Resonance and Electrostatic IEE485A Lecture10 September 2009

More fun with springsIf you yank on this mass and then let go, what happens?The fundamental resonance frequency for a system with spring constant k and mass m is given by:

Applied to MEMSW = 10 mmThe Process:Determine the spring constant of the beam.Determine the mass of the beam.Use to calculate the resonance frequencyOK, so you can treat MEMS structures as mass-spring systems, but how does that relate to an electrical signal?

Intrinsic StressAll of this beam bending stuff has assumed that the materials are deposited with no internal stress. In practice, thin films generally have internal stress, which can result in deflection even in the absence of loading.

Electrostatics IntroductionOne way to translate motion into an electrical signal.Recall the capacitance of a parallel plate capacitor:

Sensing set up structure so that A or d change and look at resulting voltage signal.Actuation apply a voltage to the capacitor that forces a change in A or d.Major advantages of electrostatic sensing and actuation:SimplicityLow powerFast response

Parallel Plate General ExpressionsDefinition of capacitance:

Energy of a capacitor:

Capacitance for a parallel plate capacitor:

Force related to change in d with constant biasing voltage:

dArea, A

ExerciseGiven a parallel plate capacitor with area 400 x 400 mm2 and an initial gap (filled with air) of 10 mm, determine the capacitance and the attractive force if V = 10 V. Repeat if the initial gap is 1 mm.

In some devices that use electrostatic sensing and actuation, charge is held constant rather than voltage. For this case, determine the force associated with changes in the plate separation, d.

Equilibrium Position of PlateElectrostatic force:

Restoring force:

Spring constant is softened by electric field, but this effect is small and well ignore it.Force balance:Vkmx0fixed platemovable plateFmechanicalFelectricx0+x

Pull-In EffectAt V= VP, there is no longer a force balance and the upper plate snaps down.x0+x|F|x0Restoring forceElectric forceElectric force at Pull InIncreasing VV = VP

Calculating the pull-in voltageRearranging force balance:

At pull-in, the slope of the two force/distance curves must be equal:Substituting expression for V2 : This is only true for x = -x0/3. So pull in occurs when the gap reaches 2/3 x0.

ExerciseVerify that x = -x0/3 is the solution for pull-in

Show that the expression for the pull-in voltage, VP, in terms of x0, km, and A is given by:

ExerciseA parallel plate capacitor is suspended by two fixed-guided cantilever beams, each with its length, width and thickness denoted by l = 100 mm, w = 10 mm, and t = 1 mm. The material is polysilicon with a Youngs modulus of 120 GPa. The initial gap is 2 mm, and the plate area is 100 mm X 100 mm.Determine the mechanical spring constant for the structure.Determine the pull-in voltageDetermine the voltage necessary for a 0.5 mm deflection.

Holding voltageAfter pull-in, the voltage required to hold the plate down is less than the pull-in voltage. VholdFmechanicalFelectricDielectric layer necessary to prevent short circuit!td

ExerciseCalculate the holding voltage for the previous example.

Dielectric chargingOne issue with electrostatic devices is the accumulation of charge in the dielectric layerCan cause pull-in voltage drift.Can cause device failure. Electrostatic actuators often use voltage waveforms that alternate in polarity to combat this problem.

*10 um: C = eA/d = 8.85e-12 F/m*400e-6^2/10e-6 = 1.416e-13 F (142 fF); Force = *142 fF*10^2/10e-6 = 708 nN1 um: C = 1.42 pF; F = 70.8 uNNote how force went up 100X when gap reduced 10XF = d/dd (1/2 Q^2/C) = Q^2 (d/dd (d/eA)) = (Q^2/eA) in this case, the force is constant!**VP = 2x0/3 * sqrt ((km * x0)/(1.5*e*A))*From appendix, d for fixed/guided beam = Fl^3/12EI, so k = F/x = 12EI/l^3 = (12*120e9*1/12(10e-6)(1e-6^3))/100e-6^3 =1.2 N/mFor two beams, k = 2X = 2.4 N/mVP = 2x0/3 * sqrt ((km * x0)/(1.5*e*A)) = 2*2e-6/3*sqrt(2.4*2e-6/(1.5*8.85e-12*100e-6^2))= 8 V0.5 um V =