Induced map on cohomology being zero impies null-homotopic?

Let $$f:A^\bullet\to B^\bullet$$ be a morphism of chain complexes (of any give abelian category). We know that if $$f$$ is homotopic to the zero map, then $$f$$ will induce zero map on cohomology. I want to know if the converse true, i.e. if $$f$$ induces zero map on cohomology, is it true that $$f$$ is homotopic to zero map?

If not, under what conditions it can be true?

Let both $$A^*$$ and $$B^*$$ be the chain complex of abelian groups $$\cdots\to0\to\mathbb Z\to\mathbb Z\to\mathbb Z/2\to0\to\cdots$$ where the map $$\mathbb Z\to\mathbb Z$$ is multiplication by 2 and the $$\cdots$$ represents just $$0$$'s. Let $$f:A^*\to B^*$$ be the identity map; it induces $$0$$ in cohomology, because the complex is an exact sequence, so its cohomology is zero. But $$f$$ is not null-homotopic. To see that, notice that the only homomorphism from $$\mathbb Z/2$$ to $$\mathbb Z$$ is zero, so a homotopy would consist entirely of zero maps except for one map from the second $$\mathbb Z$$ in $$A^*$$ to the first $$\mathbb Z$$ in $$B^*$$. That's not enough to provide the desired null-homotopy.
Let $$\mathcal A$$ be an abelian category. Your condition (zero map on cohomology implies null-homotopy) is equivalent to the condition that $$\mathcal A$$ be semisimple (i.e. any exact triple $$0\to X\to Z\to Y\to 0$$ in $$\mathcal A$$ splits, i.e. is isomorphic to $$0\to X\to X\oplus Y\to Y\to 0$$).
On the one hand, your condition implies all Yoneda Exts are zero, so that the homological dimension of $$\mathcal A$$ is zero; i.e. $$\mathcal A$$ is semisimple.
On the other hand, if $$\mathcal A$$ is semisimple, the functor $$h:(K^n,d^n)\mapsto (H^n(K),0)$$ which sends a complex $$K$$ to its complex of cohomology objects with zero differentials in all degrees, induces an equivalence between $$D(\mathcal A)$$ and the category of cyclic complexes of objects of $$\mathcal A$$ (complexes with zero differentials in all degrees).