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In fact, when the voltage input op amp is in the linear working area, there is "virtual disconnection" and "virtual short circuit" at the two input terminals. The existence of "virtual disconnection" means that the input current of the two input terminals is very small, so that it can be regarded as an open circuit.  Details Published on 2019-5-7 10:14

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Flash is not installed (or the Flash version is too low) → Fix it now. You can also use a browser other than IE browser kernel, such as Google, Sogou (be sure to use a high-speed browser, otherwise it will still be compatible with IE kernel)
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At first I wanted to paste the picture directly, but the result was like the one shown above. Later I used the method of uploading the document.
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My question is: I originally thought that if the in-phase input terminal P2 of A2 is virtual ground, then Up2=0, and Un2=0, then Un1=Up1=0; when assuming that Ui is +, then Uo is -, causing the diode to be cut off, the current flowing through is 0, according to the virtual cutoff, then In2=Ip2=0, then the current of R1 is 0, so Up2=Ui is deduced, which is inconsistent. Where is the problem?
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Originally I thought that if the non-inverting input terminal P2 of A2 is virtual ground, then Up2=0, and Un2=0, then Un1=Up1=0; In the figure below on the 6th floor, the non-inverting input terminal of A2 is originally grounded through a resistor, so where does the "P2 virtual ground" come from?  Details Published on 2019-5-6 15:11
Originally I thought that if the non-inverting input terminal P2 of A2 is virtual ground, then Up2=0, and Un2=0, then Un1=Up1=0; In the figure below on the 6th floor, the non-inverting input terminal of A2 is originally grounded through a resistor, so where does the "P2 virtual ground" come from?  Details Published on 2019-5-6 15:07
 
 
 
 

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Source document of the problem 通知.doc (488 KB, downloads: 5)
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Please help the OP to post a picture first
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Thank you  Details Published on 2019-5-6 20:10
 
 
 
 

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msddw posted on 2019-5-6 14:01 My question is: I originally thought that the non-inverting input terminal P2 of A2 is virtually grounded, then Up2=0, and Un2=0, then Un1=Up1=0; ...
I originally thought that the non-inverting input terminal P2 of A2 is virtually grounded, then Up2=0, and Un2=0, then Un1=Up1=0; In the figure below on the 6th floor, the non-inverting input terminal of A2 is originally grounded through a resistor, so where does the "P2 virtual ground" come from?
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msddw published on 2019-5-6 14:01 My question is: I originally thought that if the in-phase input terminal P2 of A2 is virtual ground, then Up2=0, and Un2=0, then Un1=Up1=0; ...
According to the virtual break, In2=Ip2=0, then the current of R1 is 0, then Up2=Ui is deduced, which is inconsistent. Only when the op amp works in the linear region will there be "virtual break" and "virtual short". When the input Ui is positive, the op amp A2 is not in the linear working area, so there is no "virtual break" and "virtual short" between the two input terminals of A2.
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What are the potentials of Up2 and Un2 respectively?  Details Published on 2019-5-6 20:49
 
 
 
 

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topwon posted on 2019-5-6 14:20 Please help the OP to post a picture first
Thank you
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maychang published on 2019-5-6 15:11 According to the virtual disconnection, In2=Ip2=0, so the current of R1 is 0, which means Up2=Ui, which is inconsistent. Only when the op amp works in the linear region, can...
What are the potentials of Up2 and Un2 respectively?
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"What are the potentials of Up2 and Un2 respectively?" Please note what I said on the 8th floor: "When the input Ui is positive, op amp A2 is not in the linear operating area, so there is no "virtual break" or "virtual short" between the two input terminals of A2." Up2 and Un2 are different for positive and negative inputs.  Details Published on 2019-5-6 20:54
 
 
 
 

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msddw posted on 2019-5-6 20:49 What are the potentials of Up2 and Un2 respectively?
"What are the potentials of Up2 and Un2 respectively?" Please note what I said on the 8th floor: "When the input Ui is positive, the operational amplifier A2 is not in the linear working area, so there is no "virtual break" or "virtual short" between the two input terminals of A2." Up2 and Un2 are different for positive and negative inputs.
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I understand that it does not work in the linear region because the negative feedback network does not exist. In the video, how did Mr. Hua say that the current in R1 is 0 and Un2 is equal to Ui? I still don't understand this part.
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When diode D is cut off, the current in the diode is zero, the current at the two op amp input terminals is also zero, and naturally the current in R1 is also zero. There is no current in R1, so naturally Un2 is equal to Ui.  Details Published on 2019-5-7 07:47
 
 
 
 

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msddw posted on 2019-5-6 21:53 I understand why it does not work in the linear region, because the negative feedback network does not exist. In the video, how did Mr. Hua say that the current in R1 is 0 and Un2 is equal to Ui...
When diode D is cut off, the current in the diode is zero, and the currents at the two op amp input terminals are also zero, so the current in R1 is also zero. There is no current in R1, so Un2 is equal to Ui.
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How can the current at the two op amp input terminals be zero?  Details Published on 2019-5-7 08:17
 
 
 
 

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maychang published on 2019-5-7 07:47 When diode D is cut off, the current in the diode is zero, the current at the two op amp input terminals is also zero, and naturally the current in R1 is also zero. There is no current in R1, so U ...
How can the current at the two op amp input terminals be zero?
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"How can the current at the two op amp inputs be zero?" For an ideal op amp, the input impedance is infinite, so of course the input current is zero. For a non-ideal op amp, the input impedance is also quite large, and the input current is usually less than 1uA, or even up to pA. Only current-input op amps have slightly larger input currents.  Details Published on 2019-5-7 09:21
 
 
 
 

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msddw published on 2019-5-7 08:17 How can the current at the two op amp inputs be zero?
"How can the current at the two op amp inputs be zero?" For an ideal op amp, the input impedance is infinite, so of course the input current is zero. For a non-ideal actual op amp, the input impedance is also quite large, and the input current is usually less than 1uA, and can even reach the order of pA. Only current-input op amps have slightly larger input currents, which are usually not negligible. But the picture clearly does not show this type of op amp. Current-input op amps can work at very high frequencies, and of course they are more expensive. Current-input op amps are generally not mentioned in general analog circuit textbooks.
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OK, I understand. Thank you, teacher.  Details Published on 2019-5-7 09:53
 
 
 
 

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I don’t understand either!
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maychang posted on 2019-5-7 09:21 "How can the current at the two op amp inputs be zero?" For an ideal op amp, the input impedance is infinite, so of course the input current is zero. For a non-ideal actual op amp...
Well, I understand. Thank you, teacher.
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In fact, when the voltage input op amp is in the linear working area, there is a "virtual disconnection" and "virtual short" at the two input terminals. The existence of a "virtual disconnection" means that the input current of the two input terminals is so small that it can be regarded as an open circuit.  Details Published on 2019-5-7 10:14
 
 
 
 

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msddw published on 2019-5-7 09:53 Yes, I understand. Thank you, teacher
In fact, when the voltage input op amp is in the linear working area, there is "virtual disconnection" and "virtual short circuit" at the two input terminals. The existence of "virtual disconnection" means that the input current of the two input terminals is very small, so that it can be regarded as an open circuit.
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