Given an increasing sequence of cardinals $\langle\kappa_\alpha\mid \alpha\in\kappa\rangle$, let $K=\prod_{\alpha\in\kappa} \kappa_\alpha$, then we call $f,g\in K$ eventually different if there exists $\alpha_0\in\kappa$ such that for all $\alpha\in[\alpha_0,\kappa)$ we have $f(\alpha)\neq g(\alpha)$. A subset $F\subseteq K$ is an almost disjoint (a.d.) family if any distinct $f,g\in F$ are eventually different.
I'm interested in finding a large a.d. family in the case where $\kappa=\sup_{\alpha\in\kappa}\kappa_\alpha$ is strongly inaccessible. It is easy to see that there exists an a.d. family of size $2^\kappa$ whevener there exists an enumeration $\langle \alpha_\xi\mid \xi\in\kappa\rangle$ of a club set of $\kappa$ such that $\kappa_{\alpha_\xi}\geq 2^{|\xi|}$ for all $\xi\in\kappa$: for each $\alpha\in\kappa$ there is $\xi$ such that $\alpha\in[\alpha_\xi,\alpha_{\xi+1})$, then we fix an injection $\iota_\alpha:2^{\xi}\to \kappa_\alpha$, and any $f:\kappa\to 2$ is mapped injectively to $f'\in K$ by letting $f'(\alpha)=\iota_\alpha(f\restriction \xi)$, then $F=\{f'\mid f:\kappa\to 2\}$ is a.d.
If we assume that the sequence of $\kappa_\alpha$'s is continuous, then there does not exist such a sequence $\langle \alpha_\xi\mid \xi\in\kappa\rangle$. My question is: if the sequence of $\kappa_\alpha$'s is continuous, for which cardinalities ${>}\kappa$ can we find an a.d. family of that size? Particularly, does there exist an a.d. family of size $2^\kappa$?
So far, I can only find references to cases where $\kappa$ is singular (e.g. in the combinatorial proof of Silver's theorem), or results where the $\kappa_\alpha$'s are all equal, and my personal attempts to find a large a.d. family fall short.
