I am starting to study the rudiments of p-adic Hodge theory and I have the following basic question. Let $\chi$ be the unramified quadratic character of $G_p = \mathrm{Gal}(\bar{\mathbb{Q}}_p/\mathbb{Q}_p)$ and let $V=\mathbb{Q}_p(\chi)$ denote the $1$-dimensional $p$-adic representation on which $G_p$ acts through $\chi$.
Then $V$ is cristalline, because $D(V) := (B_{\mathrm{cris}}\otimes V)^{G_p} = \mathbb{Q}_p\cdot (b\otimes 1)$, where $b$ is an element on the quadratic unramified extension of $\mathbb{Z}_p$ on which $G_p$ acts via $\chi$.
What is the definition of Hodge-Tate weight of $V$? Reading Berger's course one finds that $V\otimes \mathbb{C_p}$ must be isomorphic as a $G_p$-module to some Tate twist $\mathbb{C_p}(i) := \mathbb{C_p}(\varepsilon^i)$, where $\varepsilon$ is the cyclotomic character, in which case one says that the Hodge-Tate weight of $V$ is this integer $i$.
But in this simple example $V=\mathbb{Q}_p(\chi)$ and hence $V\otimes \mathbb{C_p}=\mathbb{C}_p(\chi)$, which is not isomorphic as a $G_p$-module to any $\mathbb{C_p}(i)$. What elementary thing am I missing? Is it just a matter of conventions?
Another example that I do not understand is: let $E/\mathbb{Q}_p$ be an elliptic curve with no CM with good ordinary reduction. Then the Tate module of $E$ is isomorphic to a non-trivial extension of $\mathbb{Q}_p(\psi^{-1})$ and $\mathbb{Q}_p(\psi)(1)$, where $\psi$ is the unramified character sending the frobenius to $a_p(E)$. Hence $T_p(E)\otimes \mathbb{C}_p$ is isomorphic as a $\mathbb{C}_p[G_p]$-module to a non-trivial extension of $\mathbb{C}_p(\psi^{-1})$ and $\mathbb{C}_p(\psi)(1)$. But this is not isomorphic to $\mathbb{C}_p \oplus \mathbb{C}_p(1)$, which is what according to the definitions in the literature amounts to saying that the Hodge-Tate weights are $0$ and $1$.
Can you explain me why tensoring with $\mathbb{C}_p$ allows us to forget the presence of finite order characters and allows us to work with the semisimplification of the representation?
