Modelling of piezoelectric transducers and applications

Transcription

1 st International Workshop on Smart Materials and Structures, University of Applied Science, Kiel, October 7-8, 24 Modelling of piezoelectric transducers and applications Morten Willatzen, Niels Lervad Andersen, and Yuyang Feng Mads Clausen Institute for Product Innovation University of Southern Denmark Grundtvigs Alle 5 DK-64 Sønderborg Denmark

2 OUTLINE D piezoelectric constitutive relations Transducer dynamic equations Solving transducer equations by use of Fourier analysis Numerical results and measurements i

3 D PIEZOELECTRIC CONSTITUTIVE RELATIONS D transducer constitutive equations: D = dt + ɛ T E E = β T D gt D = ɛ S E + es E = β S D hs S = de + s E T T = ee + c E S S = gd + s D T T = hd + c D S

4 TRANSDUCER DYNAMIC EQUATIONS Newton s 2nd Law: ρadz u t = (F z F z+dz ) = AT z + AT z+dz = A T z dz ρ u t = T z 2

5 Definition of strain: S = ζ z ζ: Displacement z: Position coordinate u z = ( ) ζ t z = ζ t z = S t Poisson s equation: D z = ρ free = ρ free : external or free charge density 3

6 ρ u t = T z = cd S z using: u z = S t Voltage over transmitter: Differentiate in time: using T = c D S hd V (t) = Z e I + V t = AZ 2 D e t + l 2 ɛ S D z = z z L E(z, t)dz D t h [u(z ) u(z L )] I = A D t E = D ɛ S hs D z = u z = S t 4

7 Separate equations in u and S: 2 u t 2 u 2 v2 a z = 2 2 S t 2 S 2 v2 a z = 2 where v a = c D ρ (sound speed) 5

8 SOLVING TRANSDUCER EQUATIONS BY USE OF FOURIER ANALYSIS V (t) = 2π V (ω) exp( iωt)dω V (ω) = 2π V (t) exp(iωt)dt. Solving dynamic equations for each frequency component of V (t): at a time: 2π V (ω) exp( iωt)dω u + v2 a ω 2 2 u z 2 = gives: S + v2 a ω 2 2 S z 2 = u = u A exp( ikz iωt) + u B exp(ikz iωt) S = S A exp( ikz iωt) + S B exp(ikz iωt). 6

9 Stress and displacement: T = c D S hd = c D [S A exp( ikz iωt) + S B exp(ikz iωt)] hd where: u = cd S A Z a exp( ikz iωt) cd S B Z a exp(ikz iωt) Z a = ρv a (Acoustic impedance) k = ω v a (Wave vector) u A = cd S A Z a u B = cd S B Z a (Displacement amp. - backwards prop. waves) (Displacement amp. - forwards prop. waves) Voltage equation: iωv = AZ e ω 2 D iω l ɛ SD h [u(z ) u(z L )] 7

10 lpz lac lma SB SB2 SB3 Air TL Layer Layer 2 Layer 3 SA SA2 SA3 TR Water ZL=Z Z Z2 ZR=Z3 Layer : Piezoceramic material Layer 2: Acoustic coupling layer Layer 3: Matching layer 8

11 9 unknowns: D, T L, T R, S Ai, S Bi, i =, 2, 3 Continuity of T and u at interfaces: T L = c D (S A + S B ) h D T L Z al = c D ( SA S ) B Z a Z a c D 2 (S A2 + S B2 ) = c D (S A exp( ik l ) + S B exp(ik l )) h D c D 2 ( SA2 S ) B 2 Z a2 Z a2 = c D ( SA exp( ik l ) S ) B exp(ik l ) Z a Z a c D 3 (S A 3 + S B3 ) = c D 2 (S A 2 exp( ik 2 l 2 ) + S B2 exp(ik 2 l 2 )) c D 3 ( SA3 S ) B 3 Z a3 Z a3 = c D 2 ( SA2 exp( ik 2 l 2 ) S ) B 2 exp(ik 2 l 2 ) Z a2 Z a2 T R Z ar = c D 3 T R = c D 3 (S A 3 exp( ik 3 l 3 ) + S B3 exp(ik 3 l 3 )) ( SA3 exp( ik 3 l 3 ) S ) B 3 exp(ik 3 l 3 ) Z a3 Z a3 iωv = AZ e ω 2 D iω l ɛ S D h [c D 2 ( c D SA S )] B Z a Z a ( SA2 S ) B 2 Z a2 Z a2 9

12 Solving this 9 9 matrix problem for T determines the aperture velocity: u R = T R Z ar Using the Rayleigh equation gives the incoming pressure wave at the receiving transducer: P rec ( x, ω, t) = iωρ 2π A tra da tra u R (t x x /v ar) x x For a D transducer we assume that the receiver vibrates according to the mean pressure over the receiver aperture: P rec (ω, t) = A rec A rec P rec ( x, ω, t)da rec

13 lma lac lpz Ti Water Tb SD SD2 SD3 Layer Layer 2 Layer 3 SC SC2 SC3 Tf Air ZL=Z Z Z2 ZR=Z3 Layer : Matching layer Layer 2: Acoustic coupling layer Layer 3: Piezoceramic material

14 Input to the receiver: T i = T rec (ω, t) = P rec (ω, t) The corresponding 9 equations for the receiver become: T i + T b = c D (S C + S D ) T i Z al + T b Z al = c D ( SC S ) D Z a Z a c D 2 (S C2 + S D2 ) = c D (S C exp( ik l ) + S D exp(ik l )) c D 2 ( SC2 S ) D 2 Z a2 Z a2 = c D ( SC exp( ik l ) S ) D exp(ik l ) Z a Z a c D 3 (S C3 + S D3 ) h 3 D = c D 2 (S C2 exp( ik 2 l 2 ) + S D2 exp(ik 2 l 2 )) c D 3 ( SC3 S ) D 3 Z a3 Z a3 = c D 2 ( SC2 exp( ik 2 l 2 ) S ) D 2 exp(ik 2 l 2 ) Z a2 Z a2 T f = c D 3 (S C 3 exp( ik 3 l 3 ) + S D3 exp(ik 3 l 3 )) h 3 D T f Z ar = c D 3 ( SC3 exp( ik 3 l 3 ) S ) D 3 exp(ik 3 l 3 ) Z a3 Z a3 = AZ e2 ω 2 D iω l [ 3 ɛ S D h 3 T f c D 3 Z ar ( SC3 S )] D 3 Z a3 Z a3 2

15 The time-dependent analogue to the last of the equations above is: = AZ e2 2 D t 2 + l 3 ɛ S 3 D t h [u(z R) u(z 2 )] Solving the receiver matrix equations for the electric displacement D, the induced voltage becomes: V rec(ω, t)dω = Z 2 I = iz 2 AωD Total contribution to the received voltage pulse: V rec (t) = V rec(ω, t)dω 3

16 NUMERICAL RESULTS AND MEASUREMENTS 5 5 volt. [V] time [secs] x 6 2 x 5 Im(g(w)) [sec V] frequency [Hz] x 6 4

17 8 6 4 Transmitter. Volt. [V] Time [secs] x Transm. Volt. [V] Time [secs] x 5 Transmitter voltages (a) calculated and (b) measured, 75 Ohm resistor. 5

18 6 4 2 Rec. Volt. (normalized) Time [secs] x Rec. Volt. (normalized) Time [secs] x Rec. Volt. (normalized) Time [secs] x 5 Piezoceramic layer and a steel matching layer, no bonding. (a) Calc. Z = Ohm, Z 2 = Ohm, (b) Calc. Z = Z 2 = 75 Ohm, (c) Meas. Z = Z 2 = 75 Ohm 6

19 .5.5 Rec. Volt. (normalized) Time [secs] x 5 Pure piezoceramic layer, 75 Ohm resistor. 7

20 .5.5 Rec. Volt. (normalized) Time [secs] x Rec. Volt. (normalized) Time [secs] x 5 Bonding: 5 microns (a) grease and (b) glue. Steel matching layer, 75 Ohm resistor. 8

21 .5.5 Rec. Volt. (normalized) Time [secs] x Rec. Volt. (normalized) Time [secs] x 5 Bonding: 5 microns (a) grease and (b) glue. Steel matching layer, 75 Ohm resistor. 9

22 received voltage arbitrary units diamond: measured with grease dashed: epoxy glue.6 solid: grease acoustic coupling layer thickness [m] x 5 75 Ohm resistor. 2

23 4 3 2 receiver voltage [V] time [secs] x receiver voltage [V] time [secs] x 5 Received voltages, 3 cm transmission distance. (a) Pure piezoceramic transducer. (b) PPS quarter-wave matching layer transducer. 2

24 4 3 2 receiver voltage [V] time [secs] x receiver current [A] time [secs] x 5 3 cm transmission distance. (a) Steel half-wavelength matching layer transducer (rec. voltage). (b) Steel half-wavelength matching layer transducer (rec. current). 22

25 4 3 2 receiver voltage [V] time [secs] x receiver current [A] time [secs] x 5 3 cm transmission distance. (a) Steel half-wavelength matching layer transducer (rec. voltage) with parallel inductance: L par = ω 2 C. (b) Steel half-wavelength matching layer transducer (rec. current) with parallel inductance: L par = ω 2 C. 23

26 3 x 4 2 incoming pressure receiver [Pa] time [secs] x 5 3 x 4 2 incoming pressure receiver [Pa] time [secs] x 4 Receiver mean pressure pulses, steel matching layer, no parallel inductance. (a) 2 cm transmission distance. (b) 6 cm transmission distance. 24

27 4 3 2 receiver voltage [V] time [secs] x receiver voltage [V] time [secs] x 4 Receiver voltage pulses, steel matching layer, no parallel inductance (a) 2 cm transmission distance. (b) 6 cm transmission distance. 25

28 Parameter Value Unit ω /(2π) (driving frequency) 6 Hz N (number of periods) 8 V (applied voltage amplitude). V Thickness (piezoceramic material).96 mm Transducer area 38 mm 2 Thickness (stainless steel) 2.8 mm h (piezoceramic material).42 9 F 2 V/m 3 ɛ S (piezoceramic material) 44ɛ F/m c D (piezoceramic material).9 Pa c D (stainless steel) 2.53 Pa c D (pps).58 Pa c D (water) Pa c D (air).24 5 Pa ρ (piezoceramic material) 775 kg/m 3 ρ (stainless steel) 8 kg/m 3 ρ (pps) 6 kg/m 3 ρ (water) kg/m 3 ρ (air).29 kg/m 3 α (piezoceramic material, MHz). m α (stainless steel, MHz).26 m α (pps, MHz) 9. m α (water, MHz) 25 3 m tan δ (piezoceramic material)