MODELING PIEZOELECTRIC PVDF SHEETS WITH CONDUCTIVE POLYMER ELECTRODES. Laura Marie Lediaev

Transcription

1 MODELING PIEZOELECTRIC PVDF SHEETS WITH CONDUCTIVE POLYMER ELECTRODES by Laura Marie Lediaev A thesis submitted in partial fulfillment of the requirements for the degree of Master of Science in Physics MONTANA STATE UNIVERSITY Bozeman, Montana April 2006

2 COPYRIGHT by Laura Marie Lediaev 2006 All Rights Reserved

3 ii APPROVAL of a thesis submitted by Laura Marie Lediaev This thesis has been read by each member of the thesis committee and has been found to be satisfactory regarding content, English usage, format, citations, bibliographic style, and consistency, and is ready for submission to the College of Graduate Studies. V. Hugo Schmidt, Ph.D. Approved for the Department of Physics William A. Hiscock, Ph.D. Approved for the College of Graduate Studies Joseph J. Fedock, Ph.D.

4 iii STATEMENT OF PERMISSION TO USE In presenting this thesis in partial fulfillment of the requirements for a master s degree at Montana State University, I agree that the Library shall make it availbale to borrowers under rules of the library. If I have indicated my intention to copyright this thesis by including a copyright notice page, copying is allowed only for scholarly purposes, consistent with fair use as prescribed in the U.S. Copyright Law. Requests for permission for extended quotation from or reproduction of this thesis in whole or in part may be granted only by the copyright holder. Laura Marie Lediaev April 17, 2006

5 iv TABLE OF CONTENTS 1. INTRODUCTION Piezoelectricity and PVDF Conducting Polymers and PEDOT-PSS CONSTITUTIVE FORMULAS CAPACITOR MODEL Rectangular Case Cylindrical Case MECHANICAL DEFORMATION Undamped Case Damped Case CONCLUSIONS BIBLIOGRAPHY

6 v LIST OF FIGURES Table Page 1.1 PVDF chain PVDF stretched PVDF contracted Air-brushing technique PEDOT-PSS molecular formula Liquid PEDOT-PSS Schematic of a bimorph Capacitor schematic Boundary conditions for rectangular and cylindrical geometries Amplitude vs. frequency for both electrodes at x = L Phase vs. frequency at x = L Amplitude/phase polar plot at x = L Amplitude vs. frequency for both electrodes at x = 0.5L Phase vs. frequency at x = 0.5L Amplitude/phase polar plot at x = 0.5L Amplitude vs. frequency for both electrodes at x = 0.3L Phase vs. frequency at x = 0.3L Amplitude/phase polar plot at x = 0.3L Amplitude vs. frequency for both electrodes at x = 0.1L Phase vs. frequency at x = 0.1L Amplitude/phase polar plot at x = 0.1L Amplitude vs. frequency at x = L

7 vi LIST OF FIGURES - CONTINUED Table Page 3.17 Phase vs. frequency at x = L Amplitude/phase polar plot at x = L Amplitude vs. frequency at x = 0.5L Phase vs. frequency at x = 0.5L Amplitude/phase polar plot at x = 0.5L Amplitude vs. frequency at x = 0.3L Phase vs. frequency at x = 0.3L Amplitude/phase polar plot at x = 0.3L Amplitude vs. frequency at x = 0.1L Phase vs. frequency at x = 0.1L Amplitude/phase polar plot at x = 0.1L Displacement amplitude vs. frequency Displacement amplitude vs. frequency Experimental voltage measurement Capacitor voltage solution

8 vii ABSTRACT The main concern of my research has been to find a good way to solve for the behavior of piezoelectric devices that are electroded not with metal electrodes (as has traditionally been the case) but with a conductive polymer material which has a much lower conductivity compared to metal. In this situation, if a time-varying voltage is applied at one end of the electrode, the voltage cannot be assumed to be uniform throughout the electrode because of the effects of resistivity. Determining the voltage in the electrodes as a function of time and position concurrently with the mechanical and electrical response of the piezoelectric material presents an added complexity. In this thesis the problem of the piezoelectric monomorph is considered. The piezoelectric sheet is PVDF, and the electrodes are PEDOT-PSS. As a first approximation the two problems of finding the voltage in the electrodes and the mechanical deformation in the piezoelectric material are decoupled. In order to determine the voltage distribution in the electrodes, the piezoelectric effects were neglected, which reduced the piezoelectric problem to a capacitor problem. Once the voltage function was determined the mechanical deformation of the PVDF sheet was calculated given the known voltage distribution as a function of position and time.

9 1 CHAPTER 1 INTRODUCTION Piezoelectricity and PVDF Piezoelectricity is a linear coupling between electrical and mechanical processes [1]. In the direct piezoelectric effect, when a piezoelecctric material is compressed, an electric polarization is formed across the material. In fact, the prefix piezo is derived from the Greek word for press [1]. The converse piezoelectric effect is when an applied electric field causes the piezoelectric to mechanically deform. Piezoelectricity is made possible due to certain kinds of crystal structures which lack a center of symmetry. Materials which are piezoelectric come in several forms. There are single crystals, ceramics, and polymers (semi-crystalline) [2]. Poly(vinylidene-fluoride) (PVDF) is a piezoelectric polymer, which for actuator applications often comes in the form of a thin sheet (30 microns thick). It is stretched along the x direction to align the long chain molecules, and is poled in the z direction to align the electro-negative and electro-positive parts of the molecular units (the hydrogen and fluorine atoms) to create a strong piezoelectric constant, as shown in figure 1.1. Because of the aligned ions, there is a charge polarization. When an electric field is applied across the PVDF sheet, the molecules will either stretch or contract, depending on the direction of the field, as shown in figures 1.2 and 1.3. The other dimensions of the PVDF sheet will also change. The thickness of the sheet is very small, but the length is substantial and even an elongation of only a small percent will be noticable. When an electric field is applied across two sheets that are glued

10 2 δ+ H H H H H H Electro-positive H H C C C C p C C C C δ F F F F F F F F Electro-negative Figure 1.1: PVDF chain E Figure 1.2: PVDF stretched. δ+ δ δ+ E δ Figure 1.3: PVDF contracted.

11 3 together with opposite polarization, the sheets will bend out of the plane. This is a bimorph, and the deflection is much larger than the elongation of an individual sheet. This conversion from electric field to mechanical deformation and vice versa is very useful. Conducting Polymers and PEDOT-PSS In order to apply an electric field, the PVDF sheet must have applied electrodes. In small deformation applications the electrodes can be metal, which is very good because it has very high conductivity and the voltage throughout the electrode can be assumed to be uniform. The drawback with metal is that it is stiffer than the PVDF polymer and so hinders its deformation. Polymer electrodes are more flexible, but the conductivity is much lower, and so the voltage is not uniform. The higher the frequency, the more non-uniform the voltage will be. A big problem is certainly the amplitude attenuation along the electrode. Voltage amplitude is decreased by the resistivity (Ohm s law: V = IR). One conducting polymer is PEDOT-PSS. Poly(ethylene dioxythiophene) is a conjugated polymer, and poly(styrene sulfonate) is a dopant which dramatically increases the conductivity of PEDOT [2]. See figure 1.5 for the molecular representation of PEDOT-PSS and figure 1.6 for a picture of PEDOT-PSS in it s liquid form. There are several ways to apply the polymer electrodes onto the PVDF sheets. One method that has been attempted is spraying [2]. In this method a spray gun is used to spray liquid PEDOT-PSS onto the surface of the PVDF, which then is allowed to dry. See figure 1.4 to see an example of this technique. PVDF is hydrophobic, and since PEDOT-PSS comes in a water-diluted form, if the PEDOT-PSS is applied too

12 4 Figure 1.4: Air-brushing technique. thickly it will bead up. A method that seems to work quite well is inkjet printing. In this method PEDOT-PSS is printed onto the PVDF sheets using an ordinary inkjet printer. The thickness of the applied layers can be easily controlled to produce uniform layers [3].

13 5 Figure 1.5: PEDOT-PSS molecular formula. Figure 1.6: Liquid PEDOT-PSS.

14 6 CHAPTER 2 CONSTITUTIVE FORMULAS There are four state variables which describe the electro-mechanical state of a piezoelectric actuator. The two mechanical state variables are stress (T ), and strain (S). The two electric state variables are electric field (E) and electric displacement (D). In a piezoelectric material the two sets of state variables are coupled, which means both sets must be determined simultaneously, as one affects the other. The fundamental (S-E)-type relation [1, Table 2.1(b)] is given in equation (2.1). There are other, equally valid, sets of state variables to use. For example, instead of D, the electric polarization (P ) can be used, in which case we would have an (S-P)type relation. T = c E S e E D = e S + ε S E (2.1a) (2.1b) The (S-E)type relation using full tensor index notation is given in equation (2.2). T ij = c E ijkls kl e mij E m D n = e nkl S kl + ε S nme m (2.2a) (2.2b) The stress is related to the applied and generated forces. The strain can be defined in terms of the mechanical displacements from equilibrium (u i ). Throughout this thesis the 1 direction refers to the x direction, the 2 direction to the y direction, and the 3 direction to the z direction. In every case the piezoelectric sheet is oriented so that the poled direction is the z direction, and the stretch direction is

15 7 along x. S ij 1 2 ( ui + u ) j x j x i (2.3) For tensors that have symmetric pairs of indices, those pairs can be condensed into a single index. This allows expressions to be condensed by combining equal terms. For indices which can take on values of 1, 2, or 3, a condensed pair can take on the values 1 through 6. This condensed form is call matrix index notation. 11 1, 22 2, 33 3, 23 4, 13 5, 12 6 (2.4) There is a convention for how to combine terms to form a single condensed term. Sometimes there is a factor of 2 or 4. This is done so that the final matrix index expression will look the same as the tensor index expression, without any extra factors in front of terms. The conversion for some quantities is given in equations (2.5) through (2.8). e mij e mn (2.5) c ijkl c λµ (2.6) T ij T λ (2.7) S ij S λ (i = j), 2S ij S λ (i j) (2.8) (2.9). The matrix index notation version of the constitutive relation is given in equation T λ = c E λµs µ e mλ E m D n = e nµ S µ + ε S nme m (2.9a) (2.9b) The form of the matrices for the material constants are given below. The dielectric permittivity is ε, the piezoelectric constant is e, and the elastic stiffness is c. PVDF

16 8 has mm2 symmetry, and so many entries are zero, which simplifies the resulting equations. e iλ = ε mn = c λµ = ε 1 ε 2 ε 3 e 15 e 24 e 31 e 32 e 33 c 11 c 12 c 13 c 12 c 22 c 23 c 13 c 23 c 33 c 44 c 55 c 66 (2.10) (2.11) (2.12) After explicitly writing out all the terms and substituting in the non-zero constants, we get the full general constitutive equations for mm2 symmetry. T 1 = c 11 S 1 + c 12 S 2 + c 13 S 3 e 31 E 3 T 2 = c 12 S 1 + c 22 S 2 + c 23 S 3 e 32 E 3 T 3 = c 13 S 1 + c 23 S 2 + c 33 S 3 e 33 E 3 T 4 = c 44 S 4 e 24 E 2 T 5 = c 55 S 5 e 15 E 1 T 6 = c 66 S 6 (2.13a) (2.13b) (2.13c) (2.13d) (2.13e) (2.13f) D 1 = e 15 S 5 + ε 1 E 1 D 2 = e 24 S 4 + ε 2 E 2 D 3 = e 31 S 1 + e 32 S 2 + e 33 S 3 + ε 3 E 3 (2.14a) (2.14b) (2.14c)

17 9 Equations of motion: S 1 = u 1 x 1 S 2 = u 2 x 2 S 3 = u 3 x 3 S 4 = u 2 x 3 + u 3 x 2 S 5 = u 1 x 3 + u 3 x 1 S 6 = u 1 x 2 + u 2 x 1 ρ 2 u i t 2 (2.15a) (2.15b) (2.15c) (2.15d) (2.15e) (2.15f) = T ij x j (2.16) ρ 2 u 1 t 2 = T 1 x 1 + T 6 x 2 + T 5 x 3 ρ 2 u 2 t 2 = T 6 x 1 + T 2 x 2 + T 4 x 3 ρ 2 u 3 t 2 = T 5 x 1 + T 4 x 2 + T 3 x 3 (2.17a) (2.17b) (2.17c) Maxwell equations: Electromagnetic constitutive relations: D = ρ f (2.18) B = 0 (2.19) E + B t = 0 (2.20) H D t = J f (2.21) D = ε 0 E + P (2.22) H = 1 µ 0 B M (2.23) J f = σ cond E (2.24)

18 10 Since the PVDF actuators are composed of non-magnetic materials, M = 0, and H = B/µ 0. (2.25) In the electrodes: In the piezoelectric material: E = ρ f /εε 0 (2.26) B = 0 (2.27) E + B t = 0 (2.28) 1 E B εε0 µ 0 t = σ conde (2.29) D = 0 (2.30) B = 0 (2.31) E + B t = 0 (2.32) 1 D B µ 0 t = 0 (2.33)

19 11 CHAPTER 3 CAPACITOR MODEL Electrode (PEDOT/PSS) V(t) PVDF PVDF P P glue layer Figure 3.1: Schematic of a bimorph. The voltage distribution in the electrodes of a bimorph can be quite complicated if the conductivity of the electrode material is not very high. As a first approximation the bimorph can be treated as a simple parallel-plate capacitor, as shown in figure 3.2. In this case the piezoelectric behavior is not included in the calculations. Although we have currents and a time-changing electric field, and therefore a magnetic field, we can treat the electric field quasi-electrostatically, and neglect magnetic contributions, so that the electric field is equal to the negative gradient of the electric potential. The electric field across the thickness of the capacitor is E 3 (in the z φ 0 φ t σ f E 1, J, I E 3 d σ f -φ φ b 0 Figure 3.2: Capacitor schematic.

20 12 direction), while the electric field in the electrodes is approximated as only being in the x direction, and is called E 1. For this problem there is an applied voltage applied to the top and bottom electrodes at x = 0. In order to have a symmetric solution the applied voltage is such that the voltage on the bottom electrode is the negative of the voltage on the top electrode. E = φ (3.1) E 3 = φ z = φ t φ b d = 2φ t d (3.2) The electric field inside a parallel plate capacitor is proportional to the free surface charge density on the plates, where σ t is the charge density on the top electrode, and σ b is the charge density on the bottom electrode. Because of symmetry σ b = σ t. E 3 = σ t σ b 2εε 0 = σ t εε 0 (3.3) By combining equations 3.2 and 3.3 we get an expression for the surface charge density in terms of the electric potential. σ t = 2εε 0 d φ t (3.4) We can use the charge continuity equation and the equation for the current density in a simple conductor to derive an expression for the voltage distribution in the electrodes. J = ρ f t (3.5) J = σ cond E (3.6) J = ( σ cond E ) = ( σ cond φ ) (3.7)

21 13 Figure 3.3: Boundary conditions for rectangular and cylindrical geometries. The general differential equation for the electric potential, before any further approximations are made, is given below. ( σ cond φ ) = ρ f t (3.8) In general the conductivity is a tensor, but in this case the electrode material is isotropic. In both the rectangular and cylindrical cases, shown in figure 3.3, only the 1 term survives because the current is assumed to flow only in either the x or radial direction due to symmetry. The volume charge density can be approximately expressed in terms of the surface charge density because the electrodes are so thin. The thickness of an electrode is w, its resistivity is ρ, and the thickness of the PVDF sheet is d. The permittivity of free space is ε 0, and the relative permittivity of PVDF is ε. ρ f σ w = 2εε 0 wd φ t (3.9)

22 14 Rectangular Case For the rectangular case, where the current flows only in the x direction, with boundary conditions φ t t = k 2 φ t x 2, k = wd 2ρεε 0 (3.10) φ t (0, t) = V 0 2 sin ωt, φ t x = 0, (3.11) x=l where V 0 is the amplitude of the applied voltage. For simplicity the input voltage function is sinusoidal. In general, of course, the input voltage can by any desired function. The resulting differential equation, equation 3.10, has the same form as the onedimensional diffusion equation. In the rectangular case the solution is comprised of sines and cosines. We can get the full solution by assuming a set of incident and reflected waves, and solving for the unknown coefficients. φ t = C inc f 1 (x, t) + D inc f 2 (x, t) + C ref f 3 (x, t) + D ref f 4 (x, t) (3.12) f 1 (x, t) = e αx cos (αx ωt) f 2 (x, t) = e αx sin (αx ωt) f 3 (x, t) = e αx cos (αx + ωt) f 4 (x, t) = e αx sin (αx + ωt) (3.13a) (3.13b) (3.13c) (3.13d) Some derivatives of the f i which will be useful later are listed below. f f t, f f x (3.14)

23 15 f 1 = ωf 2 f 2 = ωf 1 f 3 = ωf 4 f 4 = ωf 3 (3.15a) (3.15b) (3.15c) (3.15d) f 1 = ω 2 f 1 f 2 = ω 2 f 2 f 3 = ω 2 f 3 f 4 = ω 2 f 4 (3.16a) (3.16b) (3.16c) (3.16d) f 1 = α (f 1 + f 2 ) f 2 = α (f 1 f 2 ) f 3 = α (f 3 f 4 ) f 4 = α (f 3 + f 4 ) (3.17a) (3.17b) (3.17c) (3.17d) f 1 = 2α 2 f 2 (3.18a) f 2 = 2α 2 f 1 (3.18b) f 3 = 2α 2 f 4 (3.18c) f 4 = 2α 2 f 3 (3.18d) f 1 = αω (f 1 f 2 ) f 2 = αω (f 1 + f 2 ) f 3 = αω (f 3 + f 4 ) f 4 = αω (f 3 f 4 ) (3.19a) (3.19b) (3.19c) (3.19d)

24 16 f 1 = 2α 2 ωf 1 (3.20a) f 2 = 2α 2 ωf 2 (3.20b) f 3 = 2α 2 ωf 3 (3.20c) f 4 = 2α 2 ωf 4 (3.20d) When we put our form of the solution through the differential equation we get the following requirement for α. α = ω 2k (3.21) By applying the first boundary condition we get the following conditions on the coefficients. φ t (0, t) = (C inc + C ref ) cos (ωt) + (D ref D inc ) sin (ωt) = V 0 2 C inc + C ref = 0, D ref D inc = V 0 2 We can make the substitutions sin (ωt) (3.22) C ref = C inc, D ref = V D inc. (3.23) After applying the second boundary condition we can fully specify the coefficients. [ φ (L, t) = α [ = α = 0 C inc (f 1 + f 2 ) + D inc (f 1 f 2 ) ( V0 + C inc (f 4 f 3 ) D inc (D inc C inc ) f 1 (C inc + D inc ) f 2 ( ) V D inc C inc f 3 + ) ] (f 3 + f 4 ) x=l ( V0 2 + C inc + D inc ) f 4 ] x=l (3.24) The f i can be separated into terms of cos ωt and sin ωt, and then equation (3.24) can be but into the form C c cos ωt + C s sin ωt = 0. Because cos ωt and sin ωt are

25 17 orthogonal, both expressions C c and C s must individually equal zero. f 1 (L, t) = e αl cos (αl) cos (ωt) + e αl sin (αl) sin (ωt) f 2 (L, t) = e αl sin (αl) cos (ωt) e αl cos (αl) sin (ωt) f 3 (L, t) = e αl cos (αl) cos (ωt) e αl sin (αl) sin (ωt) f 4 (L, t) = e αl sin (αl) cos (ωt) + e αl cos (αl) sin (ωt) (3.25a) (3.25b) (3.25c) (3.25d) C c = (D inc C inc ) e αl cos αl (C inc + D inc ) e αl sin αl ( ) ( ) V D inc C inc e αl V0 cos αl C inc + D inc e αl sin αl [ ) ( )] = C inc e (cos αl αl + sin αl + e αl sin αl cos αl [ ) ( )] + D inc e (cos αl αl sin αl + e αl cos αl + sin αl + V ( ) 0 2 eαl cos αl + sin αl = 0 C s = (D inc C inc ) e αl sin αl + (C inc + D inc ) e αl cos αl ( ) ( ) V0 2 + D inc C inc e αl V0 sin αl C inc + D inc e αl cos αl [ ) ( )] = C inc e (cos αl αl sin αl + e αl cos αl + sin αl [ ) ( )] + D inc e (cos αl αl + sin αl + e αl cos αl sin αl + V ( ) 0 2 eαl cos αl sin αl = 0 (3.26) (3.27) Equations (3.26) and (3.27) can be somewhat simplified. C inc ( e αl e αl) sin αl + D inc ( e αl + e αl) cos αl = V 0 2 eαl cos αl (3.28) C inc ( e αl + e αl) cos αl + D inc ( e αl e αl) sin αl = V 0 2 eαl sin αl (3.29)

26 18 With these two equation and equation 3.23 we now have four equations for the four coefficients. The expressions for the coefficients are shown below. sin (αl) cos (αl) C inc = V 0 e 2αL + 2 ( cos 2 αl sin 2 αl ) + e 2αL = V 0 2 D inc = V 0 2 = V 0 2 sin 2αL e 2αL + 2 cos 2αL + e 2αL e αl [ e αl + e αl ( cos 2 αl sin 2 αl )] e 2αL + 2 ( cos 2 αl sin 2 αl ) + e 2αL e 2αL + cos 2αL e 2αL + 2 cos 2αL + e 2αL C ref = V 0 2 D ref = V 0 2 (3.30) (3.31) sin 2αL e 2αL + 2 cos 2αL + e 2αL (3.32) e 2αL + cos 2αL e 2αL + 2 cos 2αL + e 2αL (3.33) The full expression for the voltage on the top electrode is given below. The full potential difference across the sheet is twice φ t. φ t (x, t) = V e 2αL + 2 cos 2αL + e 2αL {e αx [ sin 2αL cos (αx ωt) ( e 2αL + cos 2αL ) sin (αx ωt) (3.34) + e [ αx sin 2αL cos (αx + ωt) + ( e 2αL + cos 2αL ) ] } sin (αx + ωt) ] The potential function is sinusoidal in time, and so it can be expressed as a sine function in t with an x dependent phase (relative to the input voltage). φ t = φ t sin (ωt + δ) = φ t sin δ cos ωt + φ t cos δ sin ωt (3.35)

27 19 φ t sin δ = V 0 2 φ t cos δ = V 0 2 sin δ = cos δ = [ 1 (e e 2αL + 2 cos 2αL + e αx e αx) sin 2αL cos αx 2αL ( (e + 2αL + cos 2αL ) e αx ( e 2αL + cos 2αL ) ) ] (3.36) e αx sin αx [ 1 (e e 2αL + 2 cos 2αL + e αx + e αx) sin 2αL sin αx 2αL ( (e + 2αL + cos 2αL ) e αx + ( e 2αL + cos 2αL ) ) ] e αx cos αx φ t = V 0 2 = V 0 2 e 2α(L x) + 2 cos (2α (L x)) + e 2α(L x) e 2αL + 2 cos 2αL + e 2αL cosh (2α (L x)) + cos (2α (L x)) cosh 2αL + cos 2αL (e αx e αx ) sin (α (2L x)) + ( e α(2l x) e α(2l x)) sin (αx) (e 2αL + 2 cos 2αL + e 2αL ) (e 2α(L x) + 2 cos (2α (L x)) + e 2α(L x) ) sinh (αx) sin (α (2L x)) + sinh (α (2L x)) sin (αx) = (cosh 2αL + cos 2αL) (cosh (2α (L x)) + cos (2α (L x))) = (e αx + e αx ) cos (α (2L x)) + ( e α(2l x) + e α(2l x)) cos (αx) (e 2αL + 2 cos 2αL + e 2αL ) (e 2α(L x) + 2 cos (2α (L x)) + e 2α(L x) ) cosh (αx) cos (α (2L x)) + cosh (α (2L x)) cos (αx) (cosh 2αL + cos 2αL) (cosh (2α (L x)) + cos (2α (L x))) (3.37) (3.38) (3.39) (3.40) Graphs of the amplitude and phase of the symmetric solution for the top and bottom electrodes are shown below at various positions along the sheet. For the polar plots, the low frequency regions are on the outside, and the high frequency region is near the center of the spiral (at low amplitudes). It s important to note that although the plots for different x look similar, the graphs for smaller x go up to a much higher frequency. The closer to x = 0, the higher the required frequency to substantially reduce the amplitude.

28 20 amplitude frequency Figure 3.4: Amplitude vs. frequency for both electrodes at x = L. δ (radians) π top electrode bottom electrode 0 frequency -π Figure 3.5: Phase vs. frequency at x = L.

29 21 top electrode bottom electrode δ=45 δ=30 δ=15 δ=180 δ=0 φ=0.1 φ=0.2 φ=0.3 φ=0.4 Figure 3.6: Amplitude/phase polar plot at x = L.

30 22 amplitude frequency Figure 3.7: Amplitude vs. frequency for both electrodes at x = 0.5L. δ (radians) π top electrode bottom electrode 0 frequency -π Figure 3.8: Phase vs. frequency at x = 0.5L.

31 23 top electrode bottom electrode δ=45 δ=30 δ=15 δ=180 δ=0 φ=0.1 φ=0.2 φ=0.3 φ=0.4 Figure 3.9: Amplitude/phase polar plot at x = 0.5L.

32 24 amplitude frequency Figure 3.10: Amplitude vs. frequency for both electrodes at x = 0.3L. δ (radians) π top electrode bottom electrode 0 frequency -π Figure 3.11: Phase vs. frequency at x = 0.3L.

33 25 top electrode bottom electrode δ=45 δ=30 δ=15 δ=180 δ=0 φ=0.1 φ=0.2 φ=0.3 φ=0.4 Figure 3.12: Amplitude/phase polar plot at x = 0.3L.

34 26 amplitude frequency Figure 3.13: Amplitude vs. frequency for both electrodes at x = 0.1L. δ (radians) π top electrode bottom electrode 0 frequency -π Figure 3.14: Phase vs. frequency at x = 0.1L.

35 27 top electrode bottom electrode δ=45 δ=30 δ=15 δ=180 δ=0 φ=0.1 φ=0.2 φ=0.3 φ=0.4 Figure 3.15: Amplitude/phase polar plot at x = 0.1L.

36 28 In the actual case the bottom electrode is grounded at x = 0. This is done because experimentally it s easier to ground one electrode instead of splitting the applied voltage. In order to get the asymmetric solution we just add (V 0 /2) sin ωt to the symmetric solution. The new boundary condition at x = 0 becomes φ t (0, t) = V 0 sin ωt and φ b (0, t) = 0. It s interesting to note that at high enough frequencies the amplitude of the grounded electrode can be greater than the top electrode for x > 0. In the expressions below φ represents the asymmetric solution, while φ represents the symmetric solution. It s good to know the asymmetric solution to verify experimental measurements, but what really matters is the potential difference across the sheet, which is the same as for the symmetric solution. φ t = V 0 2 sin ωt + φ t (3.41) φ b = V 0 2 sin ωt φ t (3.42) Just as with the symmetric solution, we can break up the potential into terms of cos ωt and sin ωt. φ t = φ t sin(ωt + δ t ) = φ t sin δ t cos ωt + φ t cos δ t sin ωt (3.43) φ b = φ b sin(ωt + δ b ) = φ b sin δ b cos ωt + φ b cos δ b sin ωt (3.44) φ t sin δ t = φ t sin δ (3.45) φ t cos δ t = V φ t cos δ (3.46) φ t = [ φ t 2 + V ] 0 2 1/2 4 + V 0 φ t cos δ (3.47) φ b sin δ b = φ t sin δ (3.48) φ b cos δ b = V 0 2 φ t cos δ (3.49) φ b = [ φ t 2 + V ] 0 2 1/2 4 V 0 φ t cos δ (3.50)

37 29 amplitude 1.0 top electrode bottom electrode frequency Figure 3.16: Amplitude vs. frequency at x = L. δ (radians) π/2 top electrode bottom electrode 0 frequency -π/4 Figure 3.17: Phase vs. frequency at x = L.

38 30 top electrode bottom electrode δ=18 δ=9 δ=0 φ=0.1 φ=0.2 φ=0.3 φ=0.4 φ=0.6 φ=0.7 φ=0.8 φ=0.9 δ=-9 δ=-18 Figure 3.18: Amplitude/phase polar plot at x = L.

39 31 amplitude 1.0 top electrode bottom electrode frequency Figure 3.19: Amplitude vs. frequency at x = 0.5L. δ (radians) π/2 top electrode bottom electrode 0 frequency -π/4 Figure 3.20: Phase vs. frequency at x = 0.5L.

40 32 top electrode bottom electrode δ=18 δ=9 δ=0 φ=0.1 φ=0.2 φ=0.3 φ=0.4 φ=0.6 φ=0.7 φ=0.8 φ=0.9 δ=-9 Figure 3.21: Amplitude/phase polar plot at x = 0.5L.

41 33 amplitude 1.0 top electrode bottom electrode frequency Figure 3.22: Amplitude vs. frequency at x = 0.3L. δ (radians) π/2 top electrode bottom electrode 0 frequency -π/4 Figure 3.23: Phase vs. frequency at x = 0.3L.

42 34 top electrode bottom electrode δ=18 δ=9 δ=0 φ=0.1 φ=0.2 φ=0.3 φ=0.4 φ=0.6 φ=0.7 φ=0.8 φ=0.9 δ=-9 δ=-18 Figure 3.24: Amplitude/phase polar plot at x = 0.3L.

43 35 amplitude 1.0 top electrode bottom electrode frequency Figure 3.25: Amplitude vs. frequency at x = 0.1L. δ (radians) π/2 top electrode bottom electrode 0 frequency -π/4 Figure 3.26: Phase vs. frequency at x = 0.1L.

44 36 top electrode bottom electrode δ=18 δ=9 δ=0 φ=0.1 φ=0.2 φ=0.3 φ=0.4 φ=0.6 φ=0.7 φ=0.8 φ=0.9 δ=-9 δ=-18 Figure 3.27: Amplitude/phase polar plot at x = 0.1L.

45 37 Cylindrical Case In the cylindrical case the solution is modified Bessel functions. The differential equation can be solved by separation of variables. The time part is a simple exponential, e iωt, but because we want the sine part, the final solution will be the imaginary part of the composite function. The solution expressed in real form is quite complicated, so here it is left in complex form. The graphs for the solutions are very nearly identical to the graphs for the rectangular solutions. [ φ t 2 t = k φ t r r ] φ t, φ t (R 1, t) = V 0 r 2 sin ωt, φ t r = 0 (3.51) r=r2 [ ( ) ( ) ( ) ( ) ] V0 I 0 iαr K1 iαr2 + I1 iαr2 K0 iαr φ t (r, t) = Im 2 eiωt ( ) ( ) ( ) ( ) I 0 iαr1 K1 iαr2 + I1 iαr2 K0 iαr1 = φ t sin (ωt + δ) = φ t sin δ cos ωt + φ t cos δ sin ωt φ t sin δ = V [ ( ) ( ) ( ) ( ) ] 0 2 Im I0 iαr K1 iαr2 + I1 iαr2 K0 iαr ( ) ( ) ( ) ( ) I 0 iαr1 K1 iαr2 + I1 iαr2 K0 iαr1 φ t cos δ = V [ ( ) ( ) ( ) ( ) ] 0 2 Re I0 iαr K1 iαr2 + I1 iαr2 K0 iαr ( ) ( ) ( ) ( ) I 0 iαr1 K1 iαr2 + I1 iαr2 K0 iαr1 φ t = (3.52) (3.53) (3.54) ( φ t sin δ) 2 + ( φ t cos δ) 2 (3.55)

46 38 CHAPTER 4 MECHANICAL DEFORMATION Now that we have solved for the electric potential, we can now include the piezoelectric effect and solve for the mechanical deformation. For the case of the monomorph the deformation will only be in the plane (no bending). For a thin-beam approximation we can say that the monomorph is also thin in the y dimension, and so we can limit our interest to the x displacement. From the constitutive relation, equation 2.1a, the piezoelectric contribution to the deformation is proportion to the electric field. The field E 3 (x, t) applied across the thickness d is 2φ t (x, t)/d, using equation (3.34) for φ(x, t). We abbreviate the notation by writing E 3 (x, t) as E 3 (x, t) = a 1 f 1 + a 2 f 2 + a 3 f 3 + a 4 f 4, (4.1) where the f i are the same as previously defined, and the a i are defined below. a 1 = 2C inc /d, a 2 = 2D inc /d a 3 = 2C ref /d, a 4 = 2D ref /d (4.2) We assume that the y and z dimensions of the monomorph (the width and thickness) are small enough so that the inertially caused stresses T 2 and T 3 are zero. T 2 = c 12 S 1 + c 22 S 2 + c 23 S 3 e 32 E 3 = 0 (4.3) T 3 = c 13 S 1 + c 23 S 2 + c 33 S 3 e 33 E 3 = 0 (4.4) From these two equations we can solve for S 1 and S 2, and then plug these expression back into the equation for T 1. [ ] [ ] [ ] c22 c 23 S2 e32 E = 3 c 12 S 1 c 23 c 33 S 3 e 33 E 3 c 13 S 1 [ ] [ ] [ S2 1 c33 c = 23 e32 E 3 c 12 S 1 S 3 c 2 23 c 22 c 33 c 23 c 22 e 33 E 3 c 13 S 1 ] (4.5) (4.6)

47 39 S 2 = (c 23e 33 c 33 e 32 ) E 3 + (c 12 c 33 c 13 c 23 ) S 1 c 2 23 c 22 c 33 (4.7) S 3 = (c 23e 32 c 22 e 33 ) E 3 + (c 13 c 22 c 12 c 23 ) S 1 c 2 23 c 22 c 33 (4.8) T 1 = c 11 S 1 + c 12 S 2 + c 13 S 3 e 31 E 3 [ = S 1 c 11 + c ] 12 (c 12 c 33 c 13 c 23 ) + c 13 (c 13 c 22 c 12 c 23 ) c 2 23 c 22 c [ 33 + E 3 e 31 + c ] 12 (c 23 e 33 c 33 e 32 ) + c 13 (c 23 e 32 c 22 e 33 ) c 2 23 c 22 c 33 (4.9) Undamped Case The constant coefficients in equation (4.9) can be abbreviated as d 1 and d 3, respectively. T 1 = d 1 S 1 + d 3 E 3 (4.10) d 1 = c 11 + c2 12c 33 + c 2 13c 22 2c 12 c 13 c 23 c 2 23 c 22 c 33 (4.11) d 3 = e 31 + (c 13c 23 c 12 c 33 ) e 32 + (c 12 c 23 c 13 c 22 ) e 33 c 2 23 c 22 c 33 (4.12) Substituting this expression for T 1 into our equation of motion for u 1, and setting T 4, T 5, and T 6 equal to zero because there are no shearing forces (only elongation in the x 1 direction), we get ρ 2 u 1 = T 1 2 u 1 E 3 = d t d x 1 x 2 3. (4.13) 1 x 1 From now on u 1 is just u and x 1 is just x. The function that satisfies equation (4.13) before applying boundary conditions is called u ih. To this function we can add

48 40 a function u h which satisfies equation (4.16), and which together with u ih will ensure that the boundary conditions are satisfied. u = u ih + u h (4.14) ρ 2 u ih t 2 2 u ih E 3 d 1 = d x 2 3 x (4.15) ρ 2 u h t d 2 u h 2 1 x = 0 (4.16) 2 Since the derivatives of the f i give back the f i with factors, the solution u ih will be a combination of the f i. u ih = c 1 f 1 (x, t) + c 2 f 2 (x, t) + c 3 f 3 (x, t) + c 4 f 4 (x, t) (4.17) After putting this form for u ih through the differential equation, we get the following requirements for the c i. [ A B B A [ A B B A ] [ c1 c 2 ] [ c3 c 4 ] [ a = C 2 a 1 (a 1 + a 2 ) ] [ ] a3 + a = C 4 a 4 a 3 ] (4.18) (4.19) A = ρω 2, B = 2α 2 d 1, C = αd 3 (4.20) The form of the coefficient matrices are simple to invert. We get the following solutions for the c i. [ c1 c 2 [ c3 c 4 ] = ] = C A 2 + B 2 C A 2 + B 2 [ A B B A [ A B B A ] [ a 2 a 1 (a 1 + a 2 ) ] [ ] a3 + a 4 a 4 a 3 ] (4.21) (4.22) A less abbreviated expression for the c i is shown below. c 1 c 2 c 3 = αd 3 ρ 2 ω 4 + 4α 4 d 2 1 c 4 ρω 2 (a 1 a 2 ) + 2α 2 d 1 (a 1 + a 2 ) ρω 2 (a 1 + a 2 ) + 2α 2 d 1 (a 2 a 1 ) ρω 2 (a 3 + a 4 ) + 2α 2 d 1 (a 4 a 3 ) ρω 2 (a 3 a 4 ) 2α 2 d 1 (a 3 + a 4 ) (4.23)

49 41 Now we need to find u h. As a guess we choose the following form for the homogeneous solution. u h = g 1 h 1 (x, t) + g 2 h 2 (x, t) + g 3 h 3 (x, t) + g 4 h 4 (x, t) (4.24) h 1 = cos (qx ωt), h 3 = cos (qx + ωt), h 2 = sin (qx ωt) h 4 = sin (qx + ωt) (4.25) After putting this form for u h through the homogeneous differential equation, we get the following requirement for q: ρ q = ω (4.26) d 1 In this problem we assume that the left end of the monomorph is clamped, so that u(0, t) = 0, and the right end is free, so that T 1 (L, t) = 0. After applying the first boundary condition we get the following conditions on the g i, the coefficients for the homogeneous solution. u (0, t) = 0 = (g 1 + g 3 + c 1 + c 3 ) cos (ωt) + (g 4 g 2 + c 4 c 2 ) sin (ωt) g 3 = (g 1 + c 1 + c 3 ), g 4 = g 2 + c 2 c 4 (4.27) Now we apply the second boundary condition. u T 1 (L, t) = 0 = d 1 + d 3 E 3 (L, t) (4.28) x x=l ] ] f 1 (L, t) [d 1 α (c 2 c 1 ) + d 3 a 1 + f 2 (L, t) [ d 1 α (c 1 + c 2 ) + d 3 a 2 ] ] + f 3 (L, t) [d 1 α (c 3 + c 4 ) + d 3 a 3 + f 4 (L, t) [d 1 α (c 4 c 3 ) + d 3 a 4 [ ] (4.29) + d 1 q g 2 h 1 (L, t) g 1 h 2 (L, t) + g 4 h 3 (L, t) g 3 h 4 (L, t) = 0 The resulting expressions are going to start getting very messy, so I m going to start abbreviating many of the constant coefficients. ζ 1 = d 1 q = ω ρd 1 (4.30)

50 42 form. γ 1 = d 1 α (c 2 c 1 ) + d 3 a 1, γ 2 = d 1 α (c 1 + c 2 ) + d 3 a 2 (4.31) γ 3 = d 1 α (c 3 + c 4 ) + d 3 a 3, γ 4 = d 1 α (c 4 c 3 ) + d 3 a 4 Since the c i aren t too complicated, the γ i can be expressed in a less abbreviated γ 1 γ 2 γ 3 γ 4 = ρω 2 d 3 ρ 2 ω 4 + 4α 4 d 2 1 ρω 2 a 1 + 2α 2 d 1 a 2 ρω 2 a 2 2α 2 d 1 a 1 ρω 2 a 3 2α 2 d 1 a 4 ρω 2 a 4 + 2α 2 d 1 a 3 (4.32) Given these abbreviated coefficients, and equation (4.62), equation (4.29) can be written as follows. γ 1 f 1 (L, t) + γ 2 f 2 (L, t) + γ 3 f 3 (L, t) + γ 4 f 4 (L, t) + ζ 1 g 2 h 1 (L, t) ζ 1 g 1 h 2 (L, t) + ζ 1 (g 2 + c 2 c 4 ) h 3 (L, t) + ζ 1 (g 1 + c 1 + c 3 ) h 4 (L, t) (4.33) = 0 We pick up a couple more constants. [ γ5 γ 6 γ 5 = ζ 1 (c 2 c 4 ), γ 6 = ζ 1 (c 1 + c 3 ) (4.34) ] = αωd [ ] 3 ρd1 ρω 2 (2a 1 + a 2 + a 4 ) 2α 2 d 1 (2a 1 a 2 a 4 ) ρ 2 ω 4 + 4α 4 d 2 ρω 2 (2a 1 1 a 2 a 4 ) + 2α 2 (4.35) d 1 (2a 2 + a 2 + a 4 ) Since the f i (L, t) and the h i (L, t) are sinusoids in time, we once again get an expression of the form C c cos ωt + C s sin ωt = 0, where in order to satisfy equation (4.33) for all time, both C c and C s must equal zero. γ 1 e αl cos αl + γ 2 e αl sin αl + γ 3 e αl cos αl + γ 4 e αl sin αl + γ 5 cos ql + γ 6 sin ql (4.36) = 2g 2 ζ 1 cos ql = K 1 γ 1 e αl sin αl γ 2 e αl cos αl γ 3 e αl sin αl + γ 4 e αl cos αl γ 5 sin ql + γ 6 cos ql (4.37) = 2g 1 ζ 1 cos ql = K 2

51 43 With equations (4.36), (4.64), and (4.62) we can now fully solve for the g i. g 1 = K 2 2ζ 1 cos ql g 2 = K 1 2ζ 1 cos ql g 3 = K 2 2γ 6 cos ql 2ζ 1 cos ql g 4 = 2γ 5 cos ql K 1 2ζ 1 cos ql (4.38a) (4.38b) (4.38c) (4.38d) We see that the solution blows up (divide by zero) when cos ql = 0. This condition occurs at the resonance frequencies, when ql = (2n + 1)π/2. (2n + 1) π d 1 ω Res =, 2L ρ n = 0, 1, 2,... (4.39a) (2n + 1) d 1 f Res =, 4L ρ n = 0, 1, 2,... (4.39b) Since the solution u is sinusoidal in time, we can find the amplitude and phase. u = u sin (ωt + δ) = u sin δ cos ωt + u cos δ sin ωt (4.40) u sin δ =c 1 e αx cos αx + c 2 e αx sin αx + c 3 e αx cos αx + c 4 e αx sin αx + g 1 cos qx + g 2 sin qx + g 3 cos qx + g 4 sin qx u cos δ =c 1 e αx sin αx c 2 e αx cos αx c 3 e αx sin αx + c 4 e αx cos αx + g 1 sin qx g 2 cos qx g 3 sin qx + g 4 cos qx (4.41) (4.42) The amplitude of the displacement versus angular frequency is shown if figure 4.1. The displacement at ω = 0 is a finite value, as shown in the graph, though it may seem at first glance that the displacement goes to infinity. This displacement was calculated for a monomorph of L = 0.05m, and applied voltage amplitude of 1V.

52 44 Figure 4.1: Displacement amplitude vs. frequency.

53 45 Damped Case As a step closer to reality we can add a damping term to the differential equation. This should prevent the solution from blowing up at the resonance frequencies. T 1 = d 1 S 1 + d 2 S 1 t + d 3E 3 (4.43) There are several choices for the form of the damping term, so it depends on the physics model. Here d 2 is a positive damping constant. Now we go through the entire process as for the undamped case. ρ 2 u ih t 2 2 u ih d 1 x d 3 u ih 2 2 x 2 t = d 3 E 3 x (4.44) c 1 c 2 c 3 c 4 = ρ 2 u h t d 2 u h 2 1 x d 3 u h 2 2 x 2 t = 0 (4.45) A (a 2 a 1 ) + B (a 1 + a 2 ) C A (a 1 + a 2 ) + B (a 2 a 1 ) A 2 + B 2 A (a 3 + a 4 ) + B (a 4 a 3 ) (4.46) A (a 4 a 3 ) B (a 3 + a 4 ) A = 2ωα 2 d 2 ρω 2, B = 2α 2 d 1, C = αd 3 The only difference for the form of the homogeneous solution is that each term is multiplied by an exponential in x. u h = g 1 h 1 (x, t) + g 2 h 2 (x, t) + g 3 h 3 (x, t) + g 4 h 4 (x, t) (4.47) h 1 = e bx cos (qx ωt), h 2 = e bx sin (qx ωt) h 3 = e bx cos (qx + ωt), h 4 = e bx sin (qx + ωt) Some derivatives of the h i which will be useful later are listed below. ḣ 1 = ωh 2 ḣ 2 = ωh 1 ḣ 3 = ωh 4 ḣ 4 = ωh 3 (4.48) (4.49a) (4.49b) (4.49c) (4.49d)

54 46 ḧ 1 = ω 2 h 1 ḧ 2 = ω 2 h 2 ḧ 3 = ω 2 h 3 ḧ 4 = ω 2 h 4 (4.50a) (4.50b) (4.50c) (4.50d) h 1 = bh 1 qh 2 h 2 = qh 1 bh 2 h 3 = bh 3 qh 4 h 4 = qh 3 + bh 4 (4.51a) (4.51b) (4.51c) (4.51d) h 1 = ( b 2 q 2) h 1 + 2bqh 2 (4.52a) h 2 = 2bqh 1 + ( b 2 q 2) h 2 (4.52b) h 3 = ( b 2 q 2) h 3 2bqh 4 (4.52c) h 4 = 2bqh 3 + ( b 2 q 2) h 4 (4.52d) ḣ 1 = ω (qh 1 bh 2 ) ḣ 2 = ω (bh 1 + qh 2 ) ḣ 3 = ω (qh 3 + bh 4 ) ḣ 4 = ω (bh 3 qh 4 ) ḣ 1 = ω ( 2bqh 1 + ( b 2 q 2) ) h 2 ( (b ḣ 2 = ω 2 q 2) ) h 1 + 2bqh 2 ḣ 3 = ω (2bqh 3 + ( b 2 q 2) ) h 4 ḣ 4 = ω ( (b 2 q 2) h 3 2bqh 4 ) (4.53a) (4.53b) (4.53c) (4.53d) (4.54a) (4.54b) (4.54c) (4.54d)

55 47 After putting our new form for u h through the damped homogeneous differential equation, we get the following conditions on the g i. [ ] [ ] [ ] F G g1 0 = G F g 2 0 [ ] [ ] [ ] F G g3 0 = G F 0 g 4 (4.55) F = ( q 2 b 2) d 1 ρω 2, G = 2bqd 1 ( q 2 b 2) ωd 2 For non-zero g i, the determinants of the coefficient matrices must be zero. F 2 + G 2 = 0 (4.56) The only way for this to be satisfied is for both F and G to both be zero. F = 0, G = 0 (4.57) This provides two equations to find b and q. bq = ρω 3 d 2 2 (d ω 2 d 2 2) q 2 b 2 = ρω2 d 1 d ω 2 d 2 2 ( ) b = ρω 2 2 d 1 ωd (d ω 2 d 2 2) ( ) q = ρω 2 2 d 1 ωd (d ω 2 d 2 2) d 1 d 1 1/2 1/2 (4.58) (4.59) (4.60) (4.61) If we set the damping constant, d 2, to zero, we get b = 0, and q reduces to the same value as for the undamped case. After applying the x = 0 boundary condition on u, we get the same condition on the g i as before. u (0, t) = 0 = (g 1 + g 3 + c 1 + c 3 ) cos (ωt) + (g 4 g 2 + c 4 c 2 ) sin (ωt) g 3 = (g 1 + c 1 + c 3 ), g 4 = g 2 + c 2 c 4 (4.62)

56 48 Now we apply the second boundary condition. u d 1 x x=l 2 u + d 2 x t x=l γ 1 f 1 (L, t) + γ 2 f 2 (L, t) + γ 3 f 3 (L, t) + γ 4 f 4 (L, t) + d 3 E 3 (L, t) = 0 (4.63) + [ ζ 2 g 1 + ζ 1 g 2 ] h1 (L, t) + [ ζ 1 g 1 + ζ 2 g 2 ] h2 (L, t) + [ ζ 2 g 1 + ζ 1 g 2 + γ 5 ] h3 (L, t) + [ ζ 1 g 1 ζ 2 g 2 + γ 6 ] h4 (L, t) (4.64) = 0 ζ 1 = d 1 q + d 2 ωb ζ 2 = d 2 ωq d 1 b (4.65a) (4.65b) γ 1 = d 1 α (c 2 c 1 ) + d 2 αω (c 1 + c 2 ) + d 3 a 1 γ 2 = d 1 α (c 1 + c 2 ) + d 2 αω (c 2 c 1 ) + d 3 a 2 γ 3 = d 1 α (c 3 + c 4 ) + d 2 αω (c 4 c 3 ) + d 3 a 3 γ 4 = d 1 α (c 4 c 3 ) d 2 αω (c 3 + c 4 ) + d 3 a 4 (4.66a) (4.66b) (4.66c) (4.66d) γ 5 = ζ 2 (c 1 + c 3 ) + ζ 1 (c 2 c 4 ) γ 6 = ζ 1 (c 1 + c 3 ) ζ 2 (c 2 c 4 ) (4.67a) (4.67b) We separate the f i (L, t) and the h i (L, t) into cos ωt and sin ωt terms, which provides two necessary conditions in order for equation (4.64) to be zero for all time. h 1 (L, t) = e bl cos (ql) cos (ωt) + e bl sin (ql) sin (ωt) h 2 (L, t) = e bl sin (ql) cos (ωt) e bl cos (ql) sin (ωt) h 3 (L, t) = e bl cos (ql) cos (ωt) e bl sin (ql) sin (ωt) h 4 (L, t) = e bl sin (ql) cos (ωt) + e bl cos (ql) sin (ωt) (4.68a) (4.68b) (4.68c) (4.68d)

57 49 [ M1 M 2 M 3 M 4 ] [ g1 g 2 ] [ K1 = K 2 ] (4.69) K 1 = γ 1 e αl cos αl + γ 2 e αl sin αl + γ 3 e αl cos αl + γ 4 e αl sin αl + γ 5 e bl cos ql + γ 6 e bl sin ql K 2 = γ 1 e αl sin αl γ 2 e αl cos αl γ 3 e αl sin αl + γ 4 e αl cos αl γ 5 e bl sin ql + γ 6 e bl cos ql (4.70a) (4.70b) M 1 = ζ 1 ( e bl e bl) sin ql + ζ 2 ( e bl + e bl) cos ql M 2 = ζ 1 ( e bl + e bl) cos ql + ζ 2 ( e bl e bl) sin ql M 3 = ζ 1 ( e bl + e bl) cos ql + ζ 2 ( e bl e bl) sin ql = M 2 M 4 = ζ 1 ( e bl e bl) sin ql ζ 2 ( e bl + e bl) cos ql = M 1 [ g1 g 2 [ ] [ ] [ ] M1 M 2 g1 K1 = M 2 M 1 g 2 K 2 ] [ ] [ ] 1 M1 M = 2 K1 M1 1 + M2 2 M 2 M 1 K 2 (4.71a) (4.71b) (4.71c) (4.71d) (4.72) (4.73) g 1 = M 1K 1 + M 2 K 2 M M 2 2 g 2 = M 1K 2 M 2 K 1 M M 2 2 (4.74a) (4.74b) g 3 = M 1K 1 + M 2 K 2 M M 2 2 c 1 c 3 (4.74c) g 4 = M 1K 2 M 2 K 1 M M c 2 c 4 (4.74d) Although the coefficients are a little bit messier, the solution for the damped case has pretty much the same form as for the undamped case. Once again we can find the amplitude and phase of u: u = u sin (ωt + δ) = u sin δ cos ωt + u cos δ sin ωt (4.75)

58 50 Figure 4.2: Displacement amplitude vs. frequency. u sin δ =c 1 e αx cos αx + c 2 e αx sin αx + c 3 e αx cos αx + c 4 e αx sin αx + g 1 e bx cos qx + g 2 e bx sin qx + g 3 e bx cos qx + g 4 e bx sin qx u cos δ =c 1 e αx sin αx c 2 e αx cos αx c 3 e αx sin αx + c 4 e αx cos αx + g 1 e bx sin qx g 2 e bx cos qx g 3 e bx sin qx + g 4 e bx cos qx (4.76) (4.77) 4.2. The amplitude of the displacement versus angular frequency is shown if figure

59 51 CHAPTER 5 CONCLUSIONS The voltage solution for the capacitor solution matches the form of the experimental measurements very well. Figure 5.1 shows an experimental measurement at 200 Hz [3]. In that plot the blue line is the voltage in the top electrode at x = L, and the red line is the voltage for the bottom electrode at x = L. Figure 5.2 shows the calculation from the capacitor model. In that plot the solid line represents the input voltage, and the dotted line represents the voltage in the top electrode at x = L. The results of the mechanical deformation calculations have not yet been verified experimentally, but the form seems reasonable. For future work I endeavor to solve the full coupled problem. Many people have solved the coupled piezoelectric problem for ideal electrodes (a uniform surface potential), but the problem of the coupled problem with an unkown electrode voltage distribution has yet to be solved. A method that has been widely employed to solve this type of problem is the finite element method, and that seems to me to be the best approach.

60 52 Figure 5.1: Experimental voltage measurement. Figure 5.2: Capacitor voltage solution

61 53 BIBLIOGRAPHY [1] T. Ikeda, Fundamentals of Piezoelectricity, Oxford University Press, New York, [2] J. I. Lorenz-Hallenberg, Application of poly(3,4-ethylenedioxytheophene)- poly(styrenesulfonate) to poly(vinylidene fluoride) as a Replacement for Traditional Electrodes, M.S. Thesis, Montana State University, August [3] J. Polasik and V. Hugo Schmidt, Conductive Polymer PEDOT/PSS Electrodes on the Piezoelectric Polymer PVDF, Proc. SPIE Vol. 5759, pp , Smart Structures and Materials 2005: Electroactive Polymer Actuators and Devices (EAPAD); Yoseph Bar-Cohen, Ed. [4] L. M. Lediaev and V. Hugo Schmidt, Modeling PVDF Actuators with Conducting Polymer Electrodes, Proc. SPIE Vol. 5759, pp , Smart Structures and Materials 2005: Electroactive Polymer Actuators and Devices (EAPAD); Yoseph Bar-Cohen, Ed.