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I am trying to find the D.E. that relates Vin to Vout.

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KCL at A: $$\frac{A-V_o}{12} + \frac{A-V_i}{6} + \frac{1}{12}*\frac{dV}{dt} = 0\tag1$$

KCL at B: $$\frac{B-0}{10} + \frac{B-V_o}{10} = 0\tag2$$

\$2B=V_o\$ or \$B = .5V_o\$

By using op-amp rules V+=V-(A=B) I found the following equation.

$$\frac{3}{2}V_o = \frac{dV_i}{dt} - 2V_i\tag3$$

I don't really know if this is 100% correct and would like to see what someone else comes up with.

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    \$\begingroup\$ The answer is wrong. Equations (1) and (2) are correct. You made some mistakes while combining them. \$\endgroup\$
    – nidhin
    Oct 14, 2015 at 9:21
  • \$\begingroup\$ As a matter of building good notational hygiene, you might want to use \$V_A\$ and \$V_B\$ respectively instead of A and B in situations like this. It's a detail, to be sure, but helps clarify your expressions. On first take I was wondering what you were doing subtracting a voltage from a gain, as \$A\$ is often used to denote the gain ("Amplification") of an op amp. No better time than while you're learning to form good habits :) \$\endgroup\$
    – scanny
    Oct 15, 2015 at 19:28

2 Answers 2

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the differential equation is wrong,please recheck your steps. The actual DE is

$$\frac{dV_o}{dt} + V_o = 4V_i$$

Update: However the differential equation is not valid as negative feedback fails to dominate here,the concept of virtual short circuit fails and op amp is in saturation!

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    \$\begingroup\$ This diff. equation applies only if both opamp input terminals are interchanged. \$\endgroup\$
    – LvW
    Oct 14, 2015 at 10:49
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By using op-amp rules V+=V-(A=B) I found the following equation....

You are not allowed to apply these rules because the equation (V+=V-) does apply only if the opamp is working linearly. However, this is not the case, because positive feedback overrules negative feedback. That means: The opamp does not work - it goes immediately into saturation.

In case of a drawing error (both opamp inputs interchanged) we have a damped Deboo integrator (1st order lowpass) with the transfer function

H(s)=4/(12sC+1)=4/(s+1)

It should not be a problem to transfer this equation into the time domain.

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  • \$\begingroup\$ At low frequencies , the cap acts as an open circuit,and since the series 6 ohm resistance is less than bridge resistor 12 ohm ,the negative feedback should dominate right? refer link \$\endgroup\$
    – Ashik Anuvar
    Oct 14, 2015 at 16:22
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    \$\begingroup\$ No - the negative feedback factor is (-6/18=-1/3) and the positive feedback factor is larger (+1/2). Hence, the circuit goes immediately into saturation. Therefore, you must exchange the opamp input terminals. \$\endgroup\$
    – LvW
    Oct 14, 2015 at 17:14

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