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 etaferrite is 0.90 at 726‘C?
e) What is die composition of the FeFegc 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 proeutectoid 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(ﬂm)'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“ nmPK. 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 ﬁm. 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 8.47.1:leimmune:3 The micronmanure 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 temperamire 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 ptype 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 crosssectional 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=nem 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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 Spring '08
 Rasmussen

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