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Electrostatic Sensors and Actuators

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  • pg 1
									                                                                           1




    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.

                                                        University of Illinois
               Gravity is generally ignored.            ECE 485/MSE 485
                                                                               3

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




                                                      University of Illinois
                                                      ECE 485/MSE 485
                                                                                                        4

                          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



                                                                               University of Illinois
                                                                               ECE 485/MSE 485
                                                                                 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”




                                                        University of Illinois
                                                        ECE 485/MSE 485
                                                                                               6

A threshold point                                       VPI




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



                                                                 X0



                                              Km



                                      Positive
                      fixed           feedback
                                      -snap, pull in




                                                                      University of Illinois
                                                                      ECE 485/MSE 485
                                                                                   7
    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.




                                                          University of Illinois
                                                          ECE 485/MSE 485
                                                                                    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




                                                           University of Illinois
                                                           ECE 485/MSE 485
                                                                                                 9
   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.


                                                                        University of Illinois
                                                                        ECE 485/MSE 485
                                                                             10
    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




                                                         University of Illinois
                                                         ECE 485/MSE 485
                                                                                11

               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%.




                                                           University of Illinois
                                                           ECE 485/MSE 485
                                                                           12
           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.




                                                       University of Illinois
                                                       ECE 485/MSE 485
                                                                  13




          How to Calculate K?


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




                                              University of Illinois
                                              ECE 485/MSE 485
                                                                           14

       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

                                                       University of Illinois
                                                       ECE 485/MSE 485
                            15




2




    1




        University of Illinois
        ECE 485/MSE 485
                               16

Examples




           University of Illinois
           ECE 485/MSE 485
                                                           17

       A Clamped-Clamped Beam

                        Fixed-guided




Fixed-guided




                                       University of Illinois
                                       ECE 485/MSE 485
                                          18

A Clamped-Free Beam




                      University of Illinois
                      ECE 485/MSE 485
                                                                      19

            Naming Conventions



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

                   (g)

(a)
      (b)

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




                                                  University of Illinois
                                                  ECE 485/MSE 485
                    20




University of Illinois
ECE 485/MSE 485
                                                                         21

       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.




                                                     University of Illinois
                                                     ECE 485/MSE 485
                                                                        22




                            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

                                                    University of Illinois
                                                    ECE 485/MSE 485
                                                               23

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




                                           University of Illinois
                                           ECE 485/MSE 485
                                                                                     24

        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



                                                                 University of Illinois
                                                                 ECE 485/MSE 485
                                                                           25

                  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.




                                                       University of Illinois
                                                       ECE 485/MSE 485
                    26




University of Illinois
ECE 485/MSE 485
                                                             27

                               Example
• A parallel plate capacitor
  suspended by two fixed-fixed
  cantilever beams, each with
  length, width and thickness
  denoted l, w and t, respectively.
  The material is made of
  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.


                                         University of Illinois
                                         ECE 485/MSE 485
                                                                                                             28

       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



                                                                                         University of Illinois
                                                                                         ECE 485/MSE 485
                                                                                                     29

          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.

                                                                                 University of Illinois
                                                                                 ECE 485/MSE 485
                                                                                    30

         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?



                                                                University of Illinois
                                                                ECE 485/MSE 485
                                                                                          31

             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



                                                                      University of Illinois
                                                                      ECE 485/MSE 485
                                                                          32

           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.




                                                      University of Illinois
                                                      ECE 485/MSE 485
                    33




University of Illinois
ECE 485/MSE 485
                    34




University of Illinois
ECE 485/MSE 485

								
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