6. Let the length of side $AC$ of $\triangle ABC$ be 2, and it satisfies $\sqrt{3} \tan A \tan C = \tan A + \tan C + \sqrt{3}$, then the range of the perimeter of $\triangle ABC$ is $\qquad$ .
$$
\begin{array}{l}
\tan B=-\tan (A+C)=\frac{\tan A+\tan C}{\tan A \tan C-1}=\sqrt{3} \Rightarrow B=60^{\circ}, \text { then } \sin A+\sin C \\
=\sin A+\sin \left(120^{\circ}-A\right)=\frac{3}{2} \sin A+\frac{\sqrt{3}}{2} \cos A=\sqrt{3} \cdot\left(\frac{\sqrt{3}}{2} \sin A+\frac{1}{2} \cos A\right) \\
=\sqrt{3} \sin \left(A+30^{\circ}\right), \text { noting that } A, C \neq 90^{\circ} \Rightarrow A \neq 30^{\circ}, 90^{\circ},
\end{array}
$$
$=\sqrt{3} \sin \left(A+30^{\circ}\right)$, noting that $A, C \neq 90^{\circ} \Rightarrow A \neq 30^{\circ}, 90^{\circ}$,
thus $\sin \left(A+30^{\circ}\right) \in\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \cup\left(\frac{\sqrt{3}}{2}, 1\right]$, and $2 R=\frac{b}{\sin B}=\frac{4}{\sqrt{3}}$,
so $a+b+c=2+4 \sin \left(A+30^{\circ}\right) \in(4,2+2 \sqrt{3}) \cup(2+2 \sqrt{3}, 6]$.
$$