# Electronic – Transfer function for inverting amplifier

operational-amplifiertransfer function

After reading this answer I wanted to know how the transfer function for an inverting amplifier was derived. I found this page, but it leaves me unsatisfied. Their proof is just two lines, and the opamp's amplification isn't even mentioned.
What's the proper derivation of the inverting amplifier's transfer function?

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 corollary of 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

"Moreover, being an ideal Op Amp, its gain is high, so the inverting input is at a virtual ground."

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 can make 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.