The gain of an ideal opamp is considered infinite and Vout=K(V+ - V-). where V+ is the non-inverting input and V- is the inverting input. The output voltage in the case of an inverting opamp amplifier is Vout=(-Rf/Ri)*Vin.

Does this means that (V+ - V-)=Vin?? If yes, then how it is possible as while calculating the Vout in inverting opamp amplifier we consider virtual shorting i'e V+=V-. Is then Vin is approximately zero??

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    \$\begingroup\$ You are mixing the open-loop configuration with a negative-feedback configuration in your question. The second equation is obtained from the first (open-loop) one when a certain feedback configuration present and K goes to infinity. \$\endgroup\$
    – Eugene Sh.
    Commented Feb 24, 2021 at 18:14
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    \$\begingroup\$ Vin in the op-amp inverting ampifier is applied in series to the voltage drop VR1 across the input resistor R1. The op-amp makes VR1 = Vin by adjusting the current through R1. As a result, V- = Vin - VR1 = 0 (virtual ground). Still an interesting question provoking an interesting answer... \$\endgroup\$ Commented Feb 24, 2021 at 18:20
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    \$\begingroup\$ Yes, it is zero. Zero times infinity in this case results in Vout. \$\endgroup\$
    – Eugene Sh.
    Commented Feb 24, 2021 at 18:33
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    \$\begingroup\$ Rf/Rin * Vout = Vin- = Vin+. While Vsource current goes thru Rin+Rf only to Vout. \$\endgroup\$ Commented Feb 24, 2021 at 18:38
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    \$\begingroup\$ Here's another way of thinking about negative feedback: When in a negative feedback configuration, the op-amp will try to drive its output to whatever voltage it takes to make \$V_\text{in}^+=V_\text{in}^-\$ \$\endgroup\$ Commented Feb 24, 2021 at 18:49

2 Answers 2


It seems you are confusing the open loop gain of the OpAmp (which is just the component) with the gain of the whole circuit.

What if you consider the OpAmp as ideal except for the infinite open loop gain:

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$$(V_{pos} = 0) \rightarrow V_{out} = -KV_{neg}$$

Replace Vneg in the first equation.

$$R_f * \frac{-V_{out}}{K} - Rf V_{in} = R_i V_{out} + \frac{R_i V_{out}}{K} $$

And isolate real gain:

$$\frac{V_{out}}{V_{in}} = \frac{-Rf K}{RiK + Ri + Rf}$$

As the open loop gain K tends to infinity, the real gain tends to the ideal formula.

$$\frac{V_{out}}{V_{in}} = \frac{-Rf}{Ri}$$

  • \$\begingroup\$ I started to write a similar answer but got tired of all of the \$\LaTeX\$ :) \$\endgroup\$
    – Eugene Sh.
    Commented Feb 24, 2021 at 21:25
  • \$\begingroup\$ :) yes, it is boring \$\endgroup\$
    – devnull
    Commented Feb 24, 2021 at 21:28

Op-amp calculations are a little weird, because they start with the assumption that the op amp output is not saturated. This is not automatically true, but in later steps we'll ensure that that is the case.

For the op-amp to not be pinned at either rail (unsaturated), the inputs to V+ and V- need to be essentially equal. This is why op-amp equations include the assumption V+ = V-.

Now lets look at the inverting amplifier through this lens. We assume v+ will be equal to v-, and v+ will be set at 0V, therefore v- must be maintained at 0v.

How will this happen? Well I like to think of it like this: If we start with the assumption that v- is 0 volts, then a certain current will flow across the input resistor to v-. If vin is 1v and the resistor is 1k, then 1mA will flow. As this happens the voltage at v- will start to rise, and the op amp output will start moving towards the negative rail (slewing). The op-amp will keep slewing until the current leaving the feedback resistor exactly equals the current in to v-. At that point the voltage at v- will be a steady 0 again and the output will be stable. The amount of output voltage swing required to create this current depends on the feedback resistor. If Rf is also 1k, then 1V will be required, creating a gain of -1. If Rf was instead 2k, then 2v would be required to remove that 1mA, and the gain would be -2.


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