Suppose $G$ is a $k$-regular graph on $n$ vertices, with least eigenvalue $\tau$.
Lovasz proved that
$$
\theta(G) \le \frac{n}{1-\frac{k}{\tau}}.
$$
Further if the automorphism group of $G$ acts arc-transitively, then equality holds.
In fact equality holds if G is a single class in a homogeneous coherent configuration,
for example, if $G$ is a strongly regular graph.

Class 1: Latin square graphs. Take the graph whose vertices are the $n^2$ cells of
and $n\times n$ Latin square, where two cells are adjacent if they are in the same row,
same column, or have the same contents. This graph is regular with valency $3(n-1)$
and least eigenvalue $-3$ and so $\theta(G)=n$. But if the Latin square is the multiplication
table of a cyclic group of even order then $\alpha(G) < n$ (because cyclic Latin squares
do not have transversals, but you can find a short proof on page 225 of one of my favorite
books on algebraic graph theory). The vertices in a given column form a clique of size
$n$ and so we see that $\chi(\bar{G})=n$ for any Latin square of order $n$.

Class 2: generalized quadrangles: A GQ with parameters $(s,t)$ is a collection of points
and lines satisfying some axioms, in particular each line contains exactly $s+1$ points.
Deem two points adjacent if they are collinear.
If $s,t>1$ this gives a strongly regular graph on $(s+1)(st+1)$ vertices with valency
$s(t+1)$ and least eigenvalue $-t-1$. Hence $\theta(G) = st+1$ and a coclique of size $st+1$
is known as an *ovoid*. The points on line forms a clique of size $s+1$, and a set
of lines that partition the point set is called a *spread*. Hence you want
GQ's with spreads but no ovoids.
For actual examples, I refer you to Payne and Thas's book on GQ's---the GQ's $Q(5,q)$ work.