MEAM 550 Modeling and Design of MEMS Spring Solution to homework #4

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1 Solution to homework #4 Problem 1 We know from the previous homework that the spring constant of the suspension is N/m. Let us now compute the electrostatic force. l g b p * 1 1 ε wl Co-energy = W = CV = V g * W wv Then, force in the length direction is = F ε = = l l. g The minus sign of the force is to be understood as the force that is in the direction of increasing value of the overlapping length. We just show the magnitude in what follows. Since every moving comb-finger has two parallel-plate capacitors on its two sides, the ε wv electrostatic force on each moving comb = F l =. Note that w is the width of the g parallel-plate, which is the thickness of the structure. From previous homework problem, we know that the thickness is µ m. We have five moving comb-fingers. So, the total electrostatic force = 5ε wv 15ε V 5 F l = =. This should be equal to the spring constant ( k = N / m ) g times the displacement ( δ = µ m ) of the shuttle mass of the comb-drive actuator. That is, 15ε V kδ kδ =. This leads to the desired voltage = V = = V 651V. 15ε It is a rather huge voltage to obtain on a chip. So, it makes sense to put more combs. The distance between two adjacent moving comb-fingers = g + b = 1 µ m. With µ m -wide space available on either side, more combs on each side, i.e., a total of 4 additional moving comb fingers can be accommodated. So, the number of comb fingers now is 45. Then, kδ the voltage comes down to V = = V. It is still large but this shows 45ε that electrostatic actuation needs large voltages. The positive side is that they don t consume much power (if you ignore any power consumed or lost in the electronics circuitry that generate high voltages). Problem In FEMLAB modeling, only the symmetric left half of the comb-drive is considered. In the figures below, note that only half of the central comb-finger is modeled as appropriate. First, it is

2 useful to check the capacitance so that we know how large an area (the bounding box) needs to be considered for doing the finite element solution of the electrostatic problem. ε A ( pt) Analytically, the capacitance due to one comb-finger is = ε =. Since, we have five g g 1ε ( ) comb-fingers, the total capacitance = pt E E E = = 5.14Eg E 6 16 Farads. The capacitance in FEMLAB can be estimated by doing boundary integration of the surface charge (nd_es) over the moving comb-finger boundaries or the stationary comb-finger boundaries. This integration gives the charge per unit thickness of the structure. By multiplying the charge per unit thickness with the thickness, we get the total charge. The total charge divided by the voltage gives the capacitance. That is, ψ lt CFEMLAB =. V In addition to checking the capacitance, we can also see how close the electric potential is to zero on the boundaries of the bounding box. A series of trials with increasing size of the bounding box is shown in the following figures. The voltage at the top boundary is far from being zero in the two cases above. So, we increase the size of the box much more in the following two trials. Note that, the linear dimensions are scaled up by 1E6 to use micron units in FEMLAB. Consequently, the voltage is multiplied by 1E1 as per the dimension of L in voltage s dimensions.

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4 Another check is to see how the charge distributions is (plotted with line plot with surface charge). See below. It is concentrated to the comb-fingers as to be expected. The last bounding box seems to be enough as the voltage at the top boundary is now small. It is still not zero because of our scale factor. Recall that the voltage is 65E1. The capacitance check works out quite well. ψ lt 1.154E 1 CFEMLAB = = = 5.6E-16 Farads. This is very close to the V 65 analytically computed value of 5.14E-16 F. Note that 1.154E-1 is the value obtained with boundary integration along the stationary comb-fingers and the mesh was at one levr finer than the default (for comparison, the default mesh gave 1.175E-1 instead of 1.154E-1). This will be Coulomb per unit thickness. When it is multiplied by (which is the thickness in micron units), we get the charge. Its dimensions are [C]. So, scale factors are necessary. The figure below shows the electric potential in the close-up view around the comb fingers.

5 Notice the linearity of the potential between the fingers and the crowding of the fringing fields around the corners. We stick with this bounding box for further calculation. As we know, the surface charge (or boundary charge) ψ s is denoted by nd_es variable in FEMLAB. In -D problems, FEMLAB assumes unit thickness. So, the nd_es is also the surface charge, we need in the force calculation. r ψ s We know that the force per unit surface area is given by f s = nˆ. Therefore, in FEMLAB, for ε the plane-stress element based elastostatic analysis, the boundary forces are given as follows. Fx = force per unit area in the x-direction = (nd_es)^//8.854e-*nx Fy = force per unit area in the y-direction = (nd_es)^//8.854e-*ny Note E_ above. This means that the scale factor has been applied to ε. This also means that the thickness of the structure does not enter the picture as both electrostatic and plane stress problems assume unit thickness. A voltage of 65 V was applied (of course as 65E1 with the scale factor).

6 According to our analytical calculation, we expect µ m as the displacement (or units in FEMLAB). But, as can be seen below, it only gave.16 µ m. Moreover, the combs seem to be swaying a little to the left. There is a good reason for this and it has to do with a little snag in FEMLAB s inconsistent sign convention for the normal of a surface. Read on. Generally, the normal is assumed to be positive if it is directed outwards from a surface. In FEMLAB, this is not consistently followed. We can check this using arrow plot by choosing boundaries and nx for x and ny for y. You may need to adjust the scale factor for the arrow sizes (also choose normalized ) in the pop-up window of the arrow plot. Then, you will see the surface normals as follows.

7 Clearly, some are directed outwards and some are directed inwards. Most significantly, we can see that on both the vertical boundaries of the moving combs, all the normals are to the left. This explains why the combs are swaying to the left. There is no easy way to fix it. We need to put the correct sign (+ or -) on each boundary so that the force outward to the surface. With this done, the result comes out much better. Now, as seen in the next figure, the displacement is about 1.5 µ m. It is still not close enough to µ m. But there is a problem still in that the nx components for the horizontal edges and ny components for the vertical edges are not what they should be. For horizontal edges, the boundary integration of nx should come to be zero. But it does not! Likewise, the boundary integration of ny should come equal to their length. This too is not the case. So, the element integration or something else is amiss. So, the results should be taken with a grain of salt! Hence, it is always a good thing to compare with analytical result (however simple it may be) or experimental data.

8 Problem Doing a -D analysis with our geometry seems to lead to memory problems with FEMLAB. So, we cannot show the levitation of the comb-fingers, unfortunately! Problem 4 The beam problem was modeled in FEMLAB by drawing two rectangles of size 5 µ m µ m (note that µ m is the thickness of the beam and the 15 µ m is width). The top rectangle shows the side-view of the beam. The bottom rectangle serves as the electrode (its height can be anything but we chose µ m ). Again, a bounding box of large enough size is created until the capacitance matches with that of the analytically computed value. A 8.854E 1 5E 6 15E 6 C analytical = ε = = 1.167E 14 F. g E 6 The capacitance obtained with FEMLAB (for 1 V) is equal to Q ψ l w 7.576E CFEMLAB = = = = 1.164E 14 F. So, it is very close to the analytical V V 1 value (% error is less than %). The figure below shows the electric potential plot. All of the bounding box is not shown in this figure so that we can clearly see the detail in between the plates.

9 Let us see if 1 V is below the pull-in voltage for this case. The equivalent spring constant for the cantilever beam with uniform distributed force is computed as follows. q δ L wl δ = i.e., 4 k = force deflection = wl /(8 8EI = ) Ewt = 4 8EI wl EI L L E k =.819 N / m. 16E9 (E 6) 15E 6 = 15 (5 6)

10 8kg = = V. 7ε A V PI Since we are not doing the relaxation to find the self-consistent solution in FEMLAB. We will simply calculate the electrostatic force and divide it by the spring constant to get the force rather than solving the cubic equation (which gives the correct self-consistent solution between the electrostatic and elastostatic equations). The electrostatic force is equal to ε AV g 8.854E 1 5E 6 15E 6 1 E 6 F e = =.84E 6 N + = + g w (E 6) 15E 6 Hence, the expected displacement = F.84E 6 e = =.544 E 6 m =.544 µ m. k.819 As seen in the figure below, the maximum displacement in FEMLAB is.5 µ m, which is reasonably close. Here too, the direction of the normals had to be taken into account in specifying the force in the plane stress problem. The scale factor for the displacement was 1 (we specify it in the deformed plot pop-window in FEMLAB).