欢迎登录材料期刊网

材料期刊网

高级检索

高热导率氮化硅陶瓷作为基板材料有着广泛的应用前景.如何在尽可能保持氮化硅陶瓷机械性能的前提下,提高其热导率是其实际应用的关键,而选择适当的烧结助剂是提高热导率的一个重要途径.本文研究了稀土氧化物种类及CaO、MgO烧结助剂对氮化硅陶瓷的热导率及电学和机械性能的影响,分别采用Y2O3-MgO,Y2O3-CaO,CeO2-MgO,CeO2-CaO,La2O3-MgO和La2O3-CaO 6种烧结助剂,采用放电等离子烧结后热处理的工艺制备氮化硅陶瓷.研究结果表明:氮化硅陶瓷的热导率随着烧结助剂稀土元素阳离子半径的增大有减小的趋势;与添加MgO助烧结相比,添加CaO助烧结不利于氮化硅柱状晶的生长,热导率及强度普遍较低,但硬度较高.采用Y2O3-MgO助烧剂和适当的烧结工艺,可以得到热导率高于80 W/m·K、抗弯强度大于1000 MPa、体电阻率大于1×1013Ωm、介电常数小于10、介电损耗小于3×10-3的氮化硅陶瓷.

参考文献

[1] Haggerty J S et al.[J].Ceramic Engineering and Science Proceedings,1995,16(04):475.
[2] Ning X S;Lin Y B;Peng R .[J].Key Engineering Materials,2001,224:2.
[3] Watari K;Hirao K;Brito M E et al.[J].Journal of Materials Research,1999,14(04):1538.
[4] Hirosaki N;Okamoto Y;Munakata F et al.[J].Journal of the European Ceramic Society,1999,19(12):2183.
[5] Wasanapiarnpong T;Wada S .[J].Journal of the Ceramic Society of Japan,2005,113(06):394.
[6] Yokota H;Yamada S .[J].Journal of the European Ceramic Society,2003,23:1175.
[7] Yang H T;Gao L;Shao G Q .[J].Ceramics International,2001,27:603.
[8] Lin Y B;Ning X S .[J].Key Engineering Materials,2002,224:813.
[9] Lu X;Ning X S;Xu W et al.[J].Materials Science Forum,2005,475:1279.
[10] Morikawa J;Hashimoto T .[J].Journal of Thermal Analysis and Calorimetry,2001,64:403.
[11] Hirao K;Toriyama M;Kanzaki S .[J].Journal of the American Ceramic Society,2001,84(02):353.
[12] Wasanaplarnpong T;Wada S .[J].Journal of the Ceramic Society of Japan,2006,114(09):733.
[13] Yang HT.;Yuan RZ.;Yang GT. .Vitrification and devitrification of MgO during sintering of Si3N4-MgO-CeO2 ceramics[J].Materials Chemistry and Physics,1998(2):178-181.
[14] Toropov N A;Fedorov N F;Khim Z N .[J].Russian Journal of Inorganic Chemistry,1965,10(03):358.
[15] David R Poirier.Transport Phenomena in Materials Processing[M].Warrendale,Pennsylvania:A Publication of the Minerals.Metals & Materials Society,1994:25.
[16] Lu X et al.[J].Key Engineering Materials,2004,280:1259.
上一张 下一张
上一张 下一张
计量
  • 下载量()
  • 访问量()
文章评分
  • 您的评分:
  • 1
    0%
  • 2
    0%
  • 3
    0%
  • 4
    0%
  • 5
    0%