GSO002 Problem 7
Problem Statement
Pull-In Effect Analysis in a MEMS Actuator
Problem Statement
In microelectromechanical systems (MEMS), electrostatic actuators that use electrostatic force to control movable parts are widely used. Consider the physical instability (pull-in effect) inherent in such a system.
A fixed lower plate of area S is placed horizontally. Directly above it, an upper plate (movable plate) of area S and mass m is suspended by a light insulating spring with spring constant k. The two plates form a parallel plate capacitor facing each other. The space between the plates is vacuum with permittivity ε0. Edge effects at the plate boundaries can be neglected.
Initially, no voltage is applied between the plates, and the movable plate is at rest in equilibrium between gravity and the spring's elastic force. At this point, the plate separation is d and the capacitance of the capacitor is C0.
From this state, a DC power supply is connected between the plates, and the applied voltage V is slowly increased from 0 (slowly enough that the plates do not vibrate, always maintaining force equilibrium). As the voltage increases, the plate separation decreases. When the voltage V exceeds a critical value V0, the spring's restoring force can no longer support the electrostatic force, and the movable plate is suddenly attracted to and collides with the lower plate. This phenomenon is called the "pull-in effect," and the critical voltage V0 is called the "pull-in voltage."
Find the value of V02, the square of the pull-in voltage.
Constraints
- Spring constant: k=1.35 N/m
- Initial plate separation: d=6.4×10−6 m
- Initial capacitance: C0=2.4×10−12 F
- Mass: m=2.0×10−6 kg
- Gravitational acceleration: g=9.8 m/s2
- All physical quantities are in SI units under ideal vacuum conditions.
Input Format
The square of the pull-in voltage V02 (in units of V2) is expressed as an irreducible fraction BA. Give the value of A+B as a positive integer.
Solution
1. Electrostatic Force and Force Equilibrium
Consider the state where voltage V is applied and the plate separation is x (0<x<d) in equilibrium.
Since the initial equilibrium (at zero voltage) already balances gravity and the spring force, we can treat the system effectively as a spring with natural length d.
The capacitance at separation x:
C=xε0SThe electrostatic attractive force (derived from virtual work principle at constant voltage):
Fe=21x2ε0SV2The spring restoring force acts upward with magnitude k(d−x). The force balance equation is:
k(d−x)=21x2ε0SV22. Limiting Condition for Equilibrium to Exist
Solving for V2:
V2=ε0S2kx2(d−x)Equilibrium is lost when the required V2 exceeds the maximum of f(x)=x2(d−x).
Differentiating:
f′(x)=2x(d−x)−x2=x(2d−3x)For 0<x<d, f′(x)=0 gives x=32d.
f(32d)=94d2⋅31d=274d3Thus, the square of the pull-in voltage:
V02=ε0S2k⋅274d3=27ε0S8kd33. Numerical Calculation
Using C0=dε0S, so ε0S=C0d:
V02=27C08kd2Substituting:
V02=27×2.4×10−128×1.35×(6.4×10−6)2=64.8×10−1210.8×40.96×10−12 V02=640.96=6004096Dividing numerator and denominator by gcd(4096,600)=8:
- 4096÷8=512
- 600÷8=75
Since 512=29 and 75=3×52 are coprime, the irreducible fraction is 75512.
A=512, B=75:
A+B=512+75=587Final Answer: 587