Is a filtered category necessarily (essentially) small? There is a result, which I have heard is due to Grothendieck, which says that a left exact functor $F : C \to \text{Set}$ is a filtered colimit of representable functors provided that $C$ is essentially small and has finite limits. When I tried proving this, I found that I didn't need to use the hypothesis that $C$ is essentially small, but the diagram that I'm taking a filtered colimit over is not necessarily essentially small as a result. 
Neither Wikipedia nor nLab require that filtered categories are essentially small. Is this in fact a requirement in the definition of filtered colimits? Or am I missing a detail in the proof? 
 A: The answer to the title question is, of course, no: the category $\mathbf{On}$ of all ordinals is filtered and not essentially small. 
If you do not assume $\mathcal{C}$ is essentially small, then it is certainly possible that $F$ is a colimit of a filtered diagram that is not essentially small. The only problem is that colimits of such diagrams need not exist – so in some sense this is an "accidental" colimit.
Here's one way of deducing your generalisation from the essentially small case: let $\mathbf{U}$ be the original set-theoretic universe and let $\mathbf{U}^+$ be a universe so large that $\mathbf{U} \in \mathbf{U}^+$ and $\mathcal{C}$ is locally $\mathbf{U}$-small and essentially $\mathbf{U}^+$-small. The inclusion $\mathbf{Set} \hookrightarrow \mathbf{Set}^+$ preserves finite limits, so $F$ still preserves finite limits considered as a functor $\mathcal{C} \to \mathbf{Set}^+$. Thus $F$ is a colimit of an essentially $\mathbf{U}^+$-small filtered diagram of representable functors $\mathcal{C} \to \mathbf{Set}^+$; but the inclusion $[\mathcal{C}, \mathbf{Set}] \hookrightarrow [\mathcal{C}, \mathbf{Set}^+]$ reflects all colimits, so $F$ is a colimit of a filtered diagram of representable functors $\mathcal{C} \to \mathbf{Set}$ as well. (Obviously, the property of being a filtered category does not depend on the choice of universe.)
