Physical explanation of an integrator's capacitor behavior

Question

I understand how the circuit behaves mathematically.

I understand mathematically that $$i_c = C \times \frac{d(-v_\text{out})}{dt}$$

What I have trouble understanding is what is happening physically to the positive and negative charges on the capacitor plates as the circuit integrates the input voltage. What is the fundamental behavior of the capacitor and why (physically, not mathematically) is the op-amp driving the output by integrating the input keeping Vx at 0V.

Feel free to reword this question as English is not my first language and I think I am having trouble being clear.

If I'm reading your question right, you're asking more about the physics of a capacitor just in general than about a capacitor in this specific case. Would that be correct? — Hearth
– Hearth, Commented Oct 30 at 1:13
Another thought crosses my mind, too. Given that math (which is idealistic in ways) isn't wanted, I suspect that an ideal opamp also isn't wanted in the explanation. Which may complicate answers quite a bit. — periblepsis
– periblepsis, Commented Oct 30 at 1:32

Simon Fitch · Accepted Answer · 2025-10-30 15:25:03Z

When you push an electron onto one plate of a capacitor, the increase in negative charge on that side repels one electron off the other plate and out the other terminal. This gives the illusion that current is flowing through the capacitor, even though that's physically impossible due to the ltage across C1 is changing, whiinsulating material between the plates. As charge increases on one plate, and decreases on the other, there's a commensurate increase in potential difference across the capacitor, which you could measure with a voltmeter. In other words, by pushing current through a capacitor, you cause the voltage across that capacitor to change at a rate proportional to that current, giving rise to the equation you wrote. If you passed constant current \$I\$ through a capacitor, voltage across it would rise or fall (depending on current direction) at constant rate. The other consequence is obedience of Kirchhoff's Current Law, applied to two-terminal devices - current \$I\$ into one terminal always equals current \$I\$ emerging out of the other.

To understand how we can control that current, you must first understand the effect of negative feedback on op-amp behaviour:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Negative feedback is implemented via capacitor C1. If \$V_{OUT}\$ rises, so does \$V_X\$. If \$V_{OUT}\$ falls, so does \$V_X\$. X potential follows OUT, different only by whatever voltage C1 has across it. Since X is the op-amp's inverting input, any perturbation of output \$V_{OUT}\$ results in a perturbation at that inverting-input, in turn producing a change of \$V_{OUT}\$ in opposition to the initial perturbation. Consequently any perturbation tends to cancel itself, and the system always settles into a stable equilibrium. The op-amp's voltage gain is so high (hundreds of thousands, or even millions), that even the tiniest potential difference between Y and X, \$V_Y - V_X\$ would result in a huge \$V_{OUT}\$, and so the only stable equilibrium that can occur is one in which that difference is tiny, very near zero. Consequently, the system settles into the state \$V_X \approx V_Y\$.

If \$V_X\$ were to rise even only slightly, by just a few microvolts, \$V_{OUT}\$ would plummet by many volts, taking X with it, which would very firmly and immediately restore the difference to \$V_Y - V_X \approx 0\$. And vice versa. This is the result of negative feedback and high op-amp voltage gain:

$$ V_X = V_Y $$

To put it another way, negative feedback causes an op-amp to adjust its output to whatever potential is necessary to maintain equality between its two input potentials. The resulting condition is \$V_X = V_Y\$, with some exceptionally important and useful consequences. One consequence here, in this example, is that since we have tied Y to ground (0V), so that \$V_Y = 0\$, negative feedback here will produce this condition:

$$ V_X = V_Y = 0 $$

Another consequence is that we know already, by Ohm's law, the value of current \$I\$, because we know that \$V_X = V_Y = 0\$. It's a little bit of maths, but it's easy:

$$ I = \frac{V_{IN}-V_X}{R_1} = \frac{V_{IN}-0}{R_1} = \frac{V_{IN}}{R_1} $$

One more important piece of information is that the op-amp's inputs draw negligible current, often quoted as being due to "extremely high input impedance". That means that none of current \$I\$ enters the op-amp's input, and all of it must therefore pass through capacitor C1. So we find ourselves able to precisely control capacitor current \$I\$ by setting \$V_{IN}\$, according to:

$$ I = \frac{V_{IN}}{R_1} $$

Knowing exactly how much current passes through C1, we can then predict the rate of change of voltage across C1, as it charges or discharges due to this current:

$$ \frac{dV}{dt} = \frac{I}{C_1} $$

We can trust that (due to negative feedback) the op-amp will always maintain \$V_X = 0\$, and so as the voltage across C1 increases or decreases, the op-amp will simply adjust its output to compensate, to keep \$V_X = 0\$. The result is that \$V_{OUT}\$ will always rise or fall at exactly the same rate that the voltage across C1 is changing, which is:

$$ \begin{aligned} \frac{d(V_{OUT})}{dt} &= \frac{I}{C_1} \\ \\ &= \frac{V_{IN}}{R_1C_1} \\ \\ \end{aligned} $$

Actually, that's not strictly true - there's a sign error, because of the polarity of the voltage developing across C1 due to \$I\$, so the real, correct relationship is:

$$ \begin{aligned} \frac{d(V_{OUT})}{dt} &= \frac{-V_{IN}}{R_1C_1} \\ \\ \end{aligned} $$

Is it fair to say that, as the negative ouput voltage augments, it is adding charges to the ''output'' plate of the capacitor to compensate for the input current adding charges to the ''input'' plate, hence keeping Vx at 0V? — Francis Poirier
– Francis Poirier, Commented 19 hours ago
@FrancisPoirier That seems fair to say, but I don't think it's a useful model, at least for me. In my mind I picture it like this: the capacitor's voltage (potential difference) is increasing, because current is flowing through it, but the absolute potentials either side of that potential difference are undefined until you define them - with the capacitor placed as it is, this system maintains an equilibrium through negative feedback, and in that equilibrium one side of the capacitor is held at 0V absolute, while the other (output) is permitted to rise or fall. — Simon Fitch
– Simon Fitch, Commented 14 hours ago

Pete W · Accepted Answer · 2025-10-30 17:20:15Z

Capacitor relates electric charge to voltage: \$Q = CV\$
Electric charge and current are related by integral or derivative: \$Q = \int{Idt}\$
Thus the capacitor's current-to-voltage relation: \$CV = \int{Idt}\$

Now in the context of the op-amp, it's the impedance within the feedback network that matter. That is in simplistic terms \$\frac{V_C}{I_C}\$ or more correctly \$\frac{\partial{V_C}}{\partial{I_C}}\$ . Usually the op-amp and surrounding circuit are analyzed with complex impedance[**], and the "magic rules" of the op-amp (the op-amp changes the output so that the + and - inputs become equal)...

But another way to think of the action of the op-amp is that it "inverts" the transfer functions of the feedback network - specifically that of the output to the inverting input of the op-amp. In this case the feedback network is a C-R network, which computers the derivative. The inverse of that is the integral, and that's what comes out of the output.

[**] We can transforms terms that involve functions of time, such as impedance or network computations (e.g. the ratio of a voltage divider), into functions of frequency. This usually gives the tremendous convenience of transforming the calculus into simple algebra. It isn't actually necessary to do the analysis, nor to answer your question. But mentioning it because it soon becomes a very efficient way to build up a mental model if you're willing to suffer through learning the abstraction.

glen_geek · Accepted Answer · 2025-10-30 03:03:39Z

Were it not for the capacitor's presence, a +ve voltage at \$V_{in}\$ would try to raise the voltage at point "X", because the op amp's "-" input ideally doesn't accept current.

The op amp attempts to maintain voltage at point "X" at zero volts...equal to the voltage at op amp's "+" input. Therefore, current \$I_{in}\$ is proportional to \$R_{in}\$ via Ohm's Law.

With capacitor present, the op amp has a means of maintaining point "X" at zero volts...by continually ramping its output negatively. The current \$I_{in}\$ flows through \$R_{in}\$, and through capacitor C to op amp output. From there, current flows to op amp's negative power supply until the op amp output voltage saturates near the negative supply voltage limit.
Since one end of the capacitor at point "X" is at zero volts, the capacitor's voltage drop is equal to the output voltage.

It should be clear that any input current \$I_{in}\$ causes a proportional voltage changing across the capacitor. Key to this understanding is the relationship of charge, current through capacitor and capacitor voltage.

As Periblepsis points out, an op amp only approaches the ideal case. Key specifications that make it non-ideal are:

input offset voltage (between "-" and "+" input pins)
Input bias current flowing into or out of "-" input pin
finite open-loop gain
speed of response
limits of op amp DC supply voltage

From the capacitor's point of view:
If charge coming in to its plate connected to point "X" were allowed to accumulate, its voltage would rise above zero volts. Only by continually moving charge from its other plate can its point "X" plate remain at zero volts.
If no charge comes into plate "X", capacitor voltage remains constant, since no charge need be removed from its other plate (attached to op amp output). Output voltage remains constant. This also requires \$V_{in}\$ to be zero volts.

There - no math.

TimWescott · Accepted Answer · 2025-10-30 03:13:02Z

You're going to get some math -- but I'll give you the physics behind it.

What I have trouble understanding is what is happening physically to the positive and negative charges on the capacitor plates as the circuit integrates the input voltage. What is the fundamental behavior of the capacitor...

Here's the really short version -- if you want to truly understand this in depth then there's more and better material on the web.

A capacitor has two conductors, called plates, that are separated and cannot conduct electricity between them. When current travels through the capacitor, charge is pulled from one plate and piles up on the other. This charge, in turn, induces a voltage between the two plates. The more charge, the more voltage.

The capacitance is a measure of the capacitor's ability to store charge -- for a capacitor with capacitance \$C\$ and a voltage \$v\$, the charge held by the capacitor must be $$q = C v \tag 1$$ So -- the higher the capacitance, the lower the voltage for a given charge.

Current is just the time derivative of charge: $$i = \frac{dq}{dt} \tag 2$$. This leads to the electronic engineer's conception of a capacitor: it's a thing that has the relationship $$i = C \frac{dv}{dt} \tag 3$$

(3) is not the physics of capacitors -- the physics is the complimentary excess and deficit of electrons on the plates causing a voltage on the capacitor terminals, as described by (1).

and why (physically, not mathematically) is the op-amp driving the output by integrating the input keeping Vx at 0V.

There's a lot of useful information left out here, but the op-amp has tons of gain -- a typical op-amp has a voltage gain of \$A_V \simeq 10^6\$. A good op-amp also has a negligible* current flowing into or out of its input pins. Because \$A_V \simeq 10^6\$ is usually negligibly different from \$A_V = + \infty \$, it's a common** trick in op-amp circuit design to say that if the circuit is stable, then the two input voltages are forced, by the op-amp, to be exactly equal.

So here's the physical reason why that circuit is an integrator:

An integrator circuit as you show is almost always stable.
Because we're assuming that \$A_V = + \infty \$, and because the + input is grounded, we can say the voltage at the - input is practically 0V.
Because we're neglecting current at the + input, any current flowing in \$\mathrm{R_{in}}\$ must go into the capacitor, to either charge it or discharge it.
Because the + input to the op-amp is practically at 0V, the current in \$\mathrm{R_{in}}\$ is \$\mathrm{I_{in} = \frac{V_{in}}{R_{in}}}\$.
Because of capacitor physics, the voltage on the capacitor is \$\frac 1 C\$ times the integral of the current going into it, with the + input needing to go higher than the output when current is flowing in, and lower when current is flowing out.
Because of all of the above, when current flows into the capacitor and the + input of the op-amp tries to go higher, the op-amp corrects, and makes the output go lower by that amount.

The overall result is a circuit that (within limits) integrates the input voltage into the output voltage.

* Different op-amps can have markedly different specifications. What's "good" for any given design has a lot to do with what properties you can ignore and what properties you have to pay attention to.

** And a common pitfall -- remember that later as you advance in your studies.

Hyp · Accepted Answer · 2025-10-30 07:14:12Z

As the op-amp has negative feedback, it keeps the inverting and non-inverting inputs at the same potential, in this case GND, i.e. 0V. This sets Ix = Vx/Rx.

No current flows into an ideal op-amp input, so If = Ix. If you imagine one scenario when Vx is constant and positive, then constant current Ix charges the capacitor linearly.

As the capacitor is being charged, voltage develops across it. The op-amp keeps setting its output voltage to values that "compensate" for the voltage across the capacitor to ensure its inverting input (-) stays at 0V. For example, when Vc is 1V at one point in time, the op-amp sets its output to -1V.

This behaviour can be generalized for any shape of the input voltage (not just constant).

Justme · Accepted Answer · 2025-10-30 08:36:26Z

It appears that the question asks the physical behaviour of how capacitors work accumulating charges, because how capacitors work does not depend on the circuit around them.

When you have a capacitor, and there is current through it, the charges accumulate on the plates and energy is stored in the electrical field between plates, as witnessed by voltage over the capacitor. The accumulated charge Q depends on current and time, and the voltage changes in proportional to capacitance. Q=I×t=C×U. That's equivalent of a bucket of water, if you fill it from the tap at some maybe varying flow rate, it integrates the result as amount of water, and the water level height will be some amount depending on the area (diameter) of the bucket and amount of water.

Now, how an op-amp works is another thing, all needed here is to know that in a valid circuit, the output of the op-amp voltage and current will do anything to keep the two inputs at equal voltage.

In the whole circuit, the op-amp keeps the two inputs at same 0V voltage, which means that the resistor converts the input voltage to current through resistor. As the only route where resistor current then flows is the capacitor, it means that it must accumulate charge at a rate defined by input voltage, and the current must go via op-amp output into an implicit virtual power supply that isn't drawn generally in ideal op-amp circuits.

The output voltage must match the charge accumulated in the capacitance, and charge accumulated was defined by the rate defined by input voltage.

Carl Rutschow · Accepted Answer · 2025-10-30 18:21:56Z

2

"What is the fundamental behavior of the capacitor and why (physically, not mathematically) is the op-amp driving the output by integrating the input keeping Vx at 0V."

Short "Cliff's Notes" no-math explanation:

The basic operation of most op amp circuits is that, due to their high open-loop gain, negative feedback causes the op amp to try to maintain the X (-) input essentially equal to the (+) input (which is at ground potential here, giving the common designation of "virtual ground" for the (-) input).

Thus the output voltage will change to maintain this across the capacitor, as determined by the input current through Rin (since X is basically being maintained at ground potential).
(This assumes the op op amp input bias current is typically negligible).

This means the capacitor is charged or discharged in response, making the capacitor current equal to the input current, with the corresponding change in voltage across the capacitor due to the change in charge, generating the integrate function.

Basically the op amp acts to convert the input voltage (and thus Rin current) into capacitor current.

Make sense?

answered 2 days ago

Carl Rutschow

4,3941 gold badge3 silver badges9 bronze badges

\$\begingroup\$ The input voltage creates the current through the resistor and capacitor. The op-amp only helps it by adding its output voltage to the input voltage. \$\endgroup\$

Circuit fantasist
– Circuit fantasist

2025-10-30 19:01:44 +00:00
Commented 2 days ago
\$\begingroup\$ @Circuit fantasist-- That is not true. The amp does not directly see or care what the input voltage is. It only responds to the input current as determined by the input voltage and input resistance value. It does not add its output voltage to the input voltage. The output voltage is also inverted from the input voltage. \$\endgroup\$

Carl Rutschow
– Carl Rutschow

2025-10-30 23:20:02 +00:00
Commented 2 days ago
\$\begingroup\$ The output voltage is added to the input voltage in series. This can be seen by following the loop Vin - R - C - Vout - Vin. So the effective current-creating voltage across R is Vin - Vc + Vout = Vin - Vc + Vc = Vin, and the current is Vin/R. \$\endgroup\$

Circuit fantasist
– Circuit fantasist

2025-10-31 07:50:22 +00:00
Commented 2 days ago
\$\begingroup\$ @Circuit fantasist -- Sorry, those "loop" voltages make little sense. You have Vin in the loop twice. You can't use such a loop to analyze an op amp circuit. \$\endgroup\$

Carl Rutschow
– Carl Rutschow

2025-10-31 14:39:20 +00:00
Commented 2 days ago
\$\begingroup\$ I am preparing an answer and in it I will illustrate what I mean. A picture is worth a thousand words. \$\endgroup\$

Circuit fantasist
– Circuit fantasist

2025-10-31 14:45:28 +00:00
Commented 2 days ago

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Whit3rd · Accepted Answer · 2025-10-30 06:39:04Z

One definition of a capacitor is C = Q/V; this refers to the charge on one capacitor plate (Q) and we usually ignore the fact that the capacitor is a two-terminal device. There's a charge +Q on one plate, and -Q on the other plate, though in a static electric scenario, a single-terminal capacitor having a net charge is also possible.

And, by definition, R= V/I or in other words R= V/(dQ/dt) In the amplifier-has-large-gain model of a perfect operational amplifier, the constant V applied on the resistor makes a constant dQ/dt in accord with the resistor characteristic, and the voltage on the op amp (+) and (-) terminals will be presumed nearly equal. That means the (-) terminal is near ground (pseudo-ground is the usual name of this state).

The situation with a variable V input is that the resistor and capacitor scale an integral of current over time which is present at the op amp output.

A one-terminal capacitor (often a sphere of metal) is presumed to be in a ground-at-infinity scenario; such capacitors are used as the inputs to the Theremin musical instrument, where they are adjusted... by hand.

Circuit fantasist · Accepted Answer · 2025-10-31 17:49:16Z

What is the fundamental behavior of the capacitor and why (physically, not mathematically) is the op-amp driving the output by integrating the input keeping Vx at 0V.

The circuit consists of three elements: a resistor, a capacitor, and an op-amp. To understand it, we must reveal the role and behavior of each one of them.

Building the circuit

I will do this here briefly in four steps, and you can find more detailed information from the links provided at the end.

C integrating circuit

A capacitor driven by a constant (step) current source is the simplest, yet perfect passive current-to-voltage integrator. It is perfect because the input current source ensures that the current remains constant.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The output voltage across the capacitor changes linearly through time.

RC integrating circuit

For a number of reasons, however, we prefer to use voltage instead of current. This necessitates converting the input (step) voltage Vin into current I using a resistor R, in accordance with Ohm's law I = V/R. So, the resistor acts as a passive voltage-to-current converter.

Thus, the resistor and the capacitor form the simplest integrator with a voltage input and output. This is how we managed to assemble the simple RC integrating circuit, which is usually given as something ready-made and complete!

schematic

^{simulate this circuit}

However, a problem appears - the capacitor voltage is subtracted from the input voltage and the current gradually decreases - Iin = (Vin - Vc)/R. This is why the capacitor voltage slows down its course and changes exponentially instead of linearly.

Active V-to-V integrator

Since voltage is lost across the capacitor (it is subtracted from the input voltage), then why not add the same voltage to compensate for it? For this purpose, we can connect a copying auxiliary voltage source -Vc in series, oriented so that its voltage is added to the input voltage.

schematic

^{simulate this circuit}

Now when the voltage Vc across the capacitor changes in a positive direction, the auxiliary voltage -Vc changes in a negative direction, and the result at the output V- is 0 V (the so-called "virtual ground"). Since the old output voltage is destroed, we can take the copy -Vc as an output. It has the advantages of being grounded and buffered.

If we draw the auxiliary source below the input source, we will see more clearly how their voltages add up in a "battery" Vinhelp = Vin + Vc.

schematic

^{simulate this circuit}

We see that the voltage of this "assisted input voltage source" is "lifted" by Vc and also changes linearly with time.

If we now group the capacitor and the auxiliary source into a "battery" , their voltages cancel out, and the total voltage is 0 V (the so-called "virtual short").

schematic

^{simulate this circuit}

Op-amp V-to-V integrator

In the practical circuit, the compensating voltage -Vc is taken from the output of an op-amp with negative feedback. In its effort to maintain zero input voltage, it keeps its output voltage equal to the voltage across the capacitor.

schematic

^{simulate this circuit}

As above, when Vc changes in a positive direction, -Vc changes in a negative direction, and the result is V- = 0 V ("virtual ground").

So, what does what?

When the input voltage source begins to push current through the resistor and the capacitor, the latter starts to charge. The voltage across it begins to rise and acts as a disturbance to the op-amp. It starts to overcome this disturbance by increasing its output voltage in the corresponding direction so as to compensate for the capacitor's voltage. In fact, the op-amp output voltage is added to the input voltage, such that the current I = Vin - Vce + Vce = Vin/R = const.

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Physical explanation of an integrator's capacitor behavior

9 Answers 9

Building the circuit

C integrating circuit

RC integrating circuit

Active V-to-V integrator

Op-amp V-to-V integrator

So, what does what?

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