I would like to find the transfer function of the above ideal opamp circuit. One of my approach was neglecting the effect of R2(assuming the voltages on the both side of R2 are the same) and applying node method to above node but the result is seem complex and wrong to me.

Since I haven't got the answer, I want to be sure if my approach is right or what would be the correct approach.


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ You are correct to neglect R2. Next, since this is a negative feedback circuit, assume the inverting input is driven to the same potential as the non-inverting inverting. \$\endgroup\$ – The Photon Jan 8 '17 at 1:33
  • \$\begingroup\$ In addition to the above recommendation you should ask yourself if the resistor R1 plays a major role - will it influence the voltage at the "+" node? \$\endgroup\$ – LvW Jan 8 '17 at 10:49
  • \$\begingroup\$ @ThePhoton, actually, you ignore R2 because no current can flow through it. \$\endgroup\$ – Scott Seidman Mar 22 '17 at 21:26
  • \$\begingroup\$ @ScottSeidman, isn't that what I said? 2nd sentence is to give a next step in analysis, not explain the first sentence. \$\endgroup\$ – The Photon Mar 22 '17 at 21:43
  • \$\begingroup\$ @ThePhoton, then yes. \$\endgroup\$ – Scott Seidman Mar 22 '17 at 23:49

General case

Assuming the voltage at the internal node is \$V\$

\$ \frac{0-V}{R_4} +\frac{V_{\text{in}}-V}{R_2} +\frac{V_{\text{out}}-V}{R_3} +\frac{V_{\text{out}}-V}{1/(Cs)}=0\$

which can be solved for \$V_\text{out}\$

\$V_{\text{out}}(\frac{1}{R_3}+C s)=-\frac{V_\text{in}}{R_2}+(\frac{1}{R_4}+\frac{1}{R_2}+\frac{1}{R_3}+Cs)V \$

\$V_\text{out} = -\frac{R_3}{R_2 \left(C R_3 s+1\right)} V_\text{in}+ \frac{R_2 R_3+R_4 R_3+R_2 R_4+C R_2 R_4 R_3 s}{R_2 R_4 \left(C R_3 s+1\right)} V\$

This is essentially a system with two inputs. The transfer function for \$\frac{V_\text{out}}{V_\text{in}}\$ is

\$-\frac{R_3}{R_2 \left(C R_3 s+1\right)}\$

Case when \$R_2\$ is neglected

The node equation now is

\$\frac{0-V_\text{in}}{R_4} +\frac{V_{\text{out}}-V_\text{in}}{R_3} +\frac{V_{\text{out}}-V_\text{in}}{1/(Cs)}=0\$

In this case solving for \$\frac{V_\text{out}}{V_\text{in}}\$ we get

\$\frac{C R_4 R_3 s+R_3+R_4}{R_4 \left(\mathcal{C} R_3 s+1\right)}\$

  • \$\begingroup\$ Whether you neglect \$R_2\$ or not should lead to the same transfer function. The first expression looks wrong to me (you have an inversion and no zero) while the second is ok. Please factor (\$R_3+R_4\$) in the numerator and you'll have \$N(s)=1+sC_1(R_3||R_4)\$ then extract \$R_4\$ from the denominator and form a leading term \$H_0=\frac{R_3+R_4}{R_4}\$ while \$D(s)=1+sC_1R_3\$. Now you have the correct low-entropy form : ) \$\endgroup\$ – Verbal Kint Aug 6 '17 at 15:50
  • \$\begingroup\$ Verbal Kint. What do you mean by 'first expression'. Could you be a bit more explicit. Thanks. \$\endgroup\$ – Suba Thomas Aug 7 '17 at 14:31
  • \$\begingroup\$ This is where you describe the general case and you write \$-\frac{R_3}{R_2 \left(C R_3 s+1\right)}\$ for the transfer function unless I misunderstood what you derived? \$\endgroup\$ – Verbal Kint Aug 7 '17 at 16:01
  • \$\begingroup\$ I just did not write \$-\frac{R_3}{R_2 \left(C R_3 s+1\right)}\$, I derived it from the very first equation in my answer. Unless there is something wrong in the first equation, or the derivation (which is unlikely since I checked it with Mathematica) the answer is right. Since you at least agree that the second case (with \$R_2=0\$) is correct - I have more confidence in my first general case as well because - now if you go ahead and substitute \$V_{in}\$ for \$V\$ in the first case you will get the answer in the second case. \$\endgroup\$ – Suba Thomas Aug 7 '17 at 16:45
  • \$\begingroup\$ Ah, this is good then: false alarm! Sorry for that and thanks for pointing this out. \$\endgroup\$ – Verbal Kint Aug 7 '17 at 19:01

The transfer function of a simple filter like this one can be solved by using the Fast Analytical Circuits Techniques or FACTs. Start with \$s=0\$ meaning you open the capacitor. The gain \$H_0\$ of that circuit is that of a non-inverting configuration considering a perfect op amp:


Then, reduce the excitation to 0 V or ground \$V_{in}\$ and "look" at the resistance "seen" from \$C_1\$ terminals when it is temporary removed from the circuit. As the input voltage is 0 V, then the left terminal is also at 0 V. The resistance "seen" is thus \$R_3\$ and we have \$\tau_1=R_3C_1\$. Now, set \$C_1\$ in its high-frequency state (a short circuit) and calculate the gain when \$R_3\$ is shorted: \$H_1=1\$. You can now apply the generalized formula for a 1st-order circuit:

\$H(s)=\frac{H_0+H_1s\tau_1}{1+s\tau_1}=H_0\frac{1+\frac{H_1}{H_0}s\tau_1}{1+s\tau_1}=H_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}\$ with \$\omega_z=\frac{1}{C_1(R_3||R_4)}\$ and \$\omega_p=\frac{1}{R_3C_1}\$

I have derived this expression without writing a single line of algebra, just drawing small sketches in simple configurations: in dc, when the cap. is open, when the excitation (the input) is reduced to 0 V and when the capacitor is replaced by a short circuit. You obtain a low-entropy expression in which you immediately see a dc gain, a zero and a pole.

If you want to learn about FACTs, check this APEC 2016 presentation


and look at the transfer functions derived in the book Linear Circuits Transfer Functions:



You can think of the capacitor as a resistor whose "resistance" varies with frequency.

That approach however will fail as the phase margin approaches zero.


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