Say we have a cohomological functor F from a triangulated category $C$ to the category $Ab$ of abelian groups, e.g. $F=Hom(x,-)$, where x is an object in $C$. By definition, such a functor transform exact triangles into long exact sequence. And $F(y[i])$ is like the i-th "cohomology group" of the object $y$ w.r.t. the functor F. However, according to the philosophy of derived category, the right thing to look at is the complex, instead of the "cohomology groups". My question is, could there be a complex that "computes" these "cohomology groups"? e.g. Given two objects $x, y$ in $C$, can we find a complex that's like $RHom(x,y)$ whose cohomology computes $Ext_C^i(x,y):=Hom_C(x,y[i])$?
I would guess in full generality the answer is negative. (Otherwise it should have been done by Verdier.) But is there any mild or restrictive assumption that makes this true?
Or just for cohomological functors of the form $Hom_C(x,-)$, can we make the triangulated category $C$ enriched over $D^b(Ab)$ (such that the cohomology of the $Hom$ complex computes the old $Ext^i$'s?)
When I was writing this I feel like triangulated categories might not a good place to play game like this... If the above question isn't too interesting, could anyone just tell me what's the nature playground of questions like this? [Do dg-categories or $A_\infty$-category have an advantage on this?] Thanks.