One interesting case, since you mention forcing, although it isn't your main case, is that ultrafilters on an atomless complete Boolean algebra are never generated by a chain (of any cardinality).

To see this, suppose that $U$ is an ultrafilter in a complete atomless Boolean algebra $\mathbb{B}$ generated by a chain. So the elements of $U$ are precisely the elements above some $b_\alpha$, where $$1=b_0\geq b_1\geq\cdots\geq b_\alpha\geq\cdots$$ is a continuous decreasing sequence in $\mathbb{B}$, for $\alpha\lt\kappa$. It is easy to see that $\bigwedge_\alpha b_\alpha=0$. Consider the corresponding difference antichain $d_\alpha= b_\alpha-b_{\alpha+1}$, and let $a$ be the supremum of $d_\alpha$ for $\alpha$ even. Since the chain is continuous, starts at $1$ and has meet $0$, it follows that the difference antichain is a maximal antichain, and $\neg a$ is the join of $d_\alpha$ for $\alpha$ odd. But neither $a$ nor $\neg a$ is above any $b_\alpha$, since each of them has nontrivial meet with cofinally many $b_\beta$, contradicting that $U$ is an ultrafilter. QED

I find this example illuminating in the case of forcing, since many generic ultrafilters are generated by a chain, such as when you force with a tree, which is a very common way to force. What the argument shows is that the ground model Boolean algebra in these cases is no longer a complete Boolean algebra in the forcing extension.