ECE 453 Assignment 1 Solutions_1[1]

ECE 453 Assignment 1 Solutions_1[1] - ECE 453 Assignment 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE 453 Assignment 1 Solutions 1.1 a) Substituting values from attached notes: q 2 (1.6e − 19 ) = = 38.6 µ S h 6.63e − 34 2 1 = 25.9k Ω 38.6 µ S 1.1 b) σW q2 M λ =s h λ+L λ+L q2 ⇒ M λ = σ sW h h σW ⇒M = 2 s qλ Substituting values from part (a) and from question: M = 25.9e3Ω × 1.6e − 3S × 0.1e − 6m = 41.44 ~ 41 100e − 9m 1.2 π vz = −π ∫ v cos θ dθ π −π ∫ dθ cos θ dθ π2 v = 2π = = π −π ∫ v 2 ∫ cos θ dθ 2π −π 2 2v π π vz2 = − ∫π v 2 cos 2 θ dθ π −π ∫ dθ 2 = v2 2π 2 π − ∫π cos ∫π θ dθ v = 2π = v 2 2 π − 1 + cos 2θ dθ 2 1.3 ɶɶ In this problem we have to find expressions for F + , F − from given differential equation and boundary conditions d ɶ+ d ɶ− F= F dz dz d ɶ+ ɶ− ⇒ F −F =0 dz ɶ ɶ ⇒ F + − F − = a constant, say = A ( ) Therefore, d ɶ+ d ɶ− A F = F =− dz dz λ …(1) ɶɶ Thus, F + , F − are linear. Now we need to apply the boundary conditions to find out A. Using equation 5.3 from the Notes (Lessons from nanoelectronics): ɶ ɶ ɶ F + ( L ) = TF + ( 0 ) + (1 − T ) F − ( L ) and applying the given boundary conditions from the question: ɶ F + ( L) = T = Now, λ λ+L ɶ ɶ F+ − F−= A ɶ ɶ ⇒ F + ( L) − F − ( L) = A ɶ ⇒ A=F + ( L ) = λ λ+L λ λ+L (as an aside) ɶ Therefore, F − ( 0 ) = 1 − ɶɶ Using the value of A and putting it back into (1) we find the expressions of F + , F − (using the equation of a line from coordinate geometry or by integration of (1)) as: λ+L−z ɶ F+ (z) = λ + L (as a check, we see that the F + L , F − 0 satisfy the solutions) ɶ()ɶ() L−z ɶ F− (z) = λ+L ...
View Full Document

This note was uploaded on 11/29/2010 for the course ECE 453 taught by Professor Supriyodatta during the Spring '10 term at Purdue.

Page1 / 3

ECE 453 Assignment 1 Solutions_1[1] - ECE 453 Assignment 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online