Does a triangulated category that possesses a subcategory $B$ of generators with no extensions of non-zero degree between them have to be isomorphic to $K^b(B)$? Suppose that a triangulated category $C$ contains a full additive subcategory $B$ of (strong) generators (i.e. there does not exist a proper strict triangulated subcategory $C'\subset C$ that contains $B$) such that: there are no non-zero $C$-morphisms between $B_1$ 
and $B_2[i]$ for any $B_1,B_2\in Obj B$ and $i\neq 0$. 
Is it true that $C\cong K^b(B)$? Is anything known about this question (in general)?
Upd. I know how to prove this statement for any 'algebraic' triangulated $C$ (i.e. if $C$ admits a differential graded enhancement); this includes all derived categories of sheaves. So, one can reformulate my question as follows: is such a $C$ necessarily algebraic? I know the proof of this fact when $C$ is an $f$-category (in the sense of Beilinson).
 A: Mikhail, if you assume more generally that $C$ is topological (in an appropriate sense) then your claim is also true. As Matthias suggests above, this is a tilting-like theorem, Theorem 5.1.1 in 'Stable model categories are categories of modules' by Schwede and Shipley. This assumption may comprise all examples of interest to you. So far, the very few known examples of exotic model categories do not satisfy your assumptions.
A: It follows that $C \cong D^b(mod-B)$. I don't think the latter is equivalent to $K^b(B)$ (by the way, what does $K^b$ denote?) unless in very special situation, like when the homological dimension of $B$ is zero. For example, if $X = Spec A$ is a smooth affine variety and $C = D^b(coh-X)$, then you can take $B = \{O_X\}$. This gives $C \cong D^b(mod-A)$ which is different from $K^b(A)$.
EDIT: This is an answer to the comment below. If you want $K^b(O_X)$ to be equivalent to $D^b(X)$ for a smooth affine $X$ you definitely should require that all projective $A$-modules have a free resolution (I guess it is equivalent to $Pic X = 0$). If this is not true, to get an equivlence you should take a pseudo-abelian envelope of $B$ by adding to it all direct summands. 
As for the equivalence $C \cong D^b(mod-B)$, I implicitly assumed that $C$ is algebraic. In this case it is well known.
