Coupled Electromechanics
Transcript of Coupled Electromechanics
Coupled ElectromechanicsElectromechanics
G. K. [email protected] EngineeringIndian Institute of ScienceBangalore, INDIABangalore, INDIA
November 2014, for ME 237/NE 211 course in IISc
2 Computing the electrostatic force in a parallel‐plate capacitor
0 = permittivity of free space
lw
g V = applied voltageC = capacitance
2 20 ( )1 12 2c
wlESE CV Vg
Electrostatic co‐energy
2012
cl
ESE wF Vl g
Force in the length direction
2012
1 1
cw
ESE lF Vw g
ESE wl A
Force in the width direction
F i th
G.K. Ananthasuresh, Indian Institute of Science
2 20 02 2
1 12 2
cg
ESE wl AF V Vg g g
Force in the gap direction
3 Computing the electrostatic force in general 3‐D problems
Conductor 1
22 V
Conductor 1Conductor 2
V 11 V
Electric potential = Electric field Charge density = charge per unit areaElectric field =
nF
ˆ21 2
eElectrostatic force = Permittivity of the i te e i ediu
g y g p
Surface normal
G.K. Ananthasuresh, Indian Institute of Science
intervening mediumIt is a surface force (traction).
4Electric field and iso‐potential lines
G.K. Ananthasuresh, Indian Institute of Science
5Basics of electrostatics
1 212 122
ˆ4
q qF rr
Coulomb’s law
0
12 122 200 0
4
ˆ ˆlim4 4q
r
Qq QE r rq r r
Electric field due to a single charge
work done P
E dlq
E
Electrical potential (voltage)
Electric field is the negative of the gradient of the potential
0r
E
D E E
Electric field is the negative of the gradient of the potential
Density of electric displacement vector
G ’ l
2
v vS V
D dS dV Q
Gauss’s law
Integral form
Diff i l f
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2v vD
Differential form
6Computing the electrostatic force
ˆ ˆn D n E Surface charge density
esf
s n D n E
snE
Surface charge density
Normal component of the electric field
2 2s s
sdF Edq dq dA
electric field
2
2 2
2
s
sdFdA
Force along the surface normal2dA
2
ˆsf n
Electrostatic force on the
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2esf n
surface per unit area
7
Computing the electrostatic force (contd.)
Governing equations to solve for the charge density in the differential equation form:
42 02
On the conductorsIn the intervening medium
Pl t ti l th d t ifi dPlus, potentials on the conductors are specified.
This is suited for FEM but sufficient intervening medium also needs to be meshed along with the interior of the conductorsneeds to be meshed along with the interior of the conductors.
Governing equations to solve for the charge density in the integral equation form:
'')()( dSxxxx
Surfaces
the integral equation form:
G.K. Ananthasuresh, Indian Institute of Science
This is suited for BEM because only conductors’ boundaries need tobe meshed.
8Static equilibrium of an elastic structure under electrostatic force
Charge distribution
‐‐ ‐ ‐ Charge distribution
causes electrostatic force of attraction between conductors
‐‐
‐ ‐‐ ‐ ‐ ‐‐‐+++
++ + + +
+
‐ ‐ ‐‐‐
Electrostatic force
‐‐
++
++
+
+ +++
‐ ‐ ‐ ‐
V
Electrostatic force deforms conductors
+++
VDeformation of conductors causes charges to re‐2 ˆ1 nFEl i f
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distribute02
F eElectrostatic force =
9 Coupled governing equations of electro‐ and elasto‐ statics
conductorsalloffor''
)()( sdSxx 'xxSurfaces
0
2 ˆ21
nf te
σ ineverywhere0
u
te
uufnσ
ononˆ
y
0
Tuuε
εEσ
1
:A self‐consistent solution is needed!
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uuε 2
solution is needed!
10Start with 1‐dof lumped model…
V
A0= plate area
= permittivity of free space
201 VAk 2
20
21 VA
k Static equilibrium
2
20
0
2V
xgkx
202 xg kx
x0g A cubic equation!
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A cubic equation!
11Lumped 1‐dof modeling of coupled electro‐ and elasto‐ static behavior
1 A 2
201 VA
0g
2
20
0
21 V
xgAkx
kx 202
Vxg
Forces
Three solutions
x
ble
ble
able
0)( PE
energy
202 11 VAkxPE
Potential energyStab
Stab
Unst 0
x
Two stable; one unstable; And, one infeasible
x
tential e 022
Vxg
kxPE
22 )( AVPE
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Po
30
20
2
2
)()(
xgAVk
xPE
Use to test stability.
12Pull‐in phenomenon
1V 2V< 3V< xgkVAVkPE 302
20
2 )(0)(
Condition for critical stability
gy
3/0g AgV
xgk
x 0
023
0
02
)(0)(
)(
)(1 0020 gxxgkVAkx
0g
al energ 322 2
0
xVxg
kx
x
Potentia
Agk
V inpull0
3
278
0
P
3/2 Vx
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3/2 0g inpullVinpullV
13 With a dielectric layer: pull‐up and hysteresis
3
inpull 0278
dtg
AkV
Dielectric layer027 rA
2
x0g
t0
up pull 0
2
r
dtgAkV
dt x
0gPull‐up voltage is found by
ti th f f i dequating the forces of spring and electrostatics at . 0gx Gilbert, J. R., Ananthasuresh, G. K., and Senturia, S. D., “3‐D Modeling and Simulation of Contact
G.K. Ananthasuresh, Indian Institute of Science
uppullV inpullV V, g
Problems and Hysteresis in Coupled Electromechanics,” presented at the IEEE‐MEMS‐96 Workshop, San Diego, CA, Feb. 11‐15, 1996.
14Pull‐in phenomenon used in a display device
www.iridigm.com (a QualComm acquisition)
Interference‐modulation by electrostatic actuation of vertically moving membranes.
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15Iridigm became mirasol…
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16 Distributed modeling of the electrostatically actuated beam
Finite element method Finite difference method
V
+ + + + + + + + + FEM or FDM could be used to solve the nonlinear equation:V
0)(2 2
0
20
4
4
ug
wVdx
udEI
V++ +
++ ++
++ Include the effects of
residual stress as well:V residual stress as well:
0)(2 2
0
20
2
2
04
4
ug
wVdx
udwtdx
udEI
G.K. Ananthasuresh, Indian Institute of Science
)(2 0 ugdxdx
A correction due to fringing field (edge and corner effects) is also included.
17
Solving the general 3‐D problem
Boundary element method for the integral equation of electrostatics
Finite element method for the differential equation of elastostatics
'')()( dSxxxx
Surfaces
elastostatics
Charge on kth panel
Discretize the boundary surfaces into n panels. eFUK
ipanel k
n
i i
ik xx
daaqp
''
1
Area of kth panel
Potential on kth panel
Assemble to get:2
l k( / ) ˆk k
kq a
f n
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qPp qpC
gon panel k 2 kf n
18Solution approaches
Relaxation‐‐ iterate between the elastic and electrostatic domains.‐‐ converges except in the vicinity of pull‐in voltage; but slow.converges except in the vicinity of pull in voltage; but slow.
Surface Newton‐‐ compute sensitivities of surface nodes.‐‐ use a Newton step to update those nodes.p p‐‐ then, re‐compute electrostatic force and internal deformations.
Direct Newton‐‐ compute all derivatives to update charges and deformations.
MM
RUR
UR Residuals in mechanical
and electrical domains
E
M
EE RR
qU
qR
UR
qU and electrical domains
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qUFor example, see: G. Li and N. R. Aluru, “Linear, non‐linear, and mixed‐regime analysis of electrostatic MEMS,” Sensors and Actuators, A 91, 2001, pp. 279‐291, and references therein.
19Comb‐drive example
(a) Tang, W. C., Nguyen, C., Judy, M. W., and Howe, R. T., “Electrostatic Comb‐Drive of LateralPolysilicon Resonators,” Sensors and Actuators A, 21 (1), 1990, pp. 328‐331
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(b) http://mems.sandia.gov/scripts/images.asp (Sandia National Laboratories, New Mexico, USA)
20Schematic of the comb‐drive
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21Lumped mechanical stiffness
3
3
3
33
12 EbhFl
bhFl
EIFl
3 3Ebh Etwk
1212
12 EbhbhEEI
3EbhFk
3 3kl l
3
l2Etwk
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3lk
total 3k
l
22Electric potential solution
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FEMLAB (old COMSOL) result
23Close‐up of the electric potential
FEMLAB (old COMSOL) result
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24Deformation of the comb‐drive
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25What about dynamic behavior?
in-pulldynamicV = dynamic pull‐in voltage
0gFrequency =
ergy
0g
tV
L d 1 d f d l
ntial ene
x2
0
20
)(2 xgAVkxxm
Lumped 1‐dof model
Poten 0 )(g
02
04
wVudEIuwt
Beam model
0)(2 2
0
04
ugdxEIuwt
tVtVVVtVVV acacdcdcacdc 22222 sinsin2)sin(
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acacdcdcacdc )(
Will contain a term!2So, the response will show two resonance at two frequencies.
26Damping: squeezed film effects
S d fil d iV Squeezed‐film damping
20 AVkxxbxm 02
20
4
4
wVudEIubuwt
Lumped 1‐dof model Beam model
20 )(2 xg
kxxbxm
)(2 20
4 ugdx
How do you obtain ?b
Use isothermal, compressible, narrow gap Reynolds equation to model the film of air beneath the beam/plate/membrane.It is widely used in lubrication theory
G.K. Ananthasuresh, Indian Institute of Science
It is widely used in lubrication theory.By analyzing this equation, we can extract the essence of damping as a lumped parameter – the so called “macromodeling”.
27 Modeling squeezed film effects: isothermal Reynolds equation
1)()(
Pressure distribution in the 2‐D x‐y plane Gap varies in the x‐y plane for a deformable structure
(beam,plate, membrane)
),(),(),(12
1),(),( 3 yxpyxgyxpt
yxgyxp
Viscosity of air
For lumped 1‐dof modeling, we have a rigid plate. So, g does not depend on . ),( yx
)(1)()(),( 2233 gggyxp
),(21
12),(),(
12),( 22 yxpgyxpyxpg
tgyxp
Assume further that pressure distribution is the same along the length of the plate soAssume further that pressure distribution is the same along the length of the plate so that it becomes a one dimensional problem.
)(1)( 22
3
ypggyp yAssumed pressure distribution
G.K. Ananthasuresh, Indian Institute of Science
)(212
ypt x y
S. D. Senturia, Microsystems Design, Kluwer, 2001.
28Behavior with small displacements
)(
21
12)( 22
3
ypgt
gyp
Linearize around :),( 00 gp
gggppp 00
Also, use non‐dimensional variables: ˆ,ˆ, ggppy
2
2
20
20
2
2
20
20 ˆ
12ˆˆ
12ˆ
ggp
wpg
tgp
wpg
tp
Also, use non dimensional variables:00
,,g
gp
pw
width01212 gwtwt
Separation of spatial and temporal components: teptp )(~),(ˆ22
teggpp
wpg
0
2
2
20
20 ~~
12
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Assume a sudden velocity impulse to the plate. Then, for t > 0, this term is zero.(with displacement )0xx
29Behavior with small displacements (contd.)
nnnn BAppp
wpg cossin~0~~
12 2
2
20
20
020
212pg
w nn
w12 00 pg
Boundary conditions and velocity‐impulse assumption give:
nn nwnpgn 2
220
20
4
,...5,3,1;12
;
n
tn
nenng
xAodd0
0 )sin(4
1 8Force on the plate =
n
tsq
negxwlpdtpwlptf
odd22
0
00
1
00 n
8),(ˆ)(
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Take the Laplace transform (continued on the next slide).
30Finally, getting to lumped approximation…
)(11961196)(33
ssXwlxwlsF
)(
1n1n)(
odd43
040
odd43
04 ssXsg
xsgsF
n
n
n
n
sq
)(1
)(1
196)( 30
4
3
ssXsbssXsg
wlsFsq
For only.1n
11cc
396 wlb 0
20
2 pg
Transfer function for general displacement input!
30
4gb
212 wc
Damping Cut‐off
bR )(),( sXtx )(),( sFtf sqsq
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coefficient frequency
cbC 1
31What does it mean mechanically?
396 wlb
k
208wpbk csq
30
4gb
x x0
2gcsq m
Thus, squeezed film effect creates two effects:qViscous damping + “air‐spring”Further analysis indicates that at low frequencies, damping d i t d i i t hi h f i
G.K. Ananthasuresh, Indian Institute of Science
dominates, and air‐spring at high frequencies.See S. D. Senturia, Microsystems Design, Kluwer, 2001, for details.
32Move up to beam modeling…
)()}({)(1)},({),,( 30
0 tyxptxugtyxptxugtyxp
),,()},({),,(12 0 tyxptxugtyxp
t
0)(),()(),( 20
42/
twVtxudEIdytyxptxuwt
w 0)},({2
),,( 20
42/
2
txugdx
EIdytyxpt
wtw
Note that this is still al h l d Note that this is still a parallel‐plate approxmation!
Solve these two coupled equations.
An approachUse FDM for pressure equation and FEM or FDM for di ti i th d i ti d i t t i ti
G.K. Ananthasuresh, Indian Institute of Science
discretizing the dynamic equation, and integrate in time using the Runge‐Kutta method.
33FD solution of Reynolds equation
3( )12 .( )Pg g P Pt
Solution submitted by N. Sajinu for ME 237 course at IISc.
t223 2 2 2
2 2
3 . .12 12 12
P g P P Pg P P Pg P g P g P Pt x y x x x x y y g t
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34
A typical beam’s response with squeezed film effect
The transverse deflection of the mid‐point of a fixed‐fixed beam under (Vdc+Vac) voltage input under the squeezed film effect:
pointmidu pointmid
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35What about this problem now?
VDiaphragm tV sin
Passive inlet valve Passive outlet valve
rate
Frequency
Flow
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A problem involving three energy domains that are strongly coupled. Furthermore, the fluids part is non‐trivial.
36Main points
MEMS are systems that tightly integrate many energetic phenomena which makesmany energetic phenomena, which makes their modeling non‐trivial.
Coupled multi‐physics equations need to be Coupled multi‐physics equations need to be solved.
Reduced order lumped “macro” models are Reduced order lumped macro models are useful for design and system‐level simulation
G.K. Ananthasuresh, Indian Institute of Science