329fall09hw5sol - ECE 329 Fall 2009 Homework5Solution Due:...

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Unformatted text preview: ECE 329 Fall 2009 Homework5Solution Due: Sep. 29, 2009 1. a) According to Page 2 in Lecture 11, the mobility is dened as qτ q = . m mν mobility = Although mν is not explicitly given in the problem, it is related to Ne and σ by (see Page 3 in Lecture 11) σ= or Ne q 2 , mν mν = Ne q 2 , σ q mν = therefore, mobility = q Ne q 2 σ = σ Ne q 5.8 × 107 8.45 × 1028 × 1.6 × 10−19 = 4.3 × 10−3 m2 /(s · V ) , = where the unit is derived from v = b) R= qτ m E. l 400 = = 0.98 (Ω) . 2 7 × π × (1.5 × 10−3 )2 σ · πr 5.8 × 10 c) On the one hand, J= I . πr2 On the other hand, J = σE. J I 3 = = 7.3 × 10−3 (V/m) . E= = 2 σ σ · πr 5.8 × 107 × π × (1.5 × 10−3 )2 ∴ d) The velocity of an electron is v= qτ E = mobility · E = 4.3 × 10−3 × 7.3 × 10−3 = 3.1 × 10−5 (m/s) , m so the time to drift from one end to the other is t= l 400 = = 1.3 × 107 (s) . v 3.1 × 10−5 2. 1 ECE 329 Fall 2009 a) Since V (0) − V (z ) = ´z 0 E (z ) dz and V (0) = 0, we have ˆz V (z ) = − E (z ) dz = 0 3 8 0 2z 3 8 0 [ 2d + 1 (z for 0 for d − d)] z z d (V) . 1 ∂ b) Inside the two regions, we have 2 V (z ) = ∂z2 V (z ) = 0, which means that Laplace's equation is satised inside the two regions. However, at the interface z = d, the derivative of V (z ) goes to innity, so Laplace's equation is not satised. c) Method (1): 2 V (z = 1) = 3 [ 2d + 80 1 (1 − d)] (V) , ρs (z = 1) = (−z ) · D (z = 1) = (−z ) · [ 2 E (z = 1)] = (−z ) · ˆ ˆ ˆ C= Q ρs (z = 1) A = = V V (z = 1) − V (z = 0) − 2 3 80 1 2A 2d + 1 (1 − d) ˆ 1z = (C/V) . Method (2): the capacitance for the two slabs are given by A 1d A 2 1−d C1 = C2 = 0 d z z d 1 therefore, the total capacitance is − − C = C1 1 + C2 1 −1 = d 1A 2 1 + 1−d 2A = 1 2A 2d + 1 (1 − d) (C/V) . 312 C/m2 , 80 ...
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This note was uploaded on 05/05/2010 for the course ECE 329 taught by Professor Franke during the Spring '08 term at University of Illinois at Urbana–Champaign.

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329fall09hw5sol - ECE 329 Fall 2009 Homework5Solution Due:...

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