**4. Dynamic behavior of a novel MEMS amplifier**

### **4.1. Inertial and geometric parameters:**

It is assumed that micro mechanism is made up of silicon having a density of 2.33 g/cm3. For short length of beams, lengths are 100 micron, widths and heights are 25 micron. The mass of short beams is;

$$\text{M} = 2.33^\*100^\*25^\*25^\*10^{\cdot15} \text{m} 145625^\*10^{\cdot18} \text{ [kg]} \tag{45}$$

For long length of beams, lengths are 800 micron, widths and heights are 25 micron. The mass of short beams is;

$$\text{M} = 2.33^\* 800^\* 25^\* 25^\* 10^{\cdot 15} \text{m} 1165000^\* 10^{\cdot 18} \text{ [kg]} \tag{46}$$

The mass of the slider is accepted as 145625\*10-18 in kilograms.

The mass moments of inertia of the beams are calculated as follows;

For short beams;

100 MATLAB – A Fundamental Tool for Scientific Computing and Engineering Applications – Volume 1

mechanism goes to infinity at zero degree crank angles.

**Figure 10.** Plot of displacement ratio according to first stage crank angle, Θ2

**Figure 11.** Plot of force amplifying according to first stage crank angle, Θ2

is claimed that the toggle position of the micro mechanism is a very crucial issue meaning that if the initial conditions such as crank angles are adjusted properly to enable both crank angle pass 0° at the same time, the ratio of the output to the input force applied to the

$$\mathbf{L} = \mathbf{M}\_{\ast} \, \text{\*} (\mathbf{L}^2 + \mathbf{a}^2) / 12 \mathbf{=} 145625 \, \mathbf{^\*} 10^{18} \text{(} 100^2 \text{+} 25^2 \text{)} / 12 \mathbf{=} 128938802.1 \text{\*} 10^{18} \text{ [} \mathbf{kg} \, \text{\*} \mu \text{m}^2\text{]} \tag{47}$$

For long beams;

$$\text{I} = \text{M}s^\*(\text{L}^2 + \text{a}^2)/12 \equiv 1165000^\*10^{\circ} 10^{\circ} \text{s}^\* (800^{\circ} + 25^{\circ})/12 \equiv 6.219401042^\*10^{\circ} \text{[kg}^\* \text{\mu m}^2\text{]} \tag{48}$$

#### **4.2. Kinematic behavior**

#### *4.2.1. Velocity analysis*

Kinematic simulation is used to calculate and to plot the velocities and acceleration of the beam of the MEMS amplifier.

To understand kinematic behavior of the mechanism, first of all, derivatives of vector loop equations derived in position analysis are taken with respect to time and the velocity equations are arranged as follows;

$$-r\_2 \, ^\ast \sin \theta\_2 \, ^\ast w\_2 - r\_3 \, ^\ast \sin \theta\_3 \, ^\ast w\_3 = \dot{r}\_1 \tag{49}$$

$$r\_2 \, ^\ast \cos \theta\_2 \, ^\ast w\_2 + r\_3 \, ^\ast \cos \theta\_3 \, ^\ast w\_3 = 0 \tag{50}$$

$$-r\_2 \, ^\ast \sin \theta\_2 \, ^\ast w\_2 - r\_5 \, ^\ast \sin \theta\_5 \, ^\ast w\_5 - r\_6 \, ^\ast \sin \theta\_6 \, ^\ast w\_6 = 0 \tag{51}$$

$$r\_2 \, ^\ast \cos \theta\_2 \, ^\ast w\_2 + r\_5 \, ^\ast \cos \theta\_5 \, ^\ast w\_5 + r\_6 \, ^\ast \cos \theta\_6 \, ^\ast w\_6 = 0 \tag{52}$$

The beam 6 are rotated at a constant speed, 0.01 rad/s, in clockwise direction and the initial conditions of w2, w3, w5, r� � are -0.059378175917485 [rad/s], 0.059378175917485 [rad/s], 0.011371580426033 [rad/s ], 2.062182408251533 [µm/s], respectively.

**Figure 12.** Plot of angular velocity of beam 2 and beam 3 versus time for simulation of force amplifier

The angular velocities of beams 2 and 3 in 1 stage are equal to each other in magnitude. As w3 rotate counter clockwise direction, w2 rotate clockwise direction and the absolute values of the changes in w3 and w2 equal to each other according to time as shown in Fig. 12.

Slider slows down until the first stage crank angle, Θ2 pass from 0°. When first stage beams are fully open, as having horizontal position, slider velocity is equal to zero. Then the slider moves to along -x direction and angular velocity of beam 5 decreases according to time as in Fig. 13.

**Figure 13.** Plot of angular velocity of beam 5 and slider velocity versus time for simulation of force amplifier

#### *4.2.2. Acceleration analysis*

102 MATLAB – A Fundamental Tool for Scientific Computing and Engineering Applications – Volume 1

**Figure 12.** Plot of angular velocity of beam 2 and beam 3 versus time for simulation of force amplifier

The angular velocities of beams 2 and 3 in 1 stage are equal to each other in magnitude. As w3 rotate counter clockwise direction, w2 rotate clockwise direction and the absolute values of the changes in w3 and w2 equal to each other according to time as shown in Fig. 12.

Slider slows down until the first stage crank angle, Θ2 pass from 0°. When first stage beams are fully open, as having horizontal position, slider velocity is equal to zero. Then the slider moves to along -x direction and angular velocity of beam 5 decreases according to time as in Fig. 13.

**Figure 13.** Plot of angular velocity of beam 5 and slider velocity versus time for simulation of force

amplifier

To analyze the acceleration of the beams, second derivatives of the terms must be handled. The second derivatives of the vector loop equations for the micro mechanism are as follows;

$$-r\_2 \, ^\ast \cos \theta\_2 \, ^\ast w\_2^2 - r\_2 \, ^\ast \sin \theta\_2 \, ^\ast a\_2 - r\_3 \, ^\ast \cos \theta\_3 \, ^\ast w\_3^2 - r\_3 \, ^\ast \sin \theta\_3 \, ^\ast a\_3 - \overset{\cdots}{r\_1} = 0 \tag{53}$$

$$-r\_2 \, ^\ast \sin \theta\_2 \, ^\ast w\_2^2 + r\_2 \, ^\ast \cos \theta\_2 \, ^\ast a\_2 - r\_3 \, ^\ast \sin \theta\_3 \, ^\ast w\_3^2 + r\_3 \, ^\ast \cos \theta\_3 \, ^\ast a\_3 = 0 \tag{54}$$

$$\begin{aligned} -r\_2 \, \prescript{\*}{}{\cos}\theta\_2 \, \prescript{2}{}{w\_2^2} - r\_2 \, \prescript{\*}{}{\sin}\theta\_2 \, \prescript{\*}{}{a\_2} - r\_5 \, \prescript{\*}{}{\cos}\theta\_5 \, \prescript{2}{}{w\_5^2} \\ -r\_5 \, \prescript{\*}{}{\sin}\theta\_5 \, \prescript{2}{}{a\_5} - r\_6 \, \prescript{\*}{}{\cos}\theta\_6 \, \prescript{2}{}{w\_6^2} - r\_6 \, \prescript{\*}{}{\sin}\theta\_6 \, \prescript{2}{}{a\_6} = 0 \end{aligned} \tag{55}$$

$$\begin{aligned} &-r\_2 \, ^\ast \sin \theta\_2 \, ^\ast w\_2^2 + r\_2 \, ^\ast \cos \theta\_2 \, ^\ast a\_2 - r\_5 \, ^\ast \sin \theta\_5 \, ^\ast w\_5^2 \\ &+ r\_5 \, ^\ast \cos \theta\_5 \, ^\ast a\_5 - r\_6 \, ^\ast \sin \theta\_6 \, ^\ast w\_6^2 - r\_6 \, ^\ast \cos \theta\_6 \, ^\ast a\_6 = 0 \end{aligned} \tag{56}$$

.

In acceleration simulation by Simulink, the velocities such as w2, w3, w5, *r*<sup>1</sup> , w6 are considered as known. The beam 6 rotates at a constant speed meaning that acceleration of beam 6 is zero.

Acceleration of beam 2 and beam 3 are shown in fig. 14. Both acceleration of beams decrease as the micro mechanism operates under constant w6, angular velocity. The magnitude of acceleration of beam 2 and beam 3 are equal to each other during simulation. Also, as seen in Fig. 15, acceleration of beam 5 and slider decrease as function of time.

**Figure 14.** Acceleration of beam 2 and beam 3 under constant angular acceleration, ߙ<sup>2</sup>

**Figure 15.** Acceleration of beam 5 and slider under constant angular acceleration, ߙ<sup>2</sup>

#### **4.3. Acceleration vector equations according to center of mass**

The linear acceleration of the center of mass equations are not present in vector loop equations that are previously derived. So, there must be equations relating to the acceleration of the center of mass of beams. Equation derivation is as follows and schematic representation of the center of mass acceleration in first and second loops is shown in Fig. 16 and Fig. 17, respectively.

**Figure 16.** The center of mass acceleration in first loop

The center of mass acceleration of beam 2 along x and y direction;

$$\bar{A}\_{c2} = \bar{R}\_{c2} \tag{57}$$

Dynamic and Quasi-Static Simulation of a Novel Compliant MEMS Force Amplifier by Matlab/Simulink 105

$$A\_{c2x} = -r\_{c2}"\sin\theta\_2"\,^\*a\_2 - r\_{c2}"\cos\theta\_2"\,^\*w\_2^2\tag{58}$$

$$A\_{c2y} = r\_{c2} \, \text{\*} \, \cos \theta\_2 \, \text{\*} \, a\_2 - r\_{c2} \, \text{\*} \, \sin \theta\_2 \, \text{\*} \, w\_2^2 \tag{59}$$

**Figure 17.** The center of mass acceleration in second loop

104 MATLAB – A Fundamental Tool for Scientific Computing and Engineering Applications – Volume 1

**Figure 15.** Acceleration of beam 5 and slider under constant angular acceleration, ߙ<sup>2</sup>

The linear acceleration of the center of mass equations are not present in vector loop equations that are previously derived. So, there must be equations relating to the acceleration of the center of mass of beams. Equation derivation is as follows and schematic representation of the center of mass acceleration in first and second loops is shown in Fig. 16

**θ3**

..

<sup>2</sup> <sup>2</sup> *<sup>c</sup> Ac R* (57)

r3

rc3

**Ac3**

**4.3. Acceleration vector equations according to center of mass** 

and Fig. 17, respectively.

**Figure 16.** The center of mass acceleration in first loop

The center of mass acceleration of beam 2 along x and y direction;

**θ<sup>2</sup>**

**Ac2**

r2

rc2

The center of mass acceleration of beam 3 along x and y direction;

$$A\_{c3} = \stackrel{\cdots}{R}\_{2} + \stackrel{\cdots}{R}\_{c3} \tag{60}$$

$$A\_{c3x} = -r\_2 \, ^\ast \sin \theta\_2 \, ^\ast a\_2 - r\_{c2} \, ^\ast \cos \theta\_2 \, ^\ast w\_2^2 - r\_{c3} \, ^\ast \sin \theta\_3 \, ^\ast a\_3 - r\_{c3} \, ^\ast \cos \theta\_3 \, ^\ast w\_3^2 \tag{61}$$

$$A\_{c3y} = r\_2 \, "\, \cos\theta\_2 \, "\, \alpha\_2 - r\_{c2} \, "\, \sin\theta\_2 \, "\, w\_2^2 + r\_{c3} \, "\, \cos\theta\_3 \, "\, \alpha\_3 - r\_{c3} \, "\, \sin\theta\_3 \, "\, w\_3^2 \tag{62}$$

The center of mass acceleration of beam 6 along x and y direction;

$$A\_{c\circ\delta} = \stackrel{\cdots}{R}\_{c\circ\delta} \tag{63}$$

$$A\_{c6y} = r\_{c6}{}^\* \sin \theta\_6{}^\* a\_6 + r\_{c6}{}^\* \cos \theta\_6{}^\* w\_6^2 \tag{64}$$

$$A\_{c\theta x} = -r\_{c\theta} \, ^\ast \cos \theta\_{\theta} \, ^\ast a\_{\theta} + r\_{c\theta} \, ^\ast \sin \theta\_{\theta} \, ^\ast w\_{\theta}^2 \tag{65}$$

The center of mass acceleration of beam 5 along x and y direction;

$$A\_{c5} = \stackrel{\cdot}{R}\_{2} + \stackrel{\cdot}{R}\_{c5} \tag{66}$$

$$A\_{c5y} = r\_6 \, ^\ast \sin \theta\_6 \, ^\ast a\_6 + r\_6 \, ^\ast \cos \theta\_6 \, ^\ast w\_6^2 + r\_{c5} \, ^\ast \sin \theta\_5 \, ^\ast a\_5 + r\_{c5} \, ^\ast \cos \theta\_5 \, ^\ast w\_5^2 \tag{67}$$

$$A\_{c5x} = -r\_6 \, ^\ast \cos \theta\_6 \, ^\ast a\_6 + r\_6 \, ^\ast \sin \theta\_2 \, ^\ast w\_2^2 - r\_{c5} \, ^\ast \cos \theta\_5 \, ^\ast a\_5 - r\_{c5} \, ^\ast \sin \theta\_5 \, ^\ast w\_5^2 \tag{68}$$

#### **4.4. Force and dynamic analysis of the micro mechanism**

The micro mechanism operates under constant angular velocity, 0.01 [rad/s], the slider crank starts increasing and reaches to its maximum value, 200 micron, meaning that the first stage slider crank is fully opened at 3.20 sec. then the first crank angle pass from 0° and slider begins to get close to its initial position and R1 decreases as shown Fig. 18. According to both crank angles, Θ2 and (90-Θ6), the output force increases or decreases. In the first section of the Foutput vs. time curve, first, both crank angles decrease, and two slider cranks start to open and at small crank angles, Foutput sharply increase and at 3.20 sec Θ2 is equal to 0.0013°

**Figure 18.** Displacement of slider and output force versus time

and at 3.25sec. Θ2 is equal to -0.0012° and at Θ2 these values, Foutput goes to its peak values such as -3.07\*104 µN at 3.20 sec. and 3.32\*104 µN at 3.25 sec. Θ2 decreases until 3.20 sec. and then it increases, whereas 90°-Θ6 decreases and gets close to small values during the simulation. The magnitude of first peak of Foutput at 3.25 sec. is higher than the magnitude of second peak of Foutput at 3.20 sec. due to the fact that (90°-Θ6) at 3.25 sec. is smaller than the value of (90°-Θ6) at 3.20 sec., meaning that small crank angle value of (90°-Θ6) contributes to get much more output force.
