# Finding the transfer function of an op amp filter circuit

I have the following third order filter circuit:

simulate this circuit – Schematic created using CircuitLab

And I know that the transfer function looks like:

$$\mathscr{H}\left(\text{s}\right)=\frac{1}{\alpha_1\cdot\text{s}^3+\alpha_2\cdot\text{s}^2+\alpha_3\cdot\text{s}+1}\tag1$$

Where:

$$\alpha_1=\text{C}_1\cdot\text{C}_2\cdot\text{C}_3\cdot\text{R}_1\cdot\frac{\text{R}_2\cdot\text{R}_3}{\text{R}_2+\text{R}_3}\cdot\left(\text{R}_1+\text{R}_2\right)\tag2$$

• $$\alpha_2=\text{C}_1\cdot\text{C}_2\cdot\text{R}_1\cdot\left(\text{R}_1+\text{R}_2\right)+\text{C}_2\cdot\text{C}_3\cdot\text{R}_3\cdot\left(\text{R}_1+\text{R}_2\right)\tag3$$
• $$\alpha_3=\text{C}_1\cdot\text{R}_1+\text{C}_2\cdot\left(\text{R}_1+\text{R}_2+\text{R}_3\right)\tag4$$

But how can I use current node analyses to find this transfer function (assuming an ideal op amp)?

My work:

I wrote the current node equations:

1. $$\text{I}_1=\text{I}_{\text{R}_1}+\text{I}_{\text{C}_1}+\text{I}_{\text{R}_2}\tag5$$
2. $$\text{I}_2=\text{I}_{\text{R}_2}+\text{I}_{\text{R}_3}+\text{I}_{\text{C}_3}\tag6$$
3. $$\text{I}_3=\text{I}_{\text{R}_3}+\text{I}_{\text{C}_2}\tag7$$
4. $$\text{I}_4=\text{I}_{\text{C}_3}\tag8$$
5. $$\text{I}_5=\text{I}_4=\text{I}_{\text{C}_3}\tag9$$
6. And of course I know that ideal op amp equation: $$\text{V}_+=\text{V}_-\tag{10}$$

But now I do not know how to continue using the voltages at those nodes:

• @TonyStewart.EEsince'75 I want to use this for frequency analysis so it is not DC (I think, or I misted the point you're trying to make. If that is so I'm sorry). – Looper Mar 14 '18 at 20:25
• Vdifferential(f)=0 is the point. or Vin+=Vin- – Tony Stewart EE75 Mar 14 '18 at 20:26
• Why don't you try the fast analytical techniques described electronics.stackexchange.com/questions/338350/… rather going for a tedious KVL-KCL classical analysis? – Verbal Kint Mar 14 '18 at 20:44
• @VerbalKint Thanks for the link. I see your work, but I also want to get it with algebra and math instead of using a program. – Looper Mar 14 '18 at 20:49
• @Looper, there is no program, just determining the circuit time constants and combining them to form the denominator coefficients. You can beat the FACTs on these applications! : ) – Verbal Kint Mar 14 '18 at 21:05

You mention "current node analysis" and your diagram shows you labeling currents at junctions. There's a fundamental problem with this currents do not exist in nodes. A node has a voltage. You picked points where currents aren't properly defined and labeled them with currents.

You either want mesh current analysis (where you write KVL equations) or node voltage equations (where you write KCL equations).

If you're doing node analysis, assign every node a voltage (V1, V2, etc.) and write the KCL equations at each node. The current in will typically be the difference of two voltages divided by some impedance. In the end you should have the same number of equations and variables.

Vin+ - Vin- =0 and both are equal to Vin=Vout at DC. Although both inputs are high impedance. Vin- is driven by a voltage source to track Vin+ so make the differential=0

• Thanks for your answer in the first place. Yes I did look, but I was trying to solve thi using current node analysis and I do not get how I can use it to get to the transfer function. So I was looking for help on that. I'm sorry if this offends you :( that is not my purpose. – Looper Mar 14 '18 at 20:27
• To get voltage you use I*Z(f) – Tony Stewart EE75 Mar 14 '18 at 21:02
• I know that, but I've to find the voltage difference over the components and that is what I do not get (and understand). – Looper Mar 14 '18 at 21:04
• But how else would you like to get the voltage drop other than V(s)=I(s)*Z(s) – Tony Stewart EE75 Mar 14 '18 at 21:06
• How do I determine what voltage differences and currents applies? – Looper Mar 14 '18 at 21:09

I would propose to use the node currents in the following manner (based on node voltages): The nodes labelled by you with I1, I2 and I3 are now: V1, V2, V3 and V4=Vout.

• At first, use only conductances Y for all parts. This gives shorter expressions. Later on you can insert Y=1/R or Y=sC.

• Example (for node V2): (V1-V2)Y2=(V2-V3)Y3+(V2-Vout)Y4 (with Y4=sC3)

• Similar equations exist for all nodes. Remeber also: V3=Vout (ideal unity gain)

• Finally, you have n equations for n unknown quantities (use Vout/Vin as one of these unknown values). This can be solved for Vout/Vin.