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I know there are at least two questions related to this on stackoverflow but neither really answer my question, and in any case, both questions got downvoted. What I am after is an operational understanding of how an op-amp integrator works. I know how a simple RC circuit can integrate, what I don't understand is how the feedback loop in an op-amp configuration helps. I understand how feedback works in a noninverting amplifier. I took the figure below from www.electronics-tutorials.ws. This web site has an explanation but I don't follow it. My understanding so far is this:

  1. Apply a positive voltage to input vin. Current flows through Rin resulting initially in a non-zero voltage at X (Correct?).

  2. Due to the high impedance of the op-amp at X, we can assume that all the current then flows to the capacitor (initial discharged).

  3. The capacitor starts to charge resulting in a voltage across the capacitor.

  4. The difference in voltage at the two op-amp inputs (the positive input is at zero, hence the difference is negative) resulting in the output, vout, going negative (we assume that vout was zero initially).

My question is what happens next? How does the feedback act to bring the difference between the two inputs back to zero? Or have I got this wrong?

I am very familiar with the proofs for showing that the configuration will integrate but they don't give any real intuition and many videos, wikpedia, and books but almost all regurgitate the proof without giving much insight. I'm after an intuitive understanding, not a mathematical proof.

enter image description here

Out of interest I also redrew the op-amp circuit next to the RC integrator shown below which gives the suggestion that the op-amp is amplifying the small vollage across C (assuming high R1) while having high impedance from the resistor/capacitor node. Not sure if that is a legitimate way to look at it.

enter image description here

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    \$\begingroup\$ Put in a square wave, the opamp strives to hold the (-) pin very close to zero volts, which imposes a constant voltage across the input resistor. This constant current has to go somewhere, and the only path is into the capacitor. Give constant current into a capacitor, you get a perfect RAMP output, which is the integral. \$\endgroup\$ – analogsystemsrf Jan 18 '19 at 3:04
  • \$\begingroup\$ To me - this small comment (if compared with all the answers) is the best explanation of the circuits behaviour in the time domain. \$\endgroup\$ – LvW Jan 18 '19 at 14:57
  • \$\begingroup\$ But it doesn't explain why the op-amp strives to hold the (-) pin close to zero. An op-amp on its own won't do that, it's the combination with the feedback that makes that happen. This is the bit I wasn't sure about. \$\endgroup\$ – rhody Jan 18 '19 at 18:47
  • \$\begingroup\$ The RC integrator is not a real integrator. It can be treated as one in a limited sense. But consider a charged capacitor and an input voltage of 0 V. According to how integrations work, the integral should not change. However, in this case the capacitor discharges => Your integral value goes to 0. This can also be seen in the Laplace transform of the two circuits: An integrator: H(s) = constant * 1/s. Your RC-circuit (essentially a lowpass): H(s) = constant * 1 / (1+RC * s). For high frequencies (!) 1 + RC * s can be assumed as the same as RC* s. In this case it is acting as an integrator \$\endgroup\$ – GNA May 27 '20 at 6:52
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This may help:

  • Remember that when current flows into the RC junction of your op-amp that the voltage at that point will tend to rise.
  • If the inverting input voltage rises the slightest bit above the non-inverting input voltage then the op-amp output will start to swing negative.
  • The output swinging negative will, through the capacitor1, tend to pull the inverting input down towards zero again where it stabilise (for the moment).

The result is that feeding current into the RC node causes the op-amp output to go negative.

Out of interest I also redrew the op-amp circuit next to the RC integrator shown below which gives the suggestion that the op-amp is amplifying the small voltage across C (assuming high R1) while having high impedance from the resistor/capacitor node. Not sure if that is a legitimate way to look at it.

That's correct. It might be better than you think. The simple RC circuit has the advantage that it's non-inverting but the disadvantage that it's non-linear. With a constant input voltage the output will be an exponential charge curve.

Putting the op-amp in as you have shown still allows the capacitor to charge up but maintains the top terminal at virtual ground. The advantage is a linear change in output. The disadvantage is that there is a minus sign on the integral obtained.


1 You can think of a capacitor as holding the voltage across it as a constant in the short term. That means that if the voltage on one side is changed the voltage on the other side will try to change by the same amount.


From the comments:

One question. what is the orientation of the capacitor in terms of conventional current? i.e. if vin goes positive the capacitor is I assume negative on its right-hand side (nearest vout). Now vout goes negative and therefore reduces the voltage across the capacitor until the potential at X is zero?

I think your understanding is correct.

If Vin goes positive then current flows into the X node charging up C. (Remember the op-amp's voltage hasn't changed yet.) This tends to increase the voltage on the inverting input and that causes the output voltage to decrease. This draws some charge from the right hand side of C. Now the inverting input is pulled back down to zero volts but there is charge on C so there is a voltage across it. Since the conventional current flowed to the right there is a negative voltage remaining on the capacitor.

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  • \$\begingroup\$ I think this the kind of answer I was looking for. One question. what is the orientation of the capacitor in terms of conventional current? ie if vin goes positive the capacitor is I assume negative on its right-hand side (nearest vout). Now vout goes negative and therefore reduces the voltage across the capacitor until the potential at X is zero? \$\endgroup\$ – rhody Jan 17 '19 at 22:44
  • \$\begingroup\$ See the update. \$\endgroup\$ – Transistor Jan 17 '19 at 23:15
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The op-amp is going to try its best to keep the voltage between it's plus and minus input the same. In an ideal op-amp, no current flows into the inputs, so the only way that it can do that is by changing its output voltage.

In the schematic below, \$v_+ = 0\mathrm{V}\$. That means that the op-amp will try to hold \$v_-\$ at zero, also.

Whatever voltage is generated by V2 gets turned into a current by R1. Because \$v_-\$ is being held at \$0\mathrm{V}\$, that same current has to flow in C1. And because \$v_-\$ is being held at \$0\mathrm{V}\$, the op-amp has to drive the output voltage such that the current in C1 matches the current in R1.

So if \$v_2\$ is constant, then the current into the node around the negative input is constant, which means that the current out of that node from the cap must be constant -- and that can only happen if the output voltage is falling at a constant rate. The end result is that the op-amp integrates the input voltage into the output voltage.

More complicated voltages at \$v_2\$ cause more complicated behavior, but the op-amp is always going to be trying to drive \$v_-\$ to \$0\mathrm{V}\$. It can only do that by satisfying \$ \frac{d}{dt} C_1 v_{out} + \frac{v_2}{R_1} = 0 \$. If you solve that differential equation, it says that $$ v_{out} = -\frac{1}{R_1 C_1} \int v_2 dt $$

HTH

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ In your comment "The op-amp is going to try its best to keep the voltage between it's plus and minus input the same.", strictly speaking, it's the op-amp combined with the next feedback that does this. This is the bit I was stuck on, how the feedback coupled with the op-amp manages to maintain the difference close to zero. \$\endgroup\$ – rhody Jan 18 '19 at 18:50
  • \$\begingroup\$ Well, I'm glad you worked through it! These are difficult concepts and hard to simplify. \$\endgroup\$ – TimWescott Jan 18 '19 at 18:57
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Rhody - have you heard about the MILLER effect? Well - the shown circuit is called "MILLER integrator" because the MILLER effect is exploited. Remember: This effect reduces the feedback impedance between an amplifier output (for example: collector) and the inverting input (example: base node of the transistor). And the factor of increase is the gain.

Here, we have the same principle. Hence, there will be a very small capacitive impedance (that means: A very large capacitor) between input and output of the opamp. And the factor of increase is the open-loop gain Aol of the opamp.

Hence, you can make a comparison with a simple RC circuit. However, because of the very large capacitor the cut-off frequency is very low (nearly DC).

Frequency domain: The transfer function between the opamps inverting node and the signal input is

Ho(s)=1/(1+sCo * R) with Co=Aol * C (MILLER effect).

Because of the very large value Aol, we can neglect the "1" in the denominator and arrive at

Ho(s)=1/(sC * Aol *R)

We are lucky and can use the low resistive opamp output (and multiply the function Ho(s) with the gain -Aol) and arrive at the final result (opamp output-to-signal input):

H(s)=Ho(s) * (-Aol) = - 1/sR*C (Transfer function of an ideal integrator)

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The inputs of the opamp don't take input current and the opamp will keep its input voltage equal as long as it is wired for negative feedback.

So effectively, the current that the input sinks against the 0V it sees goes straight into charging the capacitor. Usually when you charge a capacitor through a resistor, the charge building up on the capacitor reduces the voltage across the resistor and thus also the charging current, leading to an exponential decay of the charge current.

Here, however, the opamp output actively adjusts the voltage at the other side of the capacitor so that the resistor never gets to see the difference it makes, thus keeping the current through resistor (and into capacitor) independent from the charge across the capacitor.

It's like Útgarða-Loki handing Þorr the mouth of the ocean as a drinking horn and Þorr does not actually find himself able to drain it, not noticing that he causes the tides with his attempt.

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The Passive cap. integrator current decays with voltage as it approaches the input.

The active cap. integrator saturates after some time if Vin ≠ 0 because the output voltage drives current into Vin- to maintain 0V diff. until the output saturates at the supply rail.

So input offset is critical and you need an analog switch to discharge and initialize to 0V output.

anecdotal

I remember knowing nothing about this in 1st yr Eng and my once famous brother-in-law medical Dr of Anesthesia, Intensive Care and open heart surgery, gave me a tour around the hospital and said he needed an integrator to measure O2 content in blood for the brain, after a heart attack victim stops, to know the best treatment ( such as hypothermia )to give when no response to defib. and meds with the likelihood of success. I had no idea ! and was embarrassed to not know! (circa '75) Don't be. Just research it.

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It is a great challenge to find a new explanation for such a legendary circuit because everyone knows what an op-amp integrator is. But to know a specific circuit solution does not mean that you really understand it. To (deeply) understand a circuit means something more - to see the general idea behind it that links many specific circuit implementations (op-amp, BJT, FET, tube…) You can see it even in life in the form of many non-electrical applications...

1. Op-amp inverting integrator. The idea behind this circuit solution is extremely simple and intuitive. It may sound paradoxical... but to see it you only need to remove the symbol of the ground from the circuit diagram. As you can see in Fig. 1, I have only labeled the place of the virtual ground (1) and the place of the real ground (2)... and I have no longer used these names. You understand that there is no virtual ground because there is no real ground. But if you still miss the virtual ground, then you can talk about a virtual short between node 1 and 2.

Op-amp inverting integrator

Fig. 1. Op-amp inverting integrator (only the negative power supply V- is explicitly shown)

The current path is crucial here to see the great idea. Since the input voltage is positive, the op-amp output voltage is negative and the current enters the op-amp output... then passes through the negative power supply V- and returns to the input source. The positive source V+ is not essential in this case; so it is only hinted.

2. Electric equivalent circuit. The main question to be answered is, "What does the op-amp do here?" You know that it keeps almost zero voltage between its inputs so its output voltage is always equal to the voltage drop across the capacitor. So the op-amp output serves as a following voltage source. Then let's replace the op-amp with a variable voltage source VOA to simplify this electronic circuit - Fig. 2. By the way, I conducted such a real experiment in 2001 with my students in the laboratory when we used a capacitor with high capacity and zero indicator connected between 1 and 2.

Op-amp inverting integrator - equivalent circuit

Fig. 2. Electric equivalent circuit

This simple trick is enough to show the great idea behind the circuit. The voltage source VOA is connected in series to the capacitor C so that its voltage compensates the voltage drop VC across the capacitor and the voltage between the two nodes 1 and 2 is (almost) zero. So the conclusion is:

The op-amp in the circuit of the inverting amplifier compensates the voltage drop VC across the capacitor by adding equivalent voltage VOA = VC in series.

So, the key point of this explanation is adding, not amplifying. To think of the amplifier in a negative feedback circuit not as of an amplifier but rather as of something like integrator is a powerful technique for intuitive understanding and explaining such op-amp circuits. Indeed, here it seems a little strange (integrator inside integrator)... but works...

How simple is this "magic recipe"... You want to make the imperfect RC integrator perfect? Then connect a small variable "battery" with voltage VC in series to the capacitor and (the next brilliant idea) take its inverted "copy" voltage as an output. The load will consume current from this "helping" source... not from the input source (i.e., this is a buffered output).

The power of this intuitive explanation is that we can explain this sophisticated op-amp circuit to a "six year old" (Einstein)... and that will mean we understand it ourselves...

3. Virtual short. The total voltage across the network of two elements in series - a capacitor C and compensating voltage source (VOUT), is always zero. So this network behaves as a "piece of wire" that shorts the points 1 and 2 - Fig. 3. This is what the input source "sees" when "looking" through the resistor R at the op-amp input.

Op-amp inverting integrator - virtual short

Fig. 3. Equivalent circuit of the output part on the right

Figuratively speaking, the op-amp output acts as a "negative capacitor". While the "positive capacitor" C subtracts its voltage VC from the input voltage source, the op-amp "negative capacitor" adds its voltage VOUT to the input voltage.

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