设椭圆$C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)$的左右焦点为$F_1,F_2$，作$F_2$作$x$轴的垂线与$C$交于$A,B$两点，$F_{1}B$与$y$轴交于点$D$，若$AD\perp F_{1}B$，则椭圆$C$的离心率等于.