Which subcategory of the derived category of an algebra has a Serre functor?

Question

My question concerns which subcategory of the derived category has a Serre functor.

In Calabi-Yau algebras, Section 7.2 V. Ginzburg defines the category $D^{fproj}(A)$ of a left and right Noetherian algebra $A$ as the derived category of bounded complexes of projective finitely generated $A$-modules. By Lemma 7.2.8 the category $D^{fproj}(A)$ admits a Serre duality formula except that the Serre functor $\mathbb{S}$ does not preserve the category $D^{fproj}(A)$.

If the Noetherian algebra $A$ is well-behaved like a symmetric $R$-order over some central subring $R$ of Krull dimension $d$, then Theorem 7.2.14 of Ginzburg's paper implies that the category $D^{fproj}(A)$ is $d$-Calabi-Yau, that is, preserved by the Serre functor $\mathbb{S}$ which becomes isomorphic to the shift functor $[d]$.

My question boils down to the following:

Why is $D^{fproj}(A)$ not the same as the perfect derived category $\mathrm{perf} A$?

By $\mathrm{perf} A$ I mean the full subcategory of the big derived category $D(Mod A)$ whose objects are given by complexes which are quasi-isomorphic to bounded complexes of projective finitely generated $A$-modules.

The question becomes only interesting if $A$ is not a finite-dimensional $k$-algebra but an algebra with non-Artinian center like $A = k[x,y]/(xy)$.

Here are two additional notes:

1. Assume as above, that $A$ is a symmetric order with center of Krull dimension $d> 0$. If there was equality $D^{fproj}(A) = \mathrm{perf} A$ then any tilting complex of $A$ would be just some shift of $A$ by Serre duality. However, this cannot be true since preprojective algebras of affine type are symmetric orders which have non-trivial tilting complexes.

2. Results by Iyama and Reiten (Fomin-Zelevinsky mutation and tilting modules over Calabi-Yau algebras, Theorem 3.7) on the Serre functor suggest that $D^{fproj}(A)$ should mean the category $\mathrm{perf}_{{fl}} A$ of perfect complexes $P$ such that the cohomology $H^i(P)$ at each degree $i \in \mathbb{Z}$ has finite length. However, I cannot see that from Ginzburg's definition.