The following question came up in a discussion the other day and I have been wondering whether something is known about it. Everything below takes place over $\mathbb{C}$. I don't have the expertise to know if this is trivial or of interest. Suppose a commutative dga has a free-commutative model $(\wedge V , d)$ where V is a finite dimensional vector space. 


Recall that $T^{poly}$ is the Lie-algebra of polyvector fields on $\wedge V$ (yes, everything is superized as V will be in general graded) with Schouten bracket. Part of Kontsevich's formality theorem says that the HKR map $ T^{poly} \to HC^\*$(Hochschild cochains) is the first Taylor coefficient in an $L\_\infty$ quasi-isomorphism between the two. 


We can think of the derivation $d$ as corresponding to a vector-field $v$. It follows from a spectral sequence argument that the HKR map gives a quasi-isomorphism:
$$ (T^{poly},[v,-]) \to HC( \wedge V,d)$$


Question: Can this map be upgraded to a map of $L\_\infty$ algebras?


Certainly, the Taylor coefficients in the usual formality map must be doctored. 


A related statement that does seem to be true and standard is that there is an $L\_\infty$ quasi-isomorphism $(T^{poly}[[t]],[tv,-]) \to HC^\*(\wedge V[[t]],td )$ Thus, the question is in some reasonable sense about convergence of this isomorphism. Maybe one can prove the claim by a close inspection of Kontsevich's integral formulas. Based upon these facts, however, it seems plausible to me that that the statement is in general false, but I was unable to come up with a counterexamples or an a priori reason (I didn't try too hard however). Is it true for some more restrictive group of commutative dg algebras, for example pure Sullivan algebras? 


Update: Having finally looked at the Kontsevich formulas, I'm beginning to think there are some simple counting reasons that make the above formula converge, but am not sure that $f\_1$ stays the same (though I believe it remains a quasi-iso). Any confirmation or help would be great. Otherwise, I'll keep thinking and update again.



Extension of the formality theorem?
I think this does only work for so-called homological vector fields, i.e. vector fields of degree 1 which self-commute. Then you have $[v,v]=0$, which is the Maurer-Cartan equation in $T\_{poly}$, and which insures that the corresponding derivative $d$ squares to zero. 


Dealing with a Maurer-Cartan element you can simply use it to twist Kontsevich's formality $L\_\infty$-quasi-isomorphism. When the Maurer-Cartan element is a vector field we know the explicit form of the first Taylor component of the twisted $L\_\infty$-morphism: it is not HKR, but involve Bernoulli numbers (see e.g. <https://www.math.ethz.ch/u/calaqued/research/LecturesDufloETH.pdf>). 


Anyway, the $k$-th taylor component of the $v$-twisted $L\_\infty$-morphism will be given by the series
$$
\phi\_v^{(k)}(u\_1,\dots,u\_k):=\sum\_{l\geq0}\frac{t^{l+1}}{(l+1)!}\phi^{(k+l)}(u\_1,\dots,u\_k,\underbrace{v,\dots,v}\_{l~times})
$$
To conclude one just have to observe that $\phi^{(k)}$ preserves the grading given by the arity minus $2$ (arity means the number of arguments for poly-vectors and poly-differential operators). Therefore each time $v$ appears in the formula it decreases this degree by $1$. 


EDIT: I include a comment into the main answer about the shape of $\phi^{(1)}\_v$ in specific cases. Let me denote coordinates by $u^k$, and write $\partial\_k=\frac{\partial}{\partial u^k}$ and $v=\sum\_iv^i\partial\_i$. We consider a matrix-valued one-form $\Xi$ given by 
$$
\Xi\_i^j=\sum\_k\partial\_k\partial\_iv^jdu^k,
$$
and define 
$$
\Theta=\sum\_{n>0}c\_n\iota\_{tr(\Xi^n)}.
$$
Here $c\_n$ are rational coefficients that do not matter. 


Then $\phi\_v^{(1)}$ is given by the precompsition of HKR with $e^{\Theta}$. 


Now if I split the coordinates into even $x^i$ and odd $e^i$, and if I assume that 
$v=\sum\_iv^i(x^1,\dots,x^m)\frac{\partial}{\partial e^i}$, then the matrix $\Xi$ has non-zero entries only in the right-up block. In particular it is upper triangular, the trace of any power of it is therefore zero, and thus $\phi^{(1)}\_v=HKR$. 


