# How does opamp open-loop gain affect closed-loop gain in a circuit?

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?

• You have to make the distinction between opamp open loop gain (intrinsic to the op amp), and closed loop gain of the circuit (dependent on the opamp open loop gain and the circuit) in your question. – DKNguyen Nov 19 '19 at 22:58
• Sorry about that, I am asking about closed-loop gain, that's why I described the formula within R values above. – lateralus Nov 19 '19 at 23:03
• You can edit your question. I assume your question is that the closed loop gain is $A_v = -\frac{R_f}{R_i}$, so how does the opamp open loop gain come into play since it doesn't appear in the equation? If so, edit your question. – DKNguyen Nov 19 '19 at 23:16

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\$$

• Does this answer the question (which asks for Av=-Rf/Ri) ? The factor beta is Ri/(Ri+Rf) and we have 1/beta=1+Rf/Ri . This is, of course, the gain for a non-inverting opamp. But the question was about the inverting gain (-Rf/Ri). – LvW Nov 21 '19 at 15:41
• @LvW Yes, it explains why the open loop gain does not appear in the closed loop gain equation. It's just easier to understand than going into the actual circuit analysis to convert the negative feedback beta into the negative feedback resistors. The point is basically a large open loop gain approximates out of the equation. – DKNguyen Nov 21 '19 at 15:43
• Yes - that`s correct. But the result is not identical to the gain expression which was mentioned in the original question. I think, for a general answer (which applies to non-invering and inverting gains) we should include the forward damping block Rf/(Ri+Rf)) – LvW Nov 21 '19 at 15:49
• @LvW Oh, so you just wanted to replace beta with an the R expression without explaining where it came from. I guess I could do that but just felt it wasn't necessary if I wasn't going to go into deriving it.. – DKNguyen Nov 21 '19 at 15:52
• Please, see my detailed answer...I have explained my remark/comment. – LvW Nov 21 '19 at 16:19

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.

• I understand better now, thank you – lateralus Nov 20 '19 at 17:23

$$e_{out}= A_{OL}(e_+ -e_-)$$

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:

• This answer fails to adequately address the effect of a finite open loop gain, so it doesn't really answer the question posed. With that said, in my experience/opinion as someone who was taught this sort of model and then properly taught opamps, personifying an op-amp as an intelligent integral device only sets one up for confusion down the road, if and when they try to understand an op-amp more rigorously (i.e. at a transistor level, or in an AC analysis with stability involved). – nanofarad Nov 21 '19 at 15:17