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# Electrostatic Sensors and Actuators

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Analysis of Electrostatic Actuator

What happens to a parallel plate capacitor when
the applied voltage is gradually increased?

University of Illinois
ECE 485/MSE 485
2

An Equivalent Electromechanical Model
Fmechanical
x
If top plate
moves down-                     Felectric
Note: direction
ward, x<0.           Km                              definition of
variables

• This diagram depicts a parallel plate capacitor at
equilibrium position. The mechanical restoring spring with
spring constant Km (unit: N/m) is associated with the
suspension of the top plate.
• According to Hooke’s law, F
mechanical  Km x
• At equilibrium, the two forces, electrical force and
mechanical restoring force, must be equal. Less the plate
would move under Newton’s first law.

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Gravity is generally ignored.            ECE 485/MSE 485
3

Mechanical Spring
• Cantilever beams with various boundary conditions
• Torsional bars with various boundary conditions

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Electrical And Mechanical Forces
If the right-hand plate moves
closer to the fixed one, the magnitude
of mechanical force increases linearly.

Equilibrium:
If a constant voltage, V1, is applied
|electric force|=|mechanical force|
in between two plates, the electric force
changes as a function of distance. The
closer the two plates, the large the
force.

X0
x       Equilibrium
position

Km

fixed

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5

Pull-In Effect

• As the voltage bias increases from zero across a pair of
parallel plates, the distance between such plates would
decrease until they reach 2/3 of the original spacing, at
which point the two plates would be suddenly snapped
into contact.

• This behavior is called the pull-in effect.
– A.k.a. “snap in”

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A threshold point                                       VPI

Equilibrium:                                           X=-x0/3
|electric force|=|mechanical force|

X0

Km

Positive
fixed           feedback
-snap, pull in

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Mathematical Determination of Pull-in Voltage
Step 1 - Defining Electrical Force Constant

• Let’s define the tangent of the electric force term. It is
called electrical force constant, Ke.
F          CV 2
ke          ke  2
x          d

• When voltage is below the pull-in voltage, the magnitude
of Ke and Km are not equal at equilibrium.

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8

Review of Equations Related To Parallel Plate

• The electrostatic force is      E    1 A 2    1 CV 2
F          V 
d    2d 2
2 d

• The electric force constant is

1    A 2 A V 2    V2
K e   (2) 3 V      2
C 2
2    d      d d     d

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Mathematical Determination of Pull-in Voltage
Step 2 - Pull-in Condition
• At the pull-in voltage, there is only one intersection
between |Fe| and |Fm| curves.
• At the intersection, the gradient are the same, I.e. the two
curves intersect with same tangent.
Ke  K m
• This is on top of the condition that the magnitude of Fm
and Fe are equal.
– Force balance yields V 2   2km x( x  x0 )   2km x( x  x0 )
2

A                C
– Plug in expression of V2 into the expression for Ke,            CV 2
ke  2
• we get                                                       d
CV 2        2km x
ke              
( x  x0 ) 2 ( x  xo )
– This yield the position for the pull-in condition, x=-x0/3.
Irrespective of the magnitude of Km.

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Mathematical Determination of Pull-in Voltage
Step 3 - Pull-in Voltage Calculation

• Plug in the position of pull-in into Eq. * on previous page,
we get the voltage at pull-in as
2
4 x0
V p2       km
9C
• At pull in, C=1.5 Co
• Thus,
2 x0 km
Vp             .
3 1.5C0

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Implications of Pull-in Effect

• For electrostatic actuator, it is impossible to control the
displacement through the full gap. Only 1/3 of gap
distance can be moved reliably.

• Electrostatic micro mirrors
– reduced range of reliable position tuning

• Electrostatic tunable capacitor
– reduced range of tuning and reduced tuning range
– Tuning distance less than 1/3, tuning capacitance less than
50%.

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ECE 485/MSE 485
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Counteracting Pull-In Effect
Leveraged Bending for Full Gap Positioning

• E. Hung, S. Senturia, “Leveraged bending for full gap
positioning with electrostatic actuation”, Sensors and
Actuators Workshop, Hilton Head Island, p. 83, 2000.

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How to Calculate K?

Chang Liu
Micro and Nano Technology Research Group
University of Illinois at Urbana-Champaign

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Analysis of Mechanical Force Constants
• Concentrate on beams
Fixed-free
• Three types of most
relevant boundary
conditions
– free: max. degrees of
freedom
– fixed: rotation and                             Two fixed-
translation both                               guided beams
restricted
– guided: rotation
restricted.
• Beams with various
combination of boundary
conditions
– fixed-free, one-end-
fixed beam
Four fixed-guided beams
– fixed-fixed beam
– fixed-guided beam

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2

1

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Examples

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A Clamped-Clamped Beam

Fixed-guided

Fixed-guided

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A Clamped-Free Beam

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Naming Conventions

(c)
(j)
(i)
(h)

(g)

(a)
(b)

(f)
(e)
(d)
(k)

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Fixed-Free Beam by Sacrificial Etching

• Right anchor is fixed because its rotation is completely
restricted.
• Left anchor is free because it can translate as well as
rotate.
• Consider the beam only moves in 2D plane (paper plane).
No out-of-plane translation or rotation is encountered.

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t
2
M   dF (h)h  
A
  h dAh
w h t
2

t                         t
2        h               2

M      max    dA h  max   h 2 dA  max I
t     ( ) 
t          t                t
h                ( ) w h t      ( )
w
2      2         2        2       2

d2y
EI 2  M ( x),
dx

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Force Constants for Fixed-Free Beams
• Dimensions
– length, width, thickness
– unit in mm.
• Materials
– Young’s modulus, E
– Unit in Pa, or N/m2.

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Force Constants for Fixed-Free Beams

• Moment of inertia I (unit: m4)
– I= wt
3
for rectangular crosssection
12
Fl 2
• Maximum angular displacement
2 EI
Fl 3
• Maximum vertical displacement under F is
3EI
F    3EI Ewt 3
• Therefore, the equivalent force constant is         km        3 
Fl 3   l   4l 3
3EI

• Formula for 1st order resonant frequency               3.52   EIg
– where          is the beam weight per unit length. 2     l 4

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Modulus of Elasticity
• Names
– Young’s modulus           F
– Elastic modulus      x
• Definition           E     A
 x L
L
• Values of E for various materials can be found in notes,
text books, MEMS clearing house, etc.

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Example
• A parallel plate capacitor
suspended by two fixed-fixed
cantilever beams, each with
length, width and thickness
denoted l, w and t, respectively.
polysilicon, with a Young’s
modulus of 120GPa.
• L=400 mm, w=10 mm, and t=1
mm.
• The gap x0 between two plates
is 2 mm.
• The area is 400 mm by 400 mm.
• Calculate the amount of vertical
displacement when a voltage of
0.4 volts is applied.

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Step 1: Find mechanical force constants

• Calculate force constant of one beam first
– use model of left end guided, right end fixed.
– Under force F, the max deflection is        Fl 3
d
– The force constant is therefore            12 EI
F 12 EI Ewt 3 120  10 9  10  10 6  (1 10 6 ) 3
Km       3  3                                             0.01875 N / m
d   l    l               (400  10  6 ) 3

– This is a relatively “soft” spring.
– Note the spring constant is stiffer than fixed-free beams.

• Total force constant encountered by the parallel plate is
K m  0.0375 N / m

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Step 2: Find out the Pull-in Voltage

• Find out pull-in voltage and compare with the applied
voltage.
• First, find the static capacitance value Co
8.85  10 12 ( F / m)  (400  10 6 ) 2
C0                         6
 7.083  10 13 F
2  10

• Find the pull-in voltage value
2 x0 km    2  2  106       0.0375
Vp                                            0.25(volts)
3 1.5C0         3      1.5  7.0831013

• When the applied voltage is 0.4 volt, the beam has been
pulled-in. The displacement is therefore 2 mm.

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What if the applied voltage is 0.2 V?
• Not sufficient to pull-in
• Deformation can be solved by solving the following
equation
 2k m x( x  x0 )
2
 2k m x( x  x0 )
V2                       
A                  C
v 2A
x  2 x0 x  x x 
3          2    2
0       0
2k m
• or
x 3  4  106 x 2  4  1012 x  7.552  1019  0

• How to solve it?

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Solving Third Order Equation ...

• To solve     x 3  ax 2  bx  c  0

 a2            a     ab
• Apply y  x  a / 3                p       b, q  2( )3     c
3              3     3
• Use the following definition               3      2
 p q
Q    
 3   2
q           q
A3       Q,B  3     Q
2            2
• The only real solution is
•
y  A B
a
x  A B 
3

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Calculator … A Simple Way Out.

• Use HP calculator,
– x1=-2.45x10-7 mm
– x2=-1.2x10-6 mm
– x3=-2.5x10-6 mm

• Accept the first answer because the other two are out side
of pull-in range.

• If V=0.248 Volts, the displacement is -0.54 mm.

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