The gain of an opamp circuit is basically $$A_v=-\frac{R_f}{R_i}$$
But how come the gain of the opamp itself does not appear in the equation? How is the opamp amplifying anything in the circuit? What is the opamp doing in the circuit if it does not appear in the formula?
From Texas Instruments "Stability Analysis of Voltage-Feedback Op Amps Including Compensation Techniques"
http://www.ti.com/lit/an/sloa020a/sloa020a.pdf
A is the open loop gain (the gain of the opamp itself), and \$ \beta\$ is the feedback resistors.
If A\$\beta\$ is very large relative to 1, then the closed loop gain approximates as \$\frac{A}{A\beta}\$ which simplifies down to
$$ \frac{V_{out}}{V_{in}} =\frac{1}{\beta} $$
That means that with very large open loop gain, you can use negative feedback to produce a closed loop gain that is virtually independent of the exact open loop gain of the op-amp (which is difficult to control). It only depends on the negative feedback (the resistors) which makes the circuit easier to design, more predictable, and less tied to the specific op-amp.
The white paper I linked goes into it in more detail. It goes into more detail about how to actually do this calculation with resistors instead of just \$\beta\$
The formula you quote is not the gain of an opamp. It is the gain of a circuit containing an opamp and several resistors. That formula only holds when the open loop gain of the opamp is much larger than that given by the formula. When that is the case, the actual value of the opamp open loop gain drops out of the equation. The derivation of the formula is given in any textbook on opamps and can be found on many websites so I am not going to repeat it here. The point is that the complete formula for the circuit gain does include the open loop opamp gain but as long as the condition I stated before is true, the formula you gave is a very good approximation.
Let's start with the definition of the op amp:
$$e_{out}= A_{OL}(e_+ -e_-)$$
This is true for every op amp. The device is a differential amplifier, with a very high gain.
Now, given that the positive terminal is grounded, $$e_+ = 0\\ e_{out}= -A_{OL}e_-$$
Next, we can apply the assumption that the input impedance is infinite, thus \$i_b=0\$. This let's us apply Kirchoff's law at the negative input terminal, and algebraically manipulate stuff.
$$ \frac{e_{out} - e_-}{R_f} = \frac{e_{-} - e_{in}}{R_i}\\ \frac{-A_{OL}e_- - e_- }{R_f} = \frac{e_- - e_{in}}{R_i}\\ \frac{e_-}{R_{i}} + \frac{e_- (A_{OL}+1)}{Rf} = \frac{e_{in}}{R_i} \\ e_- + \frac{R_ie_- (A_{OL}+1)}{Rf}= e_{in}\\ $$ $$ e_- \left(\frac{R_i}{R_f}(A_{OL}+1) +1 \right) = e_{in}\\ \frac{e_-}{e_{in}}= \frac{1}{\frac{R_i}{R_f}(A_{OL}+1) +1 } \\ \frac{e_-}{e_{in}}= \frac{R_f \mathbin{/} R_i}{(A_{OL}+1) +R_f \mathbin{/} R_i } $$
If you now apply the ideal assumption that \$A_{OL}=\infty\$, you can see that \$e_- = 0\$, and that leads you to the ideal inverting op amp equations you posted. As \$A_{OL}\$ gets smaller, or if your closed loop gain gets ridiculously high, the ideal assumption starts to fail.
Note that if \$\frac{R_f}{R_i}\gg A_{OL}\$ then \$e_-=e_{in}\$! This is an absurd case, but it also demonstrates to some extent why an inverting amp with a gain of 100,000 is a loser.
In case of an inverting opamp configuration there is no negative sign at the summing junction of the block diagram (two signals - input and feedback - will be superimposed at the inverting input node) - therefore, a negative sign must be used for the open-loop gain A.
Of course, we still have negative feedback (because there is a minus sign in the feedback loop).
The block "alpha" is required because the input signal is not applied directly to the opamp input node. Both factors (alpha and beta) result from the superposition theorem applied to the inverting terminal (Voltage V-)
V- = Vin*(Rf/(Ri+Rf) + Vout*(Ri/(Ri+Rf) = Vin*(alpha) + Vout*(beta)
We have also
V- = - Vout/A (V+=0, non-inv. input grounded)
Equating both right sides gives us:
Vout/Vin= - alpha*[A/(1+beta*A)]
A visual interpretation of this transfer function leads to the following block diagram for the inverting opamp configuration:
simulate this circuit – Schematic created using CircuitLab
Lets examine how the opamp performs, over frequency. Assume Avol = 1,000,000X (120dB) and the closed loop gain is ~~10x (inverting configuration, 1K Rin, 10K Rfeedback). Oh, the opamp gain is flat to 100Hz, then rolls off with 1_pole until 10,000,000Hz.
What is the gain
Frequency --- Gain
DC ------------- 9.99990
100Hz --------- 9.99990
1KHz ---------- 9.99900
10,000Hz ----- 9.99000
100,000Hz --- 9.90000
1MHz ---------- 9.00000
10MHz -------- 0.9 1/(1 + J 1 *0.1)
Looking at the expression for the gain written in the body of the question, I guess the circuit of an op-amp inverting amplifier is considered. This is historically the first circuit with negative feedback because it could be implemented simply through an operational amplifier with a single-ended input. To compare the input and output voltages, they are subtracted by a simple resistive summing network R1-R2 (two resistors in series). Today’s op-amps do the same by a differential input; so, to realize the inverting circuit, the non-inverting input is non used (connected to ground).
I understand very well what excites the author of the question because I was in this position many years ago. I have found that the best way to understand and explain intuitively op-amp circuits with negative feedback is to consider the operating amplifier not as proportional but rather as an integral device... ie, as some "living being", which observes the voltage at its input and changes the voltage at its output.
It is also very important to show where currents flow by closed lines (current loops). For this purpose, the power supply rails and ground should be connected by continuous lines. Also, voltages should be visualized. I will demonstrate these techniques with the discussed circuit of the op-amp inverting amplifier:
1. Zero input voltage. We start this mental experiment with zero input voltage applied to the left end of the resistive network (R1). The op-amp "observes" the output voltage of the resistive network (at the common point between resistors or inverting input) and changes its output voltage applied to the right end of the resistive network (R2) until zero it. The result is zero voltage at the op-amp output... there are no voltages and currents in the whole circuit... it is "dead".
2. Positive input voltage. We change the input voltage towards the positive rail. The input current IIN begins flowing through the resistive network R1->R2... enters the op-amp output (the lower stage of the output emitter follower)... leaves the negative supply end of the op-amp... goes through the negative power supply V-... and returns to where it started - the negative terminal of the input voltage source.
The resistive network R1-R2 acts as a voltage divider driven with positive voltage from the left; so, the voltage of the common (summing) point tries to change to positive. However, the op-amp "sees" that and begins to oppose this attempt. It changes its output voltage to negative thus driving the R2-R1 voltage divider from right... until restores the zero voltage of this point. It behaves as a ground since it has zero voltage... but it is not connected to the real ground... it is not a ground... that is why it is called "virtual ground".
Note the op-amp does this by the help of the negative supply that is connected in series (through the ground line) and in the same direction with the input voltage source. Thus the whole network is driven by the sum of the input and output voltage... and the current is IIN = (VIN + VOUT)/(R1 + R2). "Copies" of the two voltages appears across the corresponding resistors: VIN - across R1, and VOUT - across R2. They are connected by the common current, so I = VI/R1 = VOUT/R2 or VOUT/VIN = -R2/R1. Note this relation does not belong to the op-amp... it belongs to the humble 2-resistor circuit.
In addition to the feedback current IIN, also bigger current IL flows through the load (if there is such). Note this current is provided entirely by the negative power supply.
3. Negative input voltage. Then we change the input voltage towards the negative rail. Now the input current IIN begins flowing in an opposite direction through the positive power supply V+... enters the positive supply end of the op-amp... leaves the op-amp output (the upper stage of the output emitter follower)... goes through the resistive network R2->R1... and returns to where it started - the negative terminal of the input voltage source.
Now the resistive network R1-R2 acts as a voltage divider driven with negative voltage from the left; so, the voltage of the common (summing) point tries to change to negative. Again, the op-amp "sees" that and begins to oppose this attempt. It changes its output voltage to positive thus driving the R2-R1 voltage divider from right... until restores the zero voltage of this point (the "virtual ground").
Now the op-amp does this by the help of the positive supply that is connected in series (through the ground line) and in the same direction with the input voltage source; so the whole network is driven by the sum of the input and output voltage.
In this case, the load current IL is provided entirely by the positive power supply.
Conclusion. Supprisingly, in this inverting configuration, the sophisticated op-amp is forced to serve a humble resistive circuit (passive voltage summer). The input voltage source drives this network from the left in one direction and the op-amp drives it from the right to the opposite direction so that to keep its output voltage in the middle point always equal to zero (virtual ground). Thus the proportion VOUT/VIN = -R2/R1 is always valid.
This proportion is a property of the resistive network, not of the op-amp. It can be derived by applying the superposition principle:
VSUM = VIN.R2/(R1 + R2) + VOUT.R1/(R1 + R2) = 0
The point of this is that, if the gain of the op-amp is high enough, the gain of the whole circuit does not depend on the op-amp gain... It is determined only by the ratio between two resistors... and it can be > 1 (inverting amplifier), = 1 (inverter) and even < 1 (inverting attenuator).
Simple scales are a very good mechanical analogy to this electronic circuit. The ratio of the weights to the left and right of the balance depends solely on the ratio of the arm lenghts l1 and l2 ("R1" and "R2"). It does not depend on the person (the "opamp") who performs the weighing procedure.
Here is an "animated" picture of a funny game where the Actor 2 represents the op-amp behavior in an inverting amplifier:
