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I have often seen (and made) circuits like this. It is a simple inverting op-amp buffer, plus a small current-limiting output resistor to protect the the op-amp if the output is shorted.

schematic

simulate this circuit – Schematic created using CircuitLab

But, lately I have noticed some people using this design instead:

schematic

simulate this circuit

With my very rudimentary electrical understanding, it seems like these circuits would be equivalent (aside from slightly higher gain and arguably more confusing layout in the second).

Is there a reason to favor the second design? If the output passes through a long cable or something else weird would having the output resistor included in the feedback loop help to keep the voltage accurate?

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  • \$\begingroup\$ Simple answer: Whe a certain part is not part of the feedback loop and when it has no effect on the signal which is fed back - it also has no influence on stabilty properties of the system. Example: R3 in the 1st figure has no influence on the loop only in case the output impedance of the opamp is negligible small. \$\endgroup\$
    – LvW
    Nov 8, 2023 at 19:34

4 Answers 4

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The second design has a constant output with resistive loading (until the op-amp saturates or current-limits). The first one would output only half the no-load voltage with a 1k ohm load.

There are subtle issues with stability- the first design is likely stable even without the 33pF capacitor (definitely with lower resistor values). That is true regardless of loading. Of course if you put a large capacitor on the output the output will roll off much faster as the frequency increases. The second design requires a capacitor as shown (and maybe a higher value) if the output has much capacitive loading- such as driving a meter of shielded cable.

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Both circuits are used when driving unknown loads which may be capacitive, such as long cable.

Op-amps do not tolerate much capacitance directly on their output, or the op-amp may become unstable and the output waveform may have ringing or other distortion, or it simply turns into an oscillator.

The circuits you posted implement two different methods of dealing with capacitive load.

The first circuit is called out-of-loop compensation where you simply make an op-amp circuit and put an isolation resistor between op-amp output and the load so it is not inside the feedback loop.

The second circuit is called in-loop compensation, as the added isolation resistor is inside the feedback loop, but still isolating the output from being connected directly to the load.

Now, each of the circuits have both upsides and downsides, and therefore which circuit to use depends on the situation. For more info, many op-amp manufacturers have application notes about dealing with capacitive loads, and they go into detail about how the system poles and zeroes of the system affect the stability and where to move them with external components to keep the system stable.

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Basic idea

In the behavior of negative feedback circuits, we can observe a few main phenomena that can help us answer the OP's question:

  1. When we affect NFB circuits with disturbances, they compensate them at the cost of additional efforts.

  2. They can do this until they reach their limits (saturate).

  3. In an effort to compensate for disturbances, negative feedback followers become amplifiers.

They are not just specific circuit ideas but much more general ones that can be seen all around us; they are concepts. My story is built on this belief that, in order to understand the particular phenomenon manifestations, we must first understand the general principles.

Things are simple once we understand why they are made the way they are, and what the point of it all is. The problem is that they don't tell us and we have to find out the truth ourselves. This is what I have illustrated in my story. I have not been able to explain in great detail the meaning of each concept because that would require even more space.

How to reveal it

Let's then investigate these phenomena by perturbing the operation of basic op-amp amplifier circuits with negative feedback through the resistor R3 (from the OP's circuit) and observing their response. As usual, I will take my favorite approach with a series of step-by-step CircuitLab experiments illustrated with schematics and graphical results. Indeed, this makes this story rather long, but that is the price of understanding. To truly understand something, it is not enough to become familiar with its final state; you have to follow the path of its evolution.

How to experiment it

In the most of the simulations below, I have used an "ideal" op-amp without power supply because, for some reason, CircuitLab has a hard time simulating an op-amp with supply pins; thus we cannot use the so attractive Live DC simulation.

Also, to simplify the schematics, in some of them, I have used a real voltmeter (with intentionally reduced internal resistance) as a "visualized load". You can change its resistance by opening the voltmeter parameters window and enteging the new value of the resistance.

When explaining basic things, the values and types of components are not particularly important. That is why, I have changed them to make some effects visible in the schematic.

The op-amp used is "generic" (not a specific type); therefore not labeled. We are discussing concepts here and the details only distract from the main idea.

Inverting configuration

Inverting follower

The most basic op-amp inverting configuration is the inverting follower (op-amp inverting amplifier with a gain of -1 or simply, inverter). In this circuit, the input Vin and the output Vout voltages are compared (subtracted) by being opposed via two identical resistances R1 and R2 (resistor summer with equal weighted inputs).

schematic

simulate this circuit – Schematic created using CircuitLab

If we set, for example, -1 V input voltage, the output adjusts its output voltage to 1 V. So, the output voltage is equal but opposite to the input one.

STEP 1.1

Disturbed inverting follower

Added disturbance: Now, let's connect the "disturbing resistor" R3 between the op-amp output and the load (a voltmeter with only 1 kΩ resistance), and take the feedback from the op-amp output.

schematic

simulate this circuit

As you can see from the VL-1k voltmeter and the graphs below, the voltage across the load is halved due to the voltage divider formed by the 1 kΩ R3 and the 1 kΩ RL, but the op-amp does not react because it does not sense the voltage drop.

STEP 1.2.1

Inserted disturbance: We know the remedy from life - we need to make the op-amp monitor the voltage after the disturbance R3 (close the feedback after it).

schematic

simulate this circuit

Now the op-amp op-amp overstrains itself and raises twice its output voltage up to 2 V - half (1 V) drops across R3, and the other half (1 V) drops across the load. So, the R3 disturbance is totally compensated, and the circuit acts as the understurbed inverting follower above (schematic 1.1). Figuratively speaking, we have turned an imperfect voltage source with 1 kΩ internal resistance into a perfect one with almost zero resistance.

STEP 1.2.2

Inverting amplifier

But cannot we use the "harm" caused by R3 for something useful (there was such an inventive principle)? Here we guess that if we use the op-amp output as an output, the circuit will become an amplifier.

schematic

simulate this circuit

STEP 1.3.1

T-bridge inverting amplifier

Thus we subtly invented the so-called "T-bridge inverting amplifier". We just need to draw the circuit like this.

schematic

simulate this circuit

STEP 1.3.2

Non-inverting configuration

Although the OP's circuit is inverting, I guess it will be of interest to them to see how this trick works in the non-inverting configuration. These are great ideas and both are worth considering.

Voltage follower

This is really the simplest op-amp negative feedback configuration. Here the two voltages - Vin and Vout, are compared (subtracted) even without any resistors, just connected in series and opposite.

schematic

simulate this circuit

"Looking" at the difference between its inputs, the op-amp adjusts its output voltage to make it equal to Vin (the two transfer curves match perfectly).

STEP 2.1

Disturbed follower

Added disturbance: As above, first we connect the "disturbing resistor" R (R3) out of the feedback loop.

schematic

simulate this circuit

The result is the same - twice less voltage across the load.

STEP 2.2.1

Inserted disturbance: But we know the "secret", and close the feedback after R.

schematic

simulate this circuit

The disturbance is eliminated and the "disturbed follower" acts as the undisturbed one (schematic 2.1).

STEP 2.2.2

Non-inverting amplifier

Then we take the output voltage from the op-amp output and thus obtain the ubiquitous non-inverting amllifier.

So, we conclude, a non-inverting amplifier is just a disturbed op-amp follower.

schematic

simulate this circuit

STEP 2.3.1

T-bridge non-inverting amplifier

Finally, we disturb the op-amp once more by inserting a second voltage divider R3-R4, and make it amplify 4 times. And of course, we transform the schematic into a "T" shape; so it is easier to memorize :-)

schematic

simulate this circuit

STEP 2.3.2

The role of R3

After all this "theory" above, it is time to clearly answer the OP's question, what exactly is the role of this "mystical" resistor R3.

Power-supplied op-amp

But first let's introduce an important complication. So far we have used an "ideal" op-amp that could produce any voltage at its output up to infinity. In this case, there would be no use for the R3 because it would never come into its role as an "automatic fuse". So let's replace the op-amp with one that has power terminals, and supply them with +-10 V.

schematic

simulate this circuit

We see that the picture becomes much more interesting because sections appear where the op-amp is "saturated" (its output voltage has reached the supply voltage). There are other interesting things, for example that the voltage at the inverting input is not (virtual) zero in these sections but starts to follow the input.

STEP 3.1

So until the op-amp output voltage has reached the power supply, the op-amp is able to cope with its task of compensating the voltage drop across R3.

Virtual conductor

Let's illustrate it with a "virtuoso" CircuitLab experiment thanks to the possibilities that the so-called "behavioral voltage sources" give us. For this purpose, let's move out of the op-amp the partial "voltage source" that compensates the drop VR3 across R3 (its voltage is an exact copy of VR3). You can consider this voltage as a "positive disturbance" for the op-amp.

schematic

simulate this circuit

The combination of the two elements in series behaves as a virtual wire with zero resistance. So, as you can see, the circuit output voltage (1 V) across the load is equal to the op-amp output voltage (1 V).

STEP 3.2

Negative resistor

As you can see in the schematic above, the voltage source -VR3 produces a voltage that is a mirror copy of the voltage drop VR3 across the resistor R3; so it acts as a "negative resistor" with resistance -R3. We can easily simulate this arrangement since CircuitLab can work with negative resistances.

schematic

simulate this circuit

This negative resistance -R3 neutralizes the positive resistance R3, and the combination of the two resistors in series behaves as a virtual wire with zero resistance. So, as above, the circuit output voltage (1 V) across the load is equal to the op-amp output voltage (1 V).

STEP 3.3

Varying load

So, while the load is drawing relatively low current, the op-amp output voltage has not reached the supply voltage and the op-amp manages to "zero" the resistance of the resistor R3 (i.e., it is disconnected from the circuit and does not affect the voltage across the load).

However, when the op-amp saturates, its output voltage "freezes" and it can no longer do this. The resistor R3 comes into action and its resistance begins to affect (decrease) the load voltage. We can observe this by starting to change the load resistance or best by sweeping it (run the DC Sweep Simulation).

schematic

simulate this circuit

As you can see from the graph, while RL < 120 Ω, R3 is "switched on" and protects the op-amp output; when RL > 120 Ω, R3 is "switched off" and does not affect the load voltage.

STEP 3.4

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  • \$\begingroup\$ The usual downvote from the anonymous downvoter... I think there should still be some mechanism against such pathological behavior - at the very least, to be warned... \$\endgroup\$ Nov 8, 2023 at 20:10
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    \$\begingroup\$ I downvoted. You don't answer OPs question. IMO this is a massive overkill info-dump that does not help someone with a "very rudimentary electrical understanding", and made me scroll several pages down to the next useful answer and that makes it "not useful". The downvote is the warning. \$\endgroup\$
    – pipe
    Nov 9, 2023 at 10:14
  • \$\begingroup\$ @pipe, Thanks for the reply. I agree that my answer is long and I have explained why at the beginning. I expect others, and especially the OP, to weigh in on the merits of the material. However, I want to ask you, isn't there at least one thing inside that impresses you? \$\endgroup\$ Nov 9, 2023 at 10:22
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    \$\begingroup\$ I think the issue is that it never really got to answer the OP's question. For one, this answer doesn't include the same components or circuit from the OPs question. All the other answers discuss compensation and capacitive loads, which answers the OPs question directly, IMO. This answer can be helpful, but for a different question than the one that was asked. \$\endgroup\$
    – KD9PDP
    Nov 9, 2023 at 12:50
  • \$\begingroup\$ @KD9PDP, Thanks for the response. When explaining basic things, the values and types of components are not particularly important. I have changed them to make some effects visible in the schematic. Capacitive loading is just one possible application of this basic idea. In conclusion, in order to understand the particular manifestations, you must first understand the general principles; my story is built on this belief. And one wish - let's not just talk about general things, but discuss specific facts, phenomena, techniques, etc. Also, it's good to show what we claim with something done. \$\endgroup\$ Nov 9, 2023 at 13:24
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The (closed loop) output impedance of the first circuit is 1kOhm, the output impedance of the second is 0Ohm: differences in voltage will get compensated, of course in either case limited by the supply voltage of the opamp and its slew rate. Indeed, C1 acts as a slew rate limiter that will help to avoid oscillation, assuming that the opamp is unity-gain stable.

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