For my opinion, the simplest solution makes use of the classical feedback formula from H. Black:

$$\frac{V_2}{V_1}=\frac{H(s)}{1-LG}$$

with:

- \$H(s)=H_1(s)H_2(s)\$=Forward transfer function for an
**open** loop (in our case: \$H_1=V_3/V_1\$ for \$R_2\$>>infinite and \$H_2=V_2/V_3\$.)
- Loop gain \$LG\$=Product of all
**three** transfer functions within the loop (with \$V_1=0\$ or \$R_1\$>>infinite).

Note that \$H(s)\$ is positive and the loop gain \$LG\$ must be negative (three inverting stages in series). The transfer functions of the three blocks are basic (inverting lowpass, inverting integrator, inverting amplifier).

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.

## Best Answer

Aah, that sucks! That's not even a proof, and the author should be sent to the salt mines!

The problem (and the reason why the "proof" is so short) is that for the proof they rely on something which is a

corollaryof the actual proof. It's a fallacy which sometimes occurs in mathematical proofs: you use the outcome of the proof as an assumption for it.In this case it's

That the inputs are equal will be a consequence of the proof, not something you may assume beforehand!

It's not a property of opamps!In fact it's very easy to have an opamp with the input voltages differing, even with feedback: think of the Schmitt-trigger.This way you're doing it completely backward.

(By the way, Wikipedia "proves" in the same way the transfer function of the non-inverting amplifier. Another miss by WP.)

Ok, let's get started (I hope I get the \$\LaTeX\$ right):

The opamp's transfer function is

\$ V_{OUT} = G \times (V_+ - V_-) \$

where \$G\$ is the opamp's gain. I could leave it as \$G\$, but I'll use an real value instead, 100 000 is a typical value. The actual value shouldn't matter, 200 000 would also be a possibility, that means that we'll probably have to get rid of it during the calculation.

\$V_+\$ is 0V, so

\$ V_{OUT} = -100 000 \times V_- \$

An assumption we

canmake is that the input current is negligible. It's one of the ideal opamp's axioms, and a datasheet will confirm it. Then, according to KCL:\$ \dfrac{V_{IN} - V_-}{R_{IN}} = \dfrac{V_- - V_{OUT}}{R_F} \$

or

\$ \left(\dfrac{1}{R_F} + \dfrac{1}{R_{IN}} \right) V_- = \dfrac{V_{IN}}{R_{IN}} + \dfrac{V_{OUT}}{R_F} \$

then

\$ V_- = V_{IN}\left(\dfrac{R_F}{R_{IN}+R_F}\right) + V_{OUT}\left(\dfrac{R_{IN}}{R_{IN}+R_F}\right) \$

Filling this in in our transfer function:

\$ V_{OUT} = -100000 \left(V_{IN}\left(\dfrac{R_F}{R_{IN}+R_F}\right) + V_{OUT}\left(\dfrac{R_{IN}}{R_{IN}+R_F}\right)\right) \$

Looks nasty, but we'll be alright in a minute!

\$ V_{OUT} \left(1 + 100 000 \left(\dfrac{R_{IN}}{R_{IN}+R_F}\right)\right) = -100000 \times V_{IN}\left(\dfrac{R_F}{R_{IN}+R_F}\right) \$

Now you see why I like to use the 100 000 value: you can easily see that the "1" can be neglected. If \$G\$ isn't much greater than 1 the whole reasoning becomes invalid.

\$ V_{OUT} = \dfrac{-100000 \times V_{IN}\left(\dfrac{R_F}{R_{IN}+R_F}\right)} {100 000 \left(\dfrac{R_{IN}}{R_{IN}+R_F}\right)} \$

Now we can cancel a lot, including the opamp's gain factor(!), and what remains is

\$ V_{OUT} = -\dfrac{R_F}{R_{IN}} V_{IN} \$

That's the inverting amplifier's transfer function!

If you replace the \$V_{OUT}\$ in the equation for \$V_-\$ by this value you'll find

\$ V_- = 0V \$

So the input voltages are indeed equal, but only as a consequence of the proof.