Finding the transfer function of an op-amp filter circuit

Question

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 analysis to find this transfer function (assuming an ideal op-amp)?

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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.

Answer 1

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.

Answer 2

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

Answer 3

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.