# How does this op-amp config work as a voltage follower?

I've checked a couple of books and the internet, but every source just tells me it's a voltage follower, and not why. Without any resistors, I'm having trouble applying the 2 golden rules of opamps (infinite input impedance and the voltage drop across the input terminals being 0). Why does this circuit behave as a voltage follower?

• The Op-Amp in this configuration is connected as non-inverted amplifier with unity gain (Gain=1) so the output follows the input. The advantage of this is you can get a lower output impedance than your input signal source.
– user208862
Commented Nov 7, 2021 at 7:12
• Try to read this electronics.stackexchange.com/questions/441184/… it may help.
– G36
Commented Nov 7, 2021 at 8:03
• If we slow down the opamp action we can see what is going on. At the beginning, we have Vin = (+) input = 1A and Vout = (-) input = 0V, Thus the opamp input will see a huge difference between "+" and "-" inputs. This will force the output and the (-) input into a positive direction. But notice that as the output voltage increases the voltage difference between V"+" and V"-" is reduced thus, the opamp "driving force" (Vd = V"+" - V"-") is reduced, the opamp output voltage will no longer move so fast toward the positive rail.
– G36
Commented Nov 7, 2021 at 8:29
• And if this voltage reaches V"+" = V"-" = Vin = Vout the opamp stops increases his output voltage. But if somehow op-amp overshoots and the output voltage will be larger than +1V. This immediately will change the sign of a voltage difference between V"+" and V"-" (the "driving force" ). So the opamp output will no longer be driven into a positive direction but due to the change in the voltage sign ( V"+" is now lower than V"-"). And the opamp will start to decrease his voltage to reach Vin = Vout.
– G36
Commented Nov 7, 2021 at 8:29
• @Soumil, I share your observations that "books and the internet" tell "it's a voltage follower" but do not reveal why it is done that way. That's why I wrote this fancy story especially for you as a thank you for the nice question. In it I have revealed what the idea of ​​this famous circuit is. A similar story could be written for the emitter follower and the source follower... and why not the cathode follower? Commented Mar 19, 2023 at 19:58

The op-amp strongly amplifies any difference it sees on its terminals.

With no feedback, an 'ideal' op-amp will take any difference and go between + and - infinity. A real op-amp will have a finite open-loop gain (Av) and amplify that difference by the gain.

Example op-amp with Av = open loop gain of 10,000V/V:

• input difference of +1mV, Vout will be +10V

Now add feedback. The follower ties Vout directly to (-) input. Net result: any change on (+) input (that is, Vin) is cancelled on (-) by Vout. So (-) input and Vout track Vin one-to-one, or nearly so:

• ideal op-amp: Vout will follow (+) exactly
• real op-amp: Vout will have a difference of Vin/Av

Again, with an Av = 10,000, a voltage follower like you've shown, a +1V input on (+) will have a difference of 1/10,000th of a volt vs. (-) due to limited gain.

More about op-amp gain here: https://www.electronics-notes.com/articles/analogue_circuits/operational-amplifier-op-amp/gain-equations.php

A slightly deeper, but very accessible paper here: https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/lecture-notes/22_op_amps1.pdf

In the model of ideal operational amplifier you find the following 2 equations:

V+= V-

I+ = 0

Please note that I- = 0 is not needed because it's implied by I+ = 0.

V+ = Vin implies V- = Vin

Now, "walk" over the wire from V- to Vout and you get:

Vout = V-

which means: Vout(t) = Vin(t)

• $I^- = 0$ most certainly is not implied by $I^+ = 0$. Real-world opamps model both input bias current and input current offset; there is nothing about the application circuit that guarantees zero current offset. Commented Nov 8, 2021 at 20:34
• I wrote "in the ideal model". Commented Nov 8, 2021 at 21:12
• Yes, in the ideal model both those quantities are zero, but because they are idealized that way not because the circuit guarantees it Commented Nov 8, 2021 at 23:39

I'm having trouble applying the 2 golden rules of opamps (infinite input impedance and the voltage drop across the input terminals being 0)

1. Infinite input Impedance (rule 1)
2. Vin- equals Vin+ (rule 2)

It's rule 2; the voltage difference between Vin+ and Vin- is zero. Given that Vin- is connected to the output (Vout), it's fairly safe to say that the voltage difference between Vout and Vin+ is also zero hence, it's a voltage follower.

1. Your sources are insufficient. I suggest you to do more diligent work. Besides, you had exact answer already in your sources, probably from every sources, for the reason below.

Why does this circuit behave as a voltage follower?

1. "voltage drop across the input terminals being 0", this alone gives the very reason why Vout == Vin.

A quick internet search for op-amp voltage follower theory will give you an explanation how a voltage follower works. The output being fed back to the negative input provides a unity gain amplifier. The voltage difference between Vin+ and Vin- being zero, Vin- will follow Vin+, and Vout = Vin- in the circuit.

Here are the first two search results I got: https://www.electrical4u.com/voltage-follower/

https://www.elprocus.com/what-is-voltage-follower-working-its-applications/

# How to explain circuits

I've checked a couple of books and the internet, but every source just tells me it's a voltage follower, and not why.

To explain a circuit, we must show not only WHAT and HOW but also WHY it is made that way. If we do not do this, those to whom we explain will know the circuit but will not understand it.

# Building an op-amp follower

The best way to show WHY a circuit is made that way is to build and "invent" it step by step. I will do it in a scenario of five consecutive steps illustrated by CircuitLab simulations with each subsequent step building on the previous one. I will reveal the circuit idea in the following sequence: non-electrical examples (analogies), passive electrical circuit, man-driven circuit, op-amp circuit.

## STEP 1: How do we copy voltage?

In life. The op-amp follower is a simple circuit based on a great idea that is not just electrical. We can see it all around us and we ourselves behave according to this idea when accomplishing our goals. This is a kind of "active copying" - we have some goal X in our head and want to materialize it as Y = X.

simulate this circuit – Schematic created using CircuitLab

In electronics, such a task is "voltage copying" - we want the same output voltage Vout as the input Vin1 to appear in a different place (the output).

simulate this circuit

## STEP 2: "Wire follower"

In life, the obvious solution is to get down to business and solve the problem ourselves.

In electronics, the first idea that comes to mind is to connect a piece of wire from the input source to the output. It works fine since there is no disturbance - the voltage source is "ideal" (with zero internal resistance), the wire is an "ideal" conductor (with zero line resistance), the ammeter is "ideal" (with zero internal resistance) and there is no load connected (open circuit). So the output voltage is equal to the input voltage (see the meters or hover the mouse over the circuit nodes and elements to see for yourself).

simulate this circuit

## STEP 3: Disturbed "wire follower"

In life, however, something always appears that prevents us from realizing our intentions - for example, we are weak or sick to do it, and the result is less than expected.

simulate this circuit

In electronics, we work with real elements that have some resistance. For example, a real (imperfect) input voltage source RVS (in a purple frame, in the schematic below) or the line can have an internal resistance Ri = 1k (we can present it by a 1 k resistor Ri). Also, an imperfect ammeter can have an internal resistance RA = 1k (we can set it in the ammeter CircuitLab parameters).

- Open circuit. But since still there is no load connected, no current flows and the output voltage is still equal to the input voltage. This goes against our human intuition; that is why I suggest you to see again the meters' readings or hover the mouse over the circuit nodes and elements so you can see for yourself.

simulate this circuit

- Loaded circuit. It is logical that we wish to use the voltage Vout for some purpose, and we connect a load RL to the output. As a result, a load current IL begins flowing through the internal resistance Ri and an unwanted voltage drop appears across it. This drop is subtracted from the input voltage and the output voltage across the load is less than the input voltage. Here it is an undesired effect but it is extremely useful when creating a voltage divider (Ri-RL). Also notice that there are now two input voltages - the internal Vin1 (the electromotive force) and the external Vin (at the source terminals).

Experiment with various Ri, RA and RL setting them in their parameters windows. Another trick can be to set some relatively small resistance RV of the voltmeter thus making it behave as a "load".

simulate this circuit

## STEP 4: Negative feedback follower

In life, we solve such problems through the ubiquitous negative feedback principle. In its perfect form, it means making one quantity Y equal to another quantity X by comparing the two quantities by subtraction, and varying the quantity Y until the difference X - Y is (almost) zero. In this way we maintain the temperature and lighting in a room, the speed and direction of a car, the state of our mind and body, the level of our voice when speaking and countless other examples... In our example, we can hire workers (we become bosses :-) and assign them to do our work.

simulate this circuit

In electronics. Since this principle is so powerful, then let's use it to make a perfect voltage follower

- Cloned voltage source. The first idea here is to supply the load by a separate voltage source Vout with the same voltage as the input one. But how do we make its voltage equal to Vin?

simulate this circuit

- Voltage comparison. We can adjust Vout = Vin by taking advantage of KVL. For this purpose, we include the two voltages in series and in reverse in a loop and apply their difference to a null voltage indicator NI (a sensitive voltmeter). Now you have to change Vout until the null indicator shows zero voltage. Practically, you have to open the Vout parameters window in CircuitLab and, while watching NI, change Vout until zero the reading.

simulate this circuit

- Current comparison. The trick here is that the load draws current from the new source and not from the input source. So no current flows through the zero indicator; as if it has an infinitely high resistance (open circuit). This phenomenon was used as far back as the 19th century to make "ideal" voltmeters. The problem then was that they did not have good voltmeters because they made them using an ammeter and resistor in series, and that is why they came up with this trick. The effect is striking - you have loaded the input source with a "bad" voltage indicator and a "bad" voltmeter in series... but the source "sees" an infinite resistance (open circuit, "nothing"). For the same purpose - artificially increasing resistance, this technique (aka "bootstrapping") is used today in modern amplifier stages.

The trick in this series arrangement is that we not just compare the output voltage with the input voltage; we subtract it. In this way, we zero the total voltage and current in the circuit and it does not matter what connects the two sources - a voltmeter or an ammeter. Let's try the latter as in the 19th century, replacing the voltmeter NI witn an ammeter.

simulate this circuit

## STEP 5: Op-amp follower

Op-amp as integrator. Everything we have done so far has been to understand what the op-amp does in the voltage follower circuit by putting ourselves in its place. Now it remains only to automate the circuit through an op-amp that performs the role of the "copying" voltage source Vout, the null indicator and, of course, the man. The op-amp must have a differential input because the difference between the two voltages is "floating" (maybe this was one of the reasons to make the op amp with a differential input). The op-amp does not need a null indicator since it directly "sees" the voltage difference Vd between its inputs and adjusts its output voltage to keep it near zero (the H&H "golden rule"). As a result, Vout = Vin = Vin1. There is only load current flowing from the op-amp output through the load.

simulate this circuit

Op-amp as amplifier. In fact, the op-amp behaves like this only when the input voltage changes. Then it is an integrator rather than an amplifier with a temporary gain Vout/Vd much less than its open-loop gain A_OL. Its output voltage rapidly increases and it tends to reach the equilibrium point where Vout/Vd = A_OL (see also my answer to a similar question yesterday). From this moment it becomes an amplifier with A_OL gain.

I'll add a couple of labels ,"Q" and "P", to the schematic, to help me explain:

simulate this circuit – Schematic created using CircuitLab

We must first agree that this op-amp multiplies the difference between its two inputs, $$\V_{P}-V_Q\$$, by some huge positive number $$\A\$$, and presents the result as $$\V_{OUT}\$$:

$$V_{OUT} = A\times (V_P-V_Q)$$

Also, because of the wire from OUT to Q, $$\V_Q=V_{OUT}\$$, and for the same reason, $$\V_P = V_{IN}\$$

That's all the maths I want to talk about.

Intuitively, if $$\V_{P}\$$ is even the tiniest bit positive with respect to $$\V_Q\$$, the output goes skyrocketing upwards, because $$\A\$$ is so large. But, if it rises, it takes $$\V_Q\$$ with it, because OUT and Q are connected together. Surely, therefore, $$\V_Q\$$ will pass through some point where it meets $$\V_P\$$, and the difference becomes zero. At that point, the op-amp output ceases to rise, because it is no longer motivated to continue; the difference in input potentials that initially provoked the rise is gone.

Conversely, if ever $$\V_Q\$$ is a tiny bit positive with respect to $$\V_P\$$, now the op-amp will make its output plummet towards a large negative value, taking Q with it, but in doing so it must pass the point where $$\V_Q\$$ and $$\V_P\$$ are equal again, zero difference.

Can you see how no matter what potential we apply to input P, the output will always change in a direction that reduces the difference between $$\V_P\$$ and $$\V_Q\$$? Even the tiniest deviation from this "equilibrium" will cause the op-amp output to re-adjust, so that $$\V_P = V_Q\$$.

Because the direction that the output changes always opposes any input difference that takes the system out of equilibrium, the feedback from OUT to Q employed here is called "negative feedback".

In the end, this circuit always produces whatever output is necessary to make its two inputs have equal potential. And since one of the inputs is under your control, $$\V_{IN}\$$, the output is incessantly re-adjusting to obtain the state $$\V_{OUT} = V_{IN}\$$, following $$\V_{IN}\$$ wherever it goes.

You are already familiar with negative feedback, because you use it all the time in your daily life. Every time you drive, you adjust your steering in a direction that reduces the difference (the "error") between where you want to go, and where you are going. Each time you adjust the room temperature, you are adjusting temperature the direction that decreases the difference between how hot it actually is, and how hot you want it to be.

That's what's happening here. You tell the op-amp what voltage you want by setting $$\V_{IN}\$$, and negative feedback causes the op-amp to adjust its output in whatever direction is necessary to minimise the difference between where $$\V_{OUT}\$$ actually is, and where you want it to be.

Very interestingly, when driving onto a new road having a completely arbitrary angle with respect to the one you're leaving, you don't need to know where North is, you don't even need to know what "degrees" are! You just need to be aware of where you should be going, and where you are going, and adjust accordingly. Same for temperature, all you need to know is "it's too hot" or "it's too cold" and you adjust up or down to fix the error. By that same logic, an op-amp doesn't need to know where "zero volts" is, or even how big a single volt is. If it's made aware of the desired output, then using negative feedback it constantly adjusts actual output, until $$\V_Q = V_P\$$.

Now, if you are asking (as somebody just did, in the comments) if $$\V_P = V_Q\$$, and $$\V_{OUT} = A\times (V_P-V_Q)\$$ then how is it possible, for $$\V_{OUT}\$$ to be anything but zero? Well, I lied a little.

In reality, those two potentials are only almost the same. $$\A\$$ is so large that you only need a tiny, tiny difference between $$\V_P\$$ and $$\V_Q\$$ to produce a very large output voltage. It's more accurate to say $$\V_P \approx V_Q\$$, but the difference is so small that we nearly always make the approximation $$\V_P = V_Q\$$. But it's a very good approximation, good for almost all intents and purposes.

It's often explained that an ideal op-amp has infinite gain, $$\A=\infty\$$, but that would lead to a similar inconsistency. It is more appropriate to say that the ideal operational amplifier has a gain approaching infinity. That avoids problems of infinities in the algebra, and the divisions by zero that crop up when you manipulation the equations.

One last thing; the fact that the op-amp's two inputs have equal potential is not a property of the op-amp, not a "golden rule" of op-amps. This behaviour, the tendency to make $$\V_{P} = V_Q\$$, is purely the result of negative feedback, and is not some innate property of op-amps.

• I agree that Vin = Vp = Vq = Vout and A is a large number, but if so, then your formula Vout = A × (Vp - Vq) can't be true. That would mean that for any Vp = Vq the Vout is zero. Commented Mar 20, 2023 at 10:34
• @Justme Yeah! I didn't want to go there, and now I have to! Commented Mar 20, 2023 at 12:10
• @Justme thank you for pushing me, I have addressed it Commented Mar 20, 2023 at 12:26
• Actually there is no inconsistency in ideal op-amps. If you take the formula and let Vp = Vq or let Vp - Vq approach zero, it just means that A must approach infinity or be infinity for the equation to hold. It usually is not worth mixing the open loop operation of op-amps confuse with the easy golden rules (where the open loop gain is infinite) where KCL and KVL can be used. Commented Mar 20, 2023 at 12:54