I want to find the transfer function of the following circuit algebraically:

enter image description here

First of all, we realize since we have ideal op amps, the voltages at node \$1\$ and \$2\$ are the same as well as the voltages at \$4\$ and \$5\$ and thus \$6\$.

We also realize that \$V_1 = \frac{R_2}{R_1+R_2} U_o\$ and that \$V_4 = V_5 = V_6 = U_o\$. So, all current going to node \$2\$ also goes through node \$3\$ and \$4\$.

At node \$4\$ we have that $$\frac{V_3-U_o}{R} = U_o sC \Leftrightarrow \\V_3 = U_o(sCR+1)$$

Finally we get that $$\frac{U_s-V_2}{R_2} = \frac{V_3-U_o}{R} $$

So, \$U_s - \frac{R_2}{R_1+R_2} U_o= U_osCR_2\$

Meaning that

$$U_o/U_s = \frac{1}{scR_2 + \frac{R_2}{R_1+R_2}}$$

Is this correct?

I would assume just looking from the circuit that we have some integrator of some sort, but the extra term in the denominator bothers me since integrators have transfer functions of the form \$\frac{1}{s}\$, or will it not matter with that extra term in?

  • 1
    \$\begingroup\$ "[...] So, all current going to node 2 also goes through node 3 and 4[...]" is wrong! The left OPAMP output will source or sink some current going through R1 and R. \$\endgroup\$ Commented Oct 7, 2022 at 15:31
  • \$\begingroup\$ @StefanWyss Is this not what's negligible for ideal op amps? \$\endgroup\$
    – Tanamas
    Commented Oct 7, 2022 at 15:41
  • \$\begingroup\$ No. Think of a simple OPAMP inverter with two resistors each of value R and connect a load of 10R to the output (Vo). Current through the feedback resistor will be Vo/R and current through the load will be Vo/10R. The difference current Vo/9R must be provided by the OPAMP output. \$\endgroup\$ Commented Oct 7, 2022 at 16:08
  • 1
    \$\begingroup\$ We assume no current is drawn into the op-amp inputs @Tanamas. We cannot assume no current is drawn in or produced by the op-amp output. If we did assume that then we might as well not connect the output in the circuit. Then where would we be? \$\endgroup\$
    – Andy aka
    Commented Oct 7, 2022 at 16:22
  • 1
    \$\begingroup\$ @Tanamas Your problem is interesting. So +1 to it. I get an integrator result, by the way. But simpler than what you write. \$\endgroup\$
    – jonk
    Commented Oct 7, 2022 at 23:10

6 Answers 6



This is an interesting circuit.

Partly, because it breaks down into parts we all recognize: an RC filter, a follower, and the 1st stage which is also a fairly familiar differential amplifier.

I really enjoyed looking over the analysis already here, as well as yours.

You wrote \$V_1 = \frac{R_2}{R_1+R_2} U_o\$ and I immediately mentally noted that since \$V_1=V_2\$ that it must also be equal to \$V_2=\frac{V_s\,\cdot\,R_1+V_3\,\cdot\,R_2}{R_1+R_2}\$. Equating them suggests that \$R_2\cdot V_o=V_s\,\cdot\,R_1+V_3\,\cdot\,R_2\$.

But then I backed off to just look at it, more.

Perhaps a way to see the circuit is to start at the capacitor. The follower buffers the capacitor voltage and feeds it immediately into the differential as the positive input. Given the resistors shown, a proportion of the input is subtracted from the capacitor voltage to set the output at \$V_3\$. The \$R\$ sets the output impedance as resistive (real axis) and the capacitor itself is orthogonal to it (imaginary axis.) This is important as it means the output will be chasing the input.

If a DC input is provided, then the difference will always have the same sign and the capacitor will simply force the opamps to ramp towards hitting their rails. But if AC is provided, then the chasing process continues ever around the circle.

Using Solver

In any case, I decided to just sit down and let the KCL stuff flow out, leaving the gnarly algebra to SymPy:

eq1 = Eq( v1/r1 + v1/r2, vo/r1 )                 # KCL node 1
eq2 = Eq( v2/r2 + v2/r1, vi/r2 + v3/r1 )         # KCL node 2
eq3 = Eq( v3/r1 + v3/r, io1 + v2/r1 + v4/r )     # KCL node 3
eq4 = Eq( v4/r + v4/(1/s/c), v3/r )              # KCL node 4
eq5 = Eq( vo/r1, io2 + v1/r1 )                   # KCL output node
eq6 = Eq( v1, v2 )                               # 1st opamp requirement
eq7 = Eq( v4, vo )                               # 2nd opamp requirement
simplify( solve( [ eq1, eq2, eq3, eq4, eq5, eq6, eq7 ], [ v1, v2, v3, v4, vo, io1, io2 ] )[vo] / vi )

So, this tells me that the transfer function is \$\frac{-R_1}{R\,\cdot \,R_2\,\cdot\,C}\cdot\frac1s\$. So it seems, anyway.

In the time domain, I believe this would translate into \$\frac{-R_1}{R\,\cdot \,R_2\,\cdot\,C}\int_{_0}^{^t}V_t\:\text{d}t\$. And if we limit ourselves to a sinusoidal \$V_t=V_{_0}\cdot\cos\left(\omega \cdot t\right)\$ then it works out to \$\frac{-R_1}{R\,\cdot \,R_2\,\cdot\,C}\cdot V_{_0}\cdot\frac1{\omega}\cdot\sin\left(\omega \cdot t\right)\$.

In short, I should expect to see a voltage gain that is inversely proportional to frequency. And I mean inversely. There's no near-DC flat-top here. As you get near to DC the gain grows without bound.

I should also tend to see the opamps hitting their rails with very low frequencies and diminishing towards 0 at very high frequencies. The difference between this and a standard low-pass RC is that in this case applying a non-zero DC voltage results in a ramp towards one of the rails where it will ultimately just sit. In contrast, a low-pass RC would just present a copy of the applied DC voltage. (Except for 0, that doesn't happen here. It just integrates.)

Apply Theory

Okay. That is what I did before attempting to see what LTspice would say to me. So I set down to specify some values to use and then proceeded to make some predictions from them.

(This is what I do in my engineering logbook -- always develop the theory first, make some explicit predictions from that theory, and then go test it and write down those results, as well. That way I can tell where I fail and where I succeed and this helps me learn. It also gives me a way to look back on where I have made mistakes.)

I'll use \$R_1=10\:\text{k}\$, \$R_2=30\:\text{k}\$, \$R=10\:\text{k}\$, \$C=15.9\:\text{nF}\$ and \$V_{_0}=1\:\text{V}\$.

I am predicting a voltage gain of \$\frac{2096.43606}{\omega}\$ for this configuration.

Test Theory

I'll now wire up a schematic in LTspice. I tend to use their LT1800 because it's just a generally good (and too expensive for me) device with rail-to-rail support and a behavioral model that seems to run well for me. So I'll use that.

Let's try a frequency of \$300\:\text{Hz}\$ first. In this case, I'd predict a voltage gain of \$\approx 1.1\$:

enter image description here

This shows that the peak-to-peak output is \$\approx 2.2\:\text{V}\$. With an input peak-to-peak of \$2\:\text{V}\$ and a voltage gain of \$\approx 1.1\$ that seems amazingly close.

Let's try a frequency of \$600\:\text{Hz}\$ next. In this case, I'd predict half the earlier voltage gain, or \$\approx 0.55\$:

enter image description here

And here we see the peak-to-peak output is \$\approx 1.1\:\text{V}\$. As expected.

One final try at a frequency of \$1\:\text{kHz}\$. In this case, I'd predict a voltage gain of \$\approx \frac13\$:

enter image description here

And there it is; the peak-to-peak output is \$\approx 650\:\text{mV}\$. As expected.

My logbook is complete on this, I guess.


The following is how I mentally visualized what I just wrote, earlier, in the Overview. I tried to use words there, but here's the mental picture that was in my mind while I was writing it:

enter image description here

For larger values of \$\omega\$ the reactance of the capacitor has a smaller value and therefore the radius of the above circle is similarly tighter/smaller, and the magnitide is therefore smaller. The differentially-arranged opamp is pushing on \$R\$, which causes the rotation. But no matter how hard it tries, \$R\$ is always positioned to be tangent to the \$C\$-axis, so all the pushing does is to just continue the rotation.

I admit that the \$R\$ and \$C\$ on your schematic was nicely arranged for me to see this circle superimposed on it when I was first reading it. I actually super-imposed the circle, mentally, then. ;)

The rest of the circuit is all resistive and fast. So it can be mentally lumped in mind as instantaneous. (Ignore any delays.)

Also, the schematic you have really will need (if you build one) a small capacitance (maybe \$22\:\text{pF}\$?) across the feedback resistor of the differentially-arranged 1st stage opamp. It's common practice, though. Just FYI.

  • 1
    \$\begingroup\$ Really beautiful answer I must say. I really like the way you literally presented every step of how you thought about the problem. I can confirm that I got the same transfer function after working with the algebra a bit again. Thank you again. \$\endgroup\$
    – Tanamas
    Commented Oct 8, 2022 at 14:15
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    \$\begingroup\$ @Tanamas What you see above is what I do in my logbook, mostly. I have to write down my thinking, as years later I can't necessarily recall anymore. So this is really just what I do, normally. And I'm very glad that it helped you. And still more than that, it's nice that you let me know, too. Makes it all worth the trouble! :) \$\endgroup\$
    – jonk
    Commented Oct 8, 2022 at 21:51

I like to separate these things into blocks to double check node voltage answers: enter image description here

I'd follow the math here, mapping your resistor values onto the other circuit you get

$$ U_{out1} = U_{s}\frac{R_1}{R_2} -U_{1}\frac{R_2}{R_1+R_2}\frac{R_1+R_2}{R_2} = U_{s}\frac{R_1}{R_2} -U_{out2}$$

Then this:
enter image description here
The voltage follower doesn't really do much from a math perspective and it's output can be considered the same voltage as the output of the RC filter (but the follower does provide an impedance buffer)

$$ U_{out2} = U_{out1}\frac{1}{1+sRC} $$

  • \$\begingroup\$ Because the final result must be negative, I think anywhere in your calculation a minus sign was "lost".....(see the 1st equation for Uout1) \$\endgroup\$
    – LvW
    Commented Oct 8, 2022 at 10:25
  • 1
    \$\begingroup\$ Thank you for your solution. I tried your strategy of dividing it into blocks, and it really made the problem alot easier to solve. As @LvW pointed out I think you missed a minus sign, but with this method I managed to get the transfer function to \$\frac{-R_1}{sR_2RC}\$. \$\endgroup\$
    – Tanamas
    Commented Oct 8, 2022 at 14:13
  • \$\begingroup\$ @LvW I never calculated the final result, just the sub-blocks \$\endgroup\$
    – Voltage Spike
    Commented Oct 9, 2022 at 5:07

The easiest way, in my opinion, is to apply superposition in this particular example and determine the transfer function first for \$s=0\$: open the capacitor and solve this simple circuit shown below by alternatively setting \$V_{in}\$ and \$V_{out}\$ to 0 V:

enter image description here

The schematic simplifies as, in this mode, \$R_5\$ plays no role and buffer \$E_2\$ can disappear for the calculations. Then apply superposition the way below to determine the dc gain \$H_0\$:

enter image description here

Then sum both results to obtain the final \$\epsilon\$ value and make it equal to zero since you assume an infinite open-loop gain for the op-amp. Factor \$H_0=\frac{V_{out}}{V_{in}}\$ and you have the dc gain \$H_0\$.

Now, by using the fast analytical circuits techniques or FACTs, I can tell there is no zero in this circuit but a pole surely. It's value is obtained by setting \$V_{in}\$ to 0 V and "looking" through the capacitor connection to obtain the resistance \$R\$ driving the capacitor. However, because of the active circuits, inspection is not obvious and the best is to place a current source \$I_T\$ across the capacitor connecting terminals and determine the voltage \$V_T\$ dropped across the current source terminals. The ratio \$\frac{V_T}{I_T}\$ is the resistance \$R\$ you want to obtain the time constant \$\tau\$ of this circuit and then its pole since \$\omega_p=\frac{1}{\tau}\$:

enter image description here

I realize it was more tricky than expected because it turns out this is a RHP pole because the resistance is negative and confirmed by simulation for this particular set of components values:

enter image description here

There you go, the complete transfer function you want is thus \$H(s)=H_0\frac{1}{1-\frac{s}{\omega_p}}\$ with the ac response shown below:

enter image description here

and finally a quick SIMetrix simulation to confirm this is correct:

enter image description here

If you consider an open-loop gain going infinite, the resistance involved in the time constant value becomes: \$R_{tau}=\frac{R_2R_5(R_3+R_4)}{R_2R_4-R_1R_3}\$ and it is easy to determine the time constant sign, leading to either a RHPP or a LHPP.

For instance, if you now set resistance \$R_1\$ to 1 kohms, the pole goes back into the left-half plane and the circuit now inverts in dc:

enter image description here

This is the fastest way I found to determine this transfer function and hope you can follow the steps.


Is this correct?

No, you've made an incorrect assumption as pointed out in your comments. That assumption forgot about the current contributed by the op-amp output driving node 3.

  • Begin with the formula for a difference amplifier: -

enter image description here

  • Then derive the node 4 voltage (your circuit) in terms of \$V_{OUT}\$ above (it's a simple RC filter so that is child's play).

  • Then, because you have a buffer amplifier on the output of node 4 it has unity gain thus, the voltage you derived for node 4 becomes \$V_2\$ in the diagram above.

Drill down through a few lines of algebra and you are done.

Image from here.

OK, @jonk has produced the right answer and it made me rethink my screwed up algebra...

enter image description here

  • \$\begingroup\$ This is an interesting question by the OP. I get the first term, just as you show. But I don't get the second. And if you isolate it and cancel out the 's', then you just have a constant and the inverse Laplace of a constant is the Dirac delta function. That's the bit I don't find in what I did in order to do a quick sanity test. \$\endgroup\$
    – jonk
    Commented Oct 7, 2022 at 22:48
  • \$\begingroup\$ @jonk yes you are right. My algebra was screwed. I get the TF to be \$\dfrac{-R_1}{sCRR_2}\$ now. I'll add my derivation with due credit to you. \$\endgroup\$
    – Andy aka
    Commented Oct 8, 2022 at 14:34

Two equations with two unknown quantities (V3 and Vo):



Can easiliy be solved.


Start with the differential amplifier.

The output is multiplied by \$\frac{R_{2}}{R_{1}+R_{2}}\$ then multiplied by \$\frac{R_{1}+R_{2}}{R_{2}}\$ so by superposition,


The differential amplifier output also is by working backwards from \$U_{O}\$,


Combining the two equations and solving for \$U_{O}\$,



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