Why is the output of the precision rectifier 0 when the diode does not conduct?

Question

From Wikipedia: "When the input voltage is negative, there is a negative voltage on the diode, so it works like an open circuit, no current flows through the load, and the output voltage is zero."

When the diode is off, shouldn't V- be equal to Vin if we are assuming infinite open loop gain? In that case, isn't Vout = Vin instead of 0 when the diode is off?

*** EDIT:
To address some of the comments, open loop gain is the gain when no feedback is used. So even if the diode is off, there is an open loop gain A on the node directly in front of the diode on the output. This of course Doesn't reach Vout, but it's there.

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

If the open loop gain is infinity, then Vout/A = 0 = V+ - V- --> V+ = V-. If the input resistance is also infinity, then there's no current going into Vout and it should be zero, but it's also Vin by virtue of there being infinite open loop gain.

Edit 3:

I've answered my own question below and removed my answer from the question.

Answer 1

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 – 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.

Answer 2

When dealing with opamps, we generally use the two rules:

1. Opamp inputs have infinite impedance

and

2. The opamp will adjust its output so that the V+ input and V- input are at the same potential.

When the diode is off, that means that it's not conducting any current from the opamp output. The only other thing connected to that node is the V- input, which is high impedance and can't source any current, so no current is flowing and Vout is at 0V (pulled down by the resistor).

Answer 3

The real problem I had was with the idealized model of the op amp that I was using.

Below the Open Loop Gain A is a property of the op amp. It does not change depending on the circuit configuration.

The problem with this model is that it does not capture the effect of the output on the input. From the diagram above, you can attain:

Using V+ = V2 and V- = V1,

Although you can manipulate the equation of the output voltage to attain V+ = V-, there is no feedback to actually have the output have any effect on the input. In this idealized model, it only captures the effect of the input on the output, but not the output on the input, so solving for the inputs given the output doesn't attain a realizable result. There is no causal link from the output to the input. There is only a link from the input to the output in this model.

The equation for the open loop gain cannot be used to determine the inputs. That's really the confusion that I had.

Given this clarification, it should be clear that with infinite input resistance in the ideal op amp, that there is no current going through R_L, and that V- is not equal to V+. Therefore, Vout (or V-) is pulled down to 0.