Given that $|a|=3$ and $|b|=2$, we can deduce the possible values of $a$ and $b$ as follows:

- Since $|a|=3$, this means $a=\pm 3$.
- Similarly, $|b|=2$ implies $b=\pm 2$.

Given the condition $a+b > 0$, we need to consider the possible combinations of $a$ and $b$ that satisfy this inequality:

1. If $a=3$ and $b=2$, then $a+b=3+2=5$, which is greater than $0$.
2. If $a=3$ and $b=-2$, then $a+b=3-2=1$, which is also greater than $0$.
3. The combinations where $a=-3$ would not satisfy $a+b > 0$ given the possible values of $b$.

Therefore, the valid combinations under the given conditions are $a=3, b=2$ and $a=3, b=-2$.

For these combinations, we calculate $b-a$ as follows:

1. If $a=3$ and $b=2$, then $b-a=2-3=-1$.
2. If $a=3$ and $b=-2$, then $b-a=-2-3=-5$.

Hence, the possible values for $b-a$ are $-1$ or $-5$, which can be encapsulated as $\boxed{-1 \text{ or } -5}$.