11. The eccentricity of the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{b^{2}}=1(b>0)$ is used as the radius, and the right focus is used as the center of a circle that is tangent to the asymptotes of the hyperbola. Then the eccentricity of the hyperbola is $\qquad$
11. $\frac{2 \sqrt{3}}{3}$.

It is known that $a=2, c=\sqrt{4+b^{2}}, e=\frac{\sqrt{4+b^{2}}}{2}$.
The equations of the asymptotes are
$$
\frac{x}{2} \pm \frac{y}{b}=0 \Rightarrow b x \pm 2 y=0 .
$$

The distance from the right focus $\left(\sqrt{4+b^{2}}, 0\right)$ to the asymptote is $=b$.
Thus, $b=\frac{\sqrt{4+b^{2}}}{2} \Rightarrow b=\frac{2 \sqrt{3}}{3}$
$$
\Rightarrow e=\frac{\sqrt{4+\frac{4}{3}}}{2}=\frac{2 \sqrt{3}}{3} .
$$