I'm trying to find the transfer function of a Sallen key filter in the s-domain and thought that cutting out the op-amp from the circuit by using the ideal op-amp laws was a smart move (see image below). However, when I attempt this I end up with the transfer function seen below.

\$ \dfrac{V_o}{V_i}=\dfrac{s(R_2+C_2)+1}{s^2(R_1R_2C_1C_2 + R_1R_2C_2^2) +s(R_1C_1 +2R_1C_2+R_2C_1 +R_2C_2) + (R_1/R_2 + 1)} \$

This transfer function does not agree with what any sources on the Internet say, and I'm sure that it's because of how I've remade the circuit.

If possible, could someone please explain to me why this would not work (if that's the case) and how would be better to approach this task?

Conversion from op-amp circuit

  • 5
    \$\begingroup\$ Just because V_o = V_p doesn't mean that current can flow from the non-inverting input to the output as can happen in your proposed equivalent circuit. \$\endgroup\$
    – The Photon
    Commented Oct 4, 2013 at 3:28

2 Answers 2


If you wish to redraw the circuit without the op-amp, try something like this:


simulate this circuit – Schematic created using CircuitLab

See the difference?

  • \$\begingroup\$ I can see the difference, but I'm unsure about what's happening with those wires connecting to nothing just to the right of VCVS1. \$\endgroup\$
    – Kadin
    Commented Oct 5, 2013 at 8:26
  • \$\begingroup\$ @user2222956, it's the funny way CircuitLab represents a VCVS; the controlling voltage has explicit connections. So, in this case, the controlling voltage for VCVS1 is just V+. Thus, Vo = V+ but V+ is not connected to Vo. \$\endgroup\$ Commented Oct 5, 2013 at 10:48
  • \$\begingroup\$ Ah, I see. So essentially, given that Vp = Vp = Vout, the dependent source will always be producing a voltage equal to Vp (or Vout). So is it safe to say that the circuit can be simplified to the following? i.imgur.com/CkWoHng.jpg \$\endgroup\$
    – Kadin
    Commented Oct 5, 2013 at 13:51

Well this is a very basic mistake I've seen my friends commit while solving circuits involving Op-amps.

Here is the equivalent structure of an Op-Amp


Ideally we have Rin=infinite, Rout=0 ohms(whereas in reality Rin=Mega Ohms and Rout=0-10 Ohms)

In Op-amps circuits having negative feedback we use the virtual short approximation i.e.

  • Vp=Vn
  • No Current is entering the Op-Amp

Now coming to your circuit. You committed a mistake by joining Vout directly to Vp in your equivalent model. Why so ? Because the Virtual Short approximation states that Vp=Vn & Zero amount of Current is entering the Op amp. When you join Vout and Vp you allow the current from C2 to pass through the Op amp, which actually is completely going into the Output terminal of the Op-amp.

As for your final result, Wiki has a good step by step derivation.

Reference for Virtual Short

  • \$\begingroup\$ As a slight aside, I imagine there may be cases where you could directly connect them to help with analysis. If you connected them and analysed the resulting circuit and no current flowed, then it would be a valid result. \$\endgroup\$
    – BeB00
    Commented Apr 24, 2017 at 13:22

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