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Since no current flows in a op amp,therefore no current flows via resistor$$R_c$$,which implies that $$V_+ =0$$Using the concept of virtual ground we can say $$V_-=0$$.Now using the fact that $$V_0=A.(V_+-V_-)$$and $$V+ ,V_=0$$ we get that the $$V_0=0$$ which is incorrect. I can't understand where I'm making a mistake. I will be thankful for even a hint.

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    \$\begingroup\$ It's not a virtual ground, it is a virtual short between the two inputs. Too many people call it virtual ground without clarifying that this is true only if the non-inverting input is grounded. \$\endgroup\$ – Elliot Alderson Jul 6 at 13:12
  • \$\begingroup\$ Indeed, i agree. \$\endgroup\$ – Yasir Sadiq Jul 6 at 14:29
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The assumption that V+ and V- are both equal to each other requires that the gain of the opamp be infinitely large and that the opamp be connected in a negative feedback configuration. This is because the negative feedback, with the help of the large gain, will try to equalize the two inputs.

Now when you try to use the standard differential amplifier equation, $$V_{o}=A(V_+-V_-)$$ where \$A\$ is the opamp gain, you make the implicit assumption that \$A\$ is not infinite. Otherwise, this equation does not mathematically make sense.

To see how the feedback network determines the operation of the opamp, consider the transfer function of an amplifier with negative feedback: $$H=\frac{A}{1+Af}$$ where \$A\$ is the amplifier gain and \$f\$ is the gain of the feedback network. If \$A\$ is extremely large, \$1+Af\approx{}Af\$ and we get $$H\approx\frac{1}{f}$$ Thus, the transfer function entirely depends on the feedback network.

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  • \$\begingroup\$ If A is extremely large, 1+Af≈f nope. If \$A\$ is extremely large then \$1+Af\approx Af\$. \$\endgroup\$ – Rohat Kılıç Jul 6 at 7:12
  • \$\begingroup\$ Whoops, thanks for the catch. \$\endgroup\$ – LetterSized Jul 6 at 7:14
  • \$\begingroup\$ @lettersized ,so basically your saying that we run into an indeterminate form 'infinity *zero' . So we attack it from another angle? Am I correct? \$\endgroup\$ – Yasir Sadiq Jul 6 at 10:46
  • \$\begingroup\$ @Yasir That's kind of what I'm getting at. A real opamp does not have infinite gain, and the equation Vo=A(V+ - V-) holds. Let's say you did have a perfect, linear differential amplifier with infinite gain with no feedback. Then that equation still holds...except now your output is infinite. But if you add negative feedback, the infinite gain forces the two inputs to be equal in order to make the output finite. That is, in order for Vo = A(V+ - V-) to be finite with A infinite, we NEED (V+ - V-)=0. And at this point, we need to redo the analysis with this new assumption of V+ = V-. \$\endgroup\$ – LetterSized Jul 6 at 19:05
  • \$\begingroup\$ @LetterSized 3,hearty thanks bro. \$\endgroup\$ – Yasir Sadiq Jul 6 at 22:47
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Using virtual ground DOES make you arrive at the correct answer. No current flows through \$R_c \$.

$$i_{R_C}=0 \text{A} \Rightarrow V_+ = 0\text{V} $$

Using virtual ground we have $$ V_- = V_+ = 0\text{V}$$

The current flowing through R1 is: $$i_{R_1}=\frac{V_{in}-V_-}{R_1}=\frac{V_{in}}{R_1} $$

The current flowing through Rf. $$ i_{R_f}=\frac{V_--V_o}{R_f}=\frac{-V_o}{R_f}$$

Those two currents are the same. Setting up the equation $$ i_{R_1}=i_{R_f} \Leftrightarrow \frac{V_{in}}{R_1}=\frac{-V_o}{R_f}$$ Recall that the gain is given as: \$A_v=\frac{V_o}{V_{in}} \$. Manipulating the equation we finally end up with $$A_v = \frac{V_o}{V_{in}}=\frac{-R_f}{R_{in}} $$

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  • \$\begingroup\$ Thanks Carl, i agree with you but my issue is what's wrong with my approach? \$\endgroup\$ – Yasir Sadiq Jul 6 at 11:21
  • \$\begingroup\$ @YasirSadiq in order to have an ideal virtual ground, A has to be infinity. What's infinity times zero? Or alternatively, if A is not infinity, then V- isn't 0 volts. \$\endgroup\$ – user253751 Jul 6 at 11:23
  • \$\begingroup\$ Infinity times zero is undefined. Now what i get from the answers is that we attack the problem from another angle, like what Carl has done. \$\endgroup\$ – Yasir Sadiq Jul 6 at 11:38
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I can't understand where I'm making a mistake.

The following formula, $$ V_o=A\ (V_+ - V_-) $$

is for open-loop (i.e. no feedback present) opamp circuits and \$A\$ is the open-loop gain of the opamp which can be extremely large. Don't mix this with \$A\$ shown in your circuit.

But your circuit introduces a feedback:

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Remember the basic feedback theory:

$$ A_V=\frac{V_o}{V_i}=\frac{A}{1+\beta A} $$

If there were no \$\beta\$ (i.e. open loop) then you'd be correct since the gain would be \$A\$ (open-loop gain).

So, with feedback, output-to-input voltage gain will be \$A_V=-R_f/R_1\$.

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  • \$\begingroup\$ Vo=A(V+ - V-) applies whether there is feedback or not (as long as the opamp isn't saturated). The difference with feedback is that V+ - V- depends on Vo itself. \$\endgroup\$ – LetterSized Jul 6 at 7:14
  • \$\begingroup\$ @LetterSized could you please show me an example where that formula applies to an opamp circuit with feedback? Remember the definition of A (open-loop gain): The gain of the opamp when there's no positive or negative feedback. \$\endgroup\$ – Rohat Kılıç Jul 6 at 7:17
  • \$\begingroup\$ If you consider the inverting amplifier in the question (and using the picture's labeling), Vo=A(V+ - V-). V+ = 0 because it is connected to ground. So now Vo = AV-. Also, V- = (Vo-V1) * R1/(R1 + Rf) + V1 using a voltage divider equation. Plugging this into the first equation, we have Vo = A(Vo-V1) * R1/(R1+Rf) + AV1. Now solving for Vo/V1, we get Vo/V1 = A(Rf/(R1+Rf)) / (1-AR1/(R1 + Rf)). Simplify and make the large A assumption and you get: Vo/V1 = -Rf/R1. Let me know if I made a mistake. \$\endgroup\$ – LetterSized Jul 6 at 7:32
  • \$\begingroup\$ Looks like I did make a mistake, I missed a minus sign. We should have Vo = -AV- and thus Vo = A(Vo-V1) * R1/(R1+Rf) + AV1 should be Vo = -A(Vo-V1) * R1/(R1+Rf) - AV1. Proceeding from there will give Vo/V1 = -A(Rf/(R1+Rf)) / (1+AR1/(R1 + Rf)). Making the large A assumption gives the correct answer. \$\endgroup\$ – LetterSized Jul 6 at 7:46
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    \$\begingroup\$ @Yasir You are correct, I did assume infinite input impedance. If you did not, then you would need to include the effect of the resistor Rc (shown in your diagram). \$\endgroup\$ – LetterSized Jul 6 at 18:58

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