KaSap Electronic Materials_Chapter_2_problem_solutions

KaSap Electronic Materials_Chapter_2_problem_solutions -...

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Chapter 2 solutions ___________________________________________________________________________________ 2.2 a. Apply the equation for temperature dependence of resistivity, ρ ( T ) = o [1 + α o ( T - T o )]. We have the temperature coefficient of resistivity, o , at T o where T o is the reference temperature. The two given reference temperatures are 0 ° C or 25 ° C, depending on choice. Taking T o = 0 ° C + 273 = 273 K, (-40 ° C + 273 = 233 K) = o [1 + o (233 K 273 K)] (25 ° C + 273 = 298 K) = o [1 + o (298 K 273 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 Ω 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 - × × Ω × ° = 2.03 × 10 -8 Ω m b. In ( T ) = o [1 + o ( T T o )] we have o at T o where T o is the reference temperature, for example, 0 ° C or 25 ° C depending on choice. We will choose T o 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 K 273 K)][1 + -40 (273 K 233 K)] = 1 or [1 40 o ][1 + 40 -40 ] = 1 Rearranging, -40 = (1 / [1 40 o ] 1)(1 / 40) -40 = o / [1 40 o ] 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 Ω m 2.03 × 10 -8 Ω m] / [(2.03 × 10 -8 Ω m)(65 K)] -40 = 5.23 × 10 -3 K -1 c. We know that 1/ = σ = en μ where is the electrical conductivity, e is the electron charge, and is the electron drift mobility. We also know that = e τ / m e , where is the mean free time between electron collisions and m e is the electron mass. Therefore, 1 / = e 2 n / m e 2.1
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τ = m e / ρ e 2 n (1) Here n is the number of conduction electrons per unit volume. But, from the density d and atomic mass M at , 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 = 3 n Al = 1.807 × 10 29 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 × × Ω × × = = n e m e = 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 μ 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 m 2 V -1 s -1 = 12.7 cm 2 V -1 s -1 d. The mean free path is l = u , where u is the mean speed. With u 2 × 10 6 m s -1 we find the mean free path: l = u = (2 × 10 6 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 = I 2 R and is proportional to resistivity , assuming the rms current level stays relatively constant. Then we have [ 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 Ω m / 2.72 × 10 -8 m) 1= -0.254 , or -25.4% (Negative sign means a reduction in the power loss).
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This note was uploaded on 04/24/2008 for the course MSE MSE 350 taught by Professor Morelli during the Spring '08 term at Michigan State University.

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KaSap Electronic Materials_Chapter_2_problem_solutions -...

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