The theory of harmonically driven, clamped silicon micro-beams

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1 The theory of harmonically driven, clamped silicon micro-beams Henrik Bruus Mikroelektronik Centret, bldg. 345, Technical University of Denmark, DK-28 yngby (Dated: June 1, 22) In this note the theory of harmonically driven, clamped silicon micro-beams is presented. The specific example of a beam of dimensions 1 µm 1 µm 1 µm is studied in detail. The beam is placed in a perpendicular magnetic field B, and it is forced to oscillate through the orentz force by passing an ac-current J cos(ωt) through it. The amplitude of the oscillations at resonance are of the order.1 µm and the amplitude of the associated, induced electromotive force E is of the order 1 µv. The whole setup is at room temperature in air. PACS numbers: PACS numbers: I. INTRODUCTION: THE MICRO-BEAM As a part of a DTU-MIC three weeks course we have set out to design, fabricate, and test a MEMS-oscillator in the form of a clamped silicon micro-beam etched out from a very pure single-crystal. This constitutes the first step at MIC to fabricate such devices. The micro-beam is forced to oscillate by utilizing the magnetic force per unit length, K(t) = BJ cos(ωt), that arises when the beam is placed in a homogeneous magnetic field perpendicular to the length axis of the beam, and an ac-current J cos(ωt) is passed through the beam. The oscillations of the beam are detected by measuring the induced electromotive force, E(t) = B A/ t, that arises when the oscillating beam sweeps through the area A(t) in the direction perpendicular to the homogeneous magnetic field B. The experiment is inspired by the one performed by Cleland and Roukes, 1 but instead of vacuum and cryogenic temperatures we plan to carry out our experiment at room temperature and in air. To be able to make a good design of the micro-beam some theory is called for. The purpose of this note is to provide the necessary theoretical understanding of the experiment. The theory presented in the following is based on andau and ifshitz. 2 II. THE BENT BEAM IN EQUIIBRIUM In this note the x axis is placed along the length axis of the beam. The magnetic field is applied along the y axis, and consequently the beam is deflected in the z direction when an ac-current J(t) is passed through the beam. The beam has the length, and a rectangular cross section h w, where h is the height in the z direction, and w the width in the y dirction. In the specific example presented in Sec. V we shall use h = 1 µm, w = 1 µm, = 1 µm. (1) Furthermore, silicon has the following Young s modulus E and mass density ρ, E = Pa, ρ = kg m 3. (2) The central goal of this note is to calculate the form of the deflected beam, z(x, t). Throughout the calculation we assume the clamped-beam boundary conditions, z() =, z () =, z() =, z () =. (3) In this section we study the bent beam in equilibirium, and find the basic differential equation for the deflection in terms of an applied force per unit length, K(x). The argument involves the internal forces F(x) acting on a cross section S(x) in the yz plane at a given position x and the associated internal force moment M(x) around the center of mass of the same cross section. Consider first a beam bent slightly downwards. In an infinitessimal neighborhood of x the shape is approximated by a circle having a radius of curvature R. On the top side the beam is stretched, on the bottom side it is compressed. In the center, denoted z =, it keeps its original length, dl. At the position z the length is stretched from dl to dl. By the simple geometry of concentric circle arches we find dl = dl (R + z)/r = (1 + z/r) dl, whence the strain in the x direction is given by u x (z) = z/r. By Hooke s law the xx component of the stress tensor σ then becomes σ xx (x, y, z) = E u x (z) = E z. (4) R(x) From this we can easily calculate the corresponding y component M y of the internal moment M, M y (x) = S dydz z σ xx = R(x), I y = wh3 12. (5) Here we have introduced the y component I y of the geometrical moment of inertia of a square cross section h w, I y = S dy dz z2 = w/2 w/2 dy h/2 h/2 dz z2. For small deviations the curvature is equal to the second derivative of the deflection, 1/R(x) = z (x), and we arrive at z (x) = M y(x). (6) While this relation constitutes a fundamental relation between curvature and the internal force momentum, we

2 2 however still need to study how the deflection is affected by variations in M(x) induced by the externally applied force per unit length K(x). First we consider the forces acting on the infinitessimal element of the beam between x and x+dx. This element is influenced by three forces. At the surface S(x) we have the internal force F, given by (F) i = S(x) dydz σ ix. At the surface S(x + dx) we have the internal force F + df. And finally we have the external force K(x) dx. In equilibrium we must require force balance, whence (F + df) F + K(x) dx = df dx = K. (7) Next we consider the force moments acting on the element. Again three contributions must be considered. In equilibrium they must add to zero. We study the moments acting on the center of mass C of the surface S(x + dx). The moment around C arising from the internal forces acting on the surface S(x + dx) is denoted M + dm. The moment around C due to the internal forces acting on the surface S(x) is found from the parallel theorem to be M dx F. Finally, the moment from the external force is neglected since it is proportional to K(x) dx 2, i.e. of second order in dx. In equilibrium we must require force moment balance, (M+dM) (M+dx F) = dm dx = e x F. (8) Now by combining Eqs. (5), (7) and (8) we arrive at the fourth-order differential equation, which describes the deflection z(x) of the beam due to the z component K z (x) of the external force per unit length, z (x) = K z (x). (9) For the static case with a constant external force K z the deflection is given by a simple fourth-order polynomium in x. We will not pause to examine the statics further, but push on to study the dynamical eigenmodes. III. THE EIGENMODES OF THE BEAM The eigenmodes are found by extending the static equation Eq. (9) to the dynamic case. When allowing for motion it follows from Newton s second law that the external force is changed by the addition of the inertial force per unit length, ρs z(x, t), z (x, t) = K z (x, t) ρs z(x, t). (1) ooking for eigenmodes we put K z (x) = and make the ansatz z(x, t) = z(x) cos(ωt). This results in z (x) = k 4 z(x), (11) where we have introduced the wavenumber k given by k 4 ρs ω 2. (12) The general solution to this homogeneous fourth-order differential equation is the sum of four linearly independent solutions, z(x) = a cos(kx) + b sin(kx) + c cosh(kx) + d sinh(kx). (13) By imposing the boundary conditions z() = and z () = we find immediately z(x) = a[cos(kx) cosh(kx)]+b[sin(kx) sinh(kx)]. (14) The boundary conditions z() = and z () = lead to a standard 2 2 secular equation, ( ) ( ) ( ) cos(k) cosh(k) sin(k) sinh(k) a =. sin(k) sinh(k) cos(k) cosh(k) b (15) Non-trivial solutions exist only when the determinant of the coefficient matrix is zero. The condition for this to happen is readily shown to be cos(k) = 1 cosh(k). (16) Generally, the wavenumber k of the eigenmodes can only be found by numerical methods. However, since 1/ cosh(k) tends to zero exponentially fast for large k, good approximate solutions can easily be found: cos(k) k n 2n + 1 π, n = 1, 2, 3, (17) The exact equation Eq. (16) have no solutions corresponding to n = in Eq. (17). The eigenfrequencies become ω n = ρs k2 2 (2n + 1)2 n π 8 E 3 ρ h 2. (18) Using the vaules from Eqs. (1) and (2) the exact solution of the lowest eigenfrequency is IV. ω 1 = 5.35 MHz f 1 =.85 MHz. (19) THE HARMONICAY DRIVEN BEAM Having identified the eigenmodes we now turn to the problem of the harmonically driven beam. As mentioned we use the magnetic force per unit length, K(t) = BJ cos(ωt) as the driving force K z (t). In this case the driving force does not depend on position. To make the calculation more realistic we add the damping force per unit length (γ/)ż due to the friction between the air and the beam. The damping coefficient γ is proportional to the viscosity η of air and, since the beam oscillates in the z direction, to the square root of the area in the xy plane, i.e. w. I have not have time to look up the numerical prefactor, but simply assume γ = w η. (2)

3 3 The differential equation Eq. (1) for z(x, t) with these assumptions becomes z (x, t) = K(t) ρs z(x, t) γ ż(x, t). (21) Due to the appearence of both z and ż it is useful to use complex functions. et ζ(x, t) = ζ(x)e iωt be the solution to the equation with a complex drive K(t) = Ke iωt. By complex conjugation we find that ζ (x, t) therefore is a solution with the complex drive K (t). Now, since the differential equation Eq. (21) is linear we find that Re[ζ(x)e iωt ] = [ζ(x, t) + ζ (x, t)]/2 is a solution with the drive [K(t)+K (t)]/2 = K cos(ωt), which is the physical drive. Hence our solution is [ z(x, t) Re ζ(x) e iωt]. (22) We therefore insert ζ(x) e iωt and K(t) = Ke iωt into Eq. (21) and obtain after dividing by e iωt ζ(x) (x) = K + ω 2 ρs ζ(x) iω γ ζ(x). (23) In analogy with Eqs. (11) and (12) we write this as ζ (x) = q 4 ζ(x) + K, (24) where we have introduced the complex wavenumber q given by q 4 ρs ω 2 γ i ω. (25) The differential equation Eq. (24) is inhomogeneous. The complete set of solutions is generated by adding one particular solution ζ (x) of the inhomogeneous equation to the general solution Eq. (13), with k = q, of the corresponding homogeneous equation, i.e. where K =. By insertion it is easily seen that the constant ζ = 1 q 4 K (26) is a particular solution to Eq. (24), and we thus arrive at ζ(x)=ζ [1+a cos(qx)+b sin(qx)+c cosh(qx)+d sinh(qx)]. (27) Note how the amplitude ζ(x) of the driven beam is fixed by the driving force through the value of ζ. This is in contrast to the case of the freely swinging beam Eq. (13), where no scale factor is provided. By imposing the boundary conditions z() = and z () = to Eq. (27) we find { ζ(x) = ζ a[cos(qx) cosh(qx)] (28) } +b[sin(qx) sinh(qx)] + [1 cosh(qx)] FIG. 1: The shape z(x) of the 1 µm 1 µm 1 µm Si micro-beam at time t = T 1/4 when driven at the fundamental resonance frequency ω = ω 1 with an ac-current of amplitude J = 1 ma in a magnetic field B = 1 mt. The damping is given by the viscosity of air, which at room temperature and standard pressure is η = Pa s. The boundary conditions z() = and z () = lead to a simple invertable 2 2 matrix equation for the coefficients a and b, ( ) cos(q) cosh(q) sin(q) sinh(q) ) ( a sin(q) sinh(q) cos(q) cosh(q) b ( cosh(q) 1 = sinh(q) In the following section we study this solution. ). (29) V. A SPECIFIC EXAMPE: THE BEAM SHAPE Once the clamped silicon beam has been specified by Eqs. (1), (2) and (3) the resonance frequencies are fixed. However, the amplitude ζ is controlled by the ac-current amplitude J and the strength B of the magnetic field. ζ = K q 4 = BJ q 4. (3) Moreover, the damping coefficient and thus the Q-factor of the oscillating beam is controlled through the viscosity η of the surrounding air. In this example we choose the following vaules of the external control parameters, which for η means room temperature and standard pressure J = 1 ma, B = 1 mt, η = Pa s. (31) In Fig. 1 is shown the shape of the beam at maximal deflection when driven at the first resonance frequency ω 1. This occurs at t = T 1 /4, where T 1 2π/ω 1 is the period of the oscillation. Thus the oscillation of the beam is π/2 radians out of phase with the driving ac-current. In Fig. 2 is shown the shape of the beam for t T 1 /2 when it suffers almost no deflection.

4 FIG. 2: The shape z(x) of the beam driven at the first resonance frequency as in Fig. 1, however, here at t T 1/2 where the beam happens to suffer a very small deflection. Eleven different shapes are shown for times ranging from t = T 1 / T 1 to t = T 1 / T FIG. 3: The shape z(x) of the beam as in Fig. 1, however, here driven at the third resonance, ω = ω 3, and at time t = T 3 /4 = 2π/(4ω 3 ) VI. THE INDUCED EMF E To detect the resonance frequency and Q-factor of the oscillating beam we plan to measure the electromotive force E(t) induced by the current carrying beam moving in the applied constant magnetic field B. According to Fraday s law of induction we have E(t, ω) = dφ dt da(t, ω) = B, (32) dt where A(t, ω) is the area of the closed current loop which the beam is a part of. et A be the area of the loop when the beam is at its equilibrium position. Then A(t, ω) = A + A(ω) e i(ωt+φ), (33) with the frequency dependent area-amplitude A given by [ ] A(ω) = max t dx z(x, t). (34) FIG. 4: The top panel shows the time dependence of the area amplitude A(t,.99 ω 1 ), i.e. at a driving frequency slightly below the first resonance frequency. The middle panel shows the time dependence of the area amplitude A(t, 1. ω 1 ), i.e. when driving the beam at the first resonance frequency. The bottom panel shows the time dependence of the area amplitude A(t, 1.1 ω 1 ), i.e. at a driving frequency slightly above the first resonance frequency. Notice the changes in amplitude and phase of the three panels. The amplitude E of the induced emf is therefore E(ω) = ω B A(ω). (35) Once the maximal area is found from calculations as the ones depicted in Fig. 4, we can find the frequency depedence of the induced emf E(ω) from Eqs. (34) and (35).

5 5 emf (uv) w/w1 From this follows ζ m directly: BJ ζ m (ω) = ρs(ω1 2 ω2 ) iω γ, (39) and consequently z m (ω) = ζ m = BJ (ρs)2 (ω 2 1 ω2 ) 2 + ω 2 (γ/) 2. (4) The simple estimate for the amplitude z m (ω 1 ) at resonance ω = ω 1 is found form this equation. The result z m (ω 1 ) = BJ =.33 µm, ω 1 γ (41) FIG. 5: The amplitude E(ω) of the induced emf of the 1 µm 1 µm 1 µm Si micro-beam. The resonance frequency is at ω 1 = 5.35 MHz or f 1 = ω 1/2π =.85 MHz. The amplitude of the driving ac-current is J = 1 ma and the magnetic field is B = 1 mt. The damping is given by the viscosity of air, which at room temperature and standard pressure is η = Pa s. In Fig. 5 is shown the resulting resonance curve of E(ω) near the first resonance frequency ω 1. It is seen that the maximal emf-amplitude is 1.21 µv. This is a fairly large signal that ought to be easy to measure. From the graph we can also determine the Q-factor Q 1 for the first resonance. It is given by Q 1 ω 1 = Hz ω Hz = , (36) where ω is the full width at half maximum of the resonance curve. This is a rather high value, so we can expect to observe well-defined resonance peaks in the emfspectrum. VII. SIMPE ESTIMATES The calculation in the previous section are straightforward but not easy to understand directly. We can, however, get a good grasp of the solution by making a simple approximation valid for the tiny damping forces present in the problem. According to Eqs. (11) and (12), when damping is neglected, the shape of the oscillating beam at resonance ω = ω 1 is given by the equation ζ (x) = ρsω 2 1 ζ(x). (37) If we let the entire beam be represented by the mid-point ζ m, we can derive an approximate equation for this point by inserting Eq. (37) into Eq. (23) ω 2 1ρS ζ m = BJ + ω 2 ρs ζ m + iω γ ζ m. (38) in good agreement with the.43 µm of Fig. 1. The estimate Q m for the Q-factor is even better. By setting z m (ω 1 ω/2) = z m (ω 1 )/2 we find [ (ρs) ω1 2 (ω ω)2] = 3 ω 1 γ, (42) and from this the perfect estimate Q m = ω 1 ω = ω 1ρwh 3 γ = (43) The area A(ω) is approximated with that of a triangle with base line and height z m (ω 1 ): A(ω 1 ) = 1 2 z m(ω 1 ) = BJ2 2ω 1 γ = 16.4 µm2, (44) which compares favorably with the 22.7 µm 2 of Fig. 4. Finally, we estimate the amplitude of the electromotive force E by inserting Eq. (44) into Eq. (35) E(ω 1 ) = ω 1 B A(ω 1 ) = B2 J 2 =.88 µv. (45) 2γ This is fairly close to the amplitude 1.21 µv of Fig. 5. VIII. CONCUSION The calculation presented here shows that if we succeed in fabricating a 1 µm 1 µm 1 µm Si micro-beam, then fairly sharp emf-resonance curves are expected with Q-factors around 1 4 and peak amplitudes of 1 µv at the first resonance frequency of.85 MHz. This is remarkable since the parameters in the calculation corresponds to a situation where the beam is oscillating in air at room temperature and standard pressure, while exposed to a moderate magnetic field of strength 1 mt and driven by an ac-current of 1 ma. The series of simple estimates of the system values presented in Sec. VII provides us with a good tool for obtaining a good intuitive feel for the system. 1 A. N. Cleland and M.. Roukes, Appl. Phys. ett., (1996) 2 andau and ifshitz, Theory of Elasticity (Pergamon Press 3rd edition, New York, 1986).