Computational Methods for MEMS - mcc.uiuc.edu Aluru/Aluru_files/Aluru... · Computational MEMS/NEMS Beckman Institute University of Illinois at Urbana-Champaign Scaling Laws ... ·By

University of Illinois at Urbana-ChampaignBeckman InstituteComputational MEMS/NEMS

Computational Methods for MEMS

N. R. AluruBeckman Institute for Advanced Science and Technology

Thanks to: Xiaozhong JinGang LiRui QiaoVaishali Shrivastava


Course Objectives

Understand how MEMS are designedUnderstand some of the computational techniques that go into the development of MEMS simulation toolsSpecific examples: electrostatic MEMS, microfluidics


Outline

Some MEMS ExamplesMixed-Domain Simulation of electrostatic MEMS and microfluidics

Techniques for interior problems (e.g. FEM)Techniques for exterior problems (e.g. BEM)Algorithms


Pressure Sensor

ApplicationsBiomedical (e.g. blood pressure)


Accelerometer


Micro Mirror

ApplicationsHigh performance projection displays


Outline



Dynamic Analysis


Is Electrostatics a good idea?


Scaling Laws

Useful to understand where macro-theories start requiring corrections with the aim of better understanding the physical consequences of downscalingDevelop an understanding of how systems are likely to behave when they are downsizedExamples

By reducing the size of a device, the structural stiffness generally increases relative to inertially imposed loadsThe mass or weight scales as l3, while the surface tension scales as l as the system size becomes smallerMore difficult to empty liquids from a capillary compared to spilling coffee from a cup because of increased surface tension in a capillaryHeat loss is proportional to l2; Heat generation is proportional to l3; As animals get smaller, a greater percentage of their intake is required to balance heat loss; Insects are cold blooded


Scaling in Electrostatics

L2Electrostatic forceLSurface TensionL3Gravity

L3Mass

LVelocity

LDistance

L3TorqueL3PowerL2Muscle force

L0Time

L1/4van der Waals

L2Friction

h

w

d

Consider a capacitor

The electrostatic P.E. stored in a capacitor is:

dhwVE br

me 2

20

,εε=

=bV electrical breakdown voltage


Scaling in Electrostatics

31

2110

, ll

llllE me →=

Assume scales linearly with d (the gap)bV

The maximum energy stored in the capacitor scales as L3

If L decreases by a factor of 10, the stored energy in the capacitor decreases by a factor of 1000

Electrostatic Force

∂∂−= 2

21 CV

xFx

∂∂−= 2

21 CV

yFy

∂∂−= 2

21 CV

zFz

23

lllFx →=

Electrostatic force scales as L2; This is an advantage because mass and inertial forces scale as L3; The electrostatic force gains over inertial forces as the size of the system decreases


Scaling Laws: Vertical Bracket NotationDifferent possible forces can be written as

=4

3

2

1

llll

Fcase where the force scales as L1

case where the force scales as L2

[ ][ ] 33 −− === FF lllmFa Flll

Fmxt −⋅⋅== 32

3400

11ll

llVt

xFVP

F

F

−=

=

→

=4

3

2

1

llll

F

=−

−

1

0

1

2

llll

a

=

12

1

2/3

ll

lt

=−

−

0.2

5.0

1

5.2

0llll

VP


Scaling Laws: Remarks

Even in the worst case when F = L4, the time required to perform a task remains constant when the system is scaled downUnder more favorable force scaling (e.g. F = L2), the time required decreases as L (a system 10 times smaller can perform an operation 10 times faster i.e. small things tend to be quick)

For Electrostatics 2ll F =0.1

0

11 −− === lVPltla

When the force scales as F = L2, the power per unit volume scales as L-1 => When the scale decreases by a factor of 10, the power that can be generated per unit volume increases by a factor of 10


Outline




Elastostatics

Finite-difference methodsFinite-element methodsMeshless methods


Key steps in FEM:

•Construct a weak or a variational form of the problem

•Obtain an approximate solution of the variationalequations through the use of finite element functions

Finite Element Method: Introduction


• Trial function q)1(u,Hu|u 1 =∈=δ

•Test function (or weighting functions) 0)1(w,Hw|w 1 =∈=ν

• Strong form

0dx)fu(w1

0xx,∫ =+

• Integrate by parts 0dxfwdxuwwu1

0

1

0x,x,

10x, ∫ ∫ =+−

1x0 0fu xx, ≤≤=+

B.C. Dirichlet q)1(u =

B.C. Neumann h)0(u x, =−

(S)

• Derive weak form

Weak formh)0(wdxfwdxuw

1

0

1

0x,x,∫ ∫ += (w) ( ) ( )WS ⇔

1-D Example: Strong and Weak Forms

0 1


•Notations: ∫=1

0x,x, dxuw)u,w(a ∫=

1

0

dxfw)f,w(

h)0(w)f,w()u,w(a +=•The weak form can be rewritten as

∑=

=N

1AAA

h CNw ∑=

=N

1BBB

h dNu

• Galerkin Approximation Method

hhu δ∈

hhw ν∈

h)0(w)f,w()u,w(a hhhh +=Galerkin form

( ) hC0Nf,CNdN,CNaN

1AAA

N

1AAA

N

1BBB

N

1AAA ∑∑∑∑

====

+

=

(G)

Apply the interpolation

1-D Example: Galerkin Form


( ) 0h0NfdNddNNCN

1BAABx,Bx,A

N

1AA =

−Ω−Ω∑∫ ∫∑=

Ω Ω=

CA’s are arbitrary, so

( )∑ ∫ ∫=

Ω Ω+Ω=Ω

N

1BAABx,Bx,A h0NfdNddNN for A=1,2, …, N

∑=

=N

1BABAB FdK for A=1,2, …, N

where

∫Ω Ω= dNNK x,Bx,AAB ( )h0NdfNF AAA +Ω= ∫Ω

The matrix form FdK = (M)

1-D Example: Matrix Form


[ ]

==

NNNN

N

N

AB

KKK

KKKKKK

KK

21

22221

11211 ......

M

==

N

A

F

FF

FFM2

1

==

N

A

d

dd

ddM2

1

Remarks:

• K is symmetric;

• For any given problem, ( ) ( ) ( ) ( )MGWS ⇔≈⇔

1-D Example: Matrix Form


( ) 211

21 xxx,

hxxxN ≤≤

−= ( ) n1n

1N

1NN xxx,

hxxxN ≤≤

−= −

−

−

1

x1 x2 xA-1 xA xA+1 xn-1 xn

N1 N1Nn

Shape functions

( )

≤≤−

≤≤−

= ++

−−

−

otherwize,0

xxx,h

xx

xxx,h

xx

xN 1AAA

1A

A1A1A

1A

A

Shape Functions


[ ]1AA x,x +Domain [ ]21 ,ξξ

global local

Nodes 1AA x,x + 21 ,ξξ

1AA d,d + 21 d,dDegree of freedom

1AA N,N + 21 N,NShape functions

Interpolation function

( ) ( )( ) 1A1A

AAh

dxNdxNxu

++

+= ( ) ( )( ) 22

11h

dNdNu

ξ+ξ=ξ

For a linear finite element

Local/Element Point of View


uh

NA NA+11 1

dA dA+1

eA xx 1= e

A xx 21 =+

N1 N21 1

d1e d2

e

12 +=ξ11 −=ξ

eξ

ex

Mapping


( )21

2N

aa

,a−

=ξ

=ξ

2xx

2hx

e1

e2

ee

,−

==ξ ( ) e

1e,

ex, h

2x ==ξ−

ξ

( ) ( ) ( ) ( ) e22

e11

2

1a

eaa

e xNxNxNx ξ+ξ=ξ=ξ ∑=

( )ξ−= 121N1 ( )ξ+= 1

21N 2

( ) ( )ξξ+=ξ aa 121N

( )A

1AA

hxxx2x −−−

=ξ

• Local Shape function

where

• Derivative of the Shape function

Mapping


∑=

=nel

1e

eKK

∑=

=nel

1e

eFF

[ ]ABee KK =

[ ]Aee FF =

( ) ∫Ω==e

dxNNN,NaK x,Bx,Ae

BAABe

( ) ( )h0Nf,NF Ae

AAe +=

[ ]2e

1ee x,x=Ωwhere is the domain of the e-th element

( ) ( )

( )( ) ( )( ) ( ) ξ=ξξξ=

=

−ξ

+

− ξξξ

+

−

Ω

∫∫∫

dxNNdxxNxN

dxxNxNK

1,

1

1 ,b,a,

1

1 x,bx,a

x,bx,aabe

e

−

−=

1111

h1K e

e

where

• Global stiffness

• Force vector

local stiffness

Matrix Assembly


( )( )

−

−=

1111

h1K 1

1 ( )( )

−

−=

1111

h1K 2

2

(1) (2)

10 11 12

−−+−

−=

MMM

MMM

LLLL

LLLL

LLLL

MMM

MMM

YYYYXX

XXK

10 11 12

1110

12

local stiffness

Global stiffness

example

Assembly Process


Divergence theorem Γ=Ω ∫∫ ΓΩdnfdf

ii,

∫∫∫ ΩΓΩΩ−Γ=Ω dfgdngfgdf i,ii,Integration by parts

Heat Conduction

Strong form: given ℜ→Γℜ→Γℜ→Ω hg :h,:g,:f

find ℜ→Ω:u such that

hii

g

i,i

onhnq

ongu infq

Γ=−

Γ=

Ω=

where j,iji uKq −= Kij are the conductivities

FEM: Multidimensional Problems


( ) 0dfqw i,i =Ω−∫Ω

0dfwdqwdnwq ii,ii =Ω−Ω−Γ ∫∫∫ ΩΩΓ

∫∫∫ ΓΩΩΓ+Ω=Ω−

h

whddfwdqw ii,

g on 0w Γ=

Notation: ( ) ∫Ω Ω= duKwu,wa j,iji,

( ) ∫Ω Ω= wfdf,w ( ) ∫ΓΓ Γ=h

dhwh,w

( ) ( ) ( )h

h,wf,wu,wa Γ+=

( ) ( ) ( )h

h,wf,wu,wa hhhhΓ+=

• Weak form

• Galerkin form

Apply the property of the weighting function

Weak Form: Heat Conduction Problem


∑=

=N

1AAA

h CNw ∑=

=N

1BBB

h dNu

Γ====

+

=

∑∑∑∑N

1AAA

N

1AAA

N

1BBB

N

1AAA h,CNf,CNdN,CNa

( ) ( ) ( )Γ=

+=∑ h,Nf,NdN,Na AAB

N

1BBA

FdK =• Matrix form

( )BAAB N,NaK =

( ) ( )Γ+= h,Nf,NF AAA

Substitute the interpolation

where

Heat Conduction Problem: Matrix Form


Strong form: given fi, gi and hi, find ui such that

hijij

giii

ij,ij

onhn

ongu

in0f

Γ=σ

Γ=

Ω=+σ

The stress is related to the strain by Hooke’s law

klijklij c ε=σ

( )j,iu2

uu i,jj,iij =

+=ε

( ) 0dfwdnwdwdfw iijijiijj,iij,iji =Ω+Γσ+Ωσ−=Ω+σ ∫∫∫∫ ΩΓΩΩ

( ) ∑ ∫∫∫=

ΓΩΩ

Γ+Ω=Ωσ

nsd

1iiiiiij dhwdfwdj,iw

Strain tensor

• Weak form

FEM for Elastostatics


( ) ijijj,i j,iww σ=σ

( ) [ ]jiwjiww ji ,,, +=

( )

[ ] symmetric) (skew 2

,

)(symmetric 2

,

,,

,,

ijji

ijji

wwjiw

wwjiw

−=

+=

So

( ) [ ]( )( ) [ ] ijij

ijijji

jiwjiw

jiwjiww

σσ

σσ

,,

,,,

+=

+=

[ ] [ ] [ ] [ ] ijjiijij jiwijwijwjiw σσσσ ,,,, −=−=−=

[ ] 0, =ijjiw σ

( ) ijijj,i j,iww σ=σWhy

where

rewrite

since

and



Notation: ( ) ( ) ( )∫Ω Ω= dl,kucj,iwu,wa ijkl

( ) ∫Ω Ω= dfwf,w ii ( ) ∑ ∫=

ΓΓ

Γ=

nsd

1iii

hi

dhwh,w

( ) ( ) ( ) vw all for h,wf,wu,wa ∈+= Γ

( ) ( ) ( )Γ+= h,wf,wu,wa hhhh

• Weak form

• Galerkin form

s)dof' 3( 1,2,3i dNun

1AiAA

hi == ∑

=

FdK =• Matrix form



4

1 2

3

ξ

η

12

3

4

( )1,1 −−

( )ee yx 22 ,

( )ee yx 33 ,( )ee yx 44 ,

( )1,1 −

( )1,1( )1,1−

ηξ

yx

eΩ

( ) ( )

( ) ( ) ea

aa

ea

aa

yNy

xNx

∑

∑

=

=

=

=

4

1

4

1

,,

,,

ηξηξ

ηξηξ

Bilinear Quadrilateral Element


( ) ( )( )

( ) ( )( )

( ) ( )( )

( ) ( )( )ηξηξ

ηξηξ

ηξηξ

ηξηξ

+−=

++=

−+=

−−=

1141,

1141,

1141,

1141,

4

3

2

1

N

N

N

N

( ) ( )( ) ( )( ) ( )( ) ( )( )

1

114111

4111

4111

41,

1

=

+−++++−++−−=∑=

ηξηξηξηξηξnen

aaN

• Shape functions

Property of the shape function

Shape Functions


( ) ( ) ( ) l

n

lll

n

ll wgRwgdg ∑∑∫ ≅+=

+

−

intint1

1ξξξξ

• Trapezoidal rule

1

1

2

2

1

int

+=

−=

=

ξ

ξ

n2,11 == lwl

( )ξξξ,32 gR −=

• Simpson’s rule

1

0

1

3

3

2

1

int

=

=

−=

=

ξ

ξ

ξ

n

34

31

2

31

=

==

w

ww

( )( )ξ4

901 gR −=

Numerical integration

Numerical Integration


( )3

20

,

11

ξ

ξ

ξξgR

w

=

==

( )( )135

13

13

1

4

21

21

ξ

ξξ

gR

ww

=

==

=−=

( )( )15750

98

95

530

53

6

231

321

ξ

ξξξ

gR

www

=

===

==−=

1int =n

2int =n

3int =n

Gaussian Quadrature Rules


( ) ( )( )

( )( )

( )

( )

( )( )

( )( )

( )( )

( )

( )

( )( )2l

n

1l

1l

2

l

1

l

1

1

n

1l

1l

1

l

1

1

1

1

2

1int

1121

1int

111

ww,g

dw,gdd,g

∑ ∑

∫ ∑∫ ∫

=

+

−=

+

−

+

−

ηξ≅

η

ηξ≅ηξηξ

ξ

η

ξ

η

ξ

η

( ) ( )0,0g4dd,g1

1

1

1=ηξηξ∫ ∫

+

−

+

−

( )

−+

+

−+

−−=ηξηξ∫ ∫

+

−

+

− 31,

31g

31,

31g

31,

31g

31,

31gdd,g

1

1

1

1

examples

Gaussian Quadrature Rules in 2-D


( )Xx ϕ= XFx dd =

( )

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

==

3

3

2

3

1

3

3

2

2

2

1

2

3

1

2

1

1

1

XXX

XXX

XXX

D

ϕϕϕ

ϕϕϕ

ϕϕϕ

XF ϕ

( ) [ ]Fdet=ϕJ

Transformation function

Deformation gradient Volume transformation JdVdv =

1E

Original configuration

Deformed configuration

xX

2E

uXx +=3E

01 ρρJ

=

T

JFP1

=σ

Area transformation

t∂∂

=ϕV

( ) ( )( ) NFSns 1−= TdJd

Density transformation

Cauchy stress

Velocity

Finite Deformation Elastodynamics


[ ] BP 02

2

0 ρρ +=∂∂ Div

tϕ

• Boundary conditions:

EcS = FFc T= [ ]IcE −=21

SFP =where

Strong form:

[ ][ ]TΓhnP

TΓ

ihiAiA

g

,0on

,0

at

aton

g

=

=ϕ

• Initial conditions:

( ) Ω=

Ω=

=

=

V

in

in0

0

00

Vt

t ϕϕ

Elastodynamics: Governing Equations


( ) [ ] ( )( )[ ] 0:, 0 =Γ⋅−⋅−= ∫∫∫ ΓΩΩ h

ddVdVDGradG hBES ηηϕϕηηϕ ρ

( ) 0int =− extFF d

• Weak form:

• Galerkin form:

Where

( ) dVT∫Ω= Sd)

BF int ∫∫ ΓΩΓ⋅+⋅=

h

ddVext NhNB0ρF

Nonlinear equations:

( ) [ ] [ ] 0:, 0 =Γ⋅−⋅−= ∫∫∫ ΓΩΩ h

ddVdVDGradG hhhhhhh hBS ηηϕηηϕ ρ

FEM for Elastodynamics


The scalar problem: r(d) is a scalar nonlinear function of d. Find such thatd

( ) 0=dr

Possibilities:

d

r

d

r

d

r

d

one unique solution several solutions no solutions

1d 2d

Solution of Nonlinear Systems: Newton Methods


( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( )( )( )ddddKdrd

ddKdr

ddrdddrdr

ddrddddr

dddrdr

∆+=

−=∆

=∆+=

∆++≅

+∆++∆++=

=

==

01

0

0

00

000

002

2

000

0

!21

ε

εε

εε

εε

εε

KK

• Solution strategy:Guess a “good” d0 close to the solution;

d0d

1d

2d

r

• Geometric interpretation

Newton’s Method


( )

( )( )

for endif end

else

if

econvergenc untilfor

'

,2,1

0

1

0

ddddrdrd

dd

toldri

ddi

ii

i

i

i

i

i

∆+=

−=∆

=

<

==

=

+

KK

The error at the i-th iteration is given by ( ) dde ii −=

( ) ( ) kii eCe ≤+1 if then we say the algorithm converges with order k.

Newton’s method has quadratic convergence, i.e. k=2

Algorithm: Newton’s Method


Advantages of Newton’s method

•Optimal (quadratic) convergence close to solution

Disadvantages of Newton’s method

•Poor or no convergence far away from the solution;

•Computation of K(di) is very expensive in the general multidimensional case (K (di) is a matrix).

Algorithm: modified Newton’s method

( )

( )( )

endif end

else

if

econvergenc untilfor

'

,2,1

0

1

0

0

ddddrdrd

dd

toldridd

i

ii

i

i

i

i

∆+=

−=∆

=

<

==

=

+

KK

Modified Newton’s Method


( )dR

Similar to the scalar case

( ) ( ) ( ) ( )

( ) ( )( ) ( ) 0

!21

00

000

002

2

000

=∆+=

∆++≅

+∆++∆++=

=

==

ddKdR

ddRdddR

ddRddddR

dddRdR

T

ε

εε

εε

εε

εε

KK

Let be a n-dimensional vector valued nonlinear function of the n-dimensional vector d

( )( )

( )nnn

n

n

dddRR

dddRR

dddRR

,,,

,,,

,,,

21

2122

2111

KK)MM

KK)

KK)

=

=

= ( )

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=

∇=+ =

n

n

n

n

nn

n

u

uu

dR

dR

dR

dR

dR

dR

uRudRdd

M

L

MOM

MOM

L

2

1

1

1

2

1

1

1

0εεε

• Directional or Frechet derivative

( )0dK T is called the tangent stiffness matrix

Newton Method: Multidimensional Case


Remarks:

• The load step is needed since the method might not converge if the entire load is applied at once. Instead the load is applied incrementally. Each increment is converged before the next step is applied.

• Likewise one might have to apply the “g” b.c.’s incrementally. This method is called “displacement control”.

• In practice, the terminology “Newton-Raphson method”is often used to denote algorithms in which a new left hand size matrix is formed for each iteration. If KT is not updated in each iteration, but kept frozen for a couple of iterations, the term “modified Newton” method is used.

( )

( )( ) ( ) ( )

( )( )( ) ( ) ( )

( ) ( )

( )

1

1

:

K

0

0

1

1

1

int11

until

loop) (iteration do

e.g) stepping load ( loop step

+←=

<

+=∆+=

−=

−=∆

=

=

=

+

+

+

++

nndd

RR

iiddd

dFF

Rdd

dd

i

n

in

ii

iii

in

extn

iiiT

ni

ε

Newton-Raphson Method: Algorithm


Outline




BEM - Introduction

What is Boundary Element Method ?Boundary discretization only Integral based method

Approaches available for solving boundary integral equations

BEM based on CollocationBEM based on Galerkin

Analysis of a turbine blade using FEM and BEM


Comparison of FEM and BEM

BEM FEM

Fewer packages availableLot of commercial packages available

Unsymmetric, dense and smaller matrices

Symmetric, sparse and large matrices

Boundary mesh (1D/2D)Domain mesh (2D/3D)

Global approach, Integral based

Local approach


Table 1

ConstantNeumann b.c.

Dirichlet b.c.

Scalar function

Problems

ResistivityElectric current

Electro-potential

Electric conduction

PermittivityElectric flowField potential Electrostatic

Stream functionIdeal fluid flow

Warping function

Elastic torsion

Thermal conductivity

Heat flowTemperatureHeat Transfer

)0( 2 =φ∇ )( φ )qn

K( =∂φ∂)( φ=φ )K(

.)DegT( ≡ )TT( =)q

nT( =∂∂

λ− )( λ

)(ψ )q)t,rcos(r( =

)sm( 12 −≡φ )( φ=φ )qn

( =∂φ∂

)voltV( ≡)VV( =

)qnV( =∂∂

ε− )( ε

)voltE( ≡)EE( =

)qnE

k1( =∂∂

)k(

rt


Boundary Integral Formulation

Laplace equation represents many problems in engineering (Table 1)

v+-

Interior problem Exterior problem

0T2 =∇ 02 =φ∇inside outsideTemperature ‘T’ known on the surface Potential known on the surfaceφ



BEM is based on the Second Theorem of Green

Problem definition

Ω

uΓ

n

x

)x(p2 =φ∇ Ω∈x

Governing equation:

Dirichlet boundary condition (b.c):

Neumann boundary condition (b.c):

)y()y( φ=φ uy Γ∈

)y(qn

)x(

yx

=∂φ∂

=

ny Γ∈

nΓ

≠=

Poisson;0Laplace;0

)x(p

y

Definition of the problem



Derivation of BIE:Multiplying with and integrating over

( ) 0dp2 =Ωφ−φ∇∫Ω∗

Integrating by parts we get (2D case),

( ) 0dpdn =+Ωφ+φ∇⋅φ∇−Γ⋅φ∇φ ∫∫ Ω

∗∗

Γ

∗

∗φ Ω)p( 2 −φ∇

Integrating by parts the second integral we get,

( ) 0dpdndn 2 =Ωφ−φ∇φ+Γ⋅φ∇φ−Γ⋅φ∇φ ∫∫∫ Ω

∗∗

Γ

∗

Γ

∗

( )1. . .

where n is the outward normal to the boundary Γ



Therefore, Equation (1) can be written as,

is the Fundamental Solution of Laplace equation. ∗φ

( ) ( ) 0dn

dn

dpdp 22 =Γ∂φ∂

φ−Γ∂φ∂

φ+Ωφ−φ∇φ=Ωφ−φ∇ ∫∫∫∫ Γ

∗

Γ

∗

Ω

∗∗

Ω

∗

( ) ( )[ ] Γ

∂φ∂

φ−∂φ∂

φ=Ωφφ∇−φφ∇⇒ ∫∫ Γ

∗∗

Ω

∗∗ dnn

d22 ( )2. . .



Fundamental Solution for Laplace equation :satisfies Laplace equationrepresents field generated by a concentrated unit charge acting at a

point ‘i’effect of this charge is propagated from ‘i’ to infinity

∗φ

0)j,i(2 =δ+φ∇ ∗ =δ )j,i( Dirac Delta function

Multiplying with and integrating we get,φ

idjid φδφφφ −=Ω−=Ω∇ ∫∫ ΩΩ

∗ )),(()( 2

Therefore,

Γφ

∂φ∂

=Γ

∂φ∂

φ+φ ∫∫ Γ

∗

Γ

∗

dn

dn

i

( )3. . .



Fundamental Solution

Equation

Convection/decay Equation

Diffusion Equation

Wave Equation

Helmholtz

Laplace

Fundamental Solutions : One Dimensional Equations

002 =δ+φ∇ ∗

0022 =δ+φλ+φ∇ ∗∗

( ) 0tt

c 02

222 =δδ+

∂φ∂

−φ∇∗

∗

( ) 0ttk

102

2 =δδ+∂φ∂

−φ∇∗

∗

( ) 0txt 0 =δδ+βφ+∂φ∂

φ+∂φ∂ ∗

∗∗

xr,2r

==φ ∗

( )rsin21

λλ

−=φ∗

( )rctHc2

1−=φ∗

H=Heaviside function

( )

−π

−=φ∗

kt4rexp

kt4tH 2

φ

−δ−=φ φβ−

∗ rter




EquationFundamental Solutions : Two Dimensional Equations

Navier’s Equation

Plate Equation

Wave Equation

D’Arcy(orthotropic case)

Helmholtz

Laplace 002 =δ+φ∇ ∗

0022 =δ+φλ+φ∇ ∗∗

0dxdk

dxdk 02

2

2

221

2

1 =δ+φ

+φ ∗∗

( ) 0tt

c 02

222 =δδ+

∂φ∂

−φ∇∗

∗

( ) 0tt 0

422

2

=δδ+φ

∇µ−

∂∂ ∗

0x l

j

jk =δ+∂σ∂ ∗

22

21 xxr,

r1ln

21

+=

π=φ ∗

( )( )rHi4

1 20 λ=φ ∗

H0 = Hankel function

+

π−=φ∗

21

2

22

1

21

21 kx

kxln

21

kk1

( )

ππµ+=φ ∗

t4rS

4tH

i

Si = Integral sine function

( )( )222 rtcc2

rctH−π

−−=φ ∗

llkk eU ∗∗ =φ




EquationFundamental Solutions : Three Dimensional Equations

Navier’s Equation(Isotropic homogenous)

Wave Equation

D’Arcy

Helmholtz

Laplace00

2 =δ+φ∇ ∗

0022 =δ+φλ+φ∇ ∗∗

0dxdk

dxdk

dxdk 02

3

2

322

2

221

2

1 =δ+φ

+φ

+φ ∗∗∗

( ) 0tt

c 02

222 =δδ+

∂φ∂

−φ∇∗

∗

0x l

j

jk =δ+∂

σ∂ ∗

23

22

21 xxxr,

r41

++=π

=φ∗

rier4

1 λ−∗

π=φ

21

3

23

2

22

1

21

321 kx

kx

kx

41

kkk1

−

∗

++

π−=φ

r4crt

π

−δ

=φ∗

llkk eU ∗∗ =φ



What happens when point ‘i’ is on ?Γ

ε=r

Boundary surface

Boundary point ‘i’

Surface

Γ

εΓ

3D case - Hemisphere around point ‘i’ 2D case - Semicircle around point ‘i’

Boundary point ‘i’

ε=r

Boundary curve Γ

Boundary curve εΓ

Augment the boundary with Hemisphere of radius in 3DSemicircle of radius in 2D

εε



Consider equation (3) before any boundary conditions have been applied,

- RHS integral easy to deal (lower order singularity),

Γ

∂φ∂

φ=Γ

∂φ∂

φ+φ ∫∫ Γ

∗

Γ

∗

dn

dn

i

≡

πεπε

∂φ∂

=Γ

πε∂φ∂

=Γφ

∂φ∂

→εΓ→εΓ

∗

→ε ∫∫εε

04

2n

limd4

1n

limdn

lim2

000

- LHS integral behaves as,

φ−=

πεπε

φ−=Γ

πεφ−=

Γ∂φ∂

φ→εΓ→εΓ

∗

→ε ∫∫εε

i2

2

0200 21

42limd

41limd

nlim



Therefore,

Γ

∂φ∂

φ=Γ

∂φ∂

φ+φ ∫∫ Γ

∗

Γ

∗

dn

dn

c i ( )4

for smooth boundaries

. . .

,21c =

πθ

=2

c for corner pointsθ P

Boundary with corner point


Boundary Integral Formulation (contd.)

Exterior Problem - Electrostatics,

A system of ‘Nc’’ ideal conductors

n

n−

∞R

jΓ

∞Γ=Γref

02 =φ∇ outside

Potential known on the surface of each conductorφ

For 3D Electrostatic problem the boundary integral equation is,

∑ ∫=

Γ

∗ Γφ∂φ∂

=φc

j

N

1j

i dn


Boundary Element MethodEquation (4) is discretized to find system of equations

Boundary is divided into N elements

Discretized form of equation (3) at point ‘i’ is given as,

Γφ∂φ∂

=Γ∂φ∂

φ+φ ∑ ∫∑ ∫=

Γ

∗

=Γ

∗

dn

dn

cN

1j

N

1j

i

jj

Constant elements Linear elements Quadratic elementsNodes Element


Boundary Element Method

In matrix form,

[ ] [ ]

∂Φ∂

=Φn

GH

where Hij and Gij are the influence coefficients given as,

‘i’ is the source point (where fundamental solution is acting) ‘j’ is the field point (any other nodes on the boundary)


Boundary Element MethodConstant Elements:

and are assumed to be constant over each element The value of and is assumed equal to that at mid-element node

The influence coefficients, Hij and Gij are given as,

∫Γ∗

Γ∂φ∂

+δ=j

dn

)j,i(21H ij

Γφ= ∫Γ∗dG

j

ij

‘i’ is the source point (where fundamental solution is acting) ‘j’ is the field point (any other nodes on the boundary)

φ

∗φφ∗φ

Node ‘i’

Element ‘j’



Evaluation of integrals:Hij and Gij can be calculated numerically, for the case

For the case , Hij and Gij are evaluated analytically

21

21

21 =Γ

∂∂

∂∂

+=Γ∂∂

+= ∫∫ Γ

∗

Γ

∗

dnr

rd

nH

j

ii φφ

+

π=Γ

π=Γφ= ∫∫ ΓΓ

∗ 12/l

1ln2l1d

r1ln

21dG

ii

ii

ji ≠ji =

Node ‘i’r

n

lElement ‘i’

0


Boundary Element MethodLinear Elements:

and are assumed to vary linearly over each element

Nodal value of Nodal value of

[ ]

φφ

=φ2

1

21 NN

[ ]∫Γ∗

Γ∂φ∂

+δ=j

dn

NN)j,i(21H 21

ij

[ ] Γφ= ∫Γ∗dNNG

j21

ij

φ ∗φ

∗φorφφ ∗φor

l

[ ]

∂φ∂∂φ∂

=∂φ∂

nn

NNn 2

1

21

Therefore,

Node ‘i’Element ‘j’



Putting all the unknowns on LHS we get,

[ ] FxA =

Note: A is a dense matrix

Dense MatrixA

(N x N) Vec

tor

x(N

x1)

= Ax

(N x

1)

DIRECTO(N3)

ITERATIVEO(N2)


Fast Integral Equation Solver

Matrix-Vector multiplication: O(N(logN)O(N(logN)22))Storage: O(N(logN)O(N(logN)22))

Results : 2-Conductor Problem


Fast Integral Equation SolverResults : Mirror Problem

Matrix-Vector multiplication: O(N(logN)O(N(logN)22))Storage: O(N(logN)O(N(logN)22))


Fast Integral Equation SolverResults : Comb-Drive Problem

Matrix-Vector multiplication: O(NlogNO(NlogN))Storage: O(NlogNO(NlogN))


References

P.K. Banerjee, The Boundary Element Method in Engineering, Mc-Graw Hill, 1994

G. Beer, Programming the Boundary Element Method, Wiley, 2001

C.A. Brebbia and J. Dominguez Boundary Elements An Introductory Course, Mc-Graw Hill, 1996

J.H. Kane, Boundary Element Analysis in Engineering Continuum Mechanics, Prentice Hall, 1994


Outline




Coupled Electromechanical Analysis

We need to self-consistently solve the coupled electrical and mechanical equations to compute the equilibrium displacements and forces. Three approaches –

Relaxation techniqueFull-Newton methodMulti-level Newton method

Solution of elastostatic equations is represented by

))(( qPRu M=

Solution of electrostatic equations is represented by

),( VuRq E=


Relaxation Technique

Simplest black-box approachData is passed back and forth between black-box electrostatic and elastostatic analysis programs until a converged solution is obtained

))(()1( kM

k qPRu =+

)( kE

k uRq =Repeat

Compute

Compute

0;1 == kuk

1+= kk

εε ≤−≤− ++ 11 kkkk qquuUntil


Relaxation Technique

AdvantagesVery quick implementation based on black-boxesExisting mechanical and electrical solvers can be used

DisadvantagesFails to converge for strong coupling between electrical and mechanical domains


Multi-Level Newton Algorithm

Matrix-free approaches: Matrix-vector product involving a Jacobian and a vector can be computed as

εε )()( uRuuRu

uR −∆+=∆∂∂

Define a new residual

−−= )(

)(),( qRuuRqquR

M

E

The Jacobian of the residual is given by

∂∂−

∂∂−

=

∂∂

∂∂

∂∂

∂∂

=I

qR

uRI

uR

qR

uR

qR

quJM

E

22

11

),(



),(),( kkkk quRuqquJ −=

δδ

Repeatsolve

0;0;1 === kk quk

1+= kk

εε ≤−≤− ++ 11 kkkk qquuuntil

set uuu kk δ+=+1

qqq kk δ+=+1set

use an iterative solver


Iterative Solution of Linear Systems

Lets say we need to solve Key steps in GMRES algorithm

pPq =

make an initial guess to the solution, 0qset k = 0do

compute the residual, kk Pqpr −=

if ,tolrk ≤ return as the solutionkqelse

choose and ins'α βk

k

j

jj

k rqq βα += ∑=

+

0

1

to minimize 1+krset 1+= kk



θθ )()( uRruRr

uR −∗+

=∗∂∂

[ ]

[ ]

−+−

−+−=

∂∂−

∂∂−

=

)()(1

)()(1

),(qRqqRu

uRuuRq

uq

Iq

Ru

RI

uqquJ

MM

EE

M

E

θδθ

δ

θδθ

δ

δδ

δδ

ru

arusign ∗∗= )(θ

)5.0,01.0(∈a



AdvantagesBlack box based approachSuperior global convergence

DisadvantagesCan be sensitive to the choice of the matrix-free parameter


Full-Newton Technique

Represent the mechanical and electrical equations as

0)()(),( int =−= qfufquR extM

0)(),( =−= VquPquRE

Let and be self-consistent solutionsu q0),( =quRM

0),( =quRE

0..),(),( 00 =+∆∂∂+∆

∂∂+= tohq

qRu

uRquRquR MM

MM

Let and be some initial guess0u 0q

0..),(),( 00 =+∆∂∂+∆

∂∂+= tohq

qRu

uRquRquR EE

EE


Full-Newton Technique

Neglecting h.o.t

−=∆

∆

∂∂

∂∂

∂∂

∂∂

),(),(

00

00quRquR

qu

qR

uR

qR

uR

E

M

EE

MM

),( 00 quRqq

Ruu

RM

MM −=∆∂∂

+∆∂∂

),( 00 quRqq

Ruu

RE

EE −=∆∂∂+∆

∂∂

In matrix form


Full Newton Algorithm

Repeat

solve

0;0;0 )()( === ii qui

1+= ii

εε ≤∆≤∆ )()( ii quuntil

set )()1()( iii uuu ∆+= −

)()1()( iii qqq ∆+= −set

−=

∆∆

∂∂

∂∂

∂∂

∂∂

−−

−−

),(),(

)1()1(

)1()1(

)(

)(

iiE

iiM

i

i

EE

MM

quRquR

qu

qR

uR

qR

uR


Full Newton Algorithm

→∂

∂→∂∂

uuf

uRM )(int

entirely elastostatic part

→=∂−∂→

∂∂ P

qVPq

qRE )( entirely electrostatic part

→∂

∂→∂∂

qqf

qR ext

M )( electrical to mechanical coupling term

→∂

∂=∂−∂→

∂∂ q

uuP

uVPq

uRE )()( mechanical to electrical coupling

term


Microfluidics: Gas Flows


Microfilter properties:Openings of various shapesThickness between 1 and 5µmOpening size as small as 2nmHigh burst pressure achieved

Introduction to Microfilters

Design issues:Flow profilesEstimation of flow rateDependence of flow rate on:

geometrysurface propertiespressure difference

Rarefaction effects observed due to smalldimensions


Characteristics of Flows in Micro-Channels

Typical Characteristics:

• Compressible

• High Kn #

• Small Re #

• Small Ma #

• Wide range of Kn #

• Reacting

Effects of high Knudsen Number:

• Slip velocity

• Thermal jump

• Strong interaction with walls


DSMC Flow Chart


Micro-Filter Elements

1x1 1x5 0.2x1 1x10 0.05x1 0.2x2hc (µm) 1 1 0.2 1 0.05 0.2lc (µm) 1 5 1 10 1 2hp (µm) 5 5 1 5 1 1lin (µm) 4 6 4 4 4 4lout (µm) 7 7 5 7 7 5Kn 0.054 0.054 0.27 0.054 1.1 0.27

hp

lc

hc

lin lout


1µmX1µm Filter Element

X Velocity Y Velocity

TemperaturePressure


Knudsen Number and Length Effects

Effect of Kn:Slip velocity increases with Kn

Effect of Length:As lc/hc increases, 2D channel approximationholds good for smaller Kn

@ y=hc/2

@ x=lc/2


Effect of Surface Accommodation1x1 µm filter

Smaller accomodation coefficients:Strong increase in slip velocityTemperature drop increases

@ x=lc/2Velocity

@ y=hc/2


Flow Rate vs. Pressure Difference

Dependence of flow rate onpressure is linear Qualitative behavior iscaptured by 2D channel formula + 1st order slip BC(Arkilic & Breuer, 1997)Good agreement for large lc/hcEffective length can be used for smaller lc/hc


Conclusions

MEMS design is still an artCritical issues

Mixed-domain simulation toolsMultiscale approachesSystem level modeling tools

Need fast and radically simpler techniques for MEMS modeling

Documents

Computational Methods for MEMS - mcc.uiuc.edu Aluru/Aluru_files/Aluru... · Computational MEMS/NEMS Beckman Institute University of Illinois at Urbana-Champaign Scaling Laws ... ·By