Let $X$ be a scheme and $U$ be an open subscheme. The proof of the Thomason-Trobaugh Theorem implies that under some mild assumptions, for any perfect complex $F$ on $U$, we have that $F\oplus F[1]$ can be extended to a perfect complex on $X$. I'm just wondering whether there exists examples where $F$ is a perfect complex on $U$ but $F$ itself cannot be extended to $X$?

Let $X$ be a scheme and $U$ be an open subscheme. The proof of the Thomason-Trobaugh Theorem implies that under some mild assumptions, for any perfect complex $F$ on $U$, we have that $F\oplus F[1]$ can be extended to a perfect complex on $X$. I'm just wondering whether there exists examples where $F$ is a perfect complex on $U$ but $F$ itself cannot be extended to $X$? I've found an example when $X$ is the cone $xy-z^{2}=0$ and $U$ is the complement of the origin. Is there an example for smooth $X$?