The ring $R/I$ is a total quotient ring iff $I$ is... what?

Conventions.

1. All my rings are commutative.
2. By a total quotient ring (TQR) , I mean a ring whose every regular element is a unit.

Now let $R$ denote a ring and $I$ denote an ideal of $R$. The following fact is well-known:

Proposition. The ring $R/I$ is a field iff $I$ is maximal among proper ideals.

I'm wondering if there is a similar theorem for total quotient rings. Therefore, I ask:

Original Question. The ring $R/I$ is a TQR iff $I$ is... what?

Edit. I'm also interested in the following, related issue:

Second Question. Is there some weakening of freeness that holds for modules over a TQR?

If so, does this help us answer the previous question?

To give you an idea of the difficulty in pinning down a condition, consider this. For a von Neumann regular ring, every ideal has the property you speak of (that is, $R/I$ is a total quotient ring for every ideal $I$. What standard properties to the ideals have? They are all idempotent, semiprime and flat and divisible (in the sense given by Lam in Lectures on modules and rings). Unfortunately, the zero ideal in a nonfield domain also has all of these properties, and yet such a domain is not a total quotient ring.
$R/I$ is total quotient iff for every $x\in R$ with $(I:x)= I$, there exists a $y$ such that $1-xy\in I$.