An associative algebra (over a field $k$) is quasi-free if dually regarded as a noncommutative scheme it is formally smooth.
Given an associative algebra $A$ let $\Omega A$ be its universal differential envelope.
An associative unital $k$-algebra $A$ is quasi-free (or formally smooth) if one of the following equivalent conditions is satisfied
Given an extension $0\to N\to E\stackrel{q}\to B\to 0$ of algebras where the ideal $N$ is nilpotent and $a:A\to B$ an algebra map. Then there exists a homomorphism $a' : A \to E$ such that $q \circ a' = a$.
$A$ has cohomological dimension $\leq 1$ with respect to Hochschild cohomology;
$\Omega^1 A$ is a projective $A$-bimodule;
the universal Hochschild 2-cocycle $c : A\otimes A\to\Omega^2 A$, $c : a\otimes b\mapsto d a d b$ is a coboundary, i.e. $c = b\phi$ for some $\phi:A\to\Omega^2 A$ satisfying the cocycle condition $\phi(a b)=a\phi(b)+\phi(a)b-d a d b$;
there exists a “right connection” $\nabla:\Omega^1 A\to \Omega^2 A$ i.e. a $k$-linear map satisfying $\nabla(a w)=a\nabla(w)$ and $\nabla (w a) = \nabla(w)a+w d a$ where $w\in\Omega^1 A$ and $a\in A$.
This is due to (CuntzQuillen).
For $A$ an associative algebra, the object $Spec A \in [Alg_k, Set]$ is formally smooth with respect to the standard infinitesimal cohesive structure over non-commutative algebras (see there for details) precisely if it is quasi-free.
Notice that the characterization via nilpotent extensions is similar to the definition of commutative formally smooth algebras as in EGAIV4 17.1.1. However most commutative formally smooth algebras are not formally smooth in the associative noncommutative sense.
Path algebras of quivers and free algebras are some of the (few classes of) examples.
J. Cuntz and D. Quillen: Cyclic homology and nonsingularity, J. Amer. Math. Soc. 8 (1995), 373-442.
Maxim Kontsevich, Alexander Rosenberg, Noncommutative smooth spaces, (arXiv:math/9812158)