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.
Best Answer
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?
Switching to the phasor domain, we have:
\$I_c = -\dfrac{A}{R_2}\dfrac{R_3}{R_4} = j \omega C V_c\$
or
\$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} \$