摘要
采用传统的固相法制备了(1x)(K0.5Na0.5NbO3-LiSbO3-BiFeO3)-xCuFe2O4(x=0.1,0.2,0.3,0.4)磁电复合陶瓷,并借助X射线衍射仪、扫描电镜和磁电耦合系数测试仪等对复合陶瓷的微结构和性能进行了分析.结果表明,复合陶瓷的K0.5Na0.5NbO3-LiSbO3-BiFeO3和CuFe2O4物相之间发生了一定的离子相互扩散作用,且两相的颗粒大小匹配性较好.随着CuFe2O4含量增加,复合陶瓷的压电系数从130pC/N减小到30pC/N,饱和磁致伸缩系数从4.5×10-6增加到12.4×10-6左右,磁电耦合系数表现出先增加后减小,在x=0.3时获得最大的磁电耦合系数9.4mV·cm-1·Oe-1.
The （1-x）（K0.5Na0.5NbO3-LiSbO3-BiFeO3）-xCuFe2O4 （x=0.1, 0.2, 0.3 and 0.4） magnetoelectric composite ceramics are prepared by the conventional solid-state reaction method. The microstructures and properties of the composite ceramics are characterized by X-ray diffractometer, scanning electron microscope and magnetoelectric coupling coefficient meter. The weak ionic interdiffusions between the phases K0.5Na0.5NbO3-LiSbO3-BiFeO3 and CuFe2O4 are observed and their particle sizes are well matched between each other. With the increase of CuFe2O4 content, the piezoelectric coefficient （d33） of the composite ceramics decreases from 130 pC/N to 30 pC/N and the magnetostriction coefficient （-λ） increases from 4.5×10-6 to 12.4×10-6. The magnetoelectric coupling coefficient （αE） of the composite ceramics first increases and then decreases with the CuFe2O4 content increasing. When the composition x=0.3, a maximum value of αE=9.4 mV·cm-1·Oe-1 is achieved.
出处
《物理学报》
SCIE
EI
CAS
CSCD
北大核心
2013年第4期452-456,共5页
Acta Physica Sinica
基金
国家自然科学基金青年基金(批准号:51102055)资助的课题~~
