I have a RC circuit named called active realization of compensator, which is used in control systems, and is given in the following link.


I want to derive its differential equation by applying KCL and KVL. So, I did apply KCL at node A and B, which results into the following equations.

$$\frac{V_i -V_A}{R_1} + \frac{V_B - V_A}{R_2} = 0$$


$$\frac{V_A -V_B}{R_2} + C\frac{d(0 - V_B)}{dt} = 0$$

I considered KCL as sum of the current flowing away from node is equal to zero.

Now, I want to solve these equations by eliminating the voltages $V_A$ and $V_B$ to obtain a differential equation in term of $V_i$ and $V_o$.

Can anybody help me out in this regards.

Thanks in advance.


I think that reason your struggling is because you did not quite realise that: $$ V_A = V_o $$ With that in mind, and after a little algebra with the first equation, we have the following identity: $$ V_B = V_o (\frac{R_2 +R_1}{R_1}) - V_i \frac{R_2}{R_1} $$ Taking the derivative of both sides: $$ \frac{dV_B}{dt} = \frac{dV_o}{dt}(\frac{R_2 +R_1}{R_1}) - \frac{dV_i}{dt} \frac{R_2}{R_1} $$ So the final step is plugging this new equations into your second one to get ride off \$ V_B \$ and its derivative: $$V_o \frac{1-R_2^2 + R_1 R_2 }{R_2^2 R_1} + V_i \frac{1}{R_1} = \frac{dV_o}{dt}(\frac{R_2 +R_1}{R_1}) - \frac{dV_i}{dt} \frac{R_2}{R_1}$$

| improve this answer | |

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.