triangulated hull and thick hull of $\mathcal{O}_X,\dots,\mathcal{O}_X(n)$

Here I am dealing with the difference of triangulated hull and thick hull. Let $$\mathcal{D}$$ be a triangulated category and $$\mathcal{E}\subset\mathcal{D}$$ be a collection of objects. The triangulated hull $$\mathcal{E}^{\star}$$ in $$\mathcal{D}$$ is the smallest triangulated subcategory of $$\mathcal{D}$$ containing $$\mathcal{E}$$, the thick hull $$\langle\mathcal{E}\rangle$$ in $$\mathcal{D}$$ is the smallest thick subcategory of $$\mathcal{D}$$ containing $$\mathcal{E}$$.

Consider $$X$$ a complex smooth projective variety of dimension $$n$$ with a polarization $$\mathcal{O}_X(1)$$ or say $$\mathcal{O}_X(1)$$ is very ample, then we know that the thick hull of $$\mathcal{E}=\{\mathcal{O}_X,\dots,\mathcal{O}_X(n)\}$$ in $$D^b(X)$$ is $$D^b(X)$$.

Is $$\mathcal{E}^{\star}=D^b(X)$$?

I know it is true in the case of $$X=\mathbb{P}^n$$.

No. The simplest counterexample is a smooth quadric surface $$X = \mathbb{P}^1 \times \mathbb{P}^1 \subset \mathbb{P}^3$$. If $$F$$ belongs to the triangulated hull of $$\mathcal{O}_X$$, $$\mathcal{O}_X(1)$$, $$\mathcal{O}_X(2)$$, then its first Chen class is proportional to $$c_1(\mathcal{O}_X(1)) = h_1 + h_2,$$ where $$h_i$$ are the classes of the rulings. In particular, the line bundle $$\mathcal{O}_X(h_1)$$ is not in this triangulated hull.