Suppose that $K$ is a complete local field and $A$ is an affinoid $K$-algebra. Is there a known way to produce an explicit description of the units of $A$?

Here is what I already know: write $A^\circ$ for the subring consisting of elements of $A$ of norm at most $1$ and $A^{\circ\circ}$ for the ideal of $A^\circ$ consisting of all elements of norm strictly less that $1$. Certainly every element of $1+A^{\circ\circ}$ is a unit in $A$. So is every non-zero element of $K$. Let's call the group generated by these units the group of standard units of $A$. I believe that if the ring $A^\circ/A^{\circ\circ}$ is prime then all units in $A$ are standard.

I also know that in general there can be non-standard units. Perhaps an easier question than the one above is `must the group of units of $A$ modulo the group of standard units be finitely generated?'.

I have a particular application in mind for a solution to this but I feel that the question is sufficiently interesting in its own right and the application sufficiently distant from the problem that it is not worth explaining it now.

Edit: given some of the comments/answers below I probably want to modify my definition of standard units to include any non-zero element of a finite field extension of $K$ inside $A$.

Edit 2: thanks for the help so far... I'm actually happy to consider as 'standard' anything in $A^\circ$ that is a unit as an element of $A^\circ$ if that makes things easier.

domain$(A_ {K'})^0$ has ord's at dvr generic pts of special fiber. By normality, two with same ords have unit ratio in $(A_ {K'})^0$. Reduced special fiber has fin. generated unit group (since $k$ finite). $\endgroup$ – BCnrd Apr 30 '10 at 16:28