Homework%203%20Solutions

# Homework%203%20Solutions - EE 215 Homework 3 You can work...

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EE 215 Homework 3 You can work on the layout problems in your design groups and just turn in one layout per group. Be sure to list the names of all the group members. Please do the non-layout problems individually . 1) Draw the cross section A-A’ of the following layout in the MUMPS process. The red material is Poly 1, the gray material is Poly 2 and the cross-hatched area is Anchor 2. What would be the function of this device once it has been released in a sacrificial HF etch? This would function as a hinge once it has been released.

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EE 215 Homework 3 2) Assume that silicon will fracture when the axial stress reaches 1 GPa. Find the maximum length of a vertical silicon rod which, under the action of its own gravitational load, will not exceed this fracture stress. Assume the density of silicon is 2331 kg/m 3 and that the acceleration due to gravity is 9.8 m/sec 2 . σ Fracture = 1 GPa, ρ = 2331 kg/m 3 σ = F/A = mg/A = ( ρ ( π /4)d 2 Lg)/ ( π /4)d 2 = ρ Lg L = σ Fracture / ρ g = (1x10 9 N/m 2 )/(2331 kg/m 3 )(9.8 m/sec 2 ) = 4.38x10 4 meters 3) A silicon cantilever of length 500 μ m, width 50 μ m, and thickness 2 μ m is subjected to a uniform distributed transverse load P. Find the tip deflection, and calculate an effective spring constant for this beam. For a load that produces a tip deflection of 2 μ m, calculate the maximum stress at the support. Assume Young’s modulus = 160 GPa. From page 186 of Kovacs: (1) y(x) = (Px 2 /24EI)(6L 2 – 4Lx + x 2 ) (2) P = F/L Substituting for P in (1) and finding the tip deflection at L: (3) y(L) = ((F/L)L 2 /24EI))(6L 2 – 4L 2 + L 2 ) = 3FL 3 /24EI = FL 3 /8EI This is the tip deflection. To find the spring constant solve for F: (4) F = (8EI/L 3 )y(L) This is now in the form of Hook’s Law; (5) F = -k δ
EE 215 Homework 3 Where k is the spring constant and δ is the spring deflection. From (4) & (5) we get: ( 6) k = -(8EI/L 3 ) (7) σ max = PL 2 t/4I = (F/L)(L 2 t/4I) = FLt/4I For a load that produces a 2 μ m tip deflection, from (3): FL 3 /8EI = y(L) = 2 μ m So that: (8) F = (2 μ m)(8EI)/L 3 Substituting (8) into (7) and setting t = 2 μ m and L = 500 μ m: σ max = [(2 μ m)(8EI)Lt/L 3 ]/4I = (2 μ m)(8E)(2 μ m)/[(4)(500 μ m) 2 ]= 8E/(500) 2 Subsituting E = 160 GPa: σ max = 8(160 GPa)/(500) 2 = 5.12 MPa 4)

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Homework%203%20Solutions - EE 215 Homework 3 You can work...

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