Other Design Issues of Other Design Issues of Microleverage Mechanisms Microleverage Mechanisms

resonant accelerometer

4.10 Other Design Issues of Other Design Issues of Microleverage Mechanisms Microleverage Mechanisms

4.10.1 4.10.1 Beam Beam column column strength strength

For most micro-scale MEMS devices, the classical beam buckling theory still For most micro-scale MEMS devices, the classical beam buckling theory still applies (Chiao and Lin, 2000). The critical load on a column fixed at one end and free at applies (Chiao and Lin, 2000). The critical load on a column fixed at one end and free at the other (Timoshenko and Goodier, 1970), is

the other (Timoshenko and Goodier, 1970), is

10 10 12 12 14 14 16 16 18 18 20 20

A A m m p p l l i i f f i i c c a a t t i i o o n n F F a a c c t t o o r r , , A A

0 2 2 4 4 6 6 8 8 10 10 12 12 14 14 16 16 18 18 2020 Lever Arm Width, micron

Lever Arm Width, micron

All dimension All dimension in microns in microns L = 210 L = 210 l = 10 l = 10 lf = 200 lf = 200 wf = 2 wf = 2 lc = 24 lc = 24 wc = 2 wc = 2 lp = 6 lp = 6 wp = 2 wp = 2

(4.41) (4.41) 12

here 12 here w w 4 ,,

4 1

1 ³³

2 2

2 ww t t

I I ll

I I E P E

==

^π^π

==

which is one quarter of the strength of a column if both ends are hinged, 1/16 of the which is one quarter of the strength of a column if both ends are hinged, 1/16 of the strength of a column if both ends are fixed. For a long and narrow beam with a width strength of a column if both ends are fixed. For a long and narrow beam with a width ww of 2

of 2

μμ

m, thicknessm, thickness t t of 2of 2

μμ

m, lengthm, length ll of 200of 200

μμ

m, and m, and E E = 1.65 x 10= 1.65 x 10¹¹¹¹ N/mN/m²²,, the criticalthe critical load is calculated to be 1.23 x 10

load is calculated to be 1.23 x 10^-5^-5 N. This determines the maximum allowable force in aN. This determines the maximum allowable force in a microleverage mechanism before buckling. In the case of a resonant accelerometer with microleverage mechanism before buckling. In the case of a resonant accelerometer with the single-stage leverage mechanism, for a total proof-mass volume of 500 x 500 x 2 the single-stage leverage mechanism, for a total proof-mass volume of 500 x 500 x 2

μμ

mm³³ and silicon density of 2.3g/cm

and silicon density of 2.3g/cm³³, the accelerometer can sustain a maximum, the accelerometer can sustain a maximum acceleration/deceleration of 6600 ms

acceleration/deceleration of 6600 ms^-2^-2 as far as the T-F beam column strength isas far as the T-F beam column strength is concerned.

concerned.

Flexure silicon beams in a microleverage mechanism usually fail by buckling Flexure silicon beams in a microleverage mechanism usually fail by buckling before

before plastic deformation. plastic deformation. Based Based on on a a yield strength yield strength of 7x1of 7x100⁹⁹ N/mN/m²², a beam width of 2, a beam width of 2

μμ

m and a beam length of 50m and a beam length of 50

μμ

m, the maximum allowable axial load is 0.7N which arem, the maximum allowable axial load is 0.7N which are orders of magnitude higher than the applied load. For instance, in the resonant orders of magnitude higher than the applied load. For instance, in the resonant accelerometer, with a 1000g input acceleration and a leverage mechanism amplification accelerometer, with a 1000g input acceleration and a leverage mechanism amplification factor of 200, the axial load on the DETF will be 2.3x10

factor of 200, the axial load on the DETF will be 2.3x10^-4^-4 N. At this load the beam willN. At this load the beam will always be in the elastic regime, which is a basic assumption for SUGAR simulations.

always be in the elastic regime, which is a basic assumption for SUGAR simulations.

Similarly, worst case for a bending moment occurs with a 1000g input acceleration. This Similarly, worst case for a bending moment occurs with a 1000g input acceleration. This acceleration yields a force of 2.3x10

acceleration yields a force of 2.3x10^-4^-4 N N and and given given a a 200 200 micron beamicron beam length m length a a momentmoment of 4.6x10

of 4.6x10^-4^-4 N N

μμ

m. The calculated allowable maximum bending moment in the beam is them. The calculated allowable maximum bending moment in the beam is the yield strength multiplied by the section modulus, which is 2.33x10

yield strength multiplied by the section modulus, which is 2.33x10^-7^-7 Nm (0.233NNm (0.233N

μμ

m),m),

orders of magnitude higher than the calculated bending moment. This shows that for orders of magnitude higher than the calculated bending moment. This shows that for most applications and typical loading conditions, the flexure beams are in the elastic most applications and typical loading conditions, the flexure beams are in the elastic region, therefore the SUGAR simulation is valid.

region, therefore the SUGAR simulation is valid.

4.10.2 4.10.2 Effect Effect of of horizontal horizontal forces forces

If both vertical and horizontal loads act simultaneously on a beam, the resultant If both vertical and horizontal loads act simultaneously on a beam, the resultant stress or displacement at a point is equal to the superposition of the two separate effects.

stress or displacement at a point is equal to the superposition of the two separate effects.

To balance an externally applied horizontal load

To balance an externally applied horizontal load F F _h_h, the pivot and the output system, the pivot and the output system would exert horizontal forces,

would exert horizontal forces, F F _h_{h p}_p and and F F _h_{h o}_o, respectively at their joint with the lever arm., respectively at their joint with the lever arm.

The values of

The values of F F _h_{h p}_p and and F F _h_{h o}_o can be calculated from the equation of static equilibrium and acan be calculated from the equation of static equilibrium and a geometrical requirement of equal horizontal displacements at beam ends of pivot and geometrical requirement of equal horizontal displacements at beam ends of pivot and output system. It is clear that the external horizontal force does not generate vertical output system. It is clear that the external horizontal force does not generate vertical (axial) forces in the pivot or output system. Consequently, the effect of an external (axial) forces in the pivot or output system. Consequently, the effect of an external horizontal force is ignored in the analysis of a single-stage microlever.

horizontal force is ignored in the analysis of a single-stage microlever.

CHAPTER 5

In document Compliant Leverage Mechanism Design for MEMs Application (Page 97-100)