EE C245 - ME C218 Introduction to MEMS Design Fall 2003€¦ · EE C245 - ME C218 Introduction to...
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1 EE C245 – ME C218 Fall 2003 Lecture 7 EE C245 - ME C218 Introduction to MEMS Design Fall 2003 Roger Howe and Thara Srinivasan Lecture 7 Microstructural Elements* * Mostly for EE’s, but there may be a few new insights for the ME’s 2 EE C245 – ME C218 Fall 2003 Lecture 7 Today’s Lecture • The cantilever beam under small deflections • Combining cantilevers in series and parallel: folded suspensions • More accurate models: large deflections, shear, … • Design implications of residual stress and stress gradients • Reading: Senturia, S. D., Microsystem Design, Kluwer Academic Publishers, 2001, Chapter 9, pp. 201-219, 222-231. J. D. Grade, H. Jerman, and T. W. Kenny, “Design of large deflection electrostatic actuators,” Journal of Microelectromechanical Systems, 12, 335-343 (2003).
Transcript of EE C245 - ME C218 Introduction to MEMS Design Fall 2003€¦ · EE C245 - ME C218 Introduction to...
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EE C245 – ME C218 Fall 2003 Lecture 7
EE C245 - ME C218Introduction to MEMS Design
Fall 2003
Roger Howe and Thara SrinivasanLecture 7
Microstructural Elements*
* Mostly for EE’s, but theremay be a few new insightsfor the ME’s
2EE C245 – ME C218 Fall 2003 Lecture 7
Today’s Lecture
• The cantilever beam under small deflections• Combining cantilevers in series and parallel:
folded suspensions• More accurate models: large deflections, shear, …• Design implications of residual stress and stress
gradients
• Reading:Senturia, S. D., Microsystem Design, Kluwer Academic
Publishers, 2001, Chapter 9, pp. 201-219, 222-231.J. D. Grade, H. Jerman, and T. W. Kenny, “Design of large
deflection electrostatic actuators,” Journal of Microelectromechanical Systems, 12, 335-343 (2003).
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3EE C245 – ME C218 Fall 2003 Lecture 7
Macro and Milli Suspensions
2000 Ford Focus(mostly 3-D steel parts andassembly-line production)
… 100,000’s per year
Hard Disk Suspensions(stamped 20 µm stainless steelwith laminated 10 µm polyimide+ 15 µm copper interconnect)
… 1,000,000’s per week
4EE C245 – ME C218 Fall 2003 Lecture 7
Springs in MEMS• Coils: 3-D is tough for planar processing!
• Flexures: straightforward to make using surface or bulk micromachining, but details of fabrication process constrain dimensions and anchors/joints
• Simplest flexure: a “clamped-free” cantilever beam … a.k.a. a diving board
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5EE C245 – ME C218 Fall 2003 Lecture 7
A Cantilever BeamClamped:
x
x = Lc
Goal: find relation between tip deflection y(x = Lc) and applied load F
Assumptions:
1. Tip deflection is small compared with beam length2. Plane sections (normal to beam’s axis) remain plane and normal
during bending … “pure bending”3. Shear stresses are negligible
F
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Checking the Assumptions
J.-A. Schweitz, Uppsala University
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7EE C245 – ME C218 Fall 2003 Lecture 7
A Beam Segment in Pure Bending
bottom is in compression
top is in tension
neutral axis (εx = 0)
εx
y
y
h/2-h/2
8EE C245 – ME C218 Fall 2003 Lecture 7
Bending Moment Mz
• Concept of moment (basic physics): force X distance• Integrate stress through thickness of beam
=− zM
Why a minus sign? See Senturia, pp. 208-210
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9EE C245 – ME C218 Fall 2003 Lecture 7
Bending Strain and Beam Curvature
=− zM
Radius of curvature à geometric connection to strain
RR + h/2
dθ
=max,xε
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Curvature and Strain (cont.)
2
21
dxyd
R =−
=− zM
Rh
x
2/max, =ε
… result from basic calculus
Combining the curvature and moment results:
and ( ) max,
3
232
2/ xzh
hEWM ε
=−
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Flexural Rigidity (Moment of Inertia) Iz
• The term Wh3/12 is defined as the flexural rigidity, Iz(Senturia uses “moment of inertia”)
• Large flexural rigidity à low curvature à small deflections à stiff
• Design implications: 1. rigidity increases as the cube of the beam ’s thickness
(in the direction of bending)2. the aspect ratio h / W determines the ratio of bending rigidity in the y
and the z directions
2
2
dxyd
EIM zz =−
12EE C245 – ME C218 Fall 2003 Lecture 7
Revisit Cantilever Deflectiondue to Residual Stress Gradients
• Model the strain by a linear profile yy resres Γ+= εε )(
z
Peter Krulevitch,Ph.D. thesis, ME, UC Berkeley, 1994
* inferred from wafer curvature after incremental thinning of poly-Si
LPCVD poly-Si:measured* stress profiles
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13EE C245 – ME C218 Fall 2003 Lecture 7
Built-in Bending Moment
• Integrate differential moment through film thickness (sign?)
( ) == ∫−
yWdyMh
hrr
2/
2/
σ
( ) Γ
+=⋅Γ+= ∫
− 120
32/
2/
EWhdyyyEWM
h
hrr ε
• Apply moment to the cantilever à constant curvature
=2
2
dxydEIz
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Tip-Deflection (Small Deflections)
• The strain gradient Γ can be found from the tip deflection ∆:
=∆== )( Lxy
• Integrate to find the tip deflection y(x =L)
=Γ
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Boundary Conditions
• A “step-up” anchor will result in the average strain causing an offset angle at y = 0
Approach tosuppressinginitial offsetangle
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The Cantilever with a Concentrated Load
Clamped: y = 0, dy/dx = 0 at x= 0
x
x = Lc
Find the tip deflection y(x = Lc) and applied load F … get effective springconstant kc
F
=− )(xM z
The moment varies linearly with x
2
2
)(dx
ydEIxM zz =−
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Tip Deflection• Integrate ODE twice and apply boundary conditions (zero
displacement, zero slope) at anchor
2)3(6
)( xxLEIF
xy cz
−
=
• Tip deflection: y(Lc)
3
3)( c
zc L
EIF
Ly
=
• Spring constant: kc (N/m) … = (µN/ µm)
=ck
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Summary of Common Loadingand Boundary Conditions
Compendium of useful results:http://www.roarksformulas.com
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Series Combinations of Cantilevers
• Springs in series à same load; deflections add
y(L) = F/k =
y(L)
#1#2
F
Lc LcL = 2Lc
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Parallel Combinations of Springs• Same displacement à load is shared and the spring
constant is the sum of the individual spring constantsy(L)
a
b
F
y(L) = F/k
F/2
F/2
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Folded-Flexure Suspension Variants
Michael Judy, Ph.D. Thesis, EECS Dept., UC Berkeley, 1994
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Overall Spring Constant• Four pairs of clamped-guided beams, each of which
bend in series (assume that trusses are inflexible)
Force is shared by each pair à Fpair =
Displacement of two legs add (springs in series) à
Fpair
leg
rigid truss
y = Fpair/kpair =
1/kleg =
y =
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Selected Goals for Suspension Design
• Compliance ratios are often required to be large(e.g., the comb drive’s maximum force is determined by lateral instability, which is in turn directly related to the lateral spring constant)
• Undesirable resonant modes of the structure are often required to be at significantly higher frequencies, which translates to stiffer spring constants
• Robustness against residual stress and stress gradients (e.g., folded flexures release most of the residual stress and cancel deflections due to gradients)
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Folded Flexure Suspension withResidual Stress Gradient
legs warp together
comb teeth areengaged
Michael Judy, Ph.D. ThesisEECS Dept., UC Berkeley, 1994
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ADXL-50 Suspension
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ADXL-05 Suspension
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ADXL-05 Suspension
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Motorola z-Axis Accelerometer
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Limits to Linearity
• Cantilever beams: stiffen as the deflection exceeds about 10% of the length of the beam
• Note: clamped-clamped beams deviate from non-linearity for much smaller deflections (next lecture)
Michael Judy, Ph.D. ThesisEECS Dept., UC Berkeley, 1994
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When is Shear Significant?
