Multiple feedback low-pass filter design

Question

I'm trying to design second order multiple feedback low-pass filter with cut-off frequency at 1500 Hz and Chebyshev approximation.

Transfer function for this scheme is defined as:

Where:

And condition for real values of resistors:

However using Matlab and cheby1 function I get extremely high values of coefficients and as a result high values of resistors and capacitors.

hertz = 1500; % passband in hertz
rad_s = hertz*2*pi; % convert to rad/s
order = 2;
ripple = 3; % dB

[b, a] = cheby1(order, ripple, rad_s, 's');

% Extract coefficients for computing sheme values
a1 = a(2);
b1 = a(3);
H0 = -b(3);
fprintf('Coefficients:\n\ta1=%d\n\tb1=%d\n\tH0=%d\r\n', a1, b1, H0);

% Condition for capacitors
KCmin = (4*b1*(1-H0))/(a1^2);
fprintf('Capacitors condition:\n\tC2 / C1 >= %d\r\n', KCmin);

% Choose satisfying capasitors
C2 = 10^(-3); % F
C1 = 10^(-12); % F
fprintf('Chosen capacitors:\n\tC1=%d\n\tC2=%d\n\tCk=%d\r\n', C1, C2, C2/C1);

R2 = (a1*C2 - sqrt((a1^2)*(C2^2) - 4*C1*C2*b1*(1-H0))) / (2*hertz*C1*C2);
R1 = -R2 / H0;
R3 = b1 / ((hertz^2)*C1*C2*R2);

fprintf('Resistors:\n\t%d\n\t%d\n\t%d\n', R1, R2, R3);

And output is:

Coefficients:
    a1=6.078036e+03
    b1=6.288448e+07
    H0=-4.451880e+07

Capacitors condition:
    C2 / C1 >= 3.031241e+08

Chosen capacitors:
    C1=1.000000e-12
    C2=1.000000e-03
    Ck=1000000000

Resistors:
    7.518520e+03
    3.347155e+11
    8.349974e+04

Also I noticed that whatever ripple I chose condition C2/C1 was always of 10^8 order.

What am I doing wrong?

UPDATE

I believe I know what the problem is: given equations are in normalized form and I need coefficients for normalized form too. However MATLAB gives coefficients which are totally ready to be used. Therefore as I understand I have to remove w from the equations but it has not helped me in some reason.

Also I found normalized coefficients in my textbook:

The last column is normalized coefficients for 2d order 3dB Chebyshev filter. But I can't understand where from I have to take H0.

Answer 1

I am afraid that your transfer function is not correct. Cancel the wp² and the wp in the denominator - and the results will be OK. Your error can be verified very easily: All three expressions in the denominator must have the unit "1" - and this is not the case with your function.

The correct equations are like this:

1/wp²=R2R3C1C2;

1/(wp*Qp)=(R2+R3+R2R3/R1)C1.

Update:

The above two equations result from a comparison between the (corrected !) transfer function of YOUR specific circuit and the general second-order transfer function:

H(s)=Ao/[1+s/(wpQp)+s²/wp²]

For Chebyshev (3 dB ripple) the pole data are: Qp=1.30656 and the normalized pole frequency is wp/wo=0.8409 (wo=cut-off).

Answer 2

First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).

Well, we are trying to analyze the following opamp-circuit:

^{simulate this circuit – Schematic created using CircuitLab}

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

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

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

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

Now, we can put equations \$(1)\$ into \$(2)\$ to get:

$$ \begin{cases} \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_1-\text{V}_3}{\text{R}_3}+\frac{\text{V}_1-\text{V}_2}{\text{R}_4}\\ \\ 0=\frac{\text{V}_1-\text{V}_3}{\text{R}_3}+\frac{\text{V}_1-\text{V}_2}{\text{R}_4}+\text{I}_5\\ \\ \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_1-\text{V}_3}{\text{R}_3}+\frac{\text{V}_2-\text{V}_3}{\text{R}_5}\\ \\ 0=\frac{\text{V}_1-\text{V}_3}{\text{R}_3}+\frac{\text{V}_2-\text{V}_3}{\text{R}_5}+\text{I}_5 \end{cases}\tag3 $$

Now, using an ideal opamp, we know that:

$$\text{V}_+=\text{V}_-=\text{V}_2=0\space\text{V}\tag4$$

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

$$ \begin{cases} \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_1-\text{V}_3}{\text{R}_3}+\frac{\text{V}_1-0}{\text{R}_4}\\ \\ 0=\frac{\text{V}_1-\text{V}_3}{\text{R}_3}+\frac{\text{V}_1-0}{\text{R}_4}+\text{I}_5\\ \\ \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_1-\text{V}_3}{\text{R}_3}+\frac{0-\text{V}_3}{\text{R}_5}\\ \\ 0=\frac{\text{V}_1-\text{V}_3}{\text{R}_3}+\frac{0-\text{V}_3}{\text{R}_5}+\text{I}_5 \end{cases}\tag5 $$

Hence, we can solve for the transfer function:

$$\mathcal{H}:=\frac{\text{V}_3}{\text{V}_\text{i}}=-\frac{\text{R}_2\text{R}_3\text{R}_5}{\text{R}_1\text{R}_2\left(\text{R}_3+\text{R}_4+\text{R}_5\right)+\text{R}_3\text{R}_4\left(\text{R}_1+\text{R}_2\right)}\tag6$$

Where I used the following Mathematica-code:

In[1]:=Clear["Global`*"];
V2 = 0;
FullSimplify[
 Solve[{I1 == I2 + I3 + I4, 0 == I3 + I4 + I5, I1 == (Vi - V1)/R1, 
   I2 == V1/R2, I3 == (V1 - V3)/R3, I4 == (V1 - V2)/R4, 
   I4 == (V2 - V3)/R5}, {I1, I2, I3, I4, I5, V1, V3}]]

Out[1]={{I1 -> ((R3 R4 + R2 (R3 + R4 + R5)) Vi)/(
   R1 R3 R4 + R2 R3 R4 + R1 R2 (R3 + R4 + R5)), 
  I2 -> (R3 R4 Vi)/(R1 R3 R4 + R2 R3 R4 + R1 R2 (R3 + R4 + R5)), 
  I3 -> (R2 (R4 + R5) Vi)/(
   R1 R3 R4 + R2 R3 R4 + R1 R2 (R3 + R4 + R5)), 
  I4 -> (R2 R3 Vi)/(R1 R3 R4 + R2 R3 R4 + R1 R2 (R3 + R4 + R5)), 
  I5 -> -((R2 (R3 + R4 + R5) Vi)/(
    R1 R3 R4 + R2 R3 R4 + R1 R2 (R3 + R4 + R5))), 
  V1 -> (R2 R3 R4 Vi)/(R1 R3 R4 + R2 R3 R4 + R1 R2 (R3 + R4 + R5)), 
  V3 -> -((R2 R3 R5 Vi)/(
    R1 R3 R4 + R2 R3 R4 + R1 R2 (R3 + R4 + R5)))}}

My equation was also confirmed using LTspice.

---

When we want to apply the derivation from above to your circuit we need to use Laplace transform (I will use lower case function names for the functions that are in the (complex) s-domain, so \$\text{y}\left(\text{s}\right)\$ is the Laplace transform of the function \$\text{Y}\left(t\right)\$):

• $$\text{R}_2=\frac{1}{\text{sC}_1}\tag7$$`
• $$\text{R}_5=\frac{1}{\text{sC}_2}\tag8$$

So, we can rewrite the transfer function as:

$$\mathscr{H}\left(\text{s}\right)=-\frac{\frac{1}{\text{sC}_1}\cdot\text{R}_3\cdot\frac{1}{\text{sC}_2}}{\text{R}_1\cdot\frac{1}{\text{sC}_1}\cdot\left(\text{R}_3+\text{R}_4+\frac{1}{\text{sC}_2}\right)+\text{R}_3\text{R}_4\left(\text{R}_1+\frac{1}{\text{sC}_1}\right)}=$$ $$-\frac{\text{R}_3}{\text{C}_1\text{C}_2\text{R}_1\text{R}_3\text{R}_4\text{s}^2+\text{C}_2\left(\text{R}_3\text{R}_4+\text{R}_1\left(\text{R}_3+\text{R}_4\right)\right)\text{s}+\text{R}_1}\tag9$$

Now, when working with sinusoidal signals we can use \$\text{s}:=\text{j}\omega\$ (where \$\text{j}^2=-1\$ and \$\omega=2\pi\text{f}\$ with \$\text{f}\$ is the frequency of the input signal in Hertz). So, we get:

$$\underline{\mathscr{H}}\left(\text{j}\omega\right)=-\frac{\text{R}_3}{\text{C}_1\text{C}_2\text{R}_1\text{R}_3\text{R}_4\left(\text{j}\omega\right)^2+\text{C}_2\left(\text{R}_3\text{R}_4+\text{R}_1\left(\text{R}_3+\text{R}_4\right)\right)\text{j}\omega+\text{R}_1}=$$ $$-\frac{\text{R}_3}{\text{R}_1-\text{C}_1\text{C}_2\text{R}_1\text{R}_3\text{R}_4\omega^2+\text{C}_2\left(\text{R}_3\text{R}_4+\text{R}_1\left(\text{R}_3+\text{R}_4\right)\right)\omega\text{j}}\tag{10}$$

So, the absolute value if given by:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\text{R}_3}{\sqrt{\left(\text{R}_1-\text{C}_1\text{C}_2\text{R}_1\text{R}_3\text{R}_4\omega^2\right)^2+\left(\text{C}_2\left(\text{R}_3\text{R}_4+\text{R}_1\left(\text{R}_3+\text{R}_4\right)\right)\omega\right)^2}}\tag{11}$$

Now, it is relatively hard to find the cut-off frequency for the transfer function in \$(11)\$. In order to make it a bit easier we assume a few things:

1. $$\text{R}_1=\text{R}_3=\text{R}_4=\text{R}\tag{12}$$
2. $$\text{C}_1=\text{C}_2=\text{C}\tag{13}$$

So, the absolute value becomes:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{1}{\sqrt{1+\left(\text{CR}\omega\right)^2\left(7+\left(\text{CR}\omega\right)^2\right)}}\tag{14}$$

Now, we can find the cut-off frequency (\$\omega_0\$) by finding:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega_0\right)\right|=\frac{1}{\sqrt{2}}\cdot\hat{\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|}\space\Longrightarrow\space\omega_0=\sqrt{\frac{\sqrt{53}-7}{2}}\cdot\frac{1}{\text{CR}}\tag{15}$$

---

So, in your case we want to have \$\text{f}_0=1500\space\text{Hz}\$ which implies \$\omega_0=3000\pi\space\text{rad/sec}\$ we get:

$$3000\pi=\sqrt{\frac{\sqrt{53}-7}{2}}\cdot\frac{1}{\text{CR}}\space\Longleftrightarrow\space$$ $$\text{CR}=\sqrt{\frac{\sqrt{53}-7}{2}}\cdot\frac{1}{3000\pi}\approx0.000039708\space\text{sec}\tag{16}$$

So, the absolute value will be:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{1}{\sqrt{1+\left(\sqrt{\frac{\sqrt{53}-7}{2}}\cdot\frac{1}{3000\pi}\cdot\omega\right)^2\left(7+\left(\sqrt{\frac{\sqrt{53}-7}{2}}\cdot\frac{1}{3000\pi}\cdot\omega\right)^2\right)}}\tag{17}$$