# Why does an OpAmp with negative Feedback not amplify the potential difference on his inputs?

An OpAmp without any feedback just takes the potential difference between his 2 inputs and amplifies it by a certain factor. I know that in an OpAmp Circuit with negative feedback, we can make the assumption that the potential difference between his 2 inputs is ideally zero. I know this is just an ideal assumption, in reality there will be a small potential difference remaining. I also know that when we have an inverting amplifier-circuit, we can basically "adjust" the amplification to a value we want by the ratio of the 2 resistors.

However I'm wondering: Once we have that circuit, what gets amplified is the Input Voltage and NOT the differential-voltage between his 2 inputs anymore. Why is that? The OpAmp doesn't "know" about the feedback surrounding him, so why does he not amplify that really small remaining potential difference between his inputs?

• it does. until it's limited by the rails. – Neil_UK Aug 22 '19 at 9:23
• "Once we have that circuit, what gets amplified is [..] NOT the differential-voltage between his 2 inputs anymore" Why do you think so? It's wrong. – Curd Aug 22 '19 at 10:05
• It does. The negative feedback reduces the input differential to a very low value, but it is still output voltage divided by A. – Bart Aug 22 '19 at 10:55

Let me answer with a simple numeric example.

• Open-loop gain (inverting) of the opamp Aol=-10³=-1000

• Feedback resistors R2=10k and R1=1k (Design goal: Closed-loop gain Acl=-10).

• Real closed-loop gain (taking Aol into account):

Acl=-[10/(1+10)]/[1/(1+10)+0.001]=-9.89

Comment: This equation results from the classical feedback formula.

• Hence, for an input voltage of 1V the output voltage is -9.89V and the diff. voltage between both opamp inputs is

Vdiff=-9.89/-1000=9.89 mV.

• Of course, when the open-loop gain is larger, the closed-loop gain is much closer to the design goal and the diff. voltage is much smaller (µV range) and can be neglected (assumed to be zero for calculation purposes).

For an ideal opamp (Aol>>infinite): Acl=-(10/11)/(1/11)=-10

• Thanks a lot. I totally overlooked the fact that the amplification set by the resistors being exactly the resulting amplification of the entire circuit is also just an ideal assumption. This is actually a really good different view on why the difference voltage is near zero. Thanks :) – Philipp317 Aug 23 '19 at 17:56

This is just an illustration for LvW's answer, with two input voltages: Note that the (small) open loop gain is the only non-ideal characteristic simulated. With this condition, the op. amp. does amplify the differential input.

As suggested in the comment below, an example with a more realistic open loop gain (100k), showing why we normally disregard the DC open loop gain, would be: For a more step-by-step analysis, assuming only the finite open loop gain as a non-ideal behavior, consider that the currents in both resistors are the same:

$$\\dfrac{(V_{in}-V_{n})}{R_2}=\dfrac{(V_{n}-V_{out})}{R_1}\$$

The differential gain is:

$$\V_{out}=(V_p-V_n) \times A_{ol}\$$

$$\V_p=0\$$

$$\V_{out}=-V_n \times A_{ol}\$$

Substituting $$\V_n\$$ in the first expression:

$$\V_n=\dfrac{-V_{out}}{A_{ol}}\$$

$$\R_1 \times V_{in} - V_n \times R_1 = V_n \times R_2 - V_{out} \times R_2\$$

$$\V_{in} \times R_1 + V_{out} \times R_2 = V_n \times (R_1 + R_2)\$$

$$\V_{in} \times R_1 + V_{out} \times R_2 = \dfrac{-V_{out}}{A_{ol}} \times (R_1 + R_2)\$$

The closed loop gain would be:

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

$$\\dfrac{V_{out}}{V_{in}} = \dfrac{-R_1}{(R_2 + R_1/A_{ol} + R_2/A_{ol})}\$$

The larger $$\A_{ol}\$$, the closer the gain will be to $$\-R_1/R_2\$$.

• It might be interesting to show the case of $A_v=1000$ (as in LvW's answer) as the first example, and $A_v=100,000$ (a typical value for a real op-amp) as the second example. – The Photon Aug 22 '19 at 16:05
• Thanks. Will do. – vangelo Aug 22 '19 at 16:25