A story about the op-amp precision rectifier

 My comments under SE EE question Understanding how the simple precision rectifier works

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This is a story about the famous circuit with a figurative name "ideal diode". I started my contribution by writing comments under the first question.

Question 1

My comments

1. Consider the circuit as a '"disturbed op-amp follower". Here is a Wikibooks story about this phenomenon observed in negative feedback circuits.
2. Try a simple experiment - apply some negative input voltage (e.g., Vin = -1 V) and put yourself in the place of the op-amp. What will you do to make the voltage of the inverting input equal to the voltage of the non-inverting input (the "golden rule")? What will be the op-amp output voltage?
3. The so-called "golden rule" is introduced in the famous book of "The Art of Electronics". It says that an op-amp with negative feedback does its best to maintain equality between its two input voltages. In your circuit, it will make Vout = Vin. The diode is not perfect (it's not just a "short circuit")... and a voltage drop of about 0.7 V is lost across it. The op-amp "senses" it and additionally increases the magnitude of its output voltage (before the diode) with 0.7 V. In this way, the op-amp compensates the undesired VF = 0.7 V. Have you understood the great idea here?
4. The op-amp compensate the voltage loss VF across the diode by an excessive output voltage VA = Vout + VF. You have to distingiish between VA and Vout - VA is the output voltage before the disturbance VF while Vout is the output voltage after the disturbance.
5. So, let's return to my example - Vin = -1 V. The op-amp will begin decreasing its output voltage VA towards the negative supply rail "observing" the difference Vin - Vout... and it will stop when Vout becomes equal to Vin (-1 V). Then the op-amp voltage VA will be -1.7 V and the circuit output voltage Vout will be -1 V.
6. When the input voltage is positive, the op-amp begins changing its output voltage towards the positive rail so the diode is backward biased and it does not conduct. The op-amp does not manage to make the voltage of the inverting input equal to the voltage of the non-inverting input... and it reaches the positive rail. However, Vout = 0 V...

Then OP showed an interest to my answer to another related question... and our conversation continued there. Here is it...

Question 2

My answer to SE EE question Confusion in precision rectifier

My answer

Razavi's book is great but I cannot agree that "when v(input)=0 op amp raises Vy such that it barely turns on the diode." OP is right - if the op-amp is ideal, its output voltage will be zero when the input voltage is zero. The diode will be turned on only if the input voltage rises at least a little.

There is a lot of philosophy in this specific circuit that illustrates the unique property of negative feedback systems to overcome disturbances. It is not only a particular circuit phenomenon; it is a ubiquitous phenomenon that can be seen everywhere around us.

I have revealed this idea below in three consecutive steps to show its evolution. Prerequisite to understand what the op amp does in such a negative feedback circuit is to think of it not as a fast-acting amplifier but as a slow-acting "being".

1. Undisturbed follower. First replace the diode by a piece of wire (i.e., connect the op-amp output directly to its inverting input) and apply an input voltage Vin = 1 V at the non-inverting input - Fig. 1. The op-amp senses the (positive) voltage difference between its inputs and begins increasing its output voltage (Y) until zeros the difference. It does this without any effort because there is no disturbance. As a result, the circuit output voltage Vout (here, the same as the op-amp output voltage) is always equal to the input voltage. This is the well-known circuit of an op-amp follower.

Fig. 1. Undisturbed follower

We all act this way in life - we set a goal and begin to realize it by constantly comparing what we have achieved with the goal. For example, in this way we maintain the level of our speech equal to some level through the acoustic feedback between our mouth and ears... and we do it effortlessly because there is nothing to hinder us.

2. Slightly disturbed follower. Now put the diode back in its place - Fig. 2. A current begins flowing from the op-amp output through the diode and resistor R1 to ground. The diode begins conducting (forward biased) and a disturbing voltage drop VF = 0.7 V appears across the diode. It is subtracted from the op-amp output voltage (Y)... so the circuit output voltage Vout drops to 0.3 V... and again a positive voltage difference appears between the op-amp inputs. The op-amp senses it and begins increasing its output voltage (Y) until reaches 1.7 V thus zeroing the difference. As a result, the circuit output voltage Vout is again equal to the input voltage. Thus, at the cost of additional "effort" (extra op-amp output voltage of 0.7 V), the op-amp overcomes the diode interference.

Fig. 2. Slightly disturbed follower

In life, if an obstacle stands in our way when we achieve our goal, we overcome it at the cost of extra effort. For example, in our "speech analogy", imagine we enter a building; so we have to put on our mask against Covid 19. Attenuation appears in the acoustic feedback path (the mask acts as the diode in the OP's circuit) and we unconsciously begin increasing the level of our voice to compensate for this disturbance. This costs us extra effort.

3. Highly disturbed follower. Now change the polarity of the input voltage Vin = -1 V. The op-amp output voltage becomes negative and the diode is cut off. The negative feedback is broken and the circuit output voltage Vout becomes zero. Now a negative voltage difference appears between the op-amp inputs. The op-amp senses it and begins changing its output voltage (Y) below ground with the hope that it will zero the difference. In an effort to do so, it drops its output voltage lower and lower... but the difference does not change... and the output voltage finally reaches the voltage V- of the negative supply rail.

Fig. 3. Highly disturbed follower

In life, when obstacles are too great, they become insurmountable... and we expend all our energy to overcome them... but fail. In our "speech analogy", imagine we enter the Covid ward of the hospital. We are afraid of the infection and put on another very thick mask. But it completely stops our voice and we do not hear our own voice (no acoustic feedback). We start talking louder and louder... and finally we begin screaming.

I suggest you visit the laboratory exercises of my students from group 64b and 66a where you will see other interesting experiments with this diode circuit.

You can download the Flash movie Strange Things can be Put into Feedback Loop (exe file with embedded Flash player) that I dedicated to Tom Hayes and his Student Manual for the Art of Electronics. Click on the diode symbol in the library on the left.

My comments

1. То the OP: Thanks for the attention to my answer. I was going to write an answer to your question but I was pretty busy finishing the semester. OK, let's begin considering the problem here. From my experience I have understood a simple though paradoxical truth: to really understand op-amp circuits with negative feedback, you have to think of the op-amp not as an amplifier but rather as an integrator (something slow-acting as a human being); if you think of it as a fast proportional device (Vout = K.Vin), you will never understand op-amp circuits because you will revolve in a vicious circle...
2. So, the best way to understand what the op-amp does is to put yourself in its place. For example, conduct such a mental experiment - change the voltage of a variable voltage source VA (immitating the op-amp output) until the voltage Vout after the diode becomes equal to Vin (you can compare the voltages by connecting a sensitive voltmeter between them). Then let's explain, thinking in this intuitive way, what the op-amp does...
3. Let's first consider the positive input half wave. The input voltage Vin begins increasing above zero. The op-amp "senses" this positive disbalance and since its job is to keep it zero, it begins increasing its output voltage VA. Until Vin < 0.7 V, the diode does not conduct and the voltage at the non-inverting input does not "move"; so, the op-amp continues increasing VA with its maximum rate of change. When Vin approaches 0.7 V, the diode begins conducting and V(+) begins "moving" up... until it becomes (almost) equal to V(-)... and the equilibrium is restored...
4. As you know, a voltage drop about 0.7 V is lost across the diode and the op-amp has to overcome this "small disturbance" by increasing its output voltage with additional 0.7 V. You can think of the diode as of a 0.7 V "battery" connected in series and contrary to VA. That is why, VA will be "lifted" with 0.7 V above Vin until the diode is forward biased... and, as a result, Vout = Vin...
5. Now let's consider the negative half wave… When Vin begins decreasing and drops below zero, the op-amp "senses" this negative disbalance and since its job is to keep it zero, it begins changes the polarity of its output voltage to negative and again tries to make Vout = Vin… but fails since the diode has cut the connection between the output and the non-inverting input… and finally reaches the negative rail... So, this was my answer to your question above. Really, it is quite long... but this is the price of understanding... the other is just knowledge.

But obviously there was something misunderstood by the OP because he asked a third question ... and our discussion continued there... Here are my comments...

Question 3

My comments to SE EE question Understanding the precision rectifier

My comments

1. The problem for me is to understand what the problem is because the way it looks, everything is fine. If this will help you, here is another circuit explanation - when it is forward biased, the diode can be considered as a floating "shifting" battery connected in series to the op-amp output and the load. In this case, its voltage is added with a positive sign to the negative output voltage (ie, its magnitude is subtracted from it). I don't think you should concentrate so much on the formal presentation...
2. Then try another experiment - connect another diode contrary in parallel (back to back) to the existing one... and analyze the circuit operation. Or replace the existing diode with a Zener diode... first, with a low voltage... then, with a higher voltage... and analyze the circuit again... I would present the circuit operation geometrically.
3. I continue wondering what the problem is... Maybe you want to describe both cases by one equation? The problem is that the diode is a nonlinear element - its IV curve depends on the voltage polarity... and it is suitable to divide it in two parts (for the two polarities)... and to describe them by different equations...
4. As I can see, you have accepted the concept of "disturbance". It is interesting that we can enlarge it more using the disturbance as an input quantity (while keeping Vin constant). An example is a resistive sensor put in the feedback loop; VA will depend on its resistance Rs. We can elaborate more this idea making VA depend on both the input voltage and disturbance.
5. Let's try to see this idea in your circuit. While Vin < 0, the op-amp output voltage VA depends only on Vin (the first "disturbance"). When Vin crosses the zero level, the second disturbance (the diode "resistance") comes into picture and begins driving VA. Since the diode rapidly changes its "dynamic resistance", Vin can be considered as constant during the positive "jump" of VA. So, VA depends only on the "diode disturbance"...
6. As you can see, I have introduced another powerful concept - we can consider the input voltage of an op-amp amplifier with negative feedback as another "disturbance". Why? This is because the real job of such an ampifier is to keep up a zero voltage difference between the op-amp inputs. In other words, any amplifier with negative feedback is a follower disturbed by the input voltage... thus becoming an ampifier.
7. Let's now return to our diode circuit... We can explain the transition from the right to left circiit (and vice versa) in terms of resistances thinking of the diode as of a "self-variable" or "dynamic" resistor that vigorously changes its resistance from zero to infinity when Vin crosses the zero level. The op-amp reacts to this change of disturbance by varying its output voltage VA.

Here the OP stated that things have already become clear to him and we finished the discussion ...