[发明专利]一种获得宽温域高介电常数的三弛豫态铁电陶瓷的方法

权利要求书

1.一种获得宽温域高介电常数的三弛豫态铁电陶瓷的方法，包括以下步骤：

S100、确定掺杂了离子的钛酸钡体系为(Ba_1-yCa_y)(Ti_1-xSn_x)O₃，其中，x、y分别指的是BaSnO₃、CaTiO₃所占(Ba_1-yCa_y)(Ti_1-xSn_x)O₃整体的物质的量的百分比，0≤x≤0.2，y＝0.1或0.22；

S200、按掺杂离子百分比从小到大的顺序设计钛酸钡系陶瓷材料样品；

S300、制备步骤S200设计的钛酸钡系陶瓷材料样品并进行Curie-Weiss拟合；

S400、对所述钛酸钡系陶瓷材料样品进行分析，找到钛酸钡系陶瓷材料样品的单斜相、正交相和菱形相的多相共存区域，当有钛酸钡系陶瓷材料样品同时具备弛豫特性和多相共存区域，此时对应弛豫相变点和三临界点的钛酸钡系陶瓷材料组成成分即为三弛豫态铁电陶瓷材料。

2.如权利要求1所述的方法，其特征在于，步骤S200中，所述钛酸钡系陶瓷材料按掺杂离子百分比从小到大的顺序为：

(Ba_0.9Ca_0.1)TiO₃、(Ba_0.9Ca_0.1)(Ti_0.96Sn_0.04)O₃、(Ba_0.9Ca_0.1)(Ti_0.92Sn_0.08)O₃、(Ba_0.9Ca_0.1)(Ti_0.9Sn_0.1)O₃、(Ba_0.9Ca_0.1)(Ti_0.89Sn_0.11)O₃、(Ba_0.9Ca_0.1)(Ti_0.88Sn_0.12)O₃、(Ba_0.9Ca_0.1)(Ti_0.87Sn_0.13)O₃、(Ba_0.9Ca_0.1)(Ti_0.84Sn_0.16)O₃、(Ba_0.9Ca_0.1)(Ti_0.81Sn_0.19)O₃、(Ba_0.78Ca_0.22)TiO₃、(Ba_0.78Ca_0.22)(Ti_0.98Sn_0.02)O₃、(Ba_0.78Ca_0.22)(Ti_0.96Sn_0.04)O₃、(Ba_0.78Ca_0.22)(Ti_0.92Sn_0.08)O₃、(Ba_0.78Ca_0.22)(Ti_0.88Sn_0.12)O₃、(Ba_0.78Ca_0.22)(Ti_0.97Sn_0.13)O₃、(Ba_0.78Ca_0.22)(Ti_0.84Sn_0.16)O₃、(Ba_0.78Ca_0.22)(Ti_0.81Sn_0.19)O₃。