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).



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 $K^b(B)$?
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.


