1 / 33

PZN

PZN. Diffuse scattering and disorder in relaxor ferroelectrics. T.R.Welberry, D.J.Goossens. PbZn 1/3 Nb 2/3 O 3 , (PZN). computer disks. Relaxor ferroelectrics PbMg 1/3 Nb 2/3 O 3 (PMN) PbZn 1/3 Nb 2/3 O 3 (PZN). high dielectric constant

raheem
Télécharger la présentation

PZN

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. PZN Diffuse scattering and disorder in relaxor ferroelectrics. T.R.Welberry, D.J.Goossens PbZn1/3Nb2/3O3, (PZN)

  2. computer disks Relaxor ferroelectricsPbMg1/3Nb2/3O3 (PMN) PbZn1/3Nb2/3O3 (PZN) • high dielectric constant • dispersion over broad range of frequencies • and wide temperature range • evidence of polar nanostructure • plays essential role in piezo-electric properties • no consensus on exact nature of polar nanostructure

  3. [001] [110] Pb O Zn/Nb Perovskite structure important to seeoxygens use neutron scattering

  4. Neutrons vs X-rays • neutron flux on SXD at ISIS • ~ 6-7  104 neutrons per sec per mm2. • X-ray flux at 1-ID beamline at APS • ~ 1  1012 photons per sec per mm2. • is it possible to do neutron diffuse scattering at all?

  5. complete t.o.f. spectrum per pixel SXD instrument at ISIS 11 detectors 6464 pixels per detector

  6. angle subtended by 90detector bank volume of reciprocal space recorded simultaneously with one detector bank. neutron time of flight geometry A-A’ and B-B’ given by detector bank B-A and B’-A’ given by time-of-flight

  7. (h k 0) apply m3m symmetry 10 crystal settings 8 detectors (h k 0.5) (h k 1) PZN diffuse scattering nb. full 3D volume

  8. 5 5 4 3 3 2 1 1 h k 0 h k 1 h k 0.5 diffraction features • diffuse lines are in fact rods not planes • azimuthal variation of intensity - displacement along <1 1 0> • all rods present in hk0 but only oddnumbered rods in hk1 • only half of spots in h k 0.5 explained by intersection of rods

  9. a rod of scattering in reciprocal space corresponds to a plane in real-space (normal to the rod) rods are parallel to the six <110> directions hence planes are normal to <110> Fourier transform theory in this case: azimuthal variation of intensity means: atomic displacements are within these planes and parallel to another <110> direction

  10. Planar defects in PZN cation displacements in planar defect are parallel to [1 1 0] Planar defect normal to [1 -1 0]

  11. Simple MC model atoms connected by springs and allowed to vibrate at given kT most successful model had force constants in ratios:- Pb-O : Nb-O : O-O : Pb-Nb 5 : 5 : 2 : 80

  12. h k 0 h k 1 h k 0.5 odd even Simple MC model Observed patterns Calculated patterns

  13. Bond valence

  14. 8,9 8,9 2,3 2,3 6 6 12 12 1 1 4,5 4,5 10,11 10,11 Bond valence Pb atoms are grossly under-bonded in average polyhedron Pb shift along [110] achieves correct valence

  15. lone-pair electrons PZN Cations displaced from centre of coordination polyhedra

  16. NbO6 octahedron Bond valence requires a = 3.955Å for Nb valence of 5.0 ZnO6 octahedron Bond valence requires a = 4.218Å for Zn valence of 2.0 Bond valence - Nb/Zn order PZN measured cell a = 4.073Å Weighted mean (2*3.955+4.218)/3 a = 4.043Å Weighted mean (3.955+4.218)/2 a = 4.087Å Strong tendency to alternate but because of 2/3 : 1/3 stoichiometry cannot be perfect alternation

  17. Peaks due to cation displacements maximal Nb/Zn ordering random Nb/Zn0 (h k 0.5) layer Extra peaks due to Nb/Zn ordering SRO of Nb/Zn • Two models tested:- • random occupancy of Nb and Zn ? • tendency to alternate? • B-site occupancy is 2/3Nb and 1/3Zn • complete alternation not possible - max corr. = -0.5 • Nb certainly follows Zn but • after Nb sometimes Zn sometimes Nb

  18. Planar defects cation displacements in planar defect are parallel to [1 1 0] random variables to represent cation displacements

  19. Displacements refer to cation displacements in a single <110> plane modeling cation displacements Monte Carlo energy random variables to represent cation displacements Total model consists of cation displacements obtained from summing the variables from the six different <110> orientations

  20. Model 1 O1 moves in phase with Pb’s Model 1 O1 moves in phase with Pb’s Model 2 O1 moves out of phase with Pb’s displacement models

  21. 5 5 4 3 3 2 1 1 comparison of models 1 and 2 1 2

  22. h k 0 h k 1 h k 0.5 random variable model obs v. calc Observed patterns Calculated patterns

  23. Summary of Gaussian Variable models planar nanodomains normal to <110> atomic displacements parallel to <110> atomic displacements within domains correlated Pb & Nb/Zn displacements in phase O1 displacements out of phase with Pb can we construct an atomistic model satisfying these criteria?

  24. E1 E2 atomistic model • assume all Pb’s displaced in 1 of 12 different ways • assume in any {110} plane Pb displacements correlated • assume no correlation with planes above and below MC energy

  25. [001] Polar nanodomains 12 different orientations [110] E1 E2 development of atomistic model Single layer normal to [1 -1 0] diffraction Pb only Note scattering around Bragg peaks as well as diffuse rods

  26. development of atomistic model two successive planes normal to [1 -1 0] Polar nanodomains 12 different orientations [001] domains do not persist in successive layers [110]

  27. development of atomistic model view down [0 0 1] [100] Linear features do persist in successive layers [010]

  28. neighbours attract or repel each other according to their mutual orientation development of atomistic model [100] Linear features do persist in successive layers [010]

  29. P [110].[110] = 2 smaller than average E = (d - d0(1 - P e))2 [110].[101] = 1 [110].[1 -1 0] = 0 size-effect parameter average [110].[-1 0 -1] =-1 [110].[-1 -1 0] =-2 bigger than average size-effect relaxation

  30. Size-effect relaxation e = 0 e = -0.02 e = +0.020 observed (h k 0)

  31. thick domains i.e. 3D double layer 2D domains Other models

  32. Acknowledgements • M.J.Gutmann (ISIS, UK) • A.P.Heerdegen(RSC, ANU) • H. Woo (Brookhaven N.L.) • G. Xu (Brookhaven N.L.) • C. Stock (Toronto) • Z-G. Ye (Simon Fraser University) • AINSE { Crystal growth}

  33. Go back to Disordered MaterialsGo to Home Page

More Related