I am learning basic circuits and I got stuck at this question: enter image description here

This is what I got:
$$V_{o}=(\frac{sC_{1}V_{o}R_{2}+V_{o}-V_{in}}{R_{1}}-sC_{1}V_{o})\cdot \frac{-1}{sC_{2}}+sC_{1}V_{o}R_{2}+V_{o} $$

And this is how:
For the ideal op amp, V_+ = V_- = V_o, so I got the current passes through C_1, and R_2, hence got the voltage after R_1, and use this to form the formula above.

I kept checking but I don't see where I got wrong, but this seems too messy to be the answer. To get H(s), I need to divide it by V_in, and make it even messier. Could you give me some hints to get a simpler form to determine the type of this filter?

I know this is not a Helpdesk, I hope this could be an example for all who are studying the same thing. Thank!

  • 1
    \$\begingroup\$ It's completely analyzable from this, where \$K=1\$ (or that \$V_\text{OUT}=V_\text{B}\$.) You should be easily able to compose and then solve correctly the two simultaneous equations and put it into a standard form of$$\begin{align*}\frac{K}{s^2+2\zeta\:\omega_{_0}\:s +\omega_{_0}^2}=\frac{K}{s^2+\frac{\omega_{_0}}{Q}\:s +\omega_{_0}^2}\end{align*}$$again where \$K=1\$. And if you'd search a little, you'd have found at least my answer here and elsewhere (that's not the only place where I develop it.) \$\endgroup\$
    – jonk
    Nov 27, 2020 at 6:05
  • 1
    \$\begingroup\$ Beside jonk's approach, you can use the FACTs to determine this transfer function without writing equations. See my answer here. \$\endgroup\$ Nov 27, 2020 at 6:46
  • \$\begingroup\$ Thank you all. I'll take a look at them. \$\endgroup\$
    – keanehui
    Nov 27, 2020 at 7:07
  • \$\begingroup\$ @keanehui are you done with this question now. Do you need any more help drilling down through the formulas I developed in my answer? \$\endgroup\$
    – Andy aka
    Nov 28, 2020 at 17:15

1 Answer 1


I'd use the complex-AC version of Millman's theorem to solve for voltage \$\color{red}{V_X}\$: -

enter image description here


$$\color{red}{\boxed{V_X}} = \dfrac{\dfrac{V_I}{R1}+\dfrac{0}{R2 + \frac{1}{sC1}}+\dfrac{V_O}{\frac{1}{sC2}}}{\dfrac{1}{R1} +\dfrac{1}{R2 + \frac{1}{sC1}} +\dfrac{1}{\frac{1}{sC2}}}$$

And, of course for a unity gain non-inverting amplifier: $$V_X\cdot\dfrac{\frac{1}{sC1}}{R2 + \frac{1}{sC1}} = V_O$$

Or, \$\hspace{5.5cm}V_X = V_O\cdot(1+sC1R2)\$


$$V_O\cdot(1+sC1R2) = \dfrac{\dfrac{V_I}{R1}+V_O\cdot sC2}{\dfrac{1}{R1} +\dfrac{sC1}{sC1R2 + 1} +sC2}$$


$$V_O\left[\dfrac{1+sC1R2}{R1} +sC1 +sC2 +s^2C1C2R2\right] = \dfrac{V_I}{R1} + V_O\cdot sC2$$


$$V_O\left[\dfrac{1+sC1R2}{R1} +sC1 +s^2C1C2R2\right] = \dfrac{V_I}{R1} $$

$$V_O\left[1+sC1R2 +sC1R1 +s^2C1C2R1R2\right] = V_I $$

Can you do the final few steps yourself? Do you need help in this any more?

  • \$\begingroup\$ Hi Andy. Thanks for your help. Since the theorem is not covered in my syllabus, I am not quite sure how it should be applied and the details. For example, for R2, isn't there more than 1 voltage source, 0V and V_o? Why can we through away V_o from V_+ from the op amp? \$\endgroup\$
    – keanehui
    Nov 29, 2020 at 8:05
  • \$\begingroup\$ The voltage appearing at Vin+ is as a result of the voltage at Vx i.e. the inputs to the op-amp don't produce a voltage that drives R2 in any respect. Millman's theorem is just a simple extension of converting several independent voltage sources in series with their respective impedances to current sources in parallel with their impedances and totalizing things i.e. it is a shortcut that avoids a few steps. If you have done norton and thevenin equivalent sources in your course, it's a simple and straightforward extension with no smoke or mirrors or cleverness involved @keanehui \$\endgroup\$
    – Andy aka
    Nov 29, 2020 at 11:19

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