The question has been asked before here: Help with ideal op-amp problem

I don't understand why we simply can't reduce \$R_2\$, \$R_3\$ and \$R_4\$ into one single feedback resistance: $$R_f=R_4+R_2 \parallel R_3$$, and then just calculate the gain: $$\frac{V_2}{V_1}=\frac{-R_f}{R_1}=-\frac{R_4+R_2||R_3}{R_1}$$

The problem is shown here:

enter image description here

I understand the "correct" solution and how to get to it, but don't know why my solution would be "wrong".

  • \$\begingroup\$ Because the current through \$R_3\$ does not reach point B. \$\endgroup\$ Mar 21, 2017 at 10:28
  • \$\begingroup\$ Because your solution does not apply to this circuit. (yeah I know: duh !). Part of the proper solution of this circuit is that node B is a virtual ground so I(R1) = I(R2). For your simplification using Rf to be correct, the same current would have to flow to node B, either using R2 or Rf. But Your Rf makes the current through Rf the same as the current R4. The current through R2 is actually somewhat lower as some current flows away (is lost) through R3. Your Rf solution does not take this into account. \$\endgroup\$ Mar 21, 2017 at 10:29
  • \$\begingroup\$ Your expression (R4+R2||R3) is nothing else than the resistance as seen from the opamp output. However, because of R3 this resistance is not identical to the effective feedback resistance BETWEEN opamp output and inverting input. \$\endgroup\$
    – LvW
    Mar 21, 2017 at 10:43
  • \$\begingroup\$ One question, are R2 and R3 in parallel or no? If they are then I don't see why the current through R3 wouldn't reach point B. \$\endgroup\$
    – eenn
    Mar 21, 2017 at 10:47
  • \$\begingroup\$ See complex feedback networks in ti.com/lit/an/slaa068b/slaa068b.pdf \$\endgroup\$ Mar 21, 2017 at 10:56

2 Answers 2


I know you have the answer already but for this type of configuration, the best is to use the Extra-Element Theorem or EET from Dr. Middlebrook (see https://en.wikipedia.org/wiki/Extra_element_theorem). This is truly the best tool which can get you straight to the answer without almost writing a line of algebra. First, you identify the element that bothers you. Here, it is clearly \$R_3\$. You have the choice to temporarily remove it from the circuit or make it a short circuit. I choose to remove it and I calculate the gain without \$R_3\$ in place: \$H_{R_{3inf}}=-\frac{R_2+R_4}{R_1}\$. Then set the input source \$V_1\$ to 0 V and determine the resistance \$R_d\$ "seen" from \$R_3\$ terminals in this configuration. Considering a perfect op amp, you can infer by a quick sketch showing how currents circulate that \$R_d=0\$. Now, bring \$V_1\$ in place and determine the resistance "seen" from \$R_3\$ terminals in this configuration while the output is nulled. Practically, imagine connecting a current source \$I_T\$ across \$R_3\$ terminals and tweak it to have \$V_2\$ equal to 0 V. This is what is called a null double injection or NDI. By inspection, if \$V_{(-)}=0\$ and if \$V_2=0\$ also, then the resistance \$R_n=R_2||R_4\$. This is it, the complete transfer function is:


The icing on the cake is if you decide to replace \$R_3\$ by a capacitor \$C_1\$ for instance, then simply replace \$R_3\$ by \$\frac{1}{sC_1}\$ and you obtain the new transfer function in a snap-shot:

\$H(s)=H_{R_{3inf}}\frac{1+\frac{R_n}{\frac{1}{sC_1}}}{1+\frac{0}{\frac{1}{sC_1}}}=-\frac{R_2+R_4}{R_1}(1+s(R_2||R_4)C_1)=H_0(1+\frac{s}{\omega_z})\$ with \$\omega_z=\frac{1}{(R_2||R_4)C_1}\$ and \$H_0=-\frac{R_2+R_4}{R_1}\$

You can't beat the Fast Analytical Circuits Techniques or FACTs in terms of simplicity and execution speed. To learn more about the technique, you can check this PPT taught at APEC in 2016


but also the FACTs at work in a series of solved examples described here


  • \$\begingroup\$ To me, the best and most simple/logical way to solve the problem is the following: Apply the star-triangle transformation for R2, R3 and R4. Then, you will see that the classical inverter feedback scheme results - and the three resistors reduce to only two effective parts (one resistor acts as a load and does not influence the gain) . \$\endgroup\$
    – LvW
    Jun 8, 2018 at 9:17
  • \$\begingroup\$ It is certainly a valuable option but I wanted to promote the EET which is really a nice tool not well-known despite its introduction in the 80's. Would the delta-wye transformation lead to the simple arrangement I came up with at the same speed? Without resorting to these tools, superposition applied to \$V_{in}\$ and \$V_{out}\$ while making \$\epsilon\$ equal to 0 V (infinite gain considered) is another viable solution, perhaps more affordable to the student. \$\endgroup\$ Jun 8, 2018 at 11:05
  • \$\begingroup\$ Yes - I agree. I know Middlebrook´s EET as well as his various contributions to the feedback therory (loop gain). And, yes - superposition of Vin and Vout related to the inverting terminal is another good option. Fazit: 100% agreement. Fine. \$\endgroup\$
    – LvW
    Jun 8, 2018 at 13:36

When you set V1=0 the current trought R1 = 0, conseguence R2=0 too, so the voltage at VA=0, and the current trought R4 and R5 is zero too. So you see or infinite resistance, zero. Now I see only zero resistance can makes the branch equation tue in the case you use a current source, so true the resistance is zero.

  • \$\begingroup\$ When u set output = null using double injection , u have the fact that current trought R4 and R5 is null to (zero means no current flowing) and VB is at zero voltage (op ap inverting configuration) : this condition makes true that equation:: V1/R1 = VA/R2 so the resistance you see is R2//R4 \$\endgroup\$
    – sonnex
    Jun 8, 2018 at 8:54

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