Precision Rectifier confusion

I answered this SE EE question 10 hours ago. Here is my answer.

Looking for the idea

This circuit solution is much more clever than it seems at first glance. It gives the illusion of being a very imperfect half-wave rectifier... and has led some to improve it (including a second diode to maintain a virtual ground on the inverting input) to make it a perfect half-wave rectifier. But I think the idea here is different:

This circuit is intended as a "slightly imperfect full-wave rectifier", and the "improvement" only destroys this idea.

Operation

Here is how the circuit works shown in a slightly different way than generally accepted.

Negative input voltage

The diode is forward biased (on) and the circuit is simply an inverting amplifier. In this configuration, we can see (with a little more imagination) two "voltage sources" - negative and positive, connected in parallel to the load.

The first of them consists of the input voltage source and the two 2k resistors in series; so it behaves as a real (imperfect) voltage source with 4k internal resistance.

The second "voltage source" consists of the op-amp output and the low-resistance diode in series. Since the negative feedback voltage is taken after the diode, the op-amp compensates for the forward voltage VF by increasing its output voltage by VF. In addition, through the mechanism of negative feedback, the op-amp compensates for the influence of the load. As a result, the circuit is perfect and behaves as an "ideal" (perfect) voltage source with (almost) zero internal resistance.

In the unequal "struggle" between the two voltage sources, the second one (the op-amp) prevails and its voltage is applied to the load.

More precisely, the voltages of the two sources are summed (subtracted) through their respective resistances and their sum is applied to the load. And because the op-amp output resistance is much less than the input source resistance, the load voltage is almost entirely determined by the op-amp output voltage.

Positive input voltage

Now the diode is backward biased (off) and the second voltage source (the op-amp) is disconnected. Only the input source remains to operate and it supplies its voltage through the network of two resistors to the load. Figuratively speaking, the feedback has become a "feedforward":-)

And because the total resistance of the resistors is relatively low (a few kohms) and the load resistance is assumed to be high (hundreds of kohms), almost all of the input voltage is transferred to the load. A straightforward solution to improve the circuit is to insert a voltage follower before the load.

Summary

Let's finally summarize the operation of this full-wave rectifier.

During the negative half-wave, the op-amp makes an inverted copy of the input voltage and passes this positive voltage through the diode to the load.

During the positive half-wave, the input voltage is applied through the feedback network directly to the load.

Simply put, the rule for making a full-wave rectifier is: the negative half-wave is inverted and the positive half-wave is not.

Generalization

Extracting general rules (principles) from specific circuit solutions is no less important than giving precise answers to specific questions.

Following this approach, we can see this configuration (only at negative input voltage here) in any negative feedback inverting circuit where the input source tries to feed its voltage directly to the load but the op-amp does not allow it and forces its output voltage to the load.

So, if the feedback resistors in op-amp inverting circuits are with too low resistance and the input source is too powerful, it will dominate and its voltage will be transferred directly to the output (undesired).

In contrast, in non-inverting negative feedback circuits this is not observed because there the input voltage source has no connection to the op-amp output.