There are two equivalent definitions of a locally convex topological vector space. Note that since the vector space is topological, addition by a fixed vector x and multiplication by a fixed scalar r are both homeomophisms, which means it is enough to give a neighborhood basis at the vector 0.
The first definition is that $0$ has a local base consisting of open sets that are convex (if $x$ and $y$ are in $S$, then $\lambda x+(1-\lambda)y$ is also in $S$ for every $\lambda\in[0,1]$), balanced, (if $x\in S$, then $r x\in S$ for every $r\in\mathbb{F}$ with $|r|=1$, where $\mathbb F$ is either the real or complex numbers), and absorbing (for any vector $y$ there exists a $\lambda\in (0,\infty)$ such that $\lambda S$ contains $y$, i.e. $S$ absorbs $y$).
The second definition is that $0$ has a local base given by the ''balls'' of radius $r$ for each $r$ of each semi-norm in some fixed collection of semi-norms (a semi-norm is just like a norm except that non-zero vectors can have a non-zero semi-norm). The two definitions are equivalent in one direction because "balls" of semi-norms are convex, balanced and absorbing, and in the other because we can define for any convex, balanced, absorbing set $S$ the semi-norm $f_S$ by assigning to $x$ the "least" (infimum) $\lambda$ for which $x$ is in $\lambda S$.
My question is: what are some examples of topological vector spaces for which $0$ does have a local base of convex open sets, but that no local base consists of convex sets that are also balanced and absorbing?