Are all countable, nonstandard models of arithmetic given by ultrapowers?

Countable models of PA fall into two categories: the standard one $(\omega, S)$ and the nonstandard ones (all the rest). The only way I've seen to construct a nonstandard model is through taking an ultraproduct or, equivalently, using the compactness theorem. My question is wether or not these are all the models there are? There are continuum many ultrafilters and continuum many nonstandard, countable models, but I don't know if there's a surjective correspondence.

• The number of ultrafilters on a countable index set is not continuum, but $2^{2^{\aleph_0}}$ (though not all of them give rise to nonisomorphic ultrapowers). – Emil Jeřábek Feb 4 '13 at 17:34

On the other hand, if you consider the Henkin construction to be constructive enough for you, then by running his construction relative to the theory in ${\mathcal L}(+,\times,0,1,c,<)$ consisting of PA together with all the assertions $c > n$ for each $n \in {\mathbb N}$, then you would obtain a nonstandard model of PA.
• The cardinality argument is a sort of red herring: every model of true arithmetic is an elementary submodel of an ultrapower of $\mathbb N$ (and in fact, the model has an ultrapower isomorphic to an ultrapower of $\mathbb N$). What is more important is that in this way one obtains only models of true arithmetic, i.e., elementarily equivalent to $\mathbb N$. You do not get e.g. models of PA satisfying ¬Con(PA). – Emil Jeřábek Feb 4 '13 at 17:32