As a shortcut to solving this problem, I tried to use the voltage divider formula on R4 and R4 to get:

\$Vx = Vo (R3/(R3+R4))\$

Then, since the same current is going through R2 and R3, I found using KCL that

\$Vx = -Vi(R2/R1)\$

I made these two equal each other and solved for Vo/Vi but I got the wrong answer.

Why doesnt the volatage divider method work here? enter image description here Correct Answer:

enter image description here

  • 1
    \$\begingroup\$ Basically, because that's not a voltage divider. But a T network. You can see it as a loaded voltage divider, where the current 'escaping it' is too big to make the simple voltage divider formula effective. And, besides, what do you mean with "the same current is going throught R2 and R3"? There's a node, there, in the middle. There is not one single current, but three. \$\endgroup\$ – Sredni Vashtar Jun 6 '17 at 22:51
  • \$\begingroup\$ Thanks for your help. Sorry, I meant to say same current thru r1 and r2 \$\endgroup\$ – gabson Jun 7 '17 at 0:26
  • \$\begingroup\$ This transfer function has been derived here electronics.stackexchange.com/questions/293799/… \$\endgroup\$ – Verbal Kint Jun 11 '17 at 10:24

You can use the Thevenin equivalent for voltage dividers, and particularly with regard to the \$R_3\$ and \$R_4\$ combination with \$V_O\$, in an intuitive way:

$$\begin{align*} R_{TH}&=\frac{R_3\cdot R_4}{R_3+R_4}\\\\ V_{TH}&=V_O\cdot\frac{R_3}{R_3+R_4}\\\\ \end{align*}$$

Now, \$V_{TH}\$ is in series with \$R_{TH}\$ and \$R_2\$ before reaching the virtual ground \$V_-\$ node. So set up the equality for the currents and proceed:

$$\begin{align*} \frac{V_{TH}}{R_{TH}+R_2}&=-\frac{V_I}{R_1}\\\\ \frac{V_O\cdot\frac{R_3}{R_3+R_4}}{\frac{R_3\cdot R_4}{R_3+R_4}+R_2}&=-\frac{V_I}{R_1}\\\\ \therefore \frac{V_O}{V_I}&=-\frac{\frac{R_3\cdot R_4}{R_3+R_4}+R_2}{R_1\cdot\frac{R_3}{R_3+R_4}}\\\\ &=-\frac{R_3\cdot R_4+R_2\cdot R_3+R_2\cdot R_4}{R_1\cdot R_3}\\\\ &=-\left[\frac{R_4}{R_1}+\frac{R_2}{R_1}+\frac{R_2\cdot R_4}{R_1\cdot R_3}\right]\\\\ &=-\frac{1}{R_1}\cdot\left[R_4+R_2+\frac{R_2\cdot R_4}{R_3}\right]\\\\ &=-\frac{R_2}{R_1}\cdot\left[1+\frac{R_4}{R_2}+\frac{R_4}{R_3}\right] \end{align*}$$

So, this means that you can use Thevenin, just fine. As always, laws just work. It's just about following the implications accurately as you proceed through.

Please note that I didn't say that \$V_X=V_{TH}\$. \$V_X\$ is a physical node in your circuit. \$V_{TH}\$ is a virtual Thevenin node with a series resistance, \$R_{TH}\$, that leads to, but is not the same as, \$V_X\$. If you want \$V_X\$, then:


Note that this isn't what you wrote. That's probably because you misunderstood and imagined that the Thevenin voltage was actually the same as \$V_X\$. However, it's not the same thing.

  • \$\begingroup\$ Using the thevenin equivalence seems a lot of work. The nodal equation is a lot simpler: Vi/R2 + Vo/R4 = Vx/(R2||R3||R4). For the ideal opamp, Vi will be 0, reducing the equation to Vo/R4 = Vx/(R2||R3||R4) \$\endgroup\$ – Bart Jun 7 '17 at 8:54
  • \$\begingroup\$ @bart I specifically chose that approach because the OP's very first equation. I use nodal all the time here and prefer it. But there is a time and place for all things. \$\endgroup\$ – jonk Jun 7 '17 at 9:08

The voltage divider equation which you have used is only good if no current at all is drawn from the junction. Since you do propose to draw current through R2, your equation is wrong (in the sense that it does not apply).

The current through R2 and R3 is equal only if the two resistors are equal, since the other end of R3 is effectively (for the purposes of this calculation) tied to ground. Then you can treat R2 and R3 as effectively being in parallel, and calculate the voltage at the junction using the parallel value, then calculate the current in R2 using that voltage.


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