My problem is that I don't know how to prove that the gain of this circuit is given by the following relation. \begin{equation} G=\frac{U_o}{U_i}=-\frac{R_2}{R_1} \end{equation} I have the feeling that I'm wrong somewhere in my reasoning, or that I don't understand a key point.

Simple inverter

Here is my reasoning so far:

I want to use the superposition theorem, therefore I'm going to analyze separately the two sources (Alternative source, Amplifier).

For the alternative source we have the following. \begin{cases} I=I_1=I_2 \\ U_i=(R_1+R_2) \cdot I \quad \text{which means that $R_1$ and $R_2$ are in series} \end{cases}

For the amplifier we get the following. \begin{cases} U_o=(R_2+R_1) \cdot I' \quad \text{where I' is not in the same direction as I} \end{cases} From this point, I don't really know where I should go, and I feel that I made a mistake somewhere but I'm not able to tell where and why.

I made some research on this subject, but I didn't find a good explanation. I have seen some people on Youtube solving this problem, but there always have been some mysterious manipulations that I didn't understand.

Thank you in advance for your responses.

P.S.: I'm not an electrical engineer, but I have an idea of what is an ideal amplificator, and I know Kirchhoff's laws.

  • \$\begingroup\$ Assume the gain block has infinite gain. From that and given a finite output voltage, the input voltage must be zero. This is called a virtual earth. Grasp then run....... \$\endgroup\$
    – Andy aka
    Apr 30, 2018 at 20:42
  • \$\begingroup\$ @Andyaka Hum... Ok, I can somehow understand that the voltage on the inputs of the amplifier is 0 [V], and that there is no current flowing through the inputs. But I don't get the reasoning on how I should proceed after these statements. \$\endgroup\$
    – Beginner
    Apr 30, 2018 at 20:48
  • \$\begingroup\$ .... so the current flowing through R1 to the (virtual) 0 volts is?.... And the current flowing into R2 (by virtue that nothing flows into the actual input) is?..... \$\endgroup\$
    – Andy aka
    Apr 30, 2018 at 20:51
  • \$\begingroup\$ @Andyaka I think that the current flowing through R1 to the virtual ground (Inverter terminal) should be 0 [A], and the current flowing through R2 should be I2=Ui/R1 by Kirchhoff's law. \$\endgroup\$
    – Beginner
    Apr 30, 2018 at 20:57
  • \$\begingroup\$ No, the current flowing into R1 is (Vin - 0volts(virtual))/R1 = Vin/R1.... take it from there and discard the word virtual if you want. I set the scene - gain infinite, output voltage finite hecne input voltage at the amps input MUST BE 0 volts (feedback makes it that way - hint). \$\endgroup\$
    – Andy aka
    Apr 30, 2018 at 21:00

1 Answer 1


In your circuit you forgot to mark non-inverting terminal to ground. The inverting terminal is therefore at ground too. No current flows into opamp. So at inverting node, apply KCL/nodal analysis :

$$\frac{V_{in}}{R_1}= -\frac{V_o}{R_2}$$ $$V_o =-V_{in} \frac{R_2}{R_1}$$

  • \$\begingroup\$ Allright, it looks like sorcery since there is no description about the flow of the current, but I will look up for the nodal analysis, which seems interesting. \$\endgroup\$
    – Beginner
    Apr 30, 2018 at 21:01
  • \$\begingroup\$ @Beginner - Look at the first equation. According to Ohm's Law, what is voltage over resistance? \$\endgroup\$ May 1, 2018 at 1:01
  • \$\begingroup\$ "I know Kirchoff's laws" -- then this cannot be sorcery to you. \$\endgroup\$
    – Mitu Raj
    May 1, 2018 at 2:39
  • \$\begingroup\$ @WhatRoughBeast According to Ohm's Law the voltage is: $V_1=V_{In}-0=R_1 \cdot I_1$. Therefore we have: $I_1=\frac{V_{In}}{R_1}$. But don't worry too, now I understand how we get the final result for the gain. It took me some time to get in my mind that the key point is the way how we define the flow of current, and the fact that the inputs of the amplifier in some cases are creating a virtual ground. \$\endgroup\$
    – Beginner
    May 1, 2018 at 14:46
  • \$\begingroup\$ @MITURAJ Yes, only if I understand how we make the calculations step by step \$\endgroup\$
    – Beginner
    May 1, 2018 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.