How to design a 2nd Order High Pass Butterworth filter with a gain of 6 dB?

Question

I have been trying to design a 2nd order filter with a gain of 6 dB (or 2 Av), using the following Sallen-Key Topology.

I have calculated R1 and R2 to give me a desired cut-off of 1.6kHz and this part is working fine. I am however not managing to get the gain up to 6 dB.

Following the example from this book, page 450, using the Butterworth co-efficients for Second-Order Filter Parameters α = 1.414 and b = 1.0.

Calculating: R4 = 2 - α = .586 ohms

Calculating: R3 = 1 + .586/1 = 1.586 ohms

This resistor ratio is providing a gain of 1.586 as per the Butterworth coefficients. Thus my circuit looks like this:

The frequency response looks like this:

The above image may not be too clear, however the cut-off freq is as desired, however the gain is not 2 AV (6 dB) but 1.586 AV as per the above calculation. I would gladly provide more information if necessary.

How can I alter the above equations to provide a gain of 2 Av while using the Butterworth coefficients?

Answer 1

First I don't understand why you are refering to page 450 of the book;

I found your circuit on page 456. Figure 11.23 "High-pass equal-component" (VCVS).

In the book they describe two types of Sallen-Key Topology filters, one is the "unit-gain"-version which as the name suggests has A_v=1. and the other is the "equal-component"-version (the one you have), this one also has a specific/fixed gain associated with it which is A=3−α this is what the book sayes about the "equal-component"-version on page 449:

"We see that the gain and damping of the filter are linked together. Indeed, for a certain damping factor, only one specific gain will work properly: A=3−α"

Since we know that for a butterworth-filter α must be sqrt(2) that determines our gain. So to answer the question;

How can I alter the above equations to provide a gain of 2 Av while using the Butterworth coefficients?

You can't without changing the basic circuit because the gain is determined by the topology and the choise of α for a butterworth-filter.

Now to answer the broader question of

How to design a 2nd Order High Pass Butterworth filter with a gain of 6 dB?

You can easily make the gain of your circuit almost anything you want by just adding a single resistor and fiddleing with the values of the existing like this;

The circuit you have can be turned into this one:

^{simulate this circuit – Schematic created using CircuitLab}

By replacing R2 and Rf with voltage-dividers

Now the gain of your new circuit is going to be (3-α)(R3+R4)/R4

To make this work the following have to be true:

R3//R4=R2 <- The thevinin equivalent of R3//R4 has to be equal to the original R2

R5//R6=Rf <- The thevinin equivalent of R5//R6 has to be equal to the original Rf

R3/R4=R5/R6 <- The two voltage-dividers have to divide the output by the same amount.

Now R6 and Ri can of course be combined, but for the sake of understanding the circuit I left them seperate.

If I was you though I would go for the "unit gain"-type and then do as I have described using R3=R4 to amplify the output by 2 to get A_v=2

EDIT:

I followed the example in the book for a unit-gain type, I chose 1kHz cutoff and simulated it in LT-spice with the results I got for the resistors and caps. here is a screenshot of the simulation in LT-spice showing cutoff at 1kHz, 0dB in-band gain and butterworth responce;

I then replaced the feedback resistors with voltage dividers as per my suggestion and simulated the results, below is a screenshot of the simulation in LT-spice, showing 6dB in-band gain, cutoff at 1kHz and butterworth-responce.

Sorry I know the pictures are hard to make out.

Answer 2

There is a straightforward solution to the problem - starting with the general transfer function of the circuit. From this function, we can derive the following expressions for an ideal opamp...

Pole frequency: $$ \omega_p=\frac{1}{R_2C_1\sqrt{k_rk_c}} $$

Pole quality factor: $$ Q_p = \frac{\sqrt{k_rk_c}}{1+k_c+k_rk_c(1-v)} $$

Where \$ k_r = R_1/R_2 \quad k_c = C_2/C_1 \quad v = 1+\frac{R_4}{R_3} \$.

These expressions can be evaluated setting \$ v=2 \$ and \$ R_3 = R_4 \$. One possible (simple) solution is to set \$ k_c = 1 \$ (i.e. \$ C_1 = C_2 \$).

For this condition we get: $$ Q_p = \frac{\sqrt{k_r}}{2-k_r} $$

For \$k_r\$ we have a quadratic solution:

$$ k_{r1,2} = 2 + \frac{1 \pm \sqrt{1+ 8 Q_p^2}}{2Q_p^2} $$

Note that only the smallest solution is valid (with the "-" sign) in order to keep \$Q_p\$ positive.

EDIT:

The transfer function for the given highpass circuit (first form) is as follows (where \$ v = 1+R_4/R_3 \$):

$$ H(s)=N(s)/D(s) $$

$$ N(s) = s^2 v R_1 R_2 C_1 C_2 $$

$$ D(s) = 1 + s[R_2(C_1+C_2)+R_1C_2(1-v)]+s^2R_1R_2C_1C_2 $$

Now we compare this circuit specific equation with the second-order general form for deriving the equations for gain, corner frequency and pole quality:

$$ H(s)=N(s)/D(s) $$

$$ N(s) = \left(\frac{s}{\omega_p}\right)^2 A_\infty $$

$$ D(s) = 1 + \frac{s}{\omega_p Q_p} + \left(\frac{s}{\omega_p}\right)^2 $$

Therefore, for \$s\$ approaching infinite values (highpass gain) we have \$ H(s) = A_\infty \$.

Comparing both forms of \$H(s)\$ we arrive at the given expressions for \$\omega_p\$, \$Q_p\$, and \$A_\infty=v\$.