EE C245 - ME C218 Introduction to MEMS Design Fall 2003...A Cantilever Beam Clamped: y = 0, dy/dx =...

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).


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


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


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


Checking the Assumptions

J.-A. Schweitz, Uppsala University


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


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


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


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

=−


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 =−


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


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


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


Boundary Conditions

• A “step-up” anchor will result in the average strain causing an offset angle at y = 0

Approach tosuppressinginitial offsetangle


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 =−


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 =

=


Summary of Common Loadingand Boundary Conditions

Compendium of useful results:http://www.roarksformulas.com


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


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


Folded-Flexure Suspension Variants

Michael Judy, Ph.D. Thesis, EECS Dept., UC Berkeley, 1994


Folded Flexure Deflection

Michael Judy, Ph.D. Thesis,EECS Dept., UC Berkeley, 1994


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


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)


Folded Flexure Suspension withResidual Stress Gradient

legs warp together

comb teeth areengaged

Michael Judy, Ph.D. ThesisEECS Dept., UC Berkeley, 1994


ADXL-50 Suspension


ADXL-05 Suspension


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


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


Motorola z-Axis Accelerometer


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


When is Shear Significant?

EE C245 - ME C218 Introduction to MEMS Design Fall 2003...A Cantilever Beam Clamped: y = 0, dy/dx =...

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