In particular, I was trying to understand the behavior of this precision rectifier given an input wave \$V\_{in}=V\_{o}sin(\omega t)\$. To my understanding, the circuit reduces to what we have in the left for \$V\_{in}>0\$ (first half-wave) and whenever \$V\_{in}<0\$ the voltage polarity on the diode becomes positive and we move to what we have in the right.

[![Precision Amplifier, Note that is should be Vy+Von instead of Vy-Von](https://i.stack.imgur.com/Gh8CH.png)](https://i.stack.imgur.com/Gh8CH.png)
So now, we're in the second half of the wave \$V\_{in}<0\$ and we will remain there so long as the voltage polarity on the diode is positive. I'm confused because given the two equations that I have the polarity on the diode will be positive no matter what; \$V\_{out} > V\_{Y}\$ always holds true due to the second equation because \$V\_{ON}\$ should be constant. Meanwhile, I think that there should be a way for us to transition to what we have in the left again once \$V\_{in}>0\$, so what am I missing?

Edit:
In the picture, \$V\_{y}-V\_{on}\$ should indeed be \$V\_{y}+V\_{on}\$