Is it always possible to construct an extension of $\mathbb{Q}_p$ as a totally ramified extension first and then an unramified extension after? Let $K = \mathbb{Q}_p$, and $L/K$ be a finite Galois extension.

Question: Is it possible to find a totally ramified extension $L'/K$ such that $L/L'$ is unramified?

I know that it is always possible to find an unramified $L'/K$ such that $L/L'$ is totally ramified, and I was wondering whether the converse is true or not.
I guess it is, and the following example demonstrates how it may be generalized. If $K= \Bbb{Q}_3$ and $L=K(\alpha)$ with $\min_K(\alpha) = x^4-3x^2 +18$, then $L/K$ has degree $4$ and ramification index $2$. The field $L$ can be written as $L=K'(\alpha)=K(\alpha,\zeta_4)$ where $K' = K(\zeta_4)$ is the maximal unramified subextension of $L/K$.
Now I could choose an automorphism $\sigma \in \operatorname{Gal}(L/K)$ such that $\sigma(\zeta_4) = \zeta_4^3$ (i.e. $\sigma$ is a lift of the Frobenius), and choose $L' = L^{\langle \sigma \rangle}$. By construction, the residue field of $L'$ is $\mathbb{F}_3$, the same residue field as of $K$, so $L'/K$ is totally ramified. This also implies that $L/L'$ is unramified.
I guess one could always construct such a subextension that way but I am not sure if it can be always generalized like that.
 A: The answer is no, even for Galois extensions. The standard example is as follows. Suppose that $L/K$ is Galois with $\mathrm{Gal}(L/K) = \mathbf{Z}/4 \mathbf{Z}$ is cyclic of order $4$ but the inertia group is cyclic of order $2$, so $f = e = 2$. By Galois theory, there is only one (proper) intermediate subfield $L/E/K$, and it will be unramified over $K$, so $L/E$ will be ramified. You can construct examples of such extensions quite explicitly. Suppose that $p \equiv 1 \bmod 4$ so that $\mathbf{Q}_p$ contains a $4$th root of unity $i$. Now let $u \in \mathbf{Z}_p$ be any element which is a quadratic non-residue. Then take $K = \mathbf{Q}_p$ and $L = \mathbf{Q}_p((up^2)^{1/4})$.
For Galois extensions, with Galois group $G$, inertia group $I$, and cyclic quotient $G/I = \langle \phi \rangle$, the general condition is that there exists a splitting $\langle \phi \rangle \rightarrow G$, that is, there is a lift of $\phi$ to $G$ which has the same order. Then you can take $E$ to be the fixed field of that lift $\phi \subset G$. You can see the example above is non-split.
