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I have confusion related to feedback in op amp. We know that the input resistance of op amp is very very high so why does the input current not flow from input to the feedback and directly to output. If so what is use of op amp we could just use very high resistance in place of it.Circuit diagram example is in the link given below.

PS- I tried to find about it and came upon the website-https://terpconnect.umd.edu/~toh/ElectroSim/CurrentFlow.html

enter image description here

according to it the way o\p signal is amplified there is no use of op amp at all!

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  • \$\begingroup\$ Please ask questions as clearly as possible: 1) All circuit questions should have a circuit diagram, 2) Define all terms. For example, what is the "i\p current"? \$\endgroup\$ – DanielSank Mar 4 '17 at 9:24
  • \$\begingroup\$ sorry i\p is input current \$\endgroup\$ – Shreya Sharma Mar 4 '17 at 9:47
  • \$\begingroup\$ input current flows to the output, the key is that not all output current comes from the input \$\endgroup\$ – Vladimir Cravero Mar 4 '17 at 10:24
  • \$\begingroup\$ Sir could you elaborate? Could you tell me how feedback returns the output and is not simply a link (like short circuit) between input and the output? \$\endgroup\$ – Shreya Sharma Mar 4 '17 at 10:31
  • \$\begingroup\$ You have TWO voltage sources - at the input and at the opamps output. Use superposition for the current. \$\endgroup\$ – LvW Mar 4 '17 at 11:17
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We know that the input resistance of op amp is very very high so why does the input current not flow from input to the feedback and directly to output.

Because, of course, the load is not only connected to the input (via the feedback resistor) but also to the op amp output, and the output has a very low output imnpedance. The output current (and voltage) is essentially independent of the input current (but not the voltage difference between the inputs). So the output supplies current to the load and also to the feedback resistor, and the circuit in no way supplies current through the input.

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Most physical op-amps approximate an ideal op-amp.

In an ideal op amp, $$ v_{out} = \infty \cdot (v_+ - v_-) $$ and the current into any input is zero.

The feedback circuit to \$v_-\$therefore keep \$v_{out}\$ at such a level that \$v_- -v_+ = 0 \$.

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An OpAmp is an Operational Amplifier. Amplifier is here the important word. Of course there is almost no current flowing into the input (\$nA\$ to \$fA\$), however this current is amplified by a factor (let's say 1e3 to 1e6) and now has the capability to drive a sink. At this point it is important to mention that the difference between positive and negative input is amplified. Looking at an ideal OpAmp this differential gain is infinite and the common mode gain is 0.

Here is an application related explanation > Understanding Operational Amplifier Specifications.

Further it is interesting to look inside an OpAmp, then it's obvious that this device is more than a resistor. Below you can see a simplified schematic of an OpAmp. enter image description here

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  • \$\begingroup\$ "....however this current is amplified by a factor (let's say 1e3 to 1e6)". Who has told you THIS? Of, course, it is the differential VOLTAGE between both inputs that gets amplified. However, due to the large (open-loop) gain up to 120 dB we operate this devive ALWAYS with negative feedback. (Did YOU have a look into the referenced document?) \$\endgroup\$ – LvW Mar 4 '17 at 12:29
  • \$\begingroup\$ You're right, I'm going to improve my answer with the hint of the amplification of the voltage difference between both inputs. But I'm not sure about the other points. 120dB are exactly 1e6. \$\endgroup\$ – auoa Mar 4 '17 at 12:40

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