ag.algebraic geometry - Generating the derived category with line bundles

The following lemma is useful and well-known:

LEMMA If $L^{pm 1}$ is ample on proper scheme over a field $k$, then some number of powers $mathcal{O},L,...,L^{m}$ generate the unbounded derived category of quasi-coherent sheaves $D(X)$ (or split generate the subcategory of perfect complexes).

QUESTION: What about a converse? Suppose that I know some number of powers of $L$ generate $D(X)$. Then can I conclude that $L^{pm 1}$ is ample?

The best I can do so far is see that the restriction of $L$ to any integral curve
$C$ in $X$ has non-zero degree. (Since by adjunction $mathcal{O},L,...,L^{m}$ generates $D(C)$, but if $L$ had degree $0$ on $C$, there would be something orthogonal $mathcal{O},L,...,L^{m}$, for instance a generic line bundle of degree $g-1$ having no cohomology.)

Something I don't know yet: does the degree of $L$ must have the same sign on all curves?
This would be useful for numerical tests of ampleness.

Note: I think that one doesn't need properness in the above lemma, but I am willing to assume it to get a converse. It makes life easier when restricting to closed subschemes.

Note 2: When saying a collection of objects generates a triangulated category with all coproducts, like $D(X)$, one usually means that you take the smallest triangulated subcategory closed under all coproducts and containing the the collection. Once you have all coproducts, then idempotents automatically split, by a standard argument called, I think, the Eilenberg swindle. If you are working with a smaller triangulated category having only finite coproducts, like perfect complexes on a scheme, then the smallest triangulated subcategory containing a collection might not be 'thick', in the sense that some idempotents might not split, so in this case one usually adds in the missing summands. To emphasize this, some people speak of 'split generation'.