# How is an inverting opamp adder circuit able to regulate its feedback?

Consider the typical opamp inverting adder as follows:

Why don't resistor Rl and R1 form a resistor divider and impose 2.5V upon inverting node and drive it to saturation. How is opamp's feedback mechanism more powerful than source?

From the answers I understand that feedback overrides it but is there a mathematical way to validate that?

• The resistive divider formula only applies if the current through the two resistors is equal, so whatever load may be attached to the middle node must be drawing much smaller current than the static bias through the resistors. Here the middle node of the possible divider created by the voltage source, RL, R1, and ground also has the RF resistor attached which changes the current at that node significantly. Commented Sep 1 at 20:43
• "How is opamp's feedback mechanism more powerful than source?" || Because we chose the values of RI, RF and R1 to achieve exactly that Commented Sep 1 at 21:34
• Note that, to a first approximation, within its design range and in typical circuits, you can treat an op-amp as an infinite gain differential amplifier. Any difference in the two inputs is fed back to attempt to eliminate that difference. This idealized model is surprisingly useful in understanding how op-amp based circuits work, and how to design and modify them. Commented Sep 2 at 4:00
• Standard plug for Jung's IC Op-Amp Coolbook, which starts by explaining the infinite gain model and then shows a wide range of practical circuits built around that concept, with compensation for the behavior of real chips. That book is quite old at this point, and there may be a better one by now, but I remain impressed with how much useful information got packed into that book. Commented Sep 2 at 4:03

This can be solved using Kirchoff's current law and a few facts about how an operational amplifier works. Begin by summing all of the currents leaving the node at Vf, assuming the op amp's inverting input to be very high impedance: $$\frac{V_f-V_1}{R_i}+\frac{V_f-V_{out}}{R_f}+\frac{V_f-0}{R_2}=0$$ We also know that Vout is directly proportional to the difference between the inverting and non-inverting inputs of the op-amp. Let's use G for the ratio: $$V_{out}=G(V_+-V_-)=G(0-V_f)=-GV_f$$ Rearranging this allows us to eliminate Vf from the first equation: $$V_f=-\frac{V_{out}}{G}$$ $$\frac{-\frac{V_{out}}{G}-V_1}{R_i}+\frac{-\frac{V_{out}}{G}-V_{out}}{R_f}+\frac{-\frac{V_{out}}{G}-0}{R_2}=0$$ This can be rearranged into terms of Vout: $$V_{out}=\frac{V_1R_1R_f}{\frac{R_iR_f}{G}+\frac{R_1R_f}{G}+\frac{R_iR_1}{G}-R_1R_2}$$ G is very large. So large in fact that any term divided by G becomes approximately zero. The equation therefore simplifies to: $$V_{out}=\frac{V_1R_1R_f}{-R_iR_1}=-\frac{R_f}{R_i}V_1$$ In the given case, we have Vcc=5V, Rf=10kΩ, and Ri=10kΩ: $$V_{out}=-\frac{10kΩ}{10kΩ}\cdot 5V=-5V$$ This process works for most inverting op-amp feedback calculations so I'll summarise the steps:

1. Use Kirchoff's voltage law to get expressions for the voltage at the inverting input
2. Substitute in the relation Vout=G(V+-V-)
3. Rearrange to find an expression for Vout
4. Take the limit of the expression as G increases to infinity (e.g. by rearranging so that G is only used as a divisor then setting those terms to zero, or by using L'Hôpital's Rule)
5. Substitute in values if needed

They would, but the feedback via RF overrides it.

Say the voltage VF1 was slightly positive (say + 1 mV). Because it has a very high gain, the opamp's gain will drive its output very negative; RF will connect this signal to VF1, and more than compensate the +1 mV.

Eventually, things will settle down where VF1 is just very slightly positive; the opamp's output is significantly negative and the signal fed through RF 'balances' everything.

As you see, VFs is 0.5 uV positive; the opamp's gain (10 million !) drives its output to -4.999 (i.e. -5 V) and things are all self-consistent.

• Is there a mathematical approach to validate which effect will override? Commented Sep 1 at 19:14
• Yes; you can calculate the VOUT needed to have RF drive the input to 0; if this value is outside the drive capability of the opamp, the assumptions will fail. e.g. if RF was 100 kΩ, the opamp would have to go to -50 V to accomplish this and this is not possible. If RF was 1.0 k, the output would only be -0.5 V Commented Sep 1 at 19:21
• Node voltage on the negative terminal, how much current from RF is going into the terminal? Commented Sep 1 at 19:22
• in an idealized opamp, zero current flows into the inverting (and the non-inverting) inputs. All the RF current is used to pull VF1 towards 0 -- it 'fights' RI and R1. Commented Sep 1 at 19:24

## Resistor summer

The three resistors R1, R2 and R3 (Rl, R1 and RF in the OP's circuit) form a 3-input resistor summer. Figuratively speaking, they fight each other like in a "tug of war" game with three teams. Ultimately, the op-amp output wins and zeroes the voltage of the common point.

From another viewpoint, we can consider the combination of R3 (RF) and the op-amp output as an "active load" that sinks a current from the R1-R2 (Rl-R1) voltage divider's output.

# Building a conceptual inverting summer

I will answer the OP's question with more details by showing in a few steps the evolution of the idea behind this famous circuit.

## Voltage divider

Indeed, as the OP notes, the resistors R1 and R2 form a voltage divider with a "gain" of 0.5 (I have rearranged them to make the diagram more clear).

simulate this circuit – Schematic created using CircuitLab

## 2-input resistor summer

We can imagine that the resistor R2 is supplied by a second input source V2 with a voltage of 0 V. So, R1 and R2 form a 2-input resistor summer with weighted inputs.

simulate this circuit

The problem of this summer above is that the output voltage V- of the midpoint between the two resistors is not zero. So, the two inputs will influence each other.

## 3-input resistor summer compensated

That is why, we decide to destroy the "harmful" voltage V- by compensating it with an additional (negative) voltage source V3 through another resistor R3. For this purpose, we adjust V3 so that V- = 0 V, and use V3 as an inverted output voltage.

simulate this circuit

## 2-input "op-amp inverting summer"

Finally, we automate the operation of this circuit using a behavioral voltage source (the op-amp in the OP's practical circuit).

simulate this circuit

# Conclusions

• In the op-amp configuration, two real voltage sources with opposite polarity - the voltage divider R1-R2 and the op-amp output through R3 - are connected in parallel to the op-amp's inverting input and compete to impose their voltage. The op-amp output wins because the op-amp monitors the voltage at the inverting input through the negative feedback mechanism and adjusts its output voltage until it nullifies it.

• So, the op-amp output is not a simple "static source" like the input sources with 10 kΩ internal resistance, but a "dynamic source" with close to zero differential resistance (although its static resistance is also 10 kΩ).

• The circuit is designed as a 2-input op-amp inverting summer, but since the second input (R2) is grounded, it effectively operates as an inverting amplifier. R2 does not affect the circuit operation since there is no voltage across it.

• Thanks for the detailed explanation. Very helpful! Commented Sep 3 at 16:04

Why don't resistor Rl and R1 form a resistor divider and impose 2.5V upon inverting node and drive it to saturation. How is opamp's feedback mechanism more powerful than source?

The circuit is behaving exactly as it should based on the 5 volts offset from V1 (in series with RI), R1 connected to ground and, the feedback resistor. End of story. On this occasion the feedback is more powerful that the net source voltage and it's Thevenin resistance.

The short answer is negative feedback - a very useful concept.

If you assume an ideal opamp with infinite gain ($$\A_{OL}\$$), infinite input impedances on the input pins, and zero output impedance; the concept of virtual ground or virtual short between the opamp input pins $$\V_P\$$ & $$\V_N\$$ can be used to show what is going on. Beware of the term virtual. Virtual short doesn't mean there's a physical short circuit between the positive and input pins, but looks like a short because of high open loop gain and negative feedback.

The output voltage is given by: $$V_O = A_{OL}(V_P - V_N)$$

When $$\A_{OL} \to \infty\$$, then $$\(V_P - V_N) \to 0\$$.
Since $$\V_P = 0\$$, $$\V_N\$$ will be zero when $$\A_{OL} = \infty\$$, i.e., a virtual short between $$\V_P\$$ and $$\V_N\$$

This says, assuming the opamp is not is saturation, the voltage at the negative input pin is zero volts for the ideal case.

This says that the current flowing through $$\R_I\$$ is the same magnitude of current flowing through $$\R_F\$$ which will keep the negative input node at zero volts. Thus, the current flowing through $$\R_1\$$ is zero, essentially, not a contributing part of the circuit.

Graphs are a good way to visualize what's going on. The opamp used in the simulation has an open loop gain ($$\A_{OL}\$$) of around 32 million which makes this nearly an ideal opamp at DC. Your simulation takes in to consideration non-ideal characteristics of the opamp: like the input bias currents which will affect the results.

The equation used for the upper graph is: $${V_O \over V1} = {-R_F \over {{{R_I+R_F} \over A_{OL}}+R_I}} \tag{1}$$

The base equations for the equation 1 are: $$V_O = A_{OL}(V_P-V_N) \tag{2}$$ $$I_I = I_F = {{V1 - V_N}\over R_I} = {{V_N-V_O}\over R_F} \tag{3}$$

Because the op-amp will do anything to keep the negative input terminal at the same potential as the positive input terminal.

It's due to the impedances if the feedback works or it doesn't. If the input summing resistors were 1 ohm then the op-amp could not work within the range of it's output voltage and current. Unless it is an ideal op-amp with infitely high supply voltages and zero output impedance with ability to drive infinitely high currents.

Here, the opamp simply needs to keep the voltage at zero with modest current set by the input summing resistors. The input needs to have 5V over 10k resistor, so the op-amp output needs to have -5V over 10 resistor to sink the current of input. That's only 0.5 mA.

Let $$\G\$$ be the (large) open-loop gain.

You have $$V_F = \frac{V_{out} + 5 + 0}{3}$$ and $$V_{out} = -GV_F$$ so $$-3\frac{V_{out}}{G} = V_{out} + 5$$ and since $$\G\$$ is very large, $$V_{out} + 5 \approxeq 0$$ In simple language, negative feedback drives the inverting input to the same voltage as the non-inverting input.

If I isolate the resistors and sources participating in feedback, we are left with this:

simulate this circuit – Schematic created using CircuitLab

I'm treating the op-amp output as a voltage source ($$\V_{OA}\$$), because that's essentially what an op-amp output is. I'll assume the op-amp has +5V and −5V supplies, and is therefore able to output any potential between those same extremes (or rather, nearly).

You could use nodal analysis or the superposition principle to analyse this circuit, and find out how voltage VF1 varies as the op-amp output goes between +5V and −5V. Other answers have touched on that, so I'll just use the simulator to examine VF1 potential as I sweep $$\V_{OA}\$$:

As you can see, VF1 varies between 0V and more than +3V, so the op-amp is clearly able to influence the potential at VF1. It's definitely "powerful enough", and negative feedback in your system is alive and well.

The biggest hint that the op-amp has strong influence upon VF1 is that RF is comparable to RL and R1. In other words, VF1's connection via RF to the op-amp output is as "strong" as its connection to +5V via RL, and to 0V via R1. Intuitively this suggests that a change in potential of any one of those three nodes will have as much influence at VF1 as the same change at any other.

• It would be good to note that in the general case (small enough input voltages) the op-amp output "wins the battle" because it monitors the VF1 voltage and tends to keep it zero. So, it is not a simple "static source" like the inputs, but a "dynamic source" with close to zero differential resistance (although its static resistance is also 10 k). Commented Sep 4 at 16:43

To understand the operation of the operational amplifier in the inverting summing configuration, you must first draw the equivalent circuit graph and perform an analysis similar to the following: