Nodal analysis -> transfer function -> step response

Question

I've been trying to get the transfer function for the circuit in the image below, but I seem to be doing something wrong, since the simulation on Circuitlab gives me a different Bode plot and I can't find the error on my own.

Here, the equations from the nodal analysis (belong to nodes A, B, C, respectively). Va=Vb=Vc=0 due to virtual ground.

The result I'm getting for the transfer function is:

which looks a little weird to me because of the large numbers, and which translates to this Bode plot:

Simulation on Circuitlab, on the other hand, outputs the following:

From this transfer funtion I should get and plot the step response, which I get calculating the inverse Laplace from H(s)/s, but since the transfer function doesn't seem to be ok I'm not getting these right either.

I'd be very thankful if anyone could point me in the right direction.

Answer 1

For my opinion, it is not necessary to go down to KCL and KVL equations. Instead you should make use of basic available gain functions.

• Between E and output node Vo there is a damped integrator (inverting lowpass)

• Between node Vo and E there is a non-inverting active block with an inverting feedback loop (local opamp with R2, R3 and C3); the resistor R1 has no influence on the gain (ideal opamp); its only purpose is stability improvement because we have two opamps in one feedback loop.

• A closer look into this block reveals that it is nothing else than a non-inverting integrator stage (integrating capacitor C3 multiplied by the driving inverting gain factor, which - in this case - is "-1")

• These building blocks are arranged in one common feedback loop. Thus, we have one version of the classical two-integrator loop (one damped inverting integrator and one non-inverting integrator).

• Following this approach, we have a system of two equations with two unknowns:

First equation: Vo=Vi * F1 + VE * F2 with F1=f(R5, C1, C2) and F2=f(R4, R5, C1)

(Comment: Both functions F1 and F2 are simple inverting gain functions)

2nd equation: VE=Vo * (1/sT) with T=R6C3

Unknowns: VE and Vo/Vi

(using this approach, I have found the transfer function - by hand calculation - within 8 minutes).

Answer 2

Nodal analysis is based on Kirchoff's Current Laws (KCL), meaning that the sum of all currents will need to sum up to exactly \$0\$.

The issue in your equations is related to the fact that outputs of opamps are voltage sources, and they will also inject a current in a node. So, for example in node D you would get instead

$$0 = (v_A - v_D)\cdot sC_3 + (v_B - v_D)\frac{1}{R_2} + i_{oa1}$$

As is, this circuit becomes unsolvable as we just inserted another unknown. In order to make the circuit solvable again, the added unknown current will need to be fixed by another (linearly independent) equation, in our case this is the virtual short-circuiting of the inputs;

$$v_B = 0$$

This problem has 6 unknowns (A through E and \$v_0\$), so you will need 6 KCL equations. There are 3 opamp outputs, so you would (when not simplifying the system of equations) ultimately have 3 extra virtual short-circuit equations and output current unknowns totaling to 9 equations and unknowns.

Answer 3

Try plotting DB(MAG(V(OUT)/V(IN))) instead of DB(MAG(V(OUT))), because you want to get the transfer function from Output to Input. Here's the schematic I used:

^{simulate this circuit – Schematic created using CircuitLab}