Well, I've the following third order filter circuit:


simulate this circuit – Schematic created using CircuitLab

And I know that the transfer function looks like:



  • $$\alpha_1=\text{C}_1\cdot\text{C}_2\cdot\text{C}_3\cdot\text{R}_1\cdot\frac{\text{R}_2\cdot\text{R}_3}{\text{R}_2+\text{R}_3}\cdot\left(\text{R}_1+\text{R}_2\right)\tag2$$
  • $$\alpha_2=\text{C}_1\cdot\text{C}_2\cdot\text{R}_1\cdot\left(\text{R}_1+\text{R}_2\right)+\text{C}_2\cdot\text{C}_3\cdot\text{R}_3\cdot\left(\text{R}_1+\text{R}_2\right)\tag3$$
  • $$\alpha_3=\text{C}_1\cdot\text{R}_1+\text{C}_2\cdot\left(\text{R}_1+\text{R}_2+\text{R}_3\right)\tag4$$

But my question is how can I use current node analyses to find this transfer function (assuming an ideal opamp)?

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 opamp equation: $$\text{V}_+=\text{V}_-\tag{10}$$

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

  • \$\begingroup\$ @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). \$\endgroup\$ – Looper Mar 14 '18 at 20:25
  • \$\begingroup\$ Vdifferential(f)=0 is the point. or Vin+=Vin- \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 14 '18 at 20:26
  • \$\begingroup\$ Why don't you try the fast analytical techniques described electronics.stackexchange.com/questions/338350/… rather going for a tedious KVL-KCL classical analysis? \$\endgroup\$ – Verbal Kint Mar 14 '18 at 20:44
  • \$\begingroup\$ @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. \$\endgroup\$ – Looper Mar 14 '18 at 20:49
  • \$\begingroup\$ @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! : ) \$\endgroup\$ – Verbal Kint Mar 14 '18 at 21:05

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

There are lots of duplicate answers here on this. Did you even look?

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  • 1
    \$\begingroup\$ 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. \$\endgroup\$ – Looper Mar 14 '18 at 20:27
  • \$\begingroup\$ To get voltage you use I*Z(f) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 14 '18 at 21:02
  • \$\begingroup\$ I know that, but I've to find the voltage difference over the components and that is what I do not get (and understand). \$\endgroup\$ – Looper Mar 14 '18 at 21:04
  • \$\begingroup\$ But how else would you like to get the voltage drop other than V(s)=I(s)*Z(s) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 14 '18 at 21:06
  • \$\begingroup\$ How do I determine what voltage differences and currents applies? \$\endgroup\$ – Looper Mar 14 '18 at 21:09

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