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I've been looking into different ways an op-amp can be used for signal filtering and came across the following circuit.

enter image description here

I would like to solve for the transfer function, \$G=\frac{v_{in}}{v_o} \$ for this circuit (we can assume an ideal op-amp for simplicity). My initial guess was to divide up the gain multiplication into two parts: \$G=\frac{v_{1}}{v_{in}}\frac{v_{o}}{v_1} \$, where \$v_1\$ is the amplified output of the op-amp.

Going with this, the first part of the transfer function, \$\frac{v_{1}}{v_{in}}\$, to my understanding would just be a standard lowpass op-amp configuration, hence:

$$\frac{v_{1}}{v_{in}}=-\frac{R_2}{R_1}\frac{1}{1+(sC_1R_2)}$$

The second part of the transfer function, \$\frac{v_{o}}{v_1}\$, is where I found some confusion. To me, it seems like a simple voltage divider between \$v_1\$ and \$v_o\$ should work, but this would not include \$R_3\$ in the equation at all. I also thought that perhaps I need to create a Thevenin equivalent circuit for this part, but I'm not really sure how you could do that for this scenario. But, assuming the voltage division here is valid, I get:

$$\frac{v_{o}}{v_{1}}=\frac{sC_2}{(1+sC_2R_4)}$$

resulting in a final transfer function of the form:

$$G(s)=-\left(\frac{R_2}{R_1}\frac{1}{1+(sC_1R_2)}\right)\left(R_4\frac{sC_2}{(1+sC_2R_4)}\right)$$

So, is this the right way to go about solving this circuit? Is splitting up the circuit essentially into two parts valid here? Is a simple voltage division enough to solve \$\frac{v_{o}}{v_1}\$, without \$R_3\$ being a factor?

Any insight would be much appreciated, please let me know if I've messed things up completely as well.

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    \$\begingroup\$ If the op-amp is acting as an ideal voltage source, R3 is simply seen as a load and has no bearing on the downstream transfer function. I didn't check your equations but your method is sound. \$\endgroup\$ Commented May 5, 2021 at 18:12
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    \$\begingroup\$ Certainly, this is a second-order lowpass - however with rather bad properties (two real poles only), The same an be achieved with two simple (passive) RC sections. There are many other (and much better) 2nd-order lowpass topologies. \$\endgroup\$
    – LvW
    Commented May 6, 2021 at 6:58
  • \$\begingroup\$ @LvW This is interesting, I agree that it should be a second order LP, but the transfer function is of the form G~s/(s^2), which if memory serves right is the generic form of a second-order bandpass. \$\endgroup\$
    – JTaft121
    Commented May 6, 2021 at 14:19
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    \$\begingroup\$ JTaft121, sorry you are right. It is a simple bandpass. I did not look carefully enough to the circuit. Anyway - my comment remains the same: A bandpass with a rather bad selectivity (two real poles). \$\endgroup\$
    – LvW
    Commented May 6, 2021 at 14:32

2 Answers 2

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First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).

Well, we are trying to analyze the following opamp-circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_1+\text{I}_2=\text{I}_3+\text{I}_4\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_1=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_2-\text{V}_3}{\text{R}_4}\\ \\ \text{I}_4=\frac{\text{V}_3}{\text{R}_5} \end{cases}\tag2 $$

Now, using an ideal opamp, we know that:

$$\text{V}_+=\text{V}_-=\text{V}_1=0\space\text{V}\tag3$$

Now, we can solve for the transfer function:

$$\mathcal{H}:=\frac{\text{V}_3}{\text{V}_\text{i}}=-\frac{\text{R}_2\text{R}_5}{\text{R}_1\left(\text{R}_4+\text{R}_5\right)}\tag4$$

Where I used the following Mathematica-code:

In[1]:=Clear["Global`*"];
V1 = 0;
FullSimplify[
 Solve[{I1 + I2 == I3 + I4, I1 == (Vi - V1)/R1, I1 == (V1 - V2)/R2, 
   I3 == V2/R3, I4 == (V2 - V3)/R4, I4 == V3/R5}, {I1, I2, I3, I4, V2,
    V3}]]

Out[1]={{I1 -> Vi/R1, 
  I2 -> -(((R3 (R4 + R5) + R2 (R3 + R4 + R5)) Vi)/(R1 R3 (R4 + R5))), 
  I3 -> -((R2 Vi)/(R1 R3)), I4 -> -((R2 Vi)/(R1 (R4 + R5))), 
  V2 -> -((R2 Vi)/R1), V3 -> -((R2 R5 Vi)/(R1 (R4 + R5)))}}

My equation was also confirmed using LTspice.


When we want to apply the derivation from above to your circuit we need to use Laplace transform (I will use lower case function names for the functions that are in the (complex) s-domain, so \$\text{y}\left(\text{s}\right)\$ is the Laplace transform of the function \$\text{Y}\left(t\right)\$):

  • $$\text{R}_2=\frac{\frac{1}{\text{sC}_1}\cdot\text{R}_x}{\frac{1}{\text{sC}_1}+\text{R}_x}=\frac{\text{R}_x}{1+\text{C}_1\text{R}_x\text{s}}\tag5$$
  • $$\text{R}_4=\frac{1}{\text{sC}_2}\tag6$$

So, we can rewrite the transfer function as:

$$\mathscr{H}\left(\text{s}\right)=-\frac{\text{R}_x}{1+\text{C}_1\text{R}_x\text{s}}\cdot\frac{\text{R}_5}{\text{R}_1\left(\frac{1}{\text{sC}_2}+\text{R}_5\right)}=$$ $$-\frac{\text{R}_x}{1+\text{C}_1\text{R}_x\text{s}}\cdot\frac{\text{C}_2\text{R}_5\text{s}}{\text{R}_1\left(1+\text{C}_2\text{R}_5\text{s}\right)}\tag7$$

Now, when working with sinusoidal signals we can use \$\text{s}:=\text{j}\omega\$ (where \$\text{j}^2=-1\$ and \$\omega=2\pi\text{f}\$ with \$\text{f}\$ is the frequency of the input signal in Hertz). So, we get:

$$\underline{\mathscr{H}}\left(\text{j}\omega\right)=-\frac{\text{R}_x}{1+\text{C}_1\text{R}_x\text{j}\omega}\cdot\frac{\text{C}_2\text{R}_5\text{j}\omega}{\text{R}_1\left(1+\text{C}_2\text{R}_5\text{j}\omega\right)}=$$ $$-\frac{\text{R}_x}{1+\text{C}_1\text{R}_x\omega\text{j}}\cdot\frac{\text{C}_2\text{R}_5\omega\text{j}}{\text{R}_1\left(1+\text{C}_2\text{R}_5\omega\text{j}\right)}\tag8$$

So, the absolute value if given by:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\text{R}_x}{\sqrt{1+\left(\text{C}_1\text{R}_x\omega\right)^2}}\cdot\frac{\text{C}_2\text{R}_5\omega}{\text{R}_1\sqrt{1+\left(\text{C}_2\text{R}_5\omega\right)^2}}\tag9$$

And the argument:

$$\arg\left(\underline{\mathscr{H}}\left(\text{j}\omega\right)\right)=\frac{\pi}{2}-\arctan\left(\text{C}_1\text{R}_x\omega\right)-\arctan\left(\text{C}_2\text{R}_5\omega\right)\tag{10}$$

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    \$\begingroup\$ Thank you for this wonderful explanation @Jan , I had not considered using Laplace transforms to solve the problem. If I am reading your solution correctly, I believe we have the same transfer function, with Rx being R2 in my answer and R5=R4. \$\endgroup\$
    – JTaft121
    Commented May 6, 2021 at 13:53
  • \$\begingroup\$ @JTaft121 correct! \$\endgroup\$ Commented May 6, 2021 at 14:30
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Your intuition is correct. An ideal op amp has zero output impedance: it behaves like a voltage source. The load resistor R3 is immaterial in that idealized model. You can of course use a more sophisticated model that includes the output impedance of the op-amp, and then it’d be obvious that R3 plays a role. But you’re not using such a model.

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