4. As shown in Figure 2, in $\triangle A B C$, it is known that $\angle B=60^{\circ}$, $\angle A C B=75^{\circ}$, and $D$ is a moving point on side $B C$. A circle $\odot O$ is constructed with $A D$ as its diameter, intersecting $A B$ and $A C$ at points $E$ and $F$ respectively. If the minimum length of chord $E F$ is 2, then $A B=$ $\qquad$
4. $\frac{4 \sqrt{6}}{3}$.

As shown in Figure 3, connect $O E$ and $O F$.
Given $\angle B=60^{\circ}, \angle A C B=75^{\circ}$, we know $\angle B A C=45^{\circ}$. Therefore, $\angle E O F=2 \angle B A C=90^{\circ}$.
Thus, $E F=\sqrt{2} O E$.
Since the minimum value of $E F$ is 2, the minimum value of $O E$ is $\sqrt{2}$. Therefore, when $A D \perp B C$, the minimum value of $A D$ is $2 \sqrt{2}$. Hence, when the minimum value of $E F$ is 2,
$$
A B=\frac{A D}{\sin 60^{\circ}}=\frac{4 \sqrt{6}}{3} \text {. }
$$