Timeline for Understanding the precision rectifier
Current License: CC BY-SA 4.0
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20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 2, 2021 at 18:07 | history | edited | Essam | CC BY-SA 4.0 |
Fixed a sign symbol.
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| Jan 23, 2021 at 21:41 | vote | accept | Essam | ||
| Jan 23, 2021 at 21:00 | history | tweeted | twitter.com/StackElectronix/status/1353085114226507776 | ||
| Jan 23, 2021 at 20:27 | answer | added | WhatRoughBeast | timeline score: 1 | |
| Jan 23, 2021 at 15:26 | comment | added | Circuit fantasist | Let us continue this discussion in chat. | |
| Jan 23, 2021 at 13:43 | comment | added | Essam | Do we solve the problem, if on supplying the positive input signal, \$V_{Y}\$ changes (becomes positive) before \$V_{out}\$ and thus we immediately transition to the highly disturbed follower in the left? | |
| Jan 23, 2021 at 13:31 | comment | added | Essam | However, for some reason, the latter doesn't seem to apply. It's like if forward bias is satisfied (According to the equations that I have) regardless to whether \$V_{in}>0\$ or \$V_{in}<0\$ meanwhile, reverse bias is only satisified if \$V_{in}>0\$ | |
| Jan 23, 2021 at 13:27 | comment | added | Essam | Umm. I think using one equation to describe the diode would be tedious to do, I'm fine with using the constant voltage model and having two separate models for the diode (open-circuit or DC source) depending on whether it's reverse or forward bias. Here, it's so easy to see that if we're in reverse bias and supply \$V_{in}<0\$ then the polarities invert and we go to forward bias. So I thought it should be also easy to see that if we're in forward bias and at some point realize \$V_{in}>0\$ then polarities should invert and we should go to reverse bias again. | |
| Jan 23, 2021 at 13:20 | answer | added | Andy aka | timeline score: 2 | |
| Jan 23, 2021 at 13:13 | comment | added | Circuit fantasist | 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... | |
| Jan 23, 2021 at 11:12 | comment | added | Circuit fantasist | 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. | |
| Jan 23, 2021 at 10:54 | comment | added | Essam | @Circuitfantasist it's just about me trying to lay it down it in equations, it's hard to see from the equations on the right that if \$V_{in}>0\$ then it must transition to the highly distributed follower in left. | |
| Jan 23, 2021 at 10:51 | comment | added | Circuit fantasist | Another related question with a lot of comments from folk is Confusion in precision rectifier. | |
| Jan 23, 2021 at 10:45 | comment | added | Circuit fantasist | 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... | |
| Jan 23, 2021 at 10:45 | comment | added | Andy aka | 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. | |
| Jan 23, 2021 at 10:42 | comment | added | Essam | Sure, I just did that, here it is: electronics.stackexchange.com/questions/543299/… | |
| Jan 23, 2021 at 10:39 | comment | added | Andy aka | 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. | |
| Jan 23, 2021 at 10:36 | comment | added | Essam | 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. | |
| Jan 23, 2021 at 10:07 | comment | added | Andy aka | Did you ask a similar question on this general theme a few days ago? | |
| Jan 23, 2021 at 9:49 | history | asked | Essam | CC BY-SA 4.0 |