Static bifurcation and primary resonance analysis of a MEMS resonator actuated by two symmetrical electrodes

Abstract

This paper investigates the static and dynamic characteristics of a doubly clamped micro-beam-based resonator driven by two electrodes. The governing equation of motion is introduced here, which is essentially nonlinear due to its cubic stiffness and electrostatic force. In order to have a deep insight into the system, static bifurcation analysis of the Hamiltonian system is first carried out to obtain the bifurcation sets and phase portraits. Static and dynamic pull-in phenomena are distinguished from the viewpoint of energy. What follows the method of multiple scales is applied to determine the response and stability of the system for small vibration amplitude and AC voltage. Two important working conditions, where the origin of the system is a stable center or an unstable saddle point, are considered, respectively, for nonlinear dynamic analysis. Results show that the resonator can exhibit hardening-type or softening-type behavior in the neighborhood of different equilibrium positions. Besides, an attractive linear-like state may also exist under certain system parameters if the resonator vibrates around its stable origin. Whereafter, the corresponding parameter relationships are deduced and then numerically verified. Moreover, the variation of the equivalent natural frequency is analyzed as well. It is found that the later working condition may increase the equivalent natural frequency of the resonator. Finally, numerical simulations are provided to illustrate the effectiveness of the theoretical results.

Appendix

The proof of $X_3 >1$ when $\Delta >0$ can be summarized as follows.

The derivative of $X_3$ with respect to $\gamma$ can be written as

$$\begin{aligned} \frac{\hbox {d}X_3 }{\hbox {d}\gamma }=\frac{\root 3 \of {\eta _1 +6\sqrt{3}\alpha \eta _2 }-\root 3 \of {\eta _1 -6\sqrt{3}\alpha \eta _2 }}{\sqrt{3}\eta _2 } \end{aligned}$$

(35)

where $\eta _1 =-(1+\alpha )^{3}+54\alpha ^{2}\gamma$ and $\eta _2 =\sqrt{-[(1+\alpha )^{3}-27\alpha ^{2}\gamma )]\gamma }$.

As $\gamma >{(1+\alpha )^{3}}/{(27\alpha ^{2})}\ge 0.25$, it is clear that $\eta _1 >(1+\alpha )^{3}>0$ and $\eta _2 >0$, leading to $\frac{\hbox {d}X_3 }{\hbox {d}\gamma }>0$. According to the continuity of $X_3$ to $\gamma$ as $\gamma \ge {(1+\alpha )^{3}}/{(27\alpha ^{2})}$, it is clear that $X_3 >\left. {X_3 } \right| _{\gamma =0.25}$, where the expression of $\left. {X_3 } \right| _{\gamma =0.25}$ can be given by

$$\begin{aligned}&\left. {X_3 } \right| _{\gamma =0.25}\nonumber \\&\quad {=}\frac{\alpha \left[ {4+2^{2/3}\left( {B_1 {+}iB_2 } \right) ^{1/3}+2^{2/3}\left( {B_1 {-}iB_2 } \right) ^{1/3}} \right] {-}2}{6\alpha }\nonumber \\ \end{aligned}$$

(36)

where $B_1 =\frac{21\alpha ^{2}-6\alpha -2\alpha ^{3}-2}{\alpha ^{3}}$ and $B_2 =3\sqrt{3}\sqrt{\frac{(\alpha -2)^{2}(1+4\alpha )}{\alpha ^{4}}}$.

If $\alpha =2,\,B_2 =0$ and $\left. {X_3 } \right| _{\gamma =0.25} =\frac{3}{2}$, else $\alpha \ne 2,\,\left. {X_3 } \right| _{\gamma =0.25}$ can be rewritten as

$$\begin{aligned} \left. {X_3 } \right| _{\gamma =0.25} =\frac{4\alpha +4(1+\alpha )\cos (\varphi )-2}{6\alpha } \end{aligned}$$

(37)

where $\varphi =\frac{1}{3}\arccos (\frac{B_1 }{\sqrt{B_1^2 +B_2^2 }}),\,\,0\le \varphi <\frac{\pi }{3}$.

The property $\frac{1}{2}<\cos (\varphi )\le 1$ leads to $1<\left. {X_3 } \right| _{\gamma =0.25} \le \frac{4\alpha +1}{3\alpha }$. It is obvious that $X_3 >1$.$\square$