Why Sn doping significantly enhances the dielectric properties in
Ba(Ti1-xSnx)O3
Tao Shi1, Lin Xie1, Lin Gu2,Jing Zhu*1
1. Beijing National Center for Electron Microscopy, School of Materials Science and
Engineering, The State Key Laboratory of New Ceramics and Fine Processing, Laboratory of
Advanced Materials , Tsinghua University, Beijing 100084, People’s Republic of China
2.
Beijing National Laboratory for Condensed Matter Physics, Institute of Physics, Chinese Academy
of Sciences, Beijing 100190, China
The ab initio calculation constants
The first-principle calculation was carried out based on a 3×3×3 supercell including 27 barium
atoms, 81 oxygen atoms, 21 titanium atoms and 6 tin atoms, while the concentration of Sn is about
22.2%. The 5s, 5p of Ba, 3s, 3p of Ti, 4d of Sn are used in the valence as semi-core states. An
energy cutoff of 600eV of the plane wave basis and a sampling of the Brillouin zone with 6×6×6
Monkhorst-Pack mesh were used. The muffin-tin radii of Ba, O, Ti, and Sn were 2.5, 1.61, 1.86
and 2.22 a.u..
The calculation of the displacements in HAADF images.
A standard peak finding algorithm based on fitting two-dimensional Gaussian functions to the
intensity maxima is employed1,2 for the HAADF image and we can get the position and brightness
of each column. Based on these data, we can calculate the off-center ion displacements between
Ti/Sn column and the center of the unit cell which is determined by the average coordinate of the
four barium columns’, according to the following formula
𝟏
𝒖𝒊𝒋 = 𝒓𝒊𝒋 − 𝟒 (𝑹𝒊𝒋 + 𝑹(𝒊+𝟏)𝒋 + 𝑹𝒊(𝒋+𝟏) + 𝑹(𝒊+𝟏)(𝒋+𝟏) )
(1)
where i/j indicate the row/column number of each atom column, rij indicates the position of Ti/Sn
atom column, while Rij indicates the position of barium atom column.
The calculation of Sn content
Since the brightness of the HAADF image is proportional to Z1.7 and the atomic number of Sn (50)
is much larger than Ti (22), a higher brightness means there are more Sn atoms in the column.
Semi-quantitatively, we assume that
𝑩𝒊 = 𝒄 ∙ 𝒁𝟏.𝟕 = 𝐜 ∙ [𝐱 𝒊 ∙ 𝟓𝟎𝟏.𝟕 + (𝟏 − 𝐱 𝒊 ) ∙ 𝟐𝟐𝟏.𝟕 ]
(2)
Where Bi indicates the brightness of column i, xi indicates the Sn content and c represents a
constant which is influenced by instrument constant, thickness, tilt and other physical factors.
In our case, the uniform contrast of the HAADF image indicates that no abrupt changes exist in
the regions and we can simply consider c to be a constant for all columns. So we can normalize all
the value of brightness into
𝑩𝒊
𝑩𝟎
𝐱 ∙𝟓𝟎𝟏.𝟕 +(𝟏−𝐱 )∙𝟐𝟐𝟏.𝟕
= 𝐱 𝒊∙𝟓𝟎𝟏.𝟕 +(𝟏−𝐱𝒊 )∙𝟐𝟐𝟏.𝟕
𝟎
𝟎
(3)
Where B0 indicates the average brightness in a large region, x0 indicates the average content of Sn
which should be 0.2 according to the chemical formula. By this equation, we can calculate the Sn
content of each column through its brightness semi-quantitatively.
The result of X-ray diffraction experiment.
Figure 1 The X-ray diffraction spectrum of Ba(Ti0.8Sn0.2)O3 comparing with pure T-phase BaTiO3
ceramics. It is obvious that the angle of the corresponding peak of Ba(Ti0.8Sn0.2)O3 is smaller than
that of the pure BaTiO3, which means the lattice constant of Ba(Ti0.8Sn0.2)O3 is larger than that of
pure BaTiO3. Similar result has been reported by Wei, et al3 and the lattice constant of the
Ba(Ti0.8Sn0.2)O3 ceramics increases with the tin content by a linear relationship.
References:
[1] Jia C. L., Lentzen M. & Urban K. Atomic-resolution imaging of oxygen in perovskite
ceramics. Science. 299, 870.(2003)
[2] Krivanek O. L. et al. Atom-by-atom structural and chemical analysis by annular dark-field
electron microscopy. Nature 464. 571(2010).
[3] Wei X. & Yao X. Preparation, structure and dielectric property of barium stannate titanate
ceramics. Mater. Sci. Eng. B. 137, 184(2007).