Strojniški vestnik - Journal of Mechanical Engineering 57(2011)6, 517-525 D01:10.5545/sv-jme.2009.151 Paper received: 03.11.2009 Paper accepted: 23.03.2011 Theoretical Analysis of Stiffness Constant and Effective Mass for a Round-Folded Beam in MEMS Accelerometer Wai Chi Wong12* - Ishak Abdul Azid1 - Burhanuddin Yeop Majlis3 1School of Mechanical Engineering, Sains University Malaysia 2Department of Mechanical Engineering, College of Engineering, Tenaga Nasional University, Malaysia 3Institute of Microengineering and Nanoelectronic, Kebangsaan University Malaysia In this paper, the governing equations of stiffness constant and effective mass for a round folded suspension beam in Micro-Electro Mechanical System (MEMS) accelerometer are derived and solved. The stiffness constant is determined by the strain energy and Castigliano s displacement theorem, whereas the effective mass is determined by the Rayleigh principle. The stiffness constant and the effective mass are solved separately by components and then combined by using the superposition method. The results obtained by the derived equations agree well when compared with the finite element results for several thickness values. The governing equations derived in this paper can be used to predict the natural frequencies and sensitivity of the MEMS-accelerometer. ©2011 Journal of Mechanical Engineering. All rights reserved. Keywords: effective mass, folded beam, MEMS-accelerometer, stiffness constant, strain energy 0 INTRODUCTION The development of MEMS devices has become increasingly important since the beginning of the 1990s. MEMS-accelerometers are the most commercially successful MEMS devices [1] and [2]. Among these, the comb finger type capacitive accelerometer is the most successful type of MEMS accelerometer due to its high sensitivity, low drift, stable dc-characteristics, low power dissipation, high bandwidth, simplicity of the fabrication process, absence of exotic materials, and low temperature sensitivity [2] to [5]. Stiffness constant and resonant frequency, which are the topics of this research, are two of the most important functionalities required in the design of any MEMS devices with moving parts. Sensors and actuators in particular, often require a specific stiffness constant and resonant frequency to guarantee successful and repeatable performance. The stiffness constant and effective mass, which incorporate both material properties and physical geometry, characterize the sensitivity and resonant frequency of the MEMS accelerometer. The sensitivity of the accelerometer is a measure of displacement with respect to acceleration. In contrast, resonant frequency characterizes the bandwidth of the accelerometer. The suspension beam can be of different designs depending on the application of the MEMS accelerometer. However, the study on the structural analysis of suspension beam in the comb finger type accelerometer has not been established well yet. Analytical derivation of these two parameters is only available for simple designs, such as straight beams [5] to [11]. For different geometric shapes, Legtenberg et al. [8] and Zhou and Dowd [9] determined the material properties of the suspension beam and the stiffness constant by using Hooke's law and the total potential energy, respectively. Tay et al. [10] used the Rayleigh's energy principle for the determination of resonant frequency as a function of effective mass. Wittwer and Howell [11] used Castigliano's displacement theorem for the analysis of the vertical deflection due to an applied moment or shear force and the geometric shape. As far as the authors know, the formula for the round folded beam has not yet been derived. Therefore, there is a need to derive the stiffness constant theoretically for a round folded beam to predict the performance of this type of design. In this paper, a detailed derivation of the stiffness constant, as well as the effective mass for the round folded beam as used for the suspension beam in MEMS accelerometer, is carried out. The principle of the strain energy and Castigliano's displacement theorem is extensively used in the derivation. From the derived equations, the stiffness constant and effective mass for various parameters are determined and compared with the finite element simulation using ANSYS® 8.1. 1 EQUIVALENT STIFFNESS CONSTANT OF THE SUSPENSION BEAM IN MEMS ACCELEROMETER A schematic design of MEMS accelerometer with the round folded beam is shown in Fig. 1a. The close up of the round folded beam with the symbol used in derivation is shown in Fig. 1b, where L is the length of the beam, r is the radius of the round part, and w is the width of the beam. Fig. 1. Round folded beam as the suspension beam in accelerometer, a) a 2D schematic diagram of MEMS accelerometer, b) the round folded beam of suspension beam In Fig. 1a, the proof mass is suspended by four suspension beams symmetrically at the four edges. The proof mass can be approximated by a central proof mass supported by four springs. The free body diagram of the typical arrangement of an accelerometer can then be approximated by a system of mass and spring as shown in Fig. 2. In Fig. 2, m is mass of the proof mass; kb k2, k3, and k4 are the stiffness constants of each suspension beam; and x is the displacement. In this spring-mass system, the mass is supported by four springs, thus the external forces are balanced by the four springs evenly and stored as the strain potential energy. The equivalent stiffness constant of the spring-mass system as shown in Fig. 2 can be determined by the following equation of equilibrium: F = mx, (1) -(( + k2 )x- - k4 ) x = mix, mx + k„x = 0, (2) where x is the acceleration and Fx is the force. Therefore, the equivalent stiffness constant, ke = kj + k2 + k3 + k4. Since the four suspension beams are in the same dimension and of the same material, then k1 = k2 = k3 = k4 = k1/4 , ke = 4k1/4 , (3) where k1/4 is the stiffness constant of a quarter system. In obtaining the governing Eqs. of the suspension beam for the stiffness constant, the following assumptions are made: • The proof mass and comb fingers of the structure are rigid. • The flexible members are attached to perfectly rigid supports. • The device only vibrates in the sensing axis. • Young's modulus is constant. • Damping is ignored. • 3D effects such as fringing field and comb finger end effects are neglected [12] to [14]. • Euler-Bernoulli beam model assumption is applied. 2 DETERMINATION OF STIFFNESS CONSTANT IN ROUND FOLDED BEAM Fig. 2. Free body diagram of typical arrangement of an accelerometer The resolved components of the round folded beam are shown in Fig. 3. The quarter Fig. 3. Resolved components of theround folded beam model of the suspension beam together with its boundary condition is shown in Fig. 3a, and its free body diagram is shown in Fig. 3b. In the analysis, the round folded beam can be resolved into three components, with two models of half fixed-fixed beam (Fig. 3 c and d) and a model of half ring (Fig. 3e). Since the device is assumed to vibrate only in the sensing axis, the total displacement of the quarter model e>1/4 is equal to the sum of the displacement of each component, ô1/4 = c>c1 + ôc2 + ^3 . Based on Hooke's law (i.e., F = M), there is k max occurs at the midspan of the beam. If this model is cut through its midspan, this part can be modeled as a half fixed-fixed beam. The reactions at both fixed ends are the bending moment M0, shear reaction force in y-direction Ry, and the axial reaction force in x-direction Ra, as shown in Fig. 4b. Since the load is transverse to the axis of the beam, the axial reaction force Ra is extremely small compared to the bending moment and shear force. Therefore, Ra is ignored in the calculation. The shear reaction force Ry, and bending moment M0 for the model of the half fixed-fixed beam is obtained as Ry = F/2 and M0 = FL/4. The maximum displacement in the model of the half fixed-fixed beam c>max is caused by both displacement due to bending moment 8bm , and the displacement due to shear 5s , or c>max = dbm + ds. Thus, the stiffness constant of the half fixed-fixed beam is taken as: 1 1 1 khalp kbm ks (5a) Therefore, the stiffness constant for components 1 and 3 is equal to the stiffness constant of the half fixed-fixed beam as given in Eq. (5a) as: kc1 kc3 1 1 1 (5) halp Hm 2.1.1 The Stiffness Constant Due to Bending Moment Fig. 4. Fixed-fixed beam For a fixed-fixed beam (Fig. 4), the maximum deflection due to a bending moment occurs at the middle of the beam. It is given by [15]: &bm = F (2L )3 FLl 192EI 24EI (6) The stiffness constant due to the bending moment kfull for the full model of the fixed-fixed beam is then derived as: kfull = ' 24EI (7) bm Thus, the stiffness constant due to the bending moment kbm for the half model of a fixed-fixed beam is: 1 k bm 12EI (8) where E is the Young's modulus, I is the second moment of the cross sectional area, and L is the length of the half model of the fixed-fixed beam. 2.1.2 The Stiffness Constant Due to Transverse Force For a rectangular cross sectional area of width b and depth d, with the total beam length L, and with the applied transverse load of F/2, as shown in Fig. 4b, the maximum deflection (at middle point) due to the shear is given as [15]: * 3 FsL os =-- (9) where G is shear modulus, G = - E , . and u is 2 (1 + ») the Poisson's ratio. By knowing Fs = F/2 and by replacing G and Fs in ôs, the maximum deflection (at middle point) due to the shear is: s = 6 (1 + V)FL s 5 bdE ' (10) Since ks = F/Ss, stiffness constant due to shear ks in the complementary form is given as 1 _ 6 (1 + m)L k ~ F ~ 5 bdE ' (11) 2.2 The Stiffness Constant for the Second Component The second component is approximated to be the model of a half ring with simple support on the left, transferred from the first and third components. The free body diagram is shown in Fig. 5a. This component is subjected to two forces, the transverse force Ry and bending moment M. The stiffness constant of the second component kc2 is the combination of the stiffness constant due to transverse force kt (Fig. 5b), and stiffness constant due to bending moment km (Fig. 5c). kt is approximated to be half of a stiffness constant in the ring kc due to the same value of transverse force, as shown in Fig. 6. a) b) c) Fig. 5. Free body diagram of the second component of round folded beam Fig. 6. Transverse force in the second component 2.2.1 The Stiffness Constant Due to Transverse Force Acting on Half a Ring The stiffness constant of the ring component (Fig. 5b) due to transverse force can be determined by using Castigliano's displacement theorem. The theorem states that deflection at a point on a member in the direction of a force applied at the specific point is given by the partial derivative of the complementary energy with respect to the external force at the point, or ôc = dU/dR, , where U is the internal strain energy given by U = J 1 dV . For a beam subjected to bending, the direct stress is a = My / I , where M is the bending moment, y is the distance from the neutral axis to the point in question, and I is the second moment of the cross sectional area I = y 2d . Thus, U = e£f m2 2,,, rL M- , -—y dAdx = I -dx Jo J a 2EI2 Jo 2EI L M2 L = jo j a 2ei2 where L is the length of the beam. Accordingly, c dU d rLM2 , oc =-= — -dx . c dR dRJ o 2EI For a full ring, the total length is given as Jo J0 the ring, and $ is the angle at any point within the iln dS is always zero, and in view of the symmetry of the ring, only one quadrant of the ring needs to be considered as shown in Fig. 7. (•2n [•% rdfi = 4 rdfi , where r is the radius of Jo Jo Fig. 7. Transverse forces at a ring sc = = ML r# c BR.. BR. I Jo 2EI = 4 J 0 'A m_ dM 0 EI BR„ r By cutting the ring at any section within the quadrant of the ring, the bending moment at any point c with an angle of $ is given as [15]: Mc = M0 - Rr (1 - cos<£), (13) where M0, the imaginary bending moment, is unknown but may be obtained from the strain energy. The strain energy for the quadrant of the ring is: •'/1 1 Uq = i /2-M2rd« = q Jo 2EI f/2 1 Jo 2EI Ryr . M0 —(1 - cos; 2 (14) r d by: The total strain energy for the ring is given 1 " Uc = 4Uq = 4, c q Jo 2EI Ryr . M0--H1 - cos4 ■dt, (15) owing to the symmetry at $ = 0, dU/dM0 = 0. Therefore, - i: ET Jo Ryr , M0 —(1 - cosq ET 0 Since 1 / EI + 0 and r + 0, r d§ = 0. (16) Ryr Ryr M0--— + —— cosé 0 2 2 dé = 0 solved as «.if )-* f-P-» ) . Therefore, M0=vi1 (17) Substituting Eq. (17) into Eq. (13), the bending moment at any point c within the full ring, is: According to Castigliano's displacement theorem, deflection of the ring at any point c is: MC = Rr| 2cos^-n" I. (18) i 0 By partial derivative with respect to Ry: dMc (1 -= r \— cos®-- dRy 12 n (19) (12), Sc = 4 Substituting Eqs. (18) and (19) into Eq. r* M >o EI ÔR = f 2 Ryr (1 cos 6 - — I ■ r (1 cos 6- — | rd6 (20) EI Jo y |2 n) |2 n) 4Ryr3 rn2( l..^ 1' ro o I — cos 6-- EI Jo I 2 n Sc =■ 4Rvr3 ( n 1 1 EI I 16 n 2n Since Ry = F, c 4Fr3 ( n 1 1 Sc =-1---+ — EI i16 n 2n (21) (22) The stiffness constant of the ring is then given by: kc ~ F ~ EI I 16 n + 2n (23) Since the stiffness constant for the second component is equal to the half of the stiffness constant of the full ring, kt = / kc . Therefore, the stiffness constant for the second component due to transverse force is given by: 1 2 8r3 (n 1 1 — = — = —\—-- + —\. (24) kt kc EI ^ 16 n 2n 2.2.2 The Stiffness Constant Due to Bending Moment The stiffness constant due to the bending moment for this component (Fig. 5c) can be determined by using the energy method. The internal strain energy is: u = f 1 de e i 5 = -2- fA f M- yidA. rd* = 1 f^ Ml ^ 2E Jo Ja I^ r E Jo Icn M r n U= EIcn 2 n Mnr ~2ëï7 ■ Since the total internal strain energy is equal to the external work performed, U = / Fd , then, n M r = 1 F§ .Substituting M = FL/2 into 2EIc2 2 (FL X nr 1 the equation, I — I -= — F0. I 2 ) 2EIC 2 2 By rearranging the above equation, the deflection due to the bending moment is: 8 = n FL1 r 4EI (26) cl Therefore, the stiffness constant due to the bending moment is given by: 1 8 n Lr kbm2 F 4EIc2 Since the width and depth of the second component is equal to the width and depth of the first component, = Ic1 = I, thus 1 n Lr Km! 4 EI (27) 2.3 The Effective Stiffness Constant of the Round Folded Beam The effective stiffness constant of the round folded beam can then be determined from Eq. (3): A - _L = I f 1 k " 4k " 41 k 11 f 1 1 1 - +--+-- — = — + —— + 1 1 i_l+jl |+4 il +_l |+4 [_l+j_ 4 V kbm kS j 4 V kt kbm2 j 4 V kbm kS , (28) —= 1 +_L+_L+ 1 Substituting Eqs. (8), (11), (24), and (27) into Eq. (28): _L = 1 f J— 1+1 [ 6 (1+^)L " ke = 21 12EI 1+ 2 5 bdE EI ! 16 n 1 f nL r 41 4EI (29) L3 3 (1 + m)L 2r3 i,i ' 24EI 5 bdE EI I 16 2n nUr 16EI ' L3 + 3(1 + ^)L + 24r3 fn_ 1 + _L 1 + 3nL2r 2w3 5w w3 V16 n 2n ) 4w3 3 DETERMINATION OF EFFECTIVE MASS IN ROUND FOLDED BEAM The effective mass of the folded beam is determined by using the Rayleigh principle. By taking a beam model with a cross sectional area A, length L, the displacement at any point x to be d(x), and velocity at any point x to be dd(x)/dt, the maximum displacement c>max is related to the distribution function N(x) as: <5(x) = N(x) 8m dS(x) , - and —y-!- = N ( dt y . dSm dt -. (30) The effective mass is given by: ■P^N1 (x )A (x )dx. (31) 3.1 Effective Mass for Half Model of Fixed-fixed Beam Since the distribution function is independent of the applied force, the distribution function can be determined by assuming the half model of the fixed-fixed beam is deflected under a concentrated force F. Displacement at any point within the beam is given by James et al. [16] as: h* )=—r v ' 12EIL 3Lx2 - 2x3 J. (32) The maximum displacement occurs at x=L thus, ^ _ FL3 12EI ' (33) Consequently, the distribution function is: 5(x) 3Lx2 - 2x3 N (x ) = S„ L (34) Effective mass of half model of fixed-fixed beam under bending moment is then given as: "b,e rL ■■ pJo N (x)A(x)dx- :pA\ J( L ( 3Lx2 - 2x31 dx = pA L6 pA L6 13 9L x 12Lx6 4x7 ---H-- 6 7 5 9 - 2+4 57 (35) x L 7 mb,e = 35 PAL . Thus, the effective mass of the first component ml is equal to that of the third component m3 or: m1 = m3 = mb,e = 35 PAL . (36) 3.2 Effective Mass for Half Model of Ring For the half model of ring under transverse force (Fig. 5), the displacement at any point within the beam is given by Eq. (20) as: H* ) = 4R„r y 1 A 1 — cos®-- EI Jo I 2 n or by expanding: S( ) 4V3 Six ) =—— v ' EI 1 sin 20 ) 1 . , -I — +-- I--sm0 + - 41 2 4 In -. (37) The maximum displacement is given by Eq. (21) as 8. 4Rvr3 ( „ 1 1 EI ^ 16 n 2n The distribution function: N (x ) = H* ) 1 ( é sin2é 1 . , è -I — +-- I--smé + ^'V 41 2 4 I n n2 1+J_ 16 n 2n is solved by: N (x ) = a io_n_n i i — n-- 16 2n (38) L 0 2 8 Effective mass of the half model of the ring under transverse force is then given as: ma,e = pf N 2 (x )A (x )dx = J 0 f 1,. 1 . =Hf 8 ^+TI6 sin M+T2 _ 1 i — n-- 16 2n As a result, the effective mass of the second component m2 is: pAr 6 2k1 + 47n5 - 32n3 - 6144n n2 (n4 - 16n2 + 64) (39) Thus, the effective mass of the device is (40) me = m1 + m2 + m3 + mpm + n m f = = 8mbe + 4mae + mpm + nmf , where mpm is the mass of proof mass, mf is the mass of finger, and n is the number of fingers. 4 COMPARISON RESULTS WITH THE FINITE ELEMENT SIMULATION The round folded beam in MEMS accelerometer is further analyzed by using the finite element (FE) package, ANSYS® 8.1. The three-dimensional 20-node structural solid element, SOLID 186, has been used in this analysis. The material properties and the physical geometries used in the FE simulation are shown in Table 1. The FE model meshed by the free meshing is shown in Fig. 8. The simulation result will be used for a comparison with the analytical results obtained by Eqs. (29) and (40). The resonant frequency is obtained directly from the modal analysis. The stiffness constant is determined by using a static analysis. Table 1. Material properties geometries for accelerometer and physical Elastic modulus 127x103 kg/^m s2 Poison's ratio 0.27 Density 2330x10-18 kg/^m3 Length of suspension beam (L) 803 mm Width of suspension beam (w) 11 mm Radius of ring, r 50 mm Length of comb finger (lfineer) 650 mm Width of comb finger (wfinger) 15 mm Number of comb finger 56 pairs Sensing gap distance (d0) 2 mm Fig. 8. Meshed model Table 2 shows the result of the comparison between the analytical and simulation analysis for stiffness constant and effective mass for several thicknesses (t). From Table 2, it can be seen that the stiffness constant and the effective mass obtained from the derived equations based on the analytical analysis show good agreement with those obtained from the finite element simulation. The difference between analytical and simulation results is below 5%. Based on this, it can be concluded that the derived analytical formulas are capable of obtaining the effective mass and the stiffness constant of the round folded beam of the suspended comb finger type accelerometer. Therefore, they are capable of predicting the resonant frequency and the sensitivity of the accelerometer. 5 CONCLUSIONS The stiffness constant and the effective mass of round folded beam in MEMS accelerometer have been derived successfully. The derivation of the stiffness constant is obtained by using the strain energy method and the Castigliano's displacement theorem, while the derivation of the effective mass is determined by using the Rayleigh principle. 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