TITLE: Does the stable category of a nice exact category embed in (the underlying category of) a derivator?
QUESTION [6 upvotes]: In Derivators, Pointed Derivators, and Stable Derivators, Moritz Groth gives as an example of a non-invertible morphism with trivial cone an inclusion $f:X\to I$. Here $X$ is an object of injective dimension $1$ in an exact category $\mathcal{E}$ with enough injectives, $I$ is an injective, and the stable category $\underline{\mathcal{E}}$ has a "suspended structure" given in the same way as the triangulation of the stable category of a Frobenius category but without the invertibility of suspension. 
This is fine as far as it goes, but it's not obvious to me that this is an example of a cone in a pointed derivator, what he's just defined, and Groth doesn't give any indication of how to see it as such.
Question: how can I realize this example as a cone in a derivator?
Having looked around, I don't see any evidence that $\underline{\mathcal{E}}$ is known to be the underlying category of a derivator, even with extra assumptions a la a paper of Stovicek to make $\mathcal{E}$ more like a Grothendieck category. Is it, in fact, such an underlying category, given some extra assumptions? Or can we, perhaps, give it an exact embedding in a homotopy category, and extend to a derivator in that way?

REPLY [2 votes]: This is not exactly answering your question but: a Frobenius category $\mathcal{E}$ (with suitable assumptions) can be given the structure of a model category whose homotopy category is the stable category $\underline{\mathcal{E}}$. See James Gillespie's http://arxiv.org/abs/1009.3574.
Now take the derivator associated with this model category structure (again, I am not so sure about the required assumptions). Its underlying category will be the stable category $\underline{\mathcal{E}}$.