If I have a quasiprojective variety $X$, and a subscheme $Z$, then the blowup 
$$f:Y = Bl(X,Z)\rightarrow X$$ 
is projective over $X$, since it is constructed by a relative Proj construction. Can I find a relatively ample bundle on $Y$ that is trivial on
$$f^{-1}(X\backslash Z)?$$


At first I thought the construction of $O(1)$ in the Proj construction guarantees this property. However, if I have a projective small contraction $Y \rightarrow X$, then it is a blowup map, but then certainly an ample bundle can't be trivial on the birational locus, since it would be trivial on $Y$.


So under what conditions can I find such a relatively ample bundle?


Edit: Karl's answer explains away my confusion: I have to blow up more than the image of the exceptional locus in case of a small contraction. It's still reasonable to ask when it's trivial on the birational locus, in which case Sandor's answer applies.



Projectivity of blowups
This can be done if $X$ has $\mathbb Q$-factorial singularities (but this is not a necessary condition!): Let $H$ be a relative ample effective divisor (not a bundle, divisor!). Then $f\_\*H$ is a Weil divisor on $X$ and if $X$ has $\mathbb Q$-factorial singularities, then some multiple of $f\_\*H$ is Cartier. Replacing $H$ with the appropriate multiple we may assume that $f\_\*H$ is Cartier. Then it can be pulled back and $f^\*(f\_\*H)$ is relatively numerically trivial, hence $H-f^\*f\_\*H$ is still relatively ample. Notice that $f^\*f\_\*H-H$ is
an effective exceptional divisor, hence trivial when restricted to the complement of the exceptional set.


Also notice that indeed if $f$ is a small contraction, then one cannot find such a divisor, but also, if $X$ has $\mathbb Q$-factorial singularities, then the exceptional set of $f$ has pure codimension $1$.


