Completion of discrete valuation ring

Question

Let $R$ be an excellent, Henselian, discrete valuation ring with algebraically closed residue field and $\hat{R}$ be the completion of $R$. If I understand correctly, the residue field of $\hat{R}$ is the same as that of $R$. Is there any classification of subrings $R'$ of $\hat{R}$ containing $R$ such that the induced morphism $\mathrm{Spec}(R') \to \mathrm{Spec}(R)$ is of finite type?

Answer 1

The transcendence degree of $\hat{R}$ over $R$ can be large. If $R$ is the Henselization of $k[t]_{(t)}$ then the transcendence degree is max(${2^{\aleph_0}}$,$|R|$), and there are many ways to construct transcendental power series. So $R'$ is rather arbitrary.