VIEWS: 18 PAGES: 34 POSTED ON: 7/2/2011
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 dAh 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 2 106 0.0375 Vp 0.25(volts) 3 1.5C0 3 1.5 7.0831013 • 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 2A x 2 x0 x x x 3 2 2 0 0 2k m • or x 3 4 106 x 2 4 1012 x 7.552 1019 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 A3 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