W) Which ofthe following can decrease the conductivity ufa Cuwire? Cold Working Small amounts of impurity elements Increased tempmmre Grain boundaries All of the above WW" WW "fly Take the Arrhenius form of this equation for two temperatures, ambient temperature and that which doubles the conductivity, and divide them: an), mew , has.) 1.. “(TJ .& Li derangenm ’3 «(12) 21: T. T. ;_i 300K T1 1 1 e 777 = 1.784x10 “ [K 300 K T1 qutmlflln.) (Anna: Tl 7 317K which of the following is an equilibrium nnuosrruclure‘? a) Manmsuc h) Bainile c) Pearlile d) Sphereiditc c) All of the above Extrinsic, elemental semiconductors: In the semiconductor industry, Al wiring is sometimes alloyed with Cu to the conductivity of a n—type semiconductor is: improve its longs/em. performance as an electrical conductor. Which of the following will likely have the highest electrical conductivity? (5 pie.) 0' : I! I e I y, 100 wt% A1 99.5 wt.% Al, 0.5 “1% Cu 99.0 wt.% Al. 1.0 wt.% Cu 98.5m%A1,15wt.%Co No diiference Phase Diagram equaiions: Ohviolmly, these mimt add up to 1.0 4) Go? is a semiconductor material with a bandga uf2.25 eV. an electron muhility of 0.03 mZNes, and a hole mobility of0.015 in N75. For intrinsic Gal), what is the approiomate inm'ease in the conductivity in going fl'nm 20"C to 60°C? The answer should be ghen as a ratio. Also, k: 8.621(10’j cV/atcme. (15 pm.) We want to use Equation (12.36), but neglect the temperature dependence of the mobility, sinee we lack information on this. E! 011‘): exp 7 MIT Thus the ratio of conductivities at the two temperatures is: El 6 , flemi fin; earl)— [E']_ pm T2 T1 exp— nor, (1(60'C) = “p[ (2.25 eV) [ 1 1 ]] a(20"C) 2(a52xnfi eVlnlant 7 Hr) 293w 7 333°K fl = 211 a'(2ll°C) and the conductivity of a p-type semiconductor is: Uznlelfll, the conductivity of an intrinsic, elemental semiconductor is: 6:”lelfle'l'1’lelfli a:nilel(fle+#h) C“ 7 C]. R + s : a The temperature dependence of the carrier concentration E E nr. xexp —7g 11ml. {—7g ZkT ZkT We can define the resistivity (p) V 2 IR ,, - g .l where A is the cross-sectional area perpendicular to current flow and l is the length. conductivity (6), another intrinsic material 1 0':— ,0 Thus the conductivity for most materials is: a=n|em Conductors: Prior 2P1+P3+P4 P: :Po+“T The increase in resistivity with ternperamre can be described according to: Pr :PG‘HJT linear relationship between 9 and impurity concentration, ...
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