Confusion in precision rectifier

I answered this SE EE question on Nov 19, 2020.

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".

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.

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.

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.

If you still have a Flash player, you can also visit the Flash movie Strange Things can be Put into Feedback Loop that I dedicated to Tom Hayes and his Student Manual for the Art of Electronics (to watch the movie, you need to add Ruffle Flash emulator to your browser). Click on the diode symbol in the library on the left.

My comments

1. 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... (to be continued...)
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. Now it's your turn to respond...