It is easy to derive from the definition of expander graphs that a $n$-vertex expander graph $G$ does not have a $o(n)$-vertex/edge separator. I was wondering if we can build a $d$-regular expander graph that is also a bipartite graph? Thus I have a sparse bipartite graph example that does not have a $o(n)$-vertex/edge separator.
Note that by definition this bipartite graph is not exactly a bipartite expander graph, which only requires that every small subset on the left side of the bipartite graph to have large neighbors on the right.