采用放电等离子火花烧结法(SPS)制备四元ZnxBi0.5Sb1.5-xTe3(x=0.05～0.4)(摩尔分数,下同)合金,得出当Zn的量为0.05时,材料的电导率出现最大值,室温附近其值为2.5×104Ω-1·m-1,大约是三元Bi0.5Sb1.5Te3合金的1.35倍.在同温度下,功率因子p值也取得最大值(1.65×10-3W·m-1·K-2),而三元Bi0.5Sb1.5Te3合金的功率因子p值为1.35×10-3W·m-1·K-2.在该合金中用Zn替代Sb元素后,合金的微结构逐渐随Zn的含量发生变化.
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