Why is the output of the precision rectifier 0 when the diode does not conduct?
If the diode doesn't conduct, then no current flows through it.

The op-amp has high input impedance, so effectively no current flows into the inverting input pin.

So no current is flowing through the resistor.

Therefore, what is the voltage across the resistor RL?


  When the diode is off, shouldn't V- be equal to Vin if we are assuming infinite open loop gain?


If the diode is not conducting, then you don't have a closed feedback loop. 


  If there is an infinite open loop gain with no feedback, as is the case in the ideal diode, then this equation must hold: Vout = A(V+ - V-)


This is simply incorrect.

V- is driven equal to V+ by having high open-loop gain, and a negative feedback connection. High open-loop gain is not a sufficient condition to force V- equal to V+.

The equivalent circuit with the diode not conducting looks like this:







simulate this circuit &ndash; Schematic created using CircuitLab

There is no way for the op-amp's output to drive the inverting input in any way, towards the noninverting input or otherwise. There is no negative feedback and there is no forcing of V- equal to V+.

What happens in reality is that the output voltage is limited by the power supplies of the op-amp, putting the op-amp in saturation mode where the open-loop gain is no longer very high.

But that isn't why the inputs aren't forced together. If the input were 1 uV, and the gain 1,000,000, then the output could be 1 V, and it still wouldn't force the inverting input to 1 uV, because there is no closed feedback loop.