lecture 6: electrostatic sensors and actuatorsmech466/mech466-lecture-6.pdf · 11 pull-in voltage...

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MECH 466Microelectromechanical Systems

University of VictoriaDept. of Mechanical Engineering

Lecture 6:Electrostatic Sensors and Actuators

© N. Dechev, University of Victoria

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Equilibrium Positions of Electrostatic Devices

Pull-in Voltage

Case-Studies of Electrostatic Devices

Overview


3© N. Dechev, University of Victoria

Equilibrium Position of Electrostatic Devices

Consider the following parallel plate capacitor system:

(a) Without electrical bias

anchored plate

moveable plate

km

anchored plate

km

(b) With electrical bias

* Note, that C is also a function of d

We can describe the mechanical force as:

We can describe the electrical force as:



From the diagram, let:

Therefore, the expression for the electrostatic force becomes:

To find an expression for the equilibrium position, we must equate:

Therefore, we have the expression:



The previous equation can be expanded, which yields the quadratic equation for x:

*Note, this equation yields two solutions



A plot of the mechanical force and electrical force shows:Fmechanical =-kmx

Felectrical =εAV2/(2d2)

- The linear mechanical function intersects with the electrical function at two points.- Note: Only the solution closest to the starting position (right side) is realizable.

As we increase voltage, the previous graph will change as follows:


Concept of ‘Pull-in’ Voltage:

Fig 4.5 Balance between mechanical forceand electrical force [Chang Liu]

Eventually, we reach a ‘Critical Point’ voltage where there is only one solution, where the mechanical force is equal to the electrical force.

This is called the ‘Pull-in’ voltage.



Fig 4.6 Pull-in Voltage [Chang Liu]

Consider the consequences of applying a voltage higher than the ‘Pull-in’ voltage.

To conclude: Electrostatic devices should be designed or operated such that the applied voltage remains below the ‘Pull-in’ voltage to avoid ‘snap-in’.

‘Snap-in’ may damage the mechanism or cause burn-out due to contact under high applied voltage.



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Review Examples 4.2 & 4.3 in Textbook

Homework:


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Pull-in Voltage

Calculate ‘Pull-in’ voltage, Vp:

We know that at equilibrium conditions:

Fe = Fm where: and

(eq. 4.12)


Moving PlateWhere: V - Voltage applied between the two plates Km - Mechanical Spring Constantxo

Anchored Plate

xFmechanical

FelectricalKm

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Upon review of the Force vs. Displacement graphs, we know that at Vpull-in the Fe curve touches the Fm curve at one point.

At this point, the tangent (slope) of Fe is equal to the slope of Fm.

The slope of Fe at this point can be defined as:

And the slope of Fm is defined as:

Therefore, this can be written as:


Pull-in Voltage

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Substituting in the expression (4.12) for have:

Therefore:

Which gives an expression for displacement, x, as:


Pull-in Voltage

Note: This means that the plate is displaced by 1/3rd the total separation distance d, when V = Vpull-in

This displacement value for x (at Vpull-in) can be substituted back into equation 4.12, which allows us to determine Vpull-in to be:


Pull-in Voltage

Note: That C is the Capacitance at the “pull-in” displacement point. Hence, it must be recalculated at that point.

Note: Therefore, it is simpler to replace C, with 1.5Co , i.e. the original capacitance times 1.5, to represent capacitance value at xo/3.

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Example of Equilibrium Position:

Given two round plates of diameter d, directly overtop one-another, separated by a 2 um gap.

Q: What is the displacement of the upper plate if a voltage of 5 V is applied between the plates?


Parallel Plate Capacitors

where: d = 100 umL = 100 umw = 5 umt = 2 um

Material: PolysiliconE: 160 GPa



Solution:

Step 1: Calculate the mechanical spring constant Km.

- This is a “fixed-guided” beam

- From Appendix B, the maximum deflection for a fixed-guided beam is given by:

- Establish the deflection formula for the beam, by determining the appropriate model in terms of end boundary conditions



Solution-Continued:

- Note that I for a beam with rectangular cross-section is:

-therefore:



Solution-Continued:

- Since

- we have for each beam.

- Since the beams are identical and loaded symmetrically, the total system stiffness is:



Solution-Continued:

Step 2: Find the displacement

- Using equation (4.12) find an expression for displacement and solve:

* However, C varies with displacement d, therefore, for parallel plates, this expression can be re-written as:

where:



Solution-Continued:

- the previous equation can be re-arranged with respect to x as:

-substituting values:

- this equation yields three possible solutions for x:

<-- physically realizable

<-- past pull-in deflection

<-- impossible



Solution-Continued:

Step 3: Check value for Pull-in Voltage, to ensure it is not exceeded.

Therefore:

Answer: Upper plate will be displaced by 0.014 um when 5 V is applied between the plates.

Where:

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Review Case Studies 4.1, 4.3 & 4.6 in Textbook

Homework:


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Example: “Case Study 4.2” from Textbook

Torsional Parallel Plate Capacitive Accelerometer


Torsional Parallel Plate Capacitors

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Example: “Case Study 4.2” from Textbook


Torsional Parallel Plate Capacitors

See Class Notes

lecture 6: electrostatic sensors and actuatorsmech466/mech466-lecture-6.pdf · 11 pull-in voltage...

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