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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).
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
5EE C245 – ME C218 Fall 2003 Lecture 7
A Cantilever BeamClamped: y = 0, dy/dx = 0 at x= 0
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
6EE C245 – ME C218 Fall 2003 Lecture 7
Checking the Assumptions
J.-A. Schweitz, Uppsala University
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
εx,max
-εx,max
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
∫∫ −−=⋅=−
2/
2/
2/
2/])[(
h
h x
h
h xz ydyEWyWdyM εσ
Why a minus sign? See Senturia, pp. 208-210
=
2/max, hy
xx εε
9EE C245 – ME C218 Fall 2003 Lecture 7
Bending Strain and Beam Curvature
( ) max,
32/
2/max, 23
22/2/ x
h
hxz
hhEW
ydyh
yEWM εε
=
=− ∫
−
Radius of curvature à geometric connection to strain
RR + h/2dθ
( )R
hRd
RddhRx
2/2/max, =
−+=
θθθ
ε
10EE C245 – ME C218 Fall 2003 Lecture 7
Curvature and Strain (cont.)
2
21
dxyd
R =−
( )( )
2
2313
3
121212/
232
2/ dxydEWh
REWhR
hhhEW
M z ==
=− −
Rh
x2/
max, =ε
… result from basic calculus
Combining the curvature and moment results:
and ( ) max,
3
232
2/ xzh
hEW
M ε
=−
11EE C245 – ME C218 Fall 2003 Lecture 7
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
13EE C245 – ME C218 Fall 2003 Lecture 7
Built-in Bending Moment
• Integrate differential moment through film thickness (sign?)
( ) dyyyEWyWdyMh
hr
h
hrr ∫∫
−−
⋅==2/
2/
2/
2/
)(εσ
( ) Γ
+=⋅Γ+= ∫
− 120
32/
2/
EWhdyyyEWM
h
hrr ε
• Apply moment to the cantilever à constant curvature
2
23
12 dxyd
EIEWh
z=Γ
2
2
dxyd
=Γ
14EE C245 – ME C218 Fall 2003 Lecture 7
Tip-Deflection (Small Deflections)
• The strain gradient Γ can be found from the tip deflection ∆:
2
21
)( LLxy Γ=∆==
• Integrate to find the tip deflection y(x =L)
22L∆
=Γ
15EE C245 – ME C218 Fall 2003 Lecture 7
Boundary Conditions
• A “step-up” anchor will result in the average strain causing an offset angle at y = 0
Approach tosuppressinginitial offsetangle
16EE C245 – ME C218 Fall 2003 Lecture 7
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
)()( xLFxM cz −=−
The moment varies linearly with x
2
2
)(dx
ydEIxM zz =−
17EE C245 – ME C218 Fall 2003 Lecture 7
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)
3
3
3 43
cc
zc L
EWhLEI
k =
=
18EE C245 – ME C218 Fall 2003 Lecture 7
Summary of Common Loadingand Boundary Conditions
Compendium of useful results:http://www.roarksformulas.com
19EE C245 – ME C218 Fall 2003 Lecture 7
Series Combinations of Cantilevers
• Springs in series à same load; deflections add
y(L) = F/k = 2 y(Lc) = 2 (F/kc) = F(1/kc + 1/kc)
y(L)
#1#2
F
Lc LcL = 2Lc
compliances add
1/k = 1/kc + 1/kc
20EE C245 – ME C218 Fall 2003 Lecture 7
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 = Fa/ka = Fb/kb = (F/2) (1/ka)
F/2
F/2
k = 2ka
21EE C245 – ME C218 Fall 2003 Lecture 7
Folded-Flexure Suspension Variants
Michael Judy, Ph.D. Thesis, EECS Dept., UC Berkeley, 1994
22EE C245 – ME C218 Fall 2003 Lecture 7
Folded Flexure Deflection
Michael Judy, Ph.D. Thesis,EECS Dept., UC Berkeley, 1994
23EE C245 – ME C218 Fall 2003 Lecture 7
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 = F/4
Displacement of two legs add (springs in series) à
Fpair
leg
rigid truss
y = Fpair/kpair = (F/4)(1/kleg + 1/kleg)
1/kleg = 1/kc + 1/kc = 2/kc
y = (F/4)(2/kc + 2/kc) = F/kc
24EE C245 – ME C218 Fall 2003 Lecture 7
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)
25EE C245 – ME C218 Fall 2003 Lecture 7
Folded Flexure Suspension withResidual Stress Gradient
legs warp together
comb teeth areengaged
Michael Judy, Ph.D. ThesisEECS Dept., UC Berkeley, 1994
26EE C245 – ME C218 Fall 2003 Lecture 7
ADXL-50 Suspension
27EE C245 – ME C218 Fall 2003 Lecture 7
ADXL-05 Suspension
28EE C245 – ME C218 Fall 2003 Lecture 7
ADXL-05 Suspension
29EE C245 – ME C218 Fall 2003 Lecture 7
BiMEMS Foundry Process• Analog Devices, 1994 – circa 2000• σr ˜ 50 MPa – [5 MPa/µm](z) à cantilevers bend down
Per Ljung,Ph.D. ME, 1995
30EE C245 – ME C218 Fall 2003 Lecture 7
Z-Axis Accelerometer Warpage• Anchor placement can be critical for balancing doming
of proof mass and warpage of folded suspension
Per Ljung,Ph.D. ME, 1995
31EE C245 – ME C218 Fall 2003 Lecture 7
Motorola z-Axis Accelerometer
32EE C245 – ME C218 Fall 2003 Lecture 7
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
33EE C245 – ME C218 Fall 2003 Lecture 7
When is Shear Significant?