5. Given $\odot O: x^{2}+y^{2}=1$ with a moving chord $A B=\sqrt{3}$, and a moving point $P$ on $\odot C:(x-2)^{2}+(y-3)^{2}=2$. Then the minimum value of $\sqrt{\overrightarrow{P A} \cdot \overrightarrow{P B}+\frac{3}{4}}$ is $\qquad$.
5. $\sqrt{13}-\frac{1}{2}-\sqrt{2}$.

Let the midpoint of $A B$ be $D$. Then
$$
\begin{array}{l}
\overrightarrow{P A} \cdot \overrightarrow{P B}=(\overrightarrow{P D}+\overrightarrow{D A}) \cdot(\overrightarrow{P D}+\overrightarrow{D B}) \\
=(\overrightarrow{P D}+\overrightarrow{D A}) \cdot(\overrightarrow{P D}-\overrightarrow{D A}) \\
=\overrightarrow{P D}^{2}-\left(\frac{|A B|}{2}\right)^{2}=\overrightarrow{P D}^{2}-\frac{3}{4} .
\end{array}
$$

Notice that, $O$ and $C$ are fixed points.
Clearly, $P$ is one of the intersection points of $O C$ and $\odot O$. When point $D$ is on $O C$, $P D$ is the shortest, at this time,
$$
\begin{array}{l}
|O D|=\frac{1}{2},|O C|=\sqrt{13},|O P|=\sqrt{13}-\sqrt{2} . \\
\text { Hence }|P D|_{\min }=\sqrt{13}-\frac{1}{2}-\sqrt{2} .
\end{array}
$$