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In particular, I was trying to understand the behavior of this precision rectifier given an input wave \$V_{in}=V_{o}sin(\omega t)\$. To my understanding, the circuit reduces to what we have in the left for \$V_{in}>0\$ (first half-wave) and whenever \$V_{in}<0\$ the voltage polarity on the diode becomes positive and we move to what we have in the right.

Precision Amplifier, Note that is should be Vy+Von instead of Vy-Von So now, we're in the second half of the wave \$V_{in}<0\$ and we will remain there so long as the voltage polarity on the diode is positive. I'm confused because given the two equations that I have the polarity on the diode will be positive no matter what; \$V_{out} > V_{Y}\$ always holds true due to the second equation because \$V_{ON}\$ should be constant. Meanwhile, I think that there should be a way for us to transition to what we have in the left again once \$V_{in}>0\$, so what am I missing?

Edit: In the picture, \$V_{y}-V_{on}\$ should indeed be \$V_{y}+V_{on}\$

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    \$\begingroup\$ Did you ask a similar question on this general theme a few days ago? \$\endgroup\$ Commented Jan 23, 2021 at 10:07
  • \$\begingroup\$ Yes, I did. But perhaps, a part of my question from last time was answered. this is what I need to get the full picture. \$\endgroup\$ Commented Jan 23, 2021 at 10:36
  • \$\begingroup\$ You should not have deleted it. Please reinstate/undelete it. I don't mind this latest question but I'd certainly like to re-read the previous one. Maybe if you don't want to reinstate it you can provide a link to it? You should be able to view your deleted posts. I'm unsure I can do that on your stuff else I'd do it myself. \$\endgroup\$ Commented Jan 23, 2021 at 10:39
  • \$\begingroup\$ Sure, I just did that, here it is: electronics.stackexchange.com/questions/543299/… \$\endgroup\$ Commented Jan 23, 2021 at 10:42
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    \$\begingroup\$ OK, that's good. I can remember it now. It's not a good idea to delete earlier questions and certainly true if they have formal answers given. However, your earlier question had a lot of comments from folk and that kind of makes it still a valuable backdrop for this latest question. \$\endgroup\$ Commented Jan 23, 2021 at 10:45

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In your right-hand diagram, the cathode of the diode is (approximately) held to ground by the resistor. When Vy becomes positive, current cannot flow to ground through the diode and resistor. If it did, that would require that the current flow through the diode from anode to cathode - and that cannot happen unless the diode is reverse-biased - and there cannot be current through a reverse-biased diode.

To put it another way, Von is not constant. It only applies when the diode is forward-biased. "Vout>VY always holds true due to the second equation because VON should be constant."

Let's look at what happens if Vin equals, let's say, 5 volts. It should be clear that, regardless of the diode condition, Vout will be less than Vin. If the diode is reverse-biased, no current can flow to the resistor, so its voltage (and Vout) must be zero. If the the diode is forward-biased, Vout must be Von less than Vy. If Vout is greater than Vin (let's say 6 volts just to illustrate), the output of the op amp must be A(Vin - V-), or A(5 - 6) or -A. This, in turn implies that Vout will be negative, and that is impossible.

So, as soon as Vin becomes greater than zero, the diode will become reverse-biased, and Von no longer applies.

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I'm confused because given the two equations that I have the polarity on the diode will be positive no matter what; \$V_{out}>V_{Y}\$.

In the left scenario, \$V_{out}\$ will be near the positive rail of the op-amp and \$V_Y\$ will be at 0 volts hence, the diode is reverse biased and not conducting.

Therefore "the polarity on the diode will be positive no matter" is incorrect.

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  • \$\begingroup\$ But once we enter the right scenario, how do we get out? I can't conclude, using the model above that we move on to the left scenario. having \$V_{in}>0\$ while on the right doesn't stop making it forward biased. \$\endgroup\$ Commented Jan 23, 2021 at 13:37
  • \$\begingroup\$ @Essam of course it does; once Vin rises positive the open-loop gain of the op-amp rapidly causes the op-amp output to go highly positive. \$\endgroup\$ Commented Jan 23, 2021 at 13:43

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