Lecture 8-2 - University of Pittsburghqiw4/Academic/MEMS1082/Lecture 8-2.pdf · Piezoelectricity...

Department of Mechanical Engineering

Piezoelectricity and Pyroelectricity

Lecture 8-2


Direct Effect

Piezoelectricity

(Electric polarization)=(Piezoelectric coefficients) (Mechanical stress)

σdP =

[C/m2]First rank vector

[C/N]Third rank tensor

[N/m2]Second rank tensor

Piezoelectric coefficients constitute a third rank polar tensor

(Mechanical strain)=(Piezoelectric coefficients) (Electric field)

[ ]

First rank vector

[m/V]

Third rank tensor

[V/m]

Second rank tensor

Converse Effect dE=ε


Direct Effect

Piezoelectricity Converse Effect

( ) ( ) ( )Ed t=ε( ) ( )( )σdP =3x1 3x6 6x1 6x1 3x16x3

Matrix form

jkijki dP σ= iijkjk Ed=εTensor form

Units:Vm

VCmN

NC

JJ

NC

NC

=

⋅⋅

=

=

Thermodynamic Proof

TdSdPEddU kkijij ++= εσFirst Law

Electric work inside materials


PiezoelectricityFree Energy

ijijkk PETSUG εσ+−−=

SdTdEPddG kkijij −−−= σε

G is function of state because system completely specified by σ, E, and T

SdTTGdE

EGdGdG

Eijk

Tkij

TEij ,,, σσ

σσ

∂∂

+

∂∂

+

∂∂

=

ijij

G εσ

−=∂∂

kk

PEG

−=∂∂ S

TG

−=∂∂

and


Various Transduction effects

Piezoelectricity

k

ij

kij EEG

∂

∂−=

∂∂∂ ε

σ

2

ij

k

ijk

PE

Gσσ ∂

∂−=

∂∂∂

=2

Converse piezoelectric effect

Direct piezoelectric effect

TTG ij

ij ∂

∂−=

∂∂∂ εσ

2

ijij

ST

Gσσ ∂∂

−=∂∂

∂=

2

Thermal expansion coefficient

Piezo-caloric effect

TP

TEG k

k ∂∂

−=∂∂

∂ 2

kk ES

ETG

∂∂

−=∂∂

∂=

2

Pyroelectric Coefficient

Electro-caloric effect


3-subscript Tensor Notation

Tensor and Matrix Coefficients

Converse Effect

jkijki dP σ=

iijkjk Ed=ε

Direct effect

331333213231131

231232212221121

1311312112111111

σσσσσσσσσ

dddddddddP

++++++=

For example:

2-subscript Matrix Notation

jiji dP σ= i=1,2,3j=1,2,3,4,5,6

6365354343332321313

6265254243232221212

6165154143132121111

σσσσσσσσσσσσ

σσσσσσ

ddddddPddddddP

ddddddP

+++++=+++++=

+++++=

iijj Ed=ε

3362261166

3352251155

3342241144

3332231313

3322221122

3312211111

EdEdEdEdEdEdEdEdEdEdEdEdEdEdEdEdEdEd

++=++=++=++=++=++=

εεεεεε

i, j, k=1, 2, 3


Tensor and Matrix Coefficients

Identify piezoelectric coefficients

13313

12212

11111

dddddd

===

11212111216

11311313115

12313212314

222

dddddddddddd

=+==+==+=

Identify equivalent stresses

432235311362112

333222111

,,,,

σσσσσσσσσσσσσσσ

=========

………………………

………………………………


For first-, second-, third- and fourth-rank polar tensors, the well-known equations of transformation from an orthogonal x1, x2, x3 system to another similarly orthogonal x’1, x’2, x’3 system are according to their definition

Transformation of Axes

jiji TaT =′

klklikij TaaT =′

lmnknjmilijk TaaaT =′

mnoplpkojnimijkl TaaaaT =′

where the aij direction cosines are the elements of the (aij) matrix. The (aij) matrix connects the original and the `new' co-ordinates according to the matrix equation



=

′′′

3

2

1

333231

232221

131211

3

2

1

xxx

aaaaaaaaa

xxx

Tensor form ∑ ==′j

jijjiji xaxaxEinstein convention

From new coordinate to old

=

′′′

3

2

1

333231

232221

131211

3

2

1

xxx

aaaaaaaaa

xxx

( ) ( ) ( )xax t ′=or

( ) ( )( )xax =′

or

transpose

( )a : Direction cosine matrix

Tensor form ∑ ′=′=j

jijjjii xaxax



Rotated Axes and angles

θφθφθϕθ

ϕφθϕφϕφϕφθ

ϕθϕφϕφθϕφϕφθ

cossinsinsinsinsinsin

sinsincoscoscoscossinsincoscos

cossinsincoscossincossinsincoscoscos

33

32

31

23

22

21

13

12

11

====

−=−−=

−=+=−=

aaaaaaaaa



First rotation φ +π/2 about X3

( )

−=

1000sincos0cossin

φφφφ

Ia

Second rotation θ about X’1

( )

−=

θθθθ

cossin0sincos0001

IIa

Third rotation π/2−θ about X”3

( )

−=

1000coscos0cossin

ϕϕϕϕ

IIIa



General rotation is product of three individual rotation

( ) ( ) ( ) ( )IIIIII aaaa =

( )

−

−

−=

1000sincos0cossin

cossin0sincos0001

1000coscos0cossin

φφφφ

θθθθϕϕ

ϕϕa

( )

−−−

−+−=

θφθφθϕθϕθφφθϕφϕφθ

ϕθϕφϕφθϕφϕφθ

cossinsincossinsinsinsincossincoscoscossinsincoscos

cossinsincoscossincossinsinsincoscosa


Symmetry Axes Parallel to X3

2-fold φ=180o Orthorhombic(Tetragonal& cubic)(Hexagonal)

( )

+−

−=

100010001

a

3-fold φ=120o

( )

+−

−−

=10002/123

0232/1

a

Trigonal(Hexagonal)

4-fold φ=90oTetragonal& cubic

( )

++

−=

100001010

a

6-fold φ=60o( )

+

−

=100

02/123

0232/1

aHexagonal


Mirror plane perpendicular to X2, and Center of symmetry=inversion center Monoclinic, orthorhombic,

tetragonal, hexagonal, cubic

( )

+−

+=

100010001

a

Triclinic, monoclinic, orthorhombic, trigonal, tetragonal, hexagonal, cubic

Mirror plane

( )

−−

−=

100010001

a

Center of symmetry


Pyroelectric tensor (Change in polarization)=(Pyroelectric coefficients)(Change in temperature)

A vectorA scalar

TpP i ∆=∆pi is pyroelectric coefficient

jiji PaP ∆=′∆ TT ∆=′∆

How does p transform?

TpaTpaPaP jijjijjiji ′∆=∆=∆=′∆new old old new

All new TpTpaP ijiji ′∆′=′∆=′∆

jiji pap =′Pyroelectricity is a first rank polar tensor (vector) property


Effect of symmetry on p

3x1 3x3 3x1

Matrix form

First rankTransformation matrix of symmetryElement of group

( ) ( )( ) ( )ppap ==′3x1

Neumann’s Principle:Property unchanged aftertransformation

Centro-symmetric crystals have inversion symmetry

( )

≡

−−−

=

−−

−=

′′′

=′

3

2

1

3

2

1

3

2

1

3

2

1

100010001

ppp

ppp

ppp

ppp

p

0321 === ppp (No pyroelectricity)

Pyroelectricity is absent in all centro-symmetric crystals (a null property)


Poled ferroelectric ceramic

Symmetry group=m//X3

-fold axis//X3 ( )

−=

1000cossin0sincos

θθθθ

a

For all θ( )

≡

+−

=

−=

′′′

=′

3

2

1

3

21

21

3

2

1

3

2

1

cossinsincos

1000cossin0sincos

ppp

ppppp

ppp

ppp

p

θθθθ

θθθθ

If θ=90o 021 == pp

03 ≠p

( )

=

3

00

pp


Direct Effect

Piezoelectricity

(Electric polarization)=(Piezoelectric coefficients) (Mechanical stress)σdP =

[C/m2]First rank vector

[C/N]

Third rank tensor

[N/m2]

Second rank tensor

kljklijjij'

i daPaP σ=='mnnlmkjklij

'i aadaP σ=

'mn

'imn

'i dP σ= jklnlmkij

'imn daaad =

Piezoelectric coefficients constitute a third rank polar tensor

Tensor form transformation


Piezoelectric-Matrix Transformations

Stress-symmetric second rank tensors

kljlikij aa σσ =′ Equivalent tensor forms

How does 6x1 matrix transform?( ) ( )( )σασ =

3x3 3x33x3 3x3( ) ( )( )( )taa σσ =′

What is (α)?

To determine (α), write out tensor and matrix forms and equate term by term

.......................222122111121212111

21111 ++++=′ σσσσσ aaaaaa

6165154143132121111 σασασασασασασ +++++=′

21111 a=α

21212 a=α 121116 2 aa=α

Other α coefficients can be determined in similar way


α Transformation

( )

++++++

+++=

211222112311211322132312231322122111

311232113311311332133312133312321131

322131223123332132233322332332223121

323131333332233

232

231

222121232322223

222

221

121111131312213

212

211

222

222

222

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

α

( )

+++++++++

=

122122113211123122313221323122211211

132123113311133132313321313321231113

231213221332331223323322333223221312

231313333323233

223

213

221212323222232

222

212

211111313121231

221

211

222

222

222

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

aaaaaaaaaaaaaaaaaaaaaaaaaaa

tα


Stress

Piezoelectric-Matrix Transformations

( ) ( )( )σασ =′6x1 6x6 6x1

new old

( ) ( ) ( ) ( )( ) ( )σσαασα ==′ −− 11

Polarization

new old( ) ( )( )PaP =′ ( ) ( ) ( ) ( )( ) ( )PPaaPa ==′ −− 113x1 3x3 3x1

Piezoelectricity ( ) ( )( ) ( )( )( ) ( )( )( ) ( )( )( )σ

σασ′′=

′===′ −

ddadaPaP 1

( ) ( )( )( ) 1−=′ αdad3x6 3x3 3x6 6x6

new old

( ) ( )( ) ( ) ( )( )( ) ( ) ( )ddaada ==′ −−− ααα 111

( ) ( ) ( )( )αdad ′= −1

old new


Matrix Method: Point Group 2

Sucrose, 2//X2( )

−+

−=

100010001

a

( ) ( )( )( ) 1−=′ αdad

( )

−

−

−+

−=′

100000010000001000000100000010000001

100010001

363534333231

262524232221

161514131211

dddddddddddd

ddddddd

( ) ( )ddddddddddddddddddd

d =

+−+−−−−+−+++

−−−−=′

363534333231

262524232221

161514131211

( )

=

3634

25232221

1614

000000

0000

dddddd

ddd

8 independent piezoelectric coefficients


Piezoelectric ceramics

Point group =m Direct inspection method

Many mirror planes//X3

Mirror X1 1 -1 2 2 3 3

111 111 112 +122 Eliminate all coefficients with odd No. of 1’s

0331321313312221

231213212133132123122111

=============

ddddddddddddd

Mirror X2 2 -2 1 1 3 3Eliminate all coefficients with odd No. of 2’s

0332323121112233211222 ======= ddddddd

Mirror at 45o to X1 and X2 1 2 2 1 3 3

322311 dd = 223113232131 dddd ===



d matrix for group =m

( )

=

0000000000000

333131

15

15

dddd

dd

For PZT, BaTiO3, PbTiO3, etc

Lecture 8-2 - University of Pittsburghqiw4/Academic/MEMS1082/Lecture 8-2.pdf · Piezoelectricity...

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Transcript of Lecture 8-2 - University of Pittsburghqiw4/Academic/MEMS1082/Lecture 8-2.pdf · Piezoelectricity...