Finding the transfer function of a opamp circuit with 3 opamps

I have the following circuit:

I know that $$\\text{sig}_+\$$ is a voltage that is $$\n\$$ times bigger than the voltage at $$\\text{sig}_-\$$.

Question: If $$\R_g\$$ is a coil with the value $$\63\space\text{mH}\$$ and $$\R_3\$$ is a capacitor with value $$\10\space\text{nF}\$$ and $$\R_f=R_1=R_2=R_4=10\space\text{k}\Omega\$$ and $$\n=10\$$ what is the gain of this circuit at $$\f=10^6\space\text{Hz}\$$?

• What have you tried so far ? Are the opamps assumed ideal ? If so, the voltage at node just above Rg is $sig^-$ and node just below Rg is $sig^+$. This will give current through Rg. The current through Rg is same as the current through both Rf. This will help you find the voltage at the outputs of the two opamps. With those two voltages, the output of the third opamp can be found. Cross check with a simulation to ensure the assumptions made are valid.
– AJN
Aug 4 '20 at 12:09
• You can edit the image and label the nodes and the currents and show your KCL / KVL / other equations. It will be easier to spot any problems that way.
– AJN
Aug 4 '20 at 12:11
• I’m voting to close this question because this is homework with no attempt to solve. This site is not a homework answering service Dec 22 '20 at 15:14

Well, I am trying to analyze the following circuit (assuming an ideal opamp):

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\begin{cases} \text{I}_8=\text{I}_1+\text{I}_4\\ \\ \text{I}_1=\text{I}_2\\ \\ \text{I}_2=\text{I}_3\\ \\ \text{I}_5=\text{I}_3+\text{I}_9\\ \\ 0=\text{I}_4+\text{I}_7\\ \\ \text{I}_5=\text{I}_6\\ \\ \text{I}_6=\text{I}_7+\text{I}_8+\text{I}_9 \end{cases}\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$\begin{cases} \text{I}_1=\frac{\text{V}_1-\text{V}_2}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_2-\text{V}_3}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_3-\text{V}_4}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_1-\text{V}_5}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_4-\text{V}_6}{\text{R}_5}\\ \\ \text{I}_6=\frac{\text{V}_6}{\text{R}_6}\\ \\ \text{I}_7=\frac{\text{V}_7-\text{V}_5}{\text{R}_7} \end{cases}\tag2$$

Substitute $$\(2)\$$ into $$\(1)\$$, in order to get:

$$\begin{cases} \text{I}_8=\frac{\text{V}_1-\text{V}_2}{\text{R}_1}+\frac{\text{V}_1-\text{V}_5}{\text{R}_4}\\ \\ \frac{\text{V}_1-\text{V}_2}{\text{R}_1}=\frac{\text{V}_2-\text{V}_3}{\text{R}_2}\\ \\ \frac{\text{V}_2-\text{V}_3}{\text{R}_2}=\frac{\text{V}_3-\text{V}_4}{\text{R}_3}\\ \\ \frac{\text{V}_4-\text{V}_6}{\text{R}_5}=\frac{\text{V}_3-\text{V}_4}{\text{R}_3}+\text{I}_9\\ \\ 0=\frac{\text{V}_1-\text{V}_5}{\text{R}_4}+\frac{\text{V}_7-\text{V}_5}{\text{R}_7}\\ \\ \frac{\text{V}_4-\text{V}_6}{\text{R}_5}=\frac{\text{V}_6}{\text{R}_6}\\ \\ \frac{\text{V}_6}{\text{R}_6}=\frac{\text{V}_7-\text{V}_5}{\text{R}_7}+\text{I}_8+\text{I}_9 \end{cases}\tag3$$

Now, using an ideal opamp, we know that:

• $$\text{V}_\text{k}=\text{V}_{+_1}=\text{V}_{-_1}=\text{V}_2$$
• $$\text{n}\cdot\text{V}_\text{k}=\text{V}_{+_2}=\text{V}_{-_2}=\text{V}_3$$
• $$\text{V}_x:=\text{V}_{+_3}=\text{V}_{-_3}=\text{V}_5=\text{V}_6$$

So we can rewrite equation $$\(3)\$$ as follows:

$$\begin{cases} \text{I}_8=\frac{\text{V}_1-\text{V}_\text{k}}{\text{R}_1}+\frac{\text{V}_1-\text{V}_x}{\text{R}_4}\\ \\ \frac{\text{V}_1-\text{V}_\text{k}}{\text{R}_1}=\frac{\text{V}_\text{k}-\text{n}\cdot\text{V}_\text{k}}{\text{R}_2}\\ \\ \frac{\text{V}_\text{k}-\text{n}\cdot\text{V}_\text{k}}{\text{R}_2}=\frac{\text{n}\cdot\text{V}_\text{k}-\text{V}_4}{\text{R}_3}\\ \\ \frac{\text{V}_4-\text{V}_x}{\text{R}_5}=\frac{\text{n}\cdot\text{V}_\text{k}-\text{V}_4}{\text{R}_3}+\text{I}_9\\ \\ 0=\frac{\text{V}_1-\text{V}_x}{\text{R}_4}+\frac{\text{V}_7-\text{V}_x}{\text{R}_7}\\ \\ \frac{\text{V}_4-\text{V}_x}{\text{R}_5}=\frac{\text{V}_x}{\text{R}_6}\\ \\ \frac{\text{V}_x}{\text{R}_6}=\frac{\text{V}_7-\text{V}_x}{\text{R}_7}+\text{I}_8+\text{I}_9 \end{cases}\tag4$$

Now, we can solve for the transfer function:

$$\displaystyle\mathcal{H}:=\frac{\text{V}_\text{o}}{\text{V}_\text{i}}=\frac{\text{V}_7}{\text{V}_\text{k}}=\frac{\text{n}\text{R}_7\left(\text{R}_6\left(\text{R}_1+\text{R}_2+\text{R}_3\right)+\text{R}_1\text{R}_5\right)+\text{n}\text{R}_4\text{R}_6\left(\text{R}_2+\text{R}_3\right)-\text{R}_7\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_5+\text{R}_6\right)-\text{R}_3\text{R}_6\left(\text{R}_4+\text{R}_7\right)}{\text{R}_2\text{R}_4\left(\text{R}_5+\text{R}_6\right)}\tag5$$

Now, applying this to your circuit we need to use (from now on I use the lower case letters for the function in the 'complex' s-domain where I used Laplace transform):

• $$\text{R}_2=\text{sL}\tag6$$
• $$\text{R}_6=\frac{1}{\text{sC}}\tag7$$
• $$\text{R}:=\text{R}_1=\text{R}_3=\text{R}_4=\text{R}_5=\text{R}_7\tag8$$

So, the transfer function becomes:

$$\mathcal{h}\left(\text{s}\right)=\frac{\text{v}_\text{o}\left(\text{s}\right)}{\text{v}_\text{i}\left(\text{s}\right)}=\frac{\text{C}\left(\text{n}-1\right)\text{R}^2\text{s}+2\text{Lns}+3\left(\text{n}-1\right)\text{R}}{\text{Ls}\left(\text{CRs}+1\right)}-1\tag9$$

Because we are working with sinusodial signals, we can write $$\\text{s}=\text{j}\omega\$$ where $$\\text{j}^2=-1\$$ (the imaginary unit) and $$\\omega=2\pi\text{f}\$$ with $$\\text{f}\$$ is the frequency in Hertz. So, we can write:

$$\underline{\mathcal{h}}\left(\text{j}\omega\right)=\frac{\text{C}\left(\text{n}-1\right)\text{R}^2\left(\text{j}\omega\right)+2\text{Ln}\left(\text{j}\omega\right)+3\left(\text{n}-1\right)\text{R}}{\text{L}\left(\text{j}\omega\right)\left(\text{CR}\left(\text{j}\omega\right)+1\right)}-1\tag{10}$$

Now, we can find the amplitude by finding the absolute value of $$\(10)\$$ and using your given values:

$$\left|\underline{\mathcal{h}}\left(\text{j}\omega\right)\right|=\frac{1}{7}\sqrt{49-\frac{767436000000000}{\omega^2+100000000}+\frac{900000000000000}{\omega^2}}\tag{11}$$

So, at $$\\text{f}=10^6\space\text{Hz}\$$ we have $$\\omega=2\pi\cdot10^6\space\text{rad/sec}\$$, we get:

$$\left|\underline{\mathcal{h}}\left(\text{j}\cdot2\pi\cdot10^6\right)\right|=\frac{1}{7}\sqrt{49-\frac{767436000000000}{\left(2\pi\cdot10^6\right)^2+100000000}+\frac{900000000000000}{\left(2\pi\cdot10^6\right)^2}}=$$ $$\frac{\sqrt{\frac{225+1325689 \pi ^2+1960000 \pi ^4}{1+40000 \pi ^2}}}{7 \pi }\approx1.0337\tag{12}$$

I used Mathematica to solve the transfer function:

In[1]:=FullSimplify[
Solve[{I8 == I1 + I4, I1 == I2, I2 == I3, I5 == I3 + I9,
0 == I4 + I7, I5 == I6, I6 == I7 + I8 + I9, I1 == (V1 - Vk)/R1,
I2 == (Vk - n*Vk)/R2, I3 == (n*Vk - V4)/R3, I4 == (V1 - Vx)/R4,
I5 == (V4 - Vx)/R5, I6 == (Vx)/R6, I7 == (V7 - Vx)/R7}, {I1, I2,
I3, I4, I5, I6, I7, I8, I9, V1, V4, Vx, V7}]]

Out[1]={{I1 -> (Vk - n Vk)/R2, I2 -> (Vk - n Vk)/R2, I3 -> (Vk - n Vk)/R2,
I4 -> (((R1 - n R1 + R2) R5 - (-1 + n) (R1 + R2 + R3) R6) Vk)/(
R2 R4 (R5 + R6)), I5 -> ((-R3 + n (R2 + R3)) Vk)/(R2 (R5 + R6)),
I6 -> ((-R3 + n (R2 + R3)) Vk)/(R2 (R5 + R6)),
I7 -> (((-1 + n) R1 R5 - R2 R5 + (-1 + n) (R1 + R2 + R3) R6) Vk)/(
R2 R4 (R5 + R6)),
I8 -> (((R1 - n R1 + R2 + R4 - n R4) R5 - (-1 + n) (R1 + R2 + R3 +
R4) R6) Vk)/(R2 R4 (R5 + R6)),
I9 -> ((-R3 - R5 - R6 + n (R2 + R3 + R5 + R6)) Vk)/(R2 (R5 + R6)),
V1 -> ((R1 - n R1 + R2) Vk)/R2, V4 -> ((-R3 + n (R2 + R3)) Vk)/R2,
Vx -> ((-R3 + n (R2 + R3)) R6 Vk)/(R2 (R5 + R6)),
V7 -> ((n (R2 + R3) R4 R6 - (R1 + R2) (R5 + R6) R7 +
n (R1 R5 + (R1 + R2 + R3) R6) R7 - R3 R6 (R4 + R7)) Vk)/(
R2 R4 (R5 + R6))}}

In[2]:=FullSimplify[(((n (R2 + R3) R4 R6 - (R1 + R2) (R5 + R6) R7 +
n (R1 R5 + (R1 + R2 + R3) R6) R7 - R3 R6 (R4 + R7)) Vk)/(
R2 R4 (R5 + R6)))/Vk]

Out[2]=(n (R2 + R3) R4 R6 - (R1 + R2) (R5 + R6) R7 +
n (R1 R5 + (R1 + R2 + R3) R6) R7 - R3 R6 (R4 + R7))/(R2 R4 (R5 + R6))


I also found the amplitude function for the transfer function using Mathematica:

In[3]:=FullSimplify[
Sqrt[(ComplexExpand[
Re[-1 + (3 (-1 + n) R + 2 L n (I*w) + C1 (-1 + n) R^2 (I*w))/(
L (I*w) (1 + C1 R (I*w)))]])^2 + (ComplexExpand[
Im[-1 + (3 (-1 + n) R + 2 L n (I*w) + C1 (-1 + n) R^2 (I*w))/(
L (I*w) (1 + C1 R (I*w)))]])^2],
Assumptions -> L > 0 && C1 > 0 && n > 0 && w > 0 && R > 0]

Out[3]=Sqrt[(L^2 (1 - 2 n)^2 w^2 + C1^2 (-1 + n)^2 R^4 w^2 +
R^2 (9 (-1 + n)^2 + 4 C1 L (-1 + n^2) w^2 +
C1^2 L^2 w^4))/(L^2 w^2 (1 + C1^2 R^2 w^2))]


Now, using your values we get:

In[4]:=L = 63*10^(-3);
C1 = 10*10^(-9);
R = 10000;
n = 10;
FullSimplify[Sqrt[(
L^2 (1 - 2 n)^2 w^2 + C1^2 (-1 + n)^2 R^4 w^2 +
R^2 (9 (-1 + n)^2 + 4 C1 L (-1 + n^2) w^2 + C1^2 L^2 w^4))/(
L^2 w^2 (1 + C1^2 R^2 w^2))]]

Out[4]=1/7 Sqrt[49 + 900000000000000/w^2 - 767436000000000/(100000000 + w^2)]


And, we can use the limit:

In[5]:=FullSimplify[
Limit[1/7 Sqrt[
49 + 900000000000000/w^2 - 767436000000000/(100000000 + w^2)],
w -> 2*Pi*10^6]]

Out[5]=Sqrt[(225 + 1325689 \[Pi]^2 + 1960000 \[Pi]^4)/(1 +
40000 \[Pi]^2)]/(7 \[Pi])

In[6]:=N[%5]

Out[6]=1.0337

• Thank you so much! Aug 4 '20 at 12:22