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resistive
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About testbanksolutionmanual
test bank solution manual textbook
resistive
electron
3rd edition kasap
device 3rd edition
eu solution manual
http testbank360 eu
solution manual principles
testbank360 eu solution
alloy
<p>Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.1 Third Edition ( 2005 McGraw-Hill) Chapter 2 ___________________________________________________________________________________ 2.1 Electrical conduction Na is a monovalent metal (BCC) with a density of 0.9712 g cm-3. Its atomic mass is 22.99 g mol-1. The drift mobility of electrons in Na is 53 cm2 V-1 s-1. a. Consider the collection of conduction electrons in the solid. If each Na atom donates one electron to the electron sea, estimate the mean separation between the electrons. (Note: if n is the concentration of particles, then the particles’ mean separation d = 1/n1/3.) b. Estimate the mean separation between an electron (e-) and a metal ion (Na+), assuming that most of the time the electron prefers to be between two neighboring Na+ ions. What is the approximate Coulombic interaction energy (in eV) between an electron and an Na+ ion? c. How does this electron/metal-ion interaction energy compare with the average thermal energy per particle, according to the kinetic molecular theory of matter? Do you expect the kinetic molecular theory to be applicable to the conduction electrons in Na? If the mean electron/metal-ion interaction energy is of the same order of magnitude as the mean KE of the electrons, what is the mean speed of electrons in Na? Why should the mean kinetic energy be comparable to the mean electron/metal-ion interaction energy? d. Calculate the electrical conductivity of Na and compare this with the experimental value of 2.1 107 Ω-1 m-1 and comment on the difference. Solution a. If D is the density, Mat is the atomic mass and NA is Avogadro's number, then the atomic concentration nat is 3 28 1 3 1 23 at m 10 544 .2 ) mol kg 10 99 . 22 ( ) mol 10 022 .6 )( kg 2 . 971 ( M DN n A at which is also the electron concentration, given that each Na atom contributes 1 conduction electron. If d is the mean separation between the electrons then d and nat are related by (see Chapter 1 Solutions, Q1.11; this is only an estimate) 3 / 1 3 28 3 / 1 ) m 10 544 .2 ( 1 1 at n d = 3.40 10-10 m or 0.34 nm b. Na is BCC with 2 atoms in the unit cell. So if a is the lattice constant (side of the cubic unit cell), the density is given by 3 at 2 cell unit of volume ) atom 1 of mass )( cell unit in atoms ( a N M D A isolate for a, 3 / 1 1 23 3 3 1 3 3 / 1 at ) mol 10 022 .6 )( m kg 10 9712 .0 ( ) mol kg 10 99 . 22 ( 2 2 A DN M a Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.2 so that a = 4.284 10-10 m or 0.4284 nm For the BCC structure, the radius of the metal ion R and the lattice parameter a are related by (4R)2 = 3a2, so that, ) (3 (1/4) 2a R = 1.855 10-10 m or 0.1855 nm If the electron is somewhere roughly between two metal ions, then the mean electron to metal ion separation delectron-ion is roughly R. If delectron-ion R, the electrostatic potential energy PE between a conduction electron and one metal ion is then ) m 10 855 .1 )( m F 10 854 .8 ( 4 ) C 10 602 .1 )( C 10 602 .1 ( 4 ) )( ( 10 1 12 19 19 ion electron 0 d e e PE (1) PE = -1.24 10-18 J or -7.76 eV c. This electron-ion PE is much larger than the average thermal energy expected from the kinetic theory for a collection of “free” particles, that is Eaverage = KEaverage = 3(kT/2) 0.039 eV at 300 K. In the case of Na, the electron-ion interaction is very strong so we cannot assume that the electrons are moving around freely as if in the case of free gas particles in a cylinder. If we assume that the mean KE is roughly the same order of magnitude as the mean PE, J 10 24 .1 2 1 18 average 2 average PE u m KE e (2) where u is the mean speed (strictly, u = root mean square velocity) and me is the electron mass. Thus, 2 / 1 31 18 2 / 1 ) kg 10 109 .9 ( J) 10 24 . 1 ( 2 2 e average m PE u (3) so that u = 1.65 106 m/s There is a theorem in classical physics called the Virial theorem which states that if the interactions between particles in a system obey the inverse square law (as in Coulombic interactions) then the magnitude of the mean KE is equal to the magnitude of the mean PE. The Virial Theorem states that: average average PE KE 2 1 - = Indeed, using this expression in Eqn. (2), we would find that u = 1.05 106 m/s. If the conduction electrons were moving around freely and obeying the kinetic theory, then we would expect (1/2)meu 2 = (3/2)kT and u = 1.1 105 m/s, a much lower mean speed. Further, kinetic theory predicts that u increases as T1/2 whereas according to Eqns. (1) and (2), u is insensitive to the temperature. The experimental linear dependence between the resistivity and the absolute temperature T for most metals (non-magnetic) can only be explained by taking u = constant as implied by Eqns. (1) and (2). d. If is the drift mobility of the conduction electrons and n is their concentration, then the electrical conductivity of Na is = en. Assuming that each Na atom donates one conduction electron (n = nat), we have ) s V m 10 53 )( m 10 544 .2 )( C 10 602 .1 ( 1 1 2 4 3 28 19 en i.e. = 2.16 107 -1 m-1 which is quite close to the experimental value. Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.3 Nota Bene: If one takes the Na+-Na+ separation 2R to be roughly the mean electron-electron separation then this is 0.37 nm and close to d = 1/(n1/3) = 0.34 nm. In any event, all calculations are only approximate to highlight the main point. The interaction PE is substantial compared with the mean thermal energy and we cannot use (3/2)kT for the mean KE! 2.2 Electrical conduction The resistivity of aluminum at 25 C has been measured to be 2.72 10-8 m. The thermal coefficient of resistivity of aluminum at 0 C is 4.29 10-3 K-1. Aluminum has a valency of 3, a density of 2.70 g cm-3, and an atomic mass of 27. a. Calculate the resistivity of aluminum at ─40ºC. b. What is the thermal coefficient of resistivity at ─40ºC? c. Estimate the mean free time between collisions for the conduction electrons in aluminum at 25 C, and hence estimate their drift mobility. d. If the mean speed of the conduction electrons is about 2 106 m s-1, calculate the mean free path and compare this with the interatomic separation in Al (Al is FCC). What should be the thickness of an Al film that is deposited on an IC chip such that its resistivity is the same as that of bulk Al? e. What is the percentage change in the power loss due to Joule heating of the aluminum wire when the temperature drops from 25 C to ─40 ºC? Solution a. Apply the equation for temperature dependence of resistivity, (T) = o[1 + o(T-To)]. We have the temperature coefficient of resistivity, o, at To where To is the reference temperature. The two given reference temperatures are 0 C or 25 C, depending on choice. Taking To = 0 C + 273 = 273 K, (-40 C + 273 = 233 K) = o[1 + o(233 K273 K)] (25 C + 273 = 298 K) = o[1 + o(298 K273 K)] Divide the above two equations to eliminate o, (-40 C)/(25 C) = [1 + o(-40 K)] / [1 + o(25 K)] Next, substitute the given values (25 C) = 2.72 10-8 m and o = 4.29 10-3 K-1 to obtain K)] )(25 K 10 (4.29 + [1 K)] )(-40 K 10 (4.29 + [1 m) 10 (2.72 = C) (-40 1 - 3 - -1 -3 8 - = 2.03 10-8 m b. In (T) = o[1 + o(T To)] we have o at To where To is the reference temperature, for example, 0 C or 25 C depending on choice. We will choose To to be first at 0 C = 273 K and then at -40 C = 233 K so that (-40 C) = (0 C)[1 + o(233K ─ 273K)] and (0 C) = (-40 C)[1 + -40(273K ─ 233K)] Multiply and simplify the two equations above to obtain [1 + o(233 K273 K)][1 + -40(273 K233 K)] = 1 or [1 40o][1 + 40-40] = 1 Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.4 Rearranging, -40 = (1 / [140o] 1)(1 / 40) -40 = o / [140o] i.e. -40 = (4.29 10-3 K-1) / [1 (40 K)(4.29 10-3 K-1)] = 5.18 10-3 K-1 Alternatively, consider, (25 C) = (-40 C)[1 + -40(298 K 233 K)] so that -40 = [(25 C)(-40 C)] / [(-40 C)(65 K)] -40 = [2.72 10-8 m2.03 10-8 m] / [(2.03 10-8 m)(65 K)] -40 = 5.23 10 -3 K-1 c. We know that 1/ = = en where is the electrical conductivity, e is the electron charge, and is the electron drift mobility. We also know that = e / me, where is the mean free time between electron collisions and me is the electron mass. Therefore, 1/ = e2n/me = me/e2n (1) Here n is the number of conduction electrons per unit volume. But, from the density d and atomic mass Mat, atomic concentration of Al is 3 - 28 3 1 - 23 at Al m 10 022 .6 = kg/mol 027 .0 kg/m 2700 mol 10 022 6. = = M d N n A so that n = 3nAl = 1.807 1029 m-3 assuming that each Al atom contributes 3 "free" conduction electrons to the metal and substituting into (1), ) m 10 (1.807 C) 10 m)(1.602 10 (2.72 kg) 10 (9.109 3 - 29 2 19 - 8 - -31 2 n e me = 7.22 10-15 s (Note: If you do not convert to meters and instead use centimeters you will not get the correct answer because seconds is an SI unit.) The relation between the drift mobility d and the mean free time is given by Equation 2.5, so that kg s C m e e d 31 15 19 10 109 .9 10 22 .7 10 602 .1 d = 1.27 10-3 m2 V-1s-1 = 12.7 cm2 V-1s-1 d. The mean free path is l = u, where u is the mean speed. With u 2 106 m s-1 we find the mean free path: l = u = (2 106 m s-1)(7.22 10-15 s) 1.44 10-8 m 14.4 nm A thin film of Al must have a much greater thickness than l to show bulk behavior. Otherwise, scattering from the surfaces will increase the resistivity by virtue of Matthiessen's rule. e. Power P = I2R and is proportional to resistivity , assuming the rms current level stays relatively constant. Then we have Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.5 [P(-40 C) P(25 C)] / P(25 C) = P(-40 C) / P(25 C)1= (-40 C) / (25 C)1 = (2.03 10-8 m / 2.72 10-8 Ωm) ─1= -0.254, or -25.4% (Negative sign means a reduction in the power loss). 2.3 Conduction in gold Gold is in the same group as Cu and Ag. Assuming that each Au atom donates one conduction electron, calculate the drift mobility of the electrons in gold at 22��� C. What is the mean free path of the conduction electrons if their mean speed is 1.4 × 106 m s−1? (Use ρo and αo in Table 2.1.) Solution The drift mobility of electrons can be obtained by using the conductivity relation = end. Resistivity of pure gold from Table 2.1 at 0C (273 K) is 0 = 22.8 n m. Resistivity at 20 C can be calculated by using Eq. 2.19. )] ( 1 [ 0 0 0 T T The TCR 0 for Au from Table 2.1 is 1/251 K-1. Therefore the resistivity for Au at 22C is (22C)=22.8 n m [1 + 1/251 K-1(293K – 273K)] = 24.62 n m Since one Au atom donates one conduction electron, the electron concentration is at A M dN n where for gold d = density = 19300 kg m-3, atomic mass Mat = 196.67 g mol -1. Substituting for d, NA, and Mat, we have n = 5.91 1028 m-3, or 5.91 1022 cm-3. ) 10 91 . 5 )( 10 6 . 1 ( ) 10 62 . 24 ( 3 28 19 1 9 m C m en d = 4.2610-3 m2 V-1 s-1 = 42.6 cm2 V-1 s-1. Given the mean speed of electron is u = 1.4 × 106 m s−1, mean free path from Equation 2.10 is C ms kg s V m e u m l e d 19 1 6 31 1 1 2 3 10 6 .1 ) 10 4 . 1 )( 10 1 . 9 )( 10 26 . 4 ( = 3.39 10-8 m = 33.9 nm. ___________________________________________________________________________________ 2.4 Effective number of conduction electrons per atom a. Electron drift mobility in tin (Sn) is 3.9 cm2 V−1 s−1. The room temperature (20 C) resistivity of Sn is about 110 n m. Atomic mass Mat and density of Sn are 118.69 g mol−1 and 7.30 g cm−3, respectively. How many “free” electrons are donated by each Sn atom in the crystal? How does this compare with the position of Sn in Group IVB of the Periodic Table? b. Consider the resistivity of few selected metals from Groups I to IV in the Periodic Table in Table 2.7. Calculate the number of conduction electrons contributed per atom and compare this with the location of the element in the Periodic Table. What is your conclusion? Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.6 NOTE: Mobility from Hall-effect measurements. Solution a. Electron concentration can be calculated from the conductivity of Sn, = end. ) 10 9 .3 )( 10 6 .1 ( ) 10 110 ( 1 1 2 4 19 1 9 s V m C m e n d e = 1.461029 electrons m3. The atomic concentration, i.e. number of Sn atoms per unit volume is ) 10 69 . 118 ( ) 10 022 .6 )( 10 3 .7 ( 1 3 1 23 3 mol kg mol kg M dN n at A at = 3.70 1028 Sn atoms m-3. Hence the number of electrons donated by each atom is (ne/nat) = 3.94 or 4 electrons per Sn atom. This is in good agreement with the position of the Sn in the Periodic Table (IVB) and its valency of 4. b. Using the same method used above, the number of electrons donated by each atom of the element are calculated and tabulated as follows: Metal Periodic Group Valency Atomic Concentration nat (m -3) Electron Concentration ne (m -3) Number of electrons ne/nat Integer (ne/nat) Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.7 Na IA 1 2.5411028 2.8081028 1.105 1 Mg IIA 2 4.3111028 8.2621028 1.916 2 Ag IB 1 5.8621028 7.0191028 1.197 1 Zn IIB 2 6.5751028 1.3201029 2.007 2 Al IIIB 3 6.0261028 1.9651029 3.262 3 Sn IVB 4 3.7031028 1.4571029 3.934 4 Pb IVB 4 3.3131028 1.3191029 3.981 4 Table 2Q4-1: Number of electrons donated by various elements As evident from the above table, the calculated number of electrons donated by one atom of the element is the same as the valency of that element and the position in the periodic table. ___________________________________________________________________________________ 2.5 TCR and Matthiessen’s rule Determine the temperature coefficient of resistivity of pure iron and of electrotechnical steel (Fe with 4% C), which are used in various electrical machinery, at two temperatures: 0 C and 500 C. Comment on the similarities and differences in the resistivity versus temperature behavior shown in Figure 2.39 for the two materials. Solution Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.8 Figure 2Q5-1: Resistivity versus temperature for pure iron and 4% C steel. The temperature coefficient of resistivity o (TCR) is defined as follows: o o T o o T dT d o at Slope 1 where the slope is d/dT at T = To and o is the resistivity at T = To. To find the slope, we draw a tangent to the curve at T = To (To = 0 C and then To = 500 C) and obtain /T d/dT. One convenient way is to define T = 400 C and find on the tangent line and then calculate /T. Iron at 0 ºC: Slopeo (0.23 10-6 m 0 m) / (400 C) = 5.75 10-10 m C -1 Since o 0.11 10-6 m, o = Slopeo/o 0.00523 C -1 Fe + 4% C at 0 ºC: Slopeo (0.57 10-6 m0.4 10-6 m) / (400 C) = 4.25 10-10 m C -1 Since o 0.53 10-6 m, o = Slopeo/o 0.00802 C -1 Iron at 500 ºC: Slopeo (0.96 10-6 m0.4 10-6 m) / (400 C) = 1.40 10-9 m C -1 Since o 0.57 10-6 m, o = Slopeo/o 0.00245 C -1 Fe + 4% C at 500 ºC: Slopeo (1.05 10-6 m 0.68 10-6 m) / (400 C) = 9.25 10-10 m C -1 Since o 0.85 10-6 m, o = Slopeo/o 0.00109 C -1 *2.6 TCR of isomorphous alloys Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.9 a. Show that for an isomorphous alloy A%-B% (B% solute in A% solvent), the temperature coefficient of resistivity AB is given by AB A A AB where AB is the resistivity of the alloy (AB) and A and A are the resistivity and TCR of pure A. What are the assumptions behind this equation? b. Determine the composition of the Cu-Ni alloy that will have a TCR of 4×10-4 K-1, that is, a TCR that is an order of magnitude less than that of Cu. Over the composition range of interest, the resistivity of the Cu-Ni alloy can be calculated from ρCuNi ≈ ρCu + Ceff X (1-X), where Ceff, the effective Nordheim coefficient, is about 1310 nΩ m. Solution a. By the Nordheim rule, the resistivity of the alloy is alloy = o + CX (1-X). We can find the TCR of the alloy from its definition ) 1 ( 1 1 alloy alloy alloy alloy X CX dT d dT d o To obtain the desired equation, we must assume that C is temperature independent (i.e. the increase in the resistivity depends on the lattice distortion induced by the impurity) so that d[CX(1X)]/dT = 0, enabling us to substitute for do/dT using the definition of the TCR: o =(do/dT)/o. Substituting into the above equation: o o o dT d alloy alloy alloy 1 1 i.e. o o alloy alloy or A A AB AB Remember that all values for the alloy and pure substance must all be taken at the same temperature, or the equation is invalid. b. Assume room temperature T = 293 K. Using values for copper from Table 2.1 in Equation 2.19, Cu = 17.1 n m and Cu = 4.0 10-3 K-1, and from Table 2.3 the Nordheim coefficient of Ni dissolved in Cu is C = 1570 n m. We want to find the composition of the alloy such that CuNi = 4 10-4 K-1. Then, m nΩ 171.0 K 0.0004 ) m nΩ 17.1 )( K 0.0040 ( 1 1 alloy Cu Cu alloy Using Nordheim’s rule: alloy = Cu + CX(1X) i.e. 171.0 n m = 17.1 n m + (1570 n m)X(1X) X2─X+ 0.0879 = 0 solving the quadratic, we find X = 0.11. Thus the composition is 89% Cu-11% Ni. However, this value is in atomic percent as the Nordheim coefficient is in atomic percent. Note that as Cu and Ni are very close in the Periodic Table this would also be the weight percentage. Note: the quadratic will produce another value, namely X = Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.10 0.86. However, using this number to obtain a composition of 11% Cu-89% Ni is incorrect because the values we used in calculations corresponded to a solution of Ni dissolved in Cu, not vice-versa (i.e. Ni was taken to be the impurity). 2.7 Resistivity of isomorphous alloys and Nordheim’s rule What are the maximum atomic and weight percentages of Cu that can be added to Au without exceeding a resistivity that is twice that of pure gold? What are the maximum atomic and weight percentages of Au that can be added to pure Cu without exceeding twice the resistivity of pure copper? (Alloys are normally prepared by mixing the elements in weight.) Solution From combined Matthiessen and Nordheim rule, Alloy =Au + Cu, with Cu= CX(1-X). In order to keep the resistivity of the alloy less than twice of pure gold, the resistivity of solute (Cu), should be less than resistivity of pure gold, i.e. I = CX(1-X) < Au. From Table 2.3, Nordheim coefficient for Cu in Au at 20C is, C = 450 n m. Resistivity of Au at 20C, using 0 = 1/251 K-1, is m n K K K m n T T 62 . 24 )] 273 293 ( 251 1 1 [ 8 . 22 )] ( 1 [ 1 0 0 0 Therefore the condition for solute (Cu) atomic fraction is CX(1-X) < 24.62 n m. X(1-X) < (24.62 n m) /(450 n m) = 0.0547 X2 – X – 0.0547 < 0 Solving the above equation, we have X < 0.0581 or 5.81%. Therefore the atomic fraction of Cu should be less than 0.0581 in order to keep the overall resistivity of the alloy less than twice the resistivity of pure Au. The weight fraction for Cu for this atomic fraction can be calculated from ) mol g 67 . 196 )( 0581 .0 1 ( ) mol g 54 . 63 )( 0581 .0 ( ) mol g 54 . 63 )( 0581 .0 ( ) 1 ( 1 1 1 Au Cu Cu Cu M X XM XM w = 0.01956 or 1.956%. Now, we discuss the case of Au in Cu, i.e. Au as solute in Cu alloy. Resistivity of Cu at 0���C is 15.7 n m. Therefore the resistivity of Cu at 20C is m nΩ 05 . 17 )] K 273 K 293 ( 232 1 1 [m nΩ 7 . 15 )] ( 1 [ 1 0 0 0 K T T Therefore the condition for solute (Au) atomic fraction is CX(1-X) < 17.03 n m. Nordheim coefficient for Au in Cu at 20C is, C = 5500 n m. X(1-X) < (17.03 n m) /(5500 n m) = 3.1010-3. X2 – X – 3.1010-3 < 0 Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.11 Solving the inequality we get the condition, X < 3.10610-3, required to keep the resistivity of alloy less than twice of pure Cu. The weight fraction for Cu for this atomic fraction can be calculated from ) mol g 54 . 63 )( 10 106 .3 1 ( ) mol g 67 . 196 )( 10 106 .3 ( ) mol g 67 . 196 )( 10 106 .3 ( ) 1 ( 1 3 1 3 1 3 Cu Au Au Au M X XM XM w = 9.55 10-3 or 0.955%. 2.8 Nordheim’s rule and brass Brass is a Cu–Zn alloy. Table 2.8 shows some typical resistivity values for various Cu–Zn compositions in which the alloy is a solid solution (up to 30% Zn). a. Plot ρ versus X(1 − X). From the slope of the best-fit line find the mean (effective) Nordheim coefficient C for Zn dissolved in Cu over this compositional range. b. Since X is the atomic fraction of Zn in brass, for each atom in the alloy, there are X Zn atoms and (1-X) Cu atoms. The conduction electrons consist of each Zn donating two electrons and each copper donating one electron. Thus, there are 2(X) + 1(1 - X) = 1 + X conduction electrons per atom. Since the conductivity is proportional to the electron concentration, the combined Nordheim-Matthiessens rule must be scaled up by (1 + X). ) 1 ( ) 1 ( 0 brass X X CX Plot the data in Table 2.8 as ρ(1 + X) versus X(1 − X). From the best-fit line find C and ρo. What is your conclusion? (Compare the correlation coefficients of the best-fit lines in your two plots). NOTE: More rigorously, ρbrass = ρmatrix + Ceff X (1−X), in which ρmatrix is the resistivity of the perfect matrix. Accounting for the extra electrons, ρmatrix ≈ ρ0/(1+X), where ρ0 is the pure metal matrix resistivity and Ceff is the Nordheim coefficient at the composition of interest, given by Ceff ≈ C/(1+X)2/3. (It is assumed that the atomic concentration does not change significantly.) As always, there are also other theories; part b is more than sufficient for most practical purposes. SOURCE: H. A. Fairbank, Phys. Rev., 66, 274, 1944. Solution a. We know the resistivity to be alloy = o + CX(1-X). We plot alloy versus X(1-X). We have a best-fit straight line of the form y = mx + b, where m is the slope of the line. The slope is Ceff, the Nordheim coefficient. Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.12 Figure 2Q8-1: Plot of alloy resistivity against X(1-X) The equation of the line is y = 225.76x + 18.523. The slope m of the best-fit line is 225.76 n m, which is the effective Nordheim coefficient Ceff for the compositional range of Zn provided. b. Figure 2Q8-2: Plot of (1+X) against X(X-1) The slope of the best-fit line is 306.67. As given in the question, the modified combined Nordheim– Matthiessens rule must be scaled up by (1 + X), ) 1 ( ) 1 ( 0 brass X X CX Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.13 or ) 1 ( ) 1 ( 0 brass X CX X The above equation is of the straight line form y = mx +b, where m is the slope of the line. Therefore from the equation of the line y = 306.67x + 17.4, we have the effective Nordheim coefficient is Ceff = 306.67 n m and 0 is 17.40 n m. If we calculate the resistivity using the values obtained above in the combined Nordheim-Mattheisen rule we obtain the following table, Zn at.% X Experimental Resistivity (n m) Case I Case II Resistivity (n m) Ceff = 225.76 n m Resistivity (n m) Ceff = 306.67 n m Scaled by (1+X) 0 17 17.00 17.00 0.34 18.1 17.76 17.98 0.5 18.84 18.12 18.43 0.93 20.7 19.08 19.64 3.06 26.8 23.70 25.32 4.65 29.9 27.01 29.24 9.66 39.1 36.70 39.91 15.6 49 46.72 49.63 19.59 54.8 52.56 54.61 29.39 63.5 63.85 62.32 Table 2Q8-1: Ceff values calculated by fitting line to experimental data and by taking into account the effect of extra valence electron For case I, the resistivity is calculated using effective Nordheim coeffcieint (Ceff) and for the second case the combined Nordheim–Matthiessens rule is scaled up by (1 + X). It is observed that the values obtained by the later method is closer to the experimental results verifying the method of scaling taking in to consideration the number of electrons donated by the solute atoms. 2.9 Resistivity of solid solution metal alloys: testing Nordheim’s rule Nordheim’s rule accounts for the increase in the resistivity from the scattering of electrons from the random distribution of impurity (solute) atoms in the host (solvent) crystal. It can nonetheless be quite useful in approximately predicting the resistivity at one composition of a solid solution metal alloy, given the value at another composition. Table 2.9 lists some solid solution metal alloys and gives the resistivity ρ at one composition X and asks for a prediction based on Nordheim’s rule at another composition X . Fill in the table for and compare the predicted values with the experimental values, and comment. Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.14 NOTE: First symbol (e.g., Ag in AgAu) is the matrix (solvent) and the second (Au) is the added solute. X is in at.%, converted from traditional weight percentages reported with alloys. Ceff is the effective Nordheim coefficient in ) 1 ( 0 X X Ceff . Solution Combined Matthiessen and Nordheim rule is ) 1 ( 0 X X Ceff alloy therefore, from the above equation effective Nordheim coefficient Ceff is ) 1 ( 0 X X C alloy eff Ag-Au: For this alloy, it is given that for X = 8.8% Au, = 44.2 n m, with 0 = 16.2 n m, the effective Nordheim coefficient Ceff is 89 . 348 ) 088 . 0 1 ( 088 . 0 m nΩ ) 2 . 16 2 . 44 ( eff C n m Now, for X = 15.4% Au, the resistivity of the alloy will be 65 . 61 ) 154 .0 1 )( 154 .0 )( m nΩ 88 . 348 ( m nΩ 2 . 16 n m Similarly, the effective Nordheim coefficient Ceff and the resistivities of the alloys at X are calculated for the various alloys and tabulated as follows, Alloy Ag-Au Au-Ag Cu-Pd Ag-Pd Au-Pd Pd-Pt Pt-Pd Cu-Ni X (at.%) 8.8% Au 8.77% Ag 6.2% Pd 10.1% Pd 8.88% Pd 7.66% Pt 7.1% Pd 2.16% Ni 0 (n m) 16.2 22.7 17 16.2 22.7 108 105.8 17 at X (n m) 44.2 54.1 70.8 59.8 54.1 188.2 146.8 50 Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.15 Ceff 348.88 392.46 925.10 480.18 388.06 1133.85 621.60 1561.51 X 15.4% Au 24.4% Ag 13% Pd 15.2% Pd 17.1% Pd 15.5% Pt 13.8% Pd 23.4% Ni at X (n m) 61.65 95.09 121.63 78.09 77.71 256.51 179.74 296.89 at X (n m) Experimental 66.3 107.2 121.6 83.8 82.2 244 181 300 Percentage Difference 7.01% less 11.29% less 0.02% more 6.81% less 5.46% less 4.88% more 0.69% less 1.04% less Table 2Q9-1: Resistivities of solid solution metal alloys Comment: From the above table, the best case has a 0.02% difference and the worst case has a 7% difference. It is clear that the Nordheim rule is very useful in predicting the approximate resistivity of a solid solution at one composition from the resistivity at a known composition. *2.10 TCR and alloy resistivity Table 2.10 shows the resistivity and TCR (α) of Cu–Ni alloys. Plot TCR versus 1/ρ, and obtain the best-fit line. What is your conclusion? Consider the Matthiessen rule, and explain why the plot should be a straight line. What is the relationship between ρCu, Cu, ρCuNi, and CuNi? Can this be generalized? NOTE: ppm-parts per million, i.e., 10-6. Solution The plot of temperature coefficient of resistivity TCR () versus 1/, is as follows Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.16 TCR ( ) versus 1/resistivity (1/) 0 500 1000 1500 2000 2500 3000 3500 4000 4500 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 1/Resistivity, 1/ (nW m -1 ) T C R , ( p p m º C -1 ) Figure 2Q10-1: TCR () versus reciprocal of resistivity1/ From Matthiessen’s rules, we have I o I matrix alloy where o is the resistivity of the matrix, determined by scattering of electrons by thermal vibrations of crystal atoms and I is the resistivity due to scattering of electrons from the impurities. Obviously, o is a function of temperature, but I shows very little temperature dependence. From the definition of temperature coefficient of resistivity, T o o o 1 or o o o T and alloy alloy alloy alloy alloy alloy alloy 1 ) ( 1 o o o I o T T T Clearly the TCR of the alloy is inversely proportional to the resistivity of the alloy. The higher the resistivity, the smaller the TCR, which is evident from the plot. 2.11 Electrical and thermal conductivity of In Electron drift mobility in indium has been measured to be 6 cm2 V-1 s-1. The room temperature (27 C) resistivity of In is 8.37 10-8 Ωm, and its atomic mass and density are 114.82 amu or g mol-1 and 7.31 g cm-3, respectively. a. Based on the resistivity value, determine how many free electrons are donated by each In atom in the crystal. How does this compare with the position of In in the Periodic Table (Group IIIB)? b. If the mean speed of conduction electrons in In is 1.74 108 cm s-1, what is the mean free path? c. Calculate the thermal conductivity of In. How does this compare with the experimental value of 81.6 W m-1 K-1? Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.17 Solution a. From = end ( is the conductivity of the metal, e is the electron charge, and d is the electron drift mobility) we can calculate the concentration of conduction electrons (n): ) s V m 10 6 )( C 10 1.602 ( ) m Ω 10 8.37 ( 1 1 2 4 19 1 8 d e n i.e. n = 1.243 1029 m-3 Atomic concentration nat is ) mol kg 10 114.82 ( ) mol 10 6.022 )( m kg 10 7.31 ( 1 3 1 23 3 3 at at M dN n A i.e. nat = 3.834 1028 m-3 Effective number of conduction electrons donated per In atom (neff) is: neff = n / nat = (1.243 1029 m-3) / (3.834 1028 m-3) = 3.24 Conclusion: There are therefore about three electrons per atom donated to the conduction-electron sea in the metal. This is in good agreement with the position of the In element in the Periodic Table (III) and its valency of 3. b. If is the mean scattering time of the conduction electrons, then from d = e/me (me = electron mass) we have: C) 10 602 . 1 ( kg) 10 .109 9 )( s V m 10 6 ( 19 31 -1 -1 2 4 e me d = 3.412 10-15 s Taking the mean speed u 1.74 106 m s-1, the mean free path (l) is given by l = u = (1.74 106 m s-1)(3.412 10-15 s) = 5.94 10-9 m or 5.94 nm c. From the Wiedemann-Franz-Lorenz law, thermal conductivity is given as: = TCWFL = (8.37 10-8 Ω m)-1(20 ºC + 273 K)(2.44 10-8 W K-2) i.e. = 85.4 W m-1 K-1 This value reasonably agrees with the experimental value. 2.12 Electrical and thermal conductivity of Ag The electron drift mobility in silver has been measured to be 56 cm2 V-1 s-1 at 27 C. The atomic mass and density of Ag are given as 107.87 amu or g mol-1 and 10.50 g cm-3, respectively. a. Assuming that each Ag atom contributes one conduction electron, calculate the resistivity of Ag at 27 C. Compare this value with the measured value of 1.6 10-8 Ω m at the same temperature and suggest reasons for the difference. b. Calculate the thermal conductivity of silver at 27 C and at 0 C. Solution Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.18 a. Atomic concentration nat is ) mol kg 10 107.87 ( mol 10 6.022 )( m kg 10 10.50 ( 1 3 1 23 3 3 at at M dN n A = 5.862 1028 m-3 If we assume there is one conduction electron per Ag atom, the concentration of conduction electrons (n) is 5.862 1028 m-3, and the conductivity is therefore: = end = (1.602 10-19 C)(5.862 1028 m-3)(56 10-4 m2 V-1s-1) = 5.259 107 -1 m-1 and the resistivity, = 1/ = 19.0 n m The experimental value of is 16 n m. We assumed that exactly 1 "free" electron per Ag atom contributes to conduction. This is not necessarily true. We need to use energy bands to describe conduction more accurately and this is addressed in Chapter 4. b. From the Wiedemann-Franz-Lorenz law at 27 C, = TCWFL = (5.259 107 Ω-1 m-1)(27 + 273 K)(2.44 10-8 W K-2) i.e. = 385 W m-1 K-1 For pure metals such as Ag this is nearly independent of temperature (same at 0 C). 2.13 Mixture rules A 70% Cu - 30% Zn brass electrical component has been made of powdered metal and contains 15 vol. % porosity. Assume that the pores are dispersed randomly. Given that the resistivity of 70% Cu-30% Zn brass is 62 n m, calculate the effective resistivity of the brass component using the simple conductivity mixture rule, Equation 2.26 and the Reynolds and Hough rule. Solution The component has 15% air pores. Apply the empirical mixture rule in Equation 2.26. The fraction of volume with air pores is = 0.15. Then, 0.15) (1 0.15) 0.5 (1 m Ω n 62 ) 1 ( ) 2 1 1 ( eff = 78.41 n m Reynolds and Hough rule is given by Equation 2.28 as alloy air alloy air alloy alloy 2 2 For the given case air = 0, alloy = (62 n m)-1. Substituting the conductivity values in the RHS of the equation we have 1 1 ) m nΩ 62 ( 2 0 ) m nΩ 62 ( 0 ) 15 . 0 ( 2 alloy air alloy air = 0.075. Solving for effective conductivity, we have = 1.2753107 -1m-1 eff = 78.41 10-9 m or 78.41 n m. Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.19 Hence the values obtained are the same. Equation 2.26 is in fact the simplified version of Reynolds and Hough rule for the case when the resistivity of dispersed phase is considerably larger than the continuous phase. 2.14 Mixture rules a. A certain carbon electrode used in electrical arcing applications is 47 percent porous. Given that the resistivity of graphite (in polycrystalline form) at room temperature is about 9.1 µ m, estimate the effective resistivity of the carbon electrode using the appropriate Reynolds and Hough rule and the simple conductivity mixture rule. Compare your estimates with the measured value of 18 µ m and comment on the differences. b. Silver particles are dispersed in a graphite paste to increase the effective conductivity of the paste. If the volume fraction of dispersed silver is 30 percent, what is the effective conductivity of this paste? Solution a. The effective conductivity of mixture can be estimated using Reynolds and Hough rule that states c d c d c c 2 2 If the conductivity of the dispersed medium is very small compared to the continuous phase, as in the given case conductivity of pores is extremely small compared to polycrystalline carbon, i.e. c>>d. Equation 2.26 is the simplified version of Reynolds and Hough rule. The volume fraction of air pores is = 0.47 and the conductivity of graphite is c = 9.1 µ m, therefore ) 47 . 0 1 ( ) 47 . 0 5 . 0 1 ( ) m Ω μ 1 .9 ( ) 1 ( ) 2 1 1 ( eff d d c = 21.21 µ m Conductivity mixture rule is based on the assumption that the two phases and are parallel to each other and the effective conductivity from Equation 2.25 is eff = + For the given situation air = 0.47, graphite = (1 - 0.47), air = 0, graphite = (9.1 µ m)-1, therefore the effective resistivity using the conductivity mixture rule is 0 m μΩ 1 .9 ) 47 . 0 1 ( 1 eff eff = 17.17 µ m b. If the dispersed phase has higher conductivity than the continuous phase, the Reynolds and Hough rule is reduced to Equation 2.27. From Table 2.1, resistivity for silver at 273 K is 14.6 µ m. Using 0 = 1/244 K-1, the resistivity at room temperature (20C) can be estimated as m nΩ 79 . 15 ) 273 293 ( 244 1 1 ) m nΩ 6 . 14 ( ) ( 1 1 0 0 0 K K K T T Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.20 Since d < 0.1c, we can use Equation 2.27. Volume fraction of dispersed silver is 30%, d = 0.3. The effective resistivity is ) 3 . 0 2 1 ( ) 3 . 0 1 ( ) m μΩ 1 .9 ( ) 2 1 ( ) 1 ( d d c eff = 3.98 µ m. If we use the Reynolds and Hough rule to calculate the effective resistivity, we obtain 3.99 µ m, which is reasonably close to the value calculated by using Equation 2.27. 2.15 Ag–Ni alloys (contact materials) and the mixture rules Silver alloys, particularly Ag alloys with the precious metals Pt, Pd, Ni, and Au, are extensively used as contact materials in various switches. Alloying Ag with other metals generally increases the hardness, wear resistance, and corrosion resistance at the expense of electrical and thermal conductivity. For example, Ag–Ni alloys are widely used as contact materials in switches in domestic appliances, control and selector switches, circuit breakers, and automotive switches up to several hundred amperes of current. Table 2.11 shows the resistivities of four Ag–Ni alloys used in make-and-break as well as disconnect contacts with current ratings up to 100 A. a. Ag–Ni is a two-phase alloy, a mixture of Ag-rich and Ni-rich phases. Using an appropriate mixture rule, predict the resistivity of the alloy and compare with the measured values in Table 2.11. Explain the difference between the predicted and experimental values. b. Compare the resistivity of Ag–10% Ni with that of Ag–10% Pd in Table 2.9. The resistivity of the Ag–Pd alloy is almost a factor of 3 greater. Ag–Pd is an isomorphous solid solution, whereas Ag–Ni is a two-phase mixture. Explain the difference in the resistivities of Ag–Ni and Ag–Pd. NOTE: Compositions are in wt.%. Ag–10% Ni means 90% Ag–10% Ni. Vickers hardness number (VHN) is a measure of the hardness or strength of the alloy, and d is density. Solution a. The Ni contents are given in wt.%. For volume fraction we use the relation Ni Ni Ni d d w where wNi is the weight fraction of Ni, dNi is the density of Ni and, d is the density of the alloy mixture. For example, for Ni-30% wt. the volume fraction of Ni in alloy will be Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.21 32 .0 ) 10 9 . 8 ( ) 10 47 . 9 )( 3 . 0 ( 3 3 3 3 kgm kgm Ni First we use Reynolds and Hough rule for mixture of dispersed phases to calculate the effective resistivity of the alloy. From Equation 2.28 we have Ag Ni Ag Ni Ni Ag Ag 2 2 Solving for Ni-30% wt., the R.H.S. of the above equation will be 1 1 1 1 1 ) m nΩ ( 1089 .0 ) m nΩ 9 . 16 ( 2 ) m nΩ 4 . 71 ( ) m nΩ 9 . 16 ( ) m nΩ 4 . 71 ( ) 32 .0 ( Substitute the calculated value in the Reynolds and Hough rule, the effective resistivity of the alloy is = 23.96 n m. Similarly the resistivity of alloy with other Ni contents is calculated and is tabulated below. Ni % in Ag-Ni d (g cm-3) Ni eff (n m) Experimental (n m) 10 10.3 0.12 19.07 20.9 15 9.76 0.16 20.11 23.6 20 9.4 0.21 21.17 25 30 9.47 0.32 23.96 31.1 Table 2Q15-1 Resistivity of Ag-Ni contact alloys for switches Now we use resistivity-mixture rule and conductivity-mixture rule to calculate the effective resistivity and compare the results. The resistivity-mixture rule or the series rule of mixtures is defined in Equation 2.24 as eff and conductivity-mixture rule is given by the Equation 2.25 eff The values obtained using these rules are as follows, Ni % in Ag- Ni Ni eff (n m) Resistivity- Mixture Rule Conductivity- Mixture Rule Reynolds & Hough Rule Experimental 10.00 0.12 23.21 18.54 19.07 20.90 15.00 0.16 25.86 19.33 20.11 23.60 20.00 0.21 28.41 20.15 21.17 25.00 30.00 0.32 34.30 22.34 23.96 31.10 Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.22 Table 2Q15-2: Resistivity of Ag-Ni contact alloys for switches calculated using different mixture rules b. 90%Ag-10% Ni, the solid is a mixture, and has two phases with an overall = 20.9 n m. 90%Ag- 10% Pd, the solid is a solid solution with = 59.8 n m, the value is roughly 3 times greater. The resistivity of a mixture is normally much lower than the resistivity of a similar solid solution. In a solid solution, the added impurities scatter electrons and increase the resistivity. In a mixture, each phase is almost like a "pure" metal, and the overall resistivity is simply an appropriate "averaging" or combination of the two resistivities. 12.16 Ag–W alloys (contact materials) and the mixture rule Silver–tungsten alloys are frequently used in heavy-duty switching applications (e.g., current-carrying contacts and oil circuit breakers) and in arcing tips. Ag–W is a two-phase alloy, a mixture of Ag-rich and W-rich phases. The measured resistivity and density for various Ag–W compositions are summarized in Table 2.12. a. Plot the resistivity and density of the Ag–W alloy against the W content (wt. %) b. Show that the density of the mixture, d, is given by 1 1 1 d w d w d where wα is the weight fraction of phase α, wβ is the weight fraction of phase β, dα is the density of phase α, and dβ is the density of phase β. c. Show that the resistivity mixture rule is d dw d dw where ρ is the resistivity of the alloy (mixture), d is the density of the alloy (mixture), and subscripts α and β refer to phases α and β, respectively. d. Calculate the density d and the resistivity ρ of the mixture for various values of W content (in wt. %) and plot the calculated values in the same graph as the experimental values. What is your conclusion? NOTE: = resistivity and d = density. Solution a. The plot of density versus W weight fraction is as follows. Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.23 Density versus Weight 0 5 10 15 20 25 0 20 40 60 80 100 wt. % W D en si ty ( g c m -3 ) Experimental Figure 2Q16-1: Experimental density for various compositions of W and the plot of resistivity versus W wt% is, Figure 2Q16-2: Experimental resistivity for various compositions of W Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.24 Dependence of resistivity in Ag-W alloy on composition as a function of wt.% W 0 10 20 30 40 50 60 70 80 90 100 0 20 40 60 80 100 W [wt.%] re s is ti v it y [ n O h m . 0 5 10 15 20 25 d e n s it y [ g /c m ^ 3 ] resistivity density Poly. (resistivity) b. The given mixture consists of two phases α, and β. Assume that the total mass of the alloy is Mmixture. If wα and wβ are the weight fractions of α, and β phases, then their respective masses in the mixture are Mα = wα Mmixture Mβ = wβ Mmixture The densities of the phases α, and β, are d and d, therefore the volume occupied by these phases can be calculated using the definition of density. i.e. density = mass / volume, we have d M w d M V mixture of density of mass d M w d M V mixture of density of mass The total volume of the alloy mixture is Vmixture = V + V d M w d M w mixture mixture . . The density of the mixture is therefore, d M w d M w M V M d mixture mixture mixture mixture mixture . . d w d w d w d w M M 1 mixture mixture Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.25 or d w d w d 1 c. The resistivity-mixture rule or the series rule of mixtures is defined in Equation 2.24 as eff where and are the volume fractions of phase and respectively. (For detailed derivation of this rule please see Example 2.13.) Volume fractions of the two phases are, mixture V V and mixture V V From part a of this problem, the volume of the phases and in the mixture are d M w V mixture and d M w V mixture and the volume of the mixture is d M V mixture , therefore the volume fraction of the two contents is d d w d M d M w V V mixture mixture and d d w d M d M w V V mixture mixture Substituting the volume fraction and , the resistivity mixture rule is d d w d d w eff d. We calculate the density and the resistivity using the relations proved in parts (b) and (c). As an example, for 30% W wt. content, the density and resistivity are, 3 3 1 1 1 cm g 1 . 19 ) 3 . 0 ( cm g 5 . 10 ) 3 . 0 1 ( W W Ag Ag d w d w d d = 12.14 g cm-3. Using resistivity-mixture rule, we have ) 1 . 19 ( ) 3 . 0 )( 14 . 12 ( ) 6 . 55 ( ) 5 . 10 ( ) 3 . 0 1 )( 14 . 12 ( ) 2 . 16 ( 3 3 3 3 cm g cm g m n cm g cm g m n d w d d w d W W W Ag Ag Ag = 23.71 n m The experimental resistivity as given in the table is 22.7 n m. Similarly, the densities and resistivities for the given W contents are calculated and listed in the Table 2Q16-1. W wt.% Volume Fraction W Experimental density Calculated density Resistivity Mixture Rule (n m) Experimental Resistivity Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.26 (g cm-3) (g cm-3) (n m) 10 0.06 10.75 11.00 18.47 18.6 15 0.09 10.95 11.26 19.68 19.7 20 0.12 11.3 11.54 20.96 20.9 30 0.19 12 12.14 23.71 22.7 40 0.26 12.35 12.81 26.77 27.6 65 0.49 14.485 14.84 36.10 35.5 70 0.55 15.02 15.33 38.34 38.3 75 0.60 15.325 15.85 40.73 40 80 0.68 16.18 16.41 43.28 46 85 0.74 16.6 17.01 46.03 47.9 90 0.81 17.25 17.65 48.98 53.9 Table 2Q16-1: Resistivity of Ag–W alloy on composition as a function of wt.% W calculated using different mixture rules 0 5 10 15 20 25 0 50 100 wt.% W experimental density [g cm^-3] calculated density [g cm^-3] Density (g /cm^3) Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.27 Figure 2Q16-3: Calculated and experimental density for various compositions of W 0 10 20 30 40 50 60 0 20 40 60 80 100 wt.% W exprimental resistivity (nOhm m) calculated resistivity (nOhm m) Resistivity (nOhm m) Figure 2Q16-4: Calculated and experimental resistivities for various compositions of W Author's Comment: The data were collected from a variety of sources (various handbooks and papers) and combined into a single table. The data are not simply from a single source. Hence the experimental values show some scatter. Given the scatter, the resistivity mixture model is in good agreement with the experimental data. By the way, the conductivity mixture rule fails in this case. 2.17 Thermal conduction Consider brass alloys with an X atomic fraction of Zn. Since Zn addition increases the number of conduction electrons, we have to scale the final alloy resistivity calculated from the simple Matthiessen-Nordheim rule in Equation 2.22 down by a factor (1+X) (see Question 2.8) so that the resistivity of the alloy is ) 1 /( )] 1 ( [ X X CX o in which C = 300 nΩ m and m nΩ 17 Cu o . a. An 80 at .% Cu─20 at. % Zn brass disk of 40 mm diameter and 5 mm thickness is used to conduct heat from a heat source to a heat sink. (1) Calculate the thermal resistance of the brass disk. (2) If the disk is conducting heat at a rate of 100 W, calculate the temperature drop along the disk. b. What should be the composition of brass if the temperature drop across the disk is to be halved? Solution a. Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.28 (1) Assume T = 20 ºC = 293 K. Apply Equation 2.22 to find the resistivity of the brass in the disk with Cu = 17.1 n m and XZn = 0.20: brass = Cu + CZn in CuXZn(1 XZn) i.e. brass = 17.1 n m + (300 n��� m)(0.20)(1 0.20) brass = 65.1 n m We know that the thermal conductivity is given by /brass = CFWLT where brass is the conductivity of the disk, CFWL is the Lorenz number and T is the temperature. This equation can also be written as brass = CFWLT so that = CFWLT/. Applying this equation, (20 C) = (2.44 10-8 W K-2)(293 K) / (6.51 10-8 m) (20 C) = 109.8 W K-1 m-1 The thermal resistance is = L/(A), where L is the thickness of the disk and A is the cross-sectional area of the disk. = L/(A) = (5 10-3 m)/[(109.8 W K-1 m-1)()(2 10-2 m)2] = 0.0362 K W-1 (2) From dQ/dt = AT/x = T/ (x can be taken to be the same as L), and dQ/dt = P (power conducted), we can substitute to obtain: T = P = (100 W)(3.62 10-2 K W-1) = 3.62 K or 3.62 C Note: Change in temperature is the same in either Kelvins or degrees Celsius, i.e. T = T1 ─ T2 = (T1 + 273) ─ (T2 + 273). b. Since T = P, to get half T, we need half or double or double or half . We thus need 1/2brass or 1/2(65.1 n m) which can be attained if the brass composition is Xnew so that new = Cu + CZn in CuXnew(1 Xnew) i.e. 1/2(65.1 n m) = 17 n m + (300 n m)Xnew(1 Xnew) Solving this quadratic equation we get Xnew = 0.0545, or 5.5% Zn. Thus we need 94.5% Cu-5.5% Zn brass. 2.18 Thermal resistance Consider a thin insulating disc made of mica to electrically insulate a semiconductor device from a conducting heat sink. Mica has = 0.75 W m-1 K-1. The disk thickness is 0.1 mm, and the diameter is 10 mm. What is the thermal resistance of the disk? What is the temperature drop across the disk if the heat current through it is 25W? Solution The thermal resistance of the mica disk can be calculated directly from Equation 2.40. 2 2 1 1 3 2 m 10 1 K m W 0.75 m 10 1 4 4 d L A L = 1.698 K W-1 The temperature drop across the disk according to Equation 2.36 (in the textbook) is Q T = 42.4 C Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.29 *2.19 Thermal resistance Consider a coaxial cable operating under steady state conditions when the current flow through the inner conductor generates Joule heat at a rate P = I2R. The heat generated per second by the core conductor flows through the dielectric; R I Q 2 . The inner conductor reaches a temperature Ti whereas the outer conductor is at To. Show that the thermal resistance of the hollow cylindrical insulation for heat flow in the radial direction is L a b Q T T o i 2 ln ' Thermal resistance of hollow cylinder where a is the inside (core conductor) radius, b is the outside radius (outer conductor), is the thermal conductivity of the insulation, and L is the cable length. Consider a coaxial cable that has a copper core conductor and polyethylene (PE) dielectric with the following properties: Core conductor resistivity = 19 n m, core radius, a = 4 mm, dielectric thickness, b-a = 3.5 mm, dielectric thermal conductivity = 0.3 W m-1 K-1. The outside temperature To is 25 C. The cable is carrying a current of 500 A. What is the temperature of the inner conductor? Solution Consider a thin cylindrical shell of thickness dr as shown in Figure 2Q12-1. The temperature difference across dr is dT. The surface area of this shell is 2rL. Thus, from Fourier’s law, dr dT rL Q ) 2 ( which we can integrate with respect to r from r = a where T = Ti to r = b where T = To, o i T T b a dT L r dr Q 2 i.e. a b L T T Q o i ln 2 ) ( Thus the thermal resistance of the hollow cylindrical insulation is L a b Q T T o i 2 ln ) ( Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.30 Inner conductor To To Ti dr r dT a b I2R L Dielectric Thin shell Outer conductor Q Figure 2Q12-1: Thermal resistance of a hollow cylindrical shell. Consider an infinitesimally thin cylindrical shell of radius r and thickness dr in the dielectric and concentrically around the inner conductor. The surface area is 2rL. The actual length of the conductor does not affect the calculations as long as the length is sufficiently long such that there is no heat transfer along the length; heat flows radially from the inner to the outer conductor. We consider a portion of length L of a very long cable and we set L = 1 m so that the calculations are per unit length. The joule heating per unit second (power) generated by the current I through the core conductor is 2 3 9 2 2 2 ) 10 4 ( ) 1 )( 10 19 ( ) 500 ( a L I Q = 94.5 W The thermal resistance of the insulation is, ) 1 )( 3 . 0 ( 2 10 4 10 ) 5 . 3 4 ( ln 2 ln 3 3 L a b = 0.33 C/W Thus, the temperature difference T due to Q flowing through is, T = Q = (94.5 W)(0.33 C/W) = 31.2 C. The inner temperature is therefore, Ti = To + T = 25 + 31.2 = 56.2 C. Note that for simplicity we assumed that the inner conductor resistivity and thermal conductivity are constant (do not change with temperature). 2.20 The Hall effect Consider a rectangular sample, a metal or an n-type semiconductor, with a length L, width W, and thickness D. A current I is passed along L, perpendicular to the cross-sectional area WD. The face W L is exposed to a magnetic field density B. A voltmeter is connected across the width, as shown in Figure 2.40, to read the Hall voltage VH. a. Show that the Hall voltage recorded by the voltmeter is Den IB VH Hall voltage Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.31 b. Consider a 1-micron-thick strip of gold layer on an insulating substrate that is a candidate for a Hall probe sensor. If the current through the film is maintained at constant 100 mA, what is the magnetic field that can be recorded per V of Hall voltage? Solution a. The Hall coefficient, RH, is related to the electron concentration, n, by RH = -1 / (en), and is defined by RH = Ey / (JB), where Ey is the electric field in the y-direction, J is the current density and B is the magnetic field. Equating these two equations: JB E en y 1 en JB Ey This electric field is in the opposite direction of the Hall field (EH) and therefore: EH = -Ey JB en (1) The current density perpendicular (going through) the plane W D (width by depth) is: WD I J JD I W (2) The Hall voltage (VH) across W is: H H WE V If we substitute expressions (1) and (2) into this equation, the following will be obtained: Den IB VH Note: this expression only depends on the thickness and not on the length of the sample. Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.32 In general, the Hall voltage will depend on the specimen shape. In the elementary treatment here, the current flow lines were assumed to be nearly parallel from one end to the other end of the sample. In an irregularly shaped sample, one has to consider the current flow lines. However, if the specimen thickness is uniform, it is then possible to carry out meaningful Hall effect measurements using the van der Pauw technique as discussed in advanced textbooks. b. We are given the depth of the film D = 1 micron = 1 m and the current through the film I = 100 mA = 0.1 A. The Hall voltage can be taken to be VH = 1 V, since we are looking for the magnetic field B per V of Hall voltage. To be able to use the equation for Hall voltage in part (a), we must find the electron concentration of gold. Appendix B in the textbook contains values for gold’s atomic mass (Mat =196.97 g mol-1) and density (d = 19.3 g/cm3 = 19300 kg/m3). Since gold has a valency of 1 electron, the concentration of free electrons is equal to the concentration of Au atoms. 3 28 1 3 -1 23 3 m 10 901 .5 mol kg 10 97 . 196 mol 10 022 .6 m kg 19300 at A M dN n Now the magnetic field B can be found using the equation for Hall voltage: Den IB VH A 0.1 m 10 5.901 C 10 1.602 m 10 1 V 10 1 3 28 19 6 6 I Den V B H B = 0.0945 T As a side note, the power (P) dissipated in the film could be found very easily. Using the value for resistivity of Au at T = 273 K, = 22.8 n m, the resistance of the film is: 228 . 0 m 10 1 m 0001 . 0 m 001 . 0 m 10 8 . 22 6 9 WD L A L R The power dissipated is then: P = I2R = (0.1 A)2(0.228 ) = 0.00228 W 2.21 The strain gauge A strain gauge is a transducer attached to a body to measure its fractional elongation L/L under an applied load (force) F. The gauge is a grid of many folded runs of a thin, resistive wire glued to a flexible backing, as depicted in Figure 2.41. The gauge is attached to the body under test such that the resistive wire length is parallel to the strain. a. Assume that the elongation does not change the resistivity and show that the change in the resis- tance R is related to the strain, = L/L by R R(1+2) Strain gauge equation where is the Poisson ratio, which is defined by l t strain al Longitudin strain Transverse Poisson ratio where l is the strain along the applied load, that is, l = L/L =, and t is the strain in the transverse direction, that is, t = D/D, where D is the diameter (thickness) of the wire. Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.33 b. Explain why a nichrome wire would be a better choice than copper for the strain gauge (consider the TCR). c. How do temperature changes affect the response of the gauge? Consider the effect of temperature on . Also consider the differential expansion of the specimen with respect to the gauge wire such that even if there is no applied load, there is still strain, which is determined by the differential expansion coefficient, specimen gauge, where is the thermal coefficient of linear expansion: L = Lo[1 + (T To)], where To is the reference temperature. d. The gauge factor for a transducer is defined as the fractional change in the measured property R/R per unit input signal (). What is the gauge factor for a metal-wire strain gauge, given that for most metals, 3 1 ? e. Consider a strain gauge that consists of a nichrome wire of resistivity 1 m, a total length of 1 m, and a diameter of 25 m. What is R for a strain of 10-3? Assume that 3 1 . f. What will R be if constantan wire with resistivity of 500 nΩ m is used? Solution a. Consider the resistance R of the gauge wire, 2 2 D L R (1) where L and D are the length and diameter of the wire, respectively. Suppose that the applied load changes L and D by L and D which change R by R. The total derivative of a function R of two variables L and D can be found by taking partial differentials (like those used for error calculations in physics labs). Assuming that is constant, D D L L D D D R L L R R ��� 3 2 4 2 4 so that the fractional change is Full file at http://testbank360.eu/solution-manual-principles-of-electronic-materials-and-devices-3rd-edition-kasap 2.34 D D L L D D L D L L D L D R R 2 4 4 2 4 4 2 3 2 2 (2) We can now use the definitions of longitudinal and transverse strain, l L L and l t D D in the expression for R/R in Eqn. (2) to obtain, ) 2 1 ( ) ( 2 l l R R (3) where = l. b. The change in R was attributed to changes in L and D due to their extension by the applied load. There are two reasons why a nichrome wire will be a better choice. First is that nichrome has a higher resistivity which means that its resistance R will be higher than that of a similar size copper wire and hence nichrome wire will exhibit a greater change in R (R is easier to measure). Secondly nichrome has a very small TCR which means that and hence R does not change si</p>