Ideal Diode

In an ideal diode circuit, the feedback “cancels” the op-amp forward voltage drop, so that we have something like:

The circuit uses an op-amp and looks like this:

For example, if we have $V_{in}=3\text{ V}$ and $V_F=0.7\text{ V}$. In order for the op-amp to make $V_{+}-V_{-}=0$, we need $V_{out}$ after the diode to be $3\text{ V}$. So, the output of the op-amp has to be $3.7\text{ V}$ in order to account for the forward voltage.