4. As shown in Figure 1, in the tetrahedron $D-ABC$, it is known that $DA \perp$ plane $ABC$, and $\triangle ABC$ is an equilateral triangle with a side length of 2. Then, when the tangent value of the dihedral angle $A-BD-C$ is 2, $V_{D-ABC}=$ $\qquad$
4.2.

Given $D A \perp$ plane $A B C$, we know
$D A \perp A B$, plane $D A B \cdot \perp$ plane $A B C$.
As shown in Figure 3, take the midpoint $O$ of $A B$.
Then, by the problem statement, $C O \perp A B, C O$
is on plane $D A B$, and
$$
C O=\sqrt{3} \text {. }
$$

Draw $O E \perp D B$ at point $E$, and connect $C E$. Then
$$
C E \perp D B \text {. }
$$

Thus, $\angle O E C$ is the plane angle of the dihedral angle $A-B D-C$.
Let $D A=x$. Then
$$
\begin{array}{l}
O E=\frac{1}{2} \times \frac{D A \cdot A B}{D B} \\
=\frac{1}{2} \times \frac{2 x}{\sqrt{x^{2}+4}}=\frac{x}{\sqrt{x^{2}+4}} .
\end{array}
$$

Given $\tan \angle O E C=\frac{O C}{O E}=2$, we have
$$
\sqrt{3} \times \sqrt{x^{2}+4}=2 x \Rightarrow x=2 \sqrt{3} \text {. }
$$

Therefore, $V_{D \rightarrow A B C}=\frac{1}{3} \times 2 \sqrt{3}\left(\frac{\sqrt{3}}{4} \times 2^{2}\right)=2$.