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I'm very new to this topic Operational Amplifiers. Just started studying for my 4th semester. Can someone please explain to me why there is a virtual ground at the node after Rin. What is the reason for that?

op

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  • \$\begingroup\$ You should find most of what you are looking for here electronics.stackexchange.com/questions/107249/… \$\endgroup\$ Aug 10 '14 at 16:49
  • \$\begingroup\$ Because of the differential input stage. \$\endgroup\$ Aug 10 '14 at 17:50
  • \$\begingroup\$ hello all, I'm sorry if this question seems like a duplicate. I checked that solution and I didn't quite get the answer to my question. I will edit this post again by this evening. Have an assignment due right now :) \$\endgroup\$
    – Shivji
    Aug 11 '14 at 2:13
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The op amp + input is ground. The output of the op amp will feed back and drive the - input to something very close to the + input.

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  • \$\begingroup\$ It's "Virtual Ground" is another way of saying "Like or same value as Ground". In fact an easy way to fault find an analog circuit(after checking the voltage rails...) is to find the virtual grounds and check they are the same as Ground (unless there is some offset but that should be obvious from the circuit) \$\endgroup\$
    – Spoon
    Aug 10 '14 at 18:54
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The output voltage of an operational amplifier wants to be \$A \cdot (V_+ - V_-) \$. Where \$A\$ is the gain of the amplifier and is a large number. \$V_+\$ is the voltage at the non inverting input and \$V_-\$ is the voltage at the inverting input.

This circuit has negative feedback so that if you take \$V_{in}\$ positive \$V_{out}\$ will go negative. Because \$V_{out}\$ goes negative this tends to pull \$V_-\$ back down towards zero.

Similarly if you take \$V_{in}\$ negative \$V_{out}\$ will go positive. Because \$V_{out}\$ goes positive this tends to pull \$V_-\$ back up towards zero.

In reality the voltage at \$V_-\$ can't be exactly ground otherwise there would be no output but if the gain is high enough the amount that \$V_-\$ moves is negligible so we say it's a virtual ground. The voltage at \$V_+\$ is ground because that's what we have connected it to, and because of the negative feedback the voltage at \$V_-\$ is virtually (almost exactly) the same as the voltage at \$V_+\$ so it's a virtual ground.

Without this negative feedback we wouldn't have a virtual ground and if \$V_{in}\$ were to go positive the amplifier output would saturate and be close to the negative supply; if \$V_{in}\$ were to go negative the amplifier output would saturate and be close to the positive supply.

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Negative feedback establishes a kind of equilibrium which leads to the following condition:

The input voltage Vn at the inverting terminal which is caused by both voltages (superposition rule) - Vin and Vout - is Vn=Vout/Aol (Aol being the open-loop gain of the opamp.) Because of the large value of Aol the voltage Vn is very small (typically in the µV range). This is very close to zero (ground) - and this is the reason we speak of "virtual ground".

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