Part (a) Write the equation for the electrostatic energy in the both capacitance

Question

Part (a) Write the equation for the electrostatic energy in the both capacitance systems shown above. Part (b) Derive the equations for the force in the r. y and directions. Part (c) Show that the restoring force per unit edge of the structure (E/W) is proportional to the square of the electrostatic field in the capacitance system Part (d) Calculate the force Fy and the restoring force F, in a MEMS structure formed by two parallel plates that have square geometry 100um on each side, separated by 2m at a potential difference of 50V. Assume that all electrostatic actuation occurs in vacuum and ignore fringe field effects.

Explanation / Answer

for the given parallel plate capacitor

a. with plate width w, length l and plate seperation d, and full overlap

capacitance, C = epsilon*lw/d

energy stored = 0.5CV^2 = 0.5*epsilon*lw*V^2/d ( where V is applied voltage)

E = 0.4*epsilon*L*w*V^2/t

(because plate seperatio is t)

for partial overlap of x = l

C = epsilon*w*x/t

energy storefd = 0.5*epsilon*w*x*V^2/t = 0.5*epsilon*w*l*V^2/t

b. now, conisdering parial overlap in x direction as shown in the figure

Fx = -dE/dx

Fx = -0.5*epsilon*w*V^2/t

Fz = 0

Fy = -dE/dt

Fy = 0.5*epsilon*w*l*V^2/t^2

c. now, Fx/W = 0.5*epsilon*V^2/t

and E = V/t

hence

Fx/W = 0.5*epsilon*E^2*t

hence Fx/W is proportional to square of Electric field E

d. for the given data

L = 100 um

d = 2 um

V = 50 V

then

Fx = -0.5*epsilon*100*10^-6*50^2/2*10^-6

Fx = -5.5385*10^-7 N

and

Fy = 0.5*epsilon*100^2*50^2/2^2 = 2.769*10^-5 N

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