I understand that the derivation of an inverting amplifier's gain is generally treated as \$G = \dfrac{-R_2}{R_1} \$, and have seen many resources supporting this claim. I understand how the proof works by solving for the current flowing to virtual ground at equilibrium, and am satisfied with this expression. However, my EE textbook showed a different, more generalized version of the expression for gain,

\$ G = \dfrac{-A(1 - B)}{(1 + AB)}, \$

where A is the open-loop gain of the op-amp and \$B = \dfrac{R_1}{R_1 + R_2}. \$

In the limit of high open-loop gain, I understand how this expression simplifies to \$G = 1 - \dfrac{1}{B} = \dfrac{-R_2}{R_1}.\$ Can someone provide an explanation as to how this more generalized transfer function is derived, and why the general derivation that follows is inaccurate in some way for low open-loop gain?

Here is a sample of the derivation that I followed - what simplifying assumptions are made?

\$ I = \dfrac{V_{in}}{R_{1}} \$

\$ V_{out} = V_{in} + IR = V_{in} - V_{in} \times \dfrac{R_2+R_1}{R_1} = V_{in} \times \dfrac{-R_2}{R_1}\$

\$ \therefore \dfrac{V_{out}}{V_{in}} = \dfrac{-R_2}{R_1}\$

Follow-up: Does whatever correction needs to be made to the closed-loop gain of the op-amp affect the input resistance of the op-amp, or is it always \$R_1\$, regardless of the op-amp gain?

  • \$\begingroup\$ In order to create a null differential input, the forward A must be much greater than the reverse B feedback. Thus the generalized form cannot assume this and applies to Power Amps with low forward gain without integrating compensation feedback caps like Op Amps and single stage transistor amps etc. This is where you start by not assuming the differential input is zero. \$\endgroup\$ Commented Dec 5, 2016 at 0:51
  • \$\begingroup\$ "R2" and "R1" mean nothing without a schematic showing where these devices are located in your circuit. \$\endgroup\$
    – The Photon
    Commented Dec 5, 2016 at 5:18
  • \$\begingroup\$ You're right, I should have included a schematic. I just assumed that they would refer to their generalized left-to-right positions in a generic inverting amplifier diagram, such as the one posted by Cabrera below. \$\endgroup\$
    – Alekxos
    Commented Dec 5, 2016 at 5:29

1 Answer 1


This is your typical inverting opamp configuration:


simulate this circuit – Schematic created using CircuitLab

The one thing you know for sure is

$$V_o=A_{ol}(V^+-V^-) $$

and notice I am not even talking about negative feedback here. The output always follows that equation and what limits the output voltage are the rails or supply voltages (\$V_{cc}\$ and \$V_{ss}\$).

Now, back to the equation provided. In order to find the gain, you want to express it in terms of \$V_{in}\$ and \$V_o\$.

You can easily see that \$V^+=0\$. You can find \$V^-\$ in several different ways. For simplicity here is what \$V^-\$ is:

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

If you plug those values into the \$V_o=A_{ol}(V^+-V^-) \$ equation, you get:

$$V_o=-A_{ol}\bigg(\frac{R_2}{R_1+R_2}V_{in}+\frac{R_1}{R_1+R_2}V_o\bigg) $$

After some algebra, you can find \$\dfrac{V_o}{V_{in}}\$ to be:

$$\frac{V_o}{V_{in}}=-\dfrac{R_2}{R_1+\dfrac{R_1+R_2}{A_{ol}} }$$

So, in order for the gain to be approximately \$\dfrac{V_o}{V_{in}}\approx -\frac{R_2}{R_1}\$, the open loop gain has to be large enough so that the \$\dfrac{R_1+R_2}{A_{ol}}\$ term in the denominator is negligible.

  • \$\begingroup\$ Thanks a ton! Is it still fair to treat the input impedance as \$R_1\$ even in the limit of low open-loop gain? \$\endgroup\$
    – Alekxos
    Commented Dec 5, 2016 at 3:31
  • 1
    \$\begingroup\$ @Alekxos If the open loop gain wasn't large enough so that \$V^+\approx V^-\$, then the input impedance of the circuit would be affected by the feedback network. What makes the input impedance \$R_1\$ is the fact that there is a virtual ground at the \$V^-\$ node, and the virtual ground is obtained when the open loop gain is large, as it is in opamps. Take a look at this great answer on input impedance of opamp: electronics.stackexchange.com/questions/56229/… \$\endgroup\$
    – Big6
    Commented Dec 5, 2016 at 3:53

Your Answer

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

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