4. For any two points $P, Q$ on the ellipse $\frac{x^{2}}{4}+y^{2}=1$, and $O$ being the origin. If $O P \perp O Q$, then the minimum value of the area of $\triangle P O Q$ is $\qquad$ .
4. $\frac{4}{5}$.

Let $P(|O P| \cos \theta,|O P| \sin \theta)$, $Q\left(|O Q| \cos \left(\theta \pm \frac{\pi}{2}\right),|O Q| \sin \left(\theta \pm \frac{\pi}{2}\right)\right)$.
Since points $P$ and $Q$ are on the ellipse, we have
$\frac{1}{|O P|^{2}}=\frac{1}{4} \cos ^{2} \theta+\sin ^{2} \theta$,
$$
\frac{1}{|O Q|^{2}}=\frac{1}{4} \sin ^{2} \theta+\cos ^{2} \theta \text {. }
$$

Adding the two equations gives
$$
\frac{5}{4}=\frac{1}{|O P|^{2}}+\frac{1}{|O Q|^{2}} \geqslant 2 \cdot \frac{1}{|O P||O Q|} \text {. }
$$

Thus, $S_{\triangle P O Q}=\frac{1}{2}|O P||O Q| \geqslant \frac{4}{5}$.
The minimum value of $S_{\triangle P O Q}$ is $\frac{4}{5}$, which is achieved when $|O P|=|O Q|=\frac{2 \sqrt{10}}{5}$.