I am trying to prove that the inductance of a given circuit (between the source A, and ground) (see image below) is equal to (R1*R2*R4*C)/R3.

Under the assumption that it is an ideal op-amp, I know that the input to each of the op-amps is equal to the input voltage (A), and that there is no current entering.

I've attempted nodal analysis, but my main issue is that I don't know the current leaving either of the op-amps, and I can't find any information regarding a circuit like this one on the Internet.

On top of that, I also cannot figure out how to tackle the capacitor, so any advice regarding that would also be greatly appreciated.

Finally, the circuit is supposed to result in being equivalent to an inductor, L = (R1*R2*R4*C)/R3 , and that is what is required to prove.

Circuit in question

  • \$\begingroup\$ Are the crossed wires in the middle connected? (the node between R2 and C) \$\endgroup\$
    – Justin
    Aug 29 '13 at 11:57
  • \$\begingroup\$ Yes, they're connected, sorry for the lack of clarity there. \$\endgroup\$
    – Kadin
    Aug 29 '13 at 12:19
  • \$\begingroup\$ Your units don't result in inductance. \$\endgroup\$ Aug 29 '13 at 12:45
  • \$\begingroup\$ We don't mind hand-drawn schematics, but to avoid confusion next time please put a dot on each node, even the T's. \$\endgroup\$ Aug 29 '13 at 12:45
  • \$\begingroup\$ where is the frequency dependent factor of s in your equation? \$\endgroup\$
    – Andy aka
    Aug 29 '13 at 12:55

You know the current through R4:

\$i_4 = \dfrac{A}{R_4}\$

Thus, you know the current through R3:

\$i_3 = i_4 \$

Thus, you know the output voltage of the 2nd op-amp:

\$v_{O2} = i_4(R_4 + R_3) = A(1 + \dfrac{R_3}{R_4})\$

Thus, you know the voltage across R2:

\$v_{R2} = A - v_{O2} = -A\dfrac{R_3}{R_4} \$

Thus, you know the current through R2 which is identical to the current through the capacitor:

\$i_C = i_{R2} = -\dfrac{A}{R_2}\dfrac{R_3}{R_4}\$

Now recall:

\$i_C = C \dfrac{dv_C}{dt}\$

Can you take it from here?

I can't find an error in my calcs, so I think I'm just misinterpreting what Vs and Is actually are.

Switching to the phasor domain, we have:

\$I_c = -\dfrac{A}{R_2}\dfrac{R_3}{R_4} = j \omega C V_c\$


\$V_c = -\dfrac{A}{j \omega}\dfrac{R_3}{R_2R_4C}\$

Thus, the output voltage of the first op-amp is:

\$V_{o1} = A + V_c = A - \dfrac{A}{j \omega}\dfrac{R_3}{R_2R_4C}\$

And the voltage across R1 is:

\$V_{r1} = A - V_{o1} = \dfrac{A}{j \omega}\dfrac{R_3}{R_2R_4C}\$

Finally, the current through R1 is:

\$I_{r1} = \dfrac{V_{r1}}{R_1} = \dfrac{A}{j \omega}\dfrac{R_3}{R_1R_2R_4C} \$

But the source current is identical to \$I_{r1}\$, thus:

\$\dfrac{V_s}{I_s} = \dfrac{A}{I_{r1}} = j\omega \dfrac{R_1R_2R_4C}{R_3} = j \omega L_{eq} \$

  • \$\begingroup\$ I can understand all that you've done there, but unfortunately I can't really nail that last step (I think). Should I rearrange to get: C(dV_c/dt) = -(AR_3)/(R_2*R_4) dV_c/dt = -(AR_3)/(R_2*R_4*C) V_c = integral [-(AR_3)/(R_2*R_4*C)] dt V_c = -(AR_3*t)/(R_2*R_4*C) Which then leaves me with the voltage across the capacitor, and it varies with time. So then, the current across R_1, i_1 = (A-V_c)/R_1 Solve all of that.. And then (I think) all currents and voltages are found.. Right? \$\endgroup\$
    – Kadin
    Aug 29 '13 at 13:51
  • \$\begingroup\$ Shoot, and I forgot to mention in the original question, it's supposed to result in being equivalent to an inductor, L = (R1*R2*R3*C)/R4 :/ probably an important information. \$\endgroup\$
    – Kadin
    Aug 29 '13 at 13:55
  • 1
    \$\begingroup\$ @user2222956, switch to the phasor domain to make it easier: \$I_c = j\omega C V_c\$. Then the voltage across the capacitor is \$V_c = -\dfrac{A}{j\omega}\dfrac{R_3}{R_2 R_4 C}\$ \$\endgroup\$ Aug 29 '13 at 14:24
  • 1
    \$\begingroup\$ @user2222956, if you know the voltage across the capacitor, it is an elementary step to find the voltage across R1. Having found that, you now know the current through R1. Having found that, you now know the current through the source providing the voltage A. Now, you have the source voltage and the source current. That's all you need to find the equivalent inductance seen by the source. Recall, \$V_s = j\omega L_{eq}I_s \$ \$\endgroup\$ Aug 30 '13 at 1:57
  • 1
    \$\begingroup\$ @user2222956, see the edit to my answer. \$\endgroup\$ Aug 30 '13 at 13:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.