Algebras with Calabi-Yau derived categories Let $A$ be a finite dimensional algebra over a field $k$ and assume that global dimension of $A$ is finite. I want to describe such algebras $A$ with Calabi-Yau derived category.
A triangulated category with a Serre functor $S$ is called Calabi-Yau if $S(-) \cong (-)[n]$, where $n$ is an integer.
For the triangulated category $D^b(A)$ Serre's functor is Nakayama's functor
$$
S(M) = D(A) \otimes^{\mathbb{L}} M,
$$
where $M \in D^b(A)$ and $D(-)=\operatorname{Hom}_k(-,k)$
If we assume that $S(M)=M[n]$ and take $M=A$ we get $D(A) \cong A[n]$, so Calabi-Yau dimension $n=0$ and $D(A) \cong A$ as left modules. So, $A$ is a Frobenius algebra. 
The conclusion is: algebra $A$ is Frobenius if and only if $D^b(A)$ is Calabi-Yau. In this case Calabi-Yau dimension is $0$.
Is all the above correct?
 A: Almost correct, except that "Frobenius" is not sufficient. In order that $D(A)\otimes -$ is isomorphic to the identity functor (as functors from left $A$-modules to left $A$-modules) you need not only that $D(A)\cong A$ as left $A$-modules, but that they are isomorphic as $A$-bimodules. Algebras with this stronger property are called "symmetric".
Edit: Actually, I answered in a hurry and misread exactly what you asked, so let me clarify.
Unless it is semisimple, a symmetric algebra has infinite global dimension, and the derived category $D^b(A)$ is not Calabi-Yau. So the only examples are the semisimple symmetric $k$-algebras, which are precisely the separable $k$-algebras.
The sense in which more general symmetric algebras are $0$-Calabi-Yau is that the category $K^b(P_A)$ of perfect complexes (i.e., those isomorphic to bounded complexes of finitely generated projective modules) is $0$-Calabi-Yau. More generally, there is a natural isomorphism $\text{Hom}(P,X)\cong D\text{Hom}(X,P)$ whenever $P$ and $X$ are objects of $D^b(A)$ and $P$ is perfect.
