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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.

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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.

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Calculating: R4 = 2 - α = .586 ohms

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

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

enter image description here

The frequency response looks like this:

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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?

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  • \$\begingroup\$ Should R3 and 4 calculations result in kΩ rather than Ω? It's "dB" - not "Db". 'd' for deci and 'B' for bel. \$\endgroup\$ – Transistor Dec 28 '17 at 13:18
  • \$\begingroup\$ R3 and R4 calculation result is simply the ratio so units do not matter. They are in ohms since 2 - 1.414 is 0.586 Ω and not 586 Ω or 0.586 kΩ. In the circuit I provided the units are kΩ as you suggest. \$\endgroup\$ – Rrz0 Dec 28 '17 at 13:21
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    \$\begingroup\$ They'll matter if you try to drive them with an op-amp. As long as you are aware that's fine. \$\endgroup\$ – Transistor Dec 28 '17 at 13:24
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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).

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In the book they describe two types of Sallen-Key Topology filters, one is the "unit-gain"-version which as the name suggests has Av=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:

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"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:

schematic

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 Av=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;

enter image description here

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.

enter image description here

Sorry I know the pictures are hard to make out.

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  • \$\begingroup\$ Vinzent - I dont agree with your conclusion that we "cant without changing...the circuit" realize Butterworth with a gain of two. Of course, we can (see my conntribution). \$\endgroup\$ – LvW Jan 6 '18 at 17:08
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    \$\begingroup\$ @LvW I might be wrong in that conclusion, I don't know enough of the math behind to follow your example I just based my conclusion on the comment in the book. However I know that you can make the output have any gain you want by doing as I described in the end of my answer, without changing the filter itself. \$\endgroup\$ – Vinzent Jan 6 '18 at 17:17
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    \$\begingroup\$ @Vinzent, in the circuit you provided above, shouldn't the capacitors be going in the positive rail of the op-amp? \$\endgroup\$ – Rrz0 Jan 7 '18 at 6:25
  • \$\begingroup\$ (3-α)(R3+R4)/R4 does not seem to hold true on simulation since i achieved a perfect result with R£ and R4 being both equal to give in parallel 7k (14k and 14k) \$\endgroup\$ – Rrz0 Jan 7 '18 at 10:24
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    \$\begingroup\$ The (3-a) part is the original gain of the equal-component-version remember, the gain of the unit-gain version is of course 1, so the equation for the gain of the unit-gain version (compensated using my method) is of course not (3-a)*(r3+r4)/r4 but rather 1(ONE)*(r3+r4)/r4 which is indeed =2 if you make r3=r4. Otherwise your calculations are correct. Basically just take the 7kohm resistor in your unit-gain circuit and replace it with a voltage divider consisting of 2*14kohm and the same with the negative-feedback resistor (the 14kohm one) and you should be good to go \$\endgroup\$ – Vinzent Jan 7 '18 at 12:48
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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\$.

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  • \$\begingroup\$ Many thanks for your answer. I have an issue with R3 = R4, since this is the unity gain version without utilizing the butter worth coefficients. Is this still considered a Butterworth response? \$\endgroup\$ – Rrz0 Jan 6 '18 at 15:00
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    \$\begingroup\$ For Butterworth we have wp=wc (3 dB) - that means: b1=1. More than that, the pole Q Qp=0.7071 contains the value of a1 because Qp=1/a1. That means: The equations contain, of course, the coefficients - however, in another form, which is more convenient. \$\endgroup\$ – LvW Jan 6 '18 at 15:34
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    \$\begingroup\$ OK - I will edit my detailed answer. \$\endgroup\$ – LvW Jan 6 '18 at 15:38
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    \$\begingroup\$ The frequency wc is the 3dB-corner frequency which is related to the pole frequecy wp by the factor b1: b1=wc/wp. And for the Butterworth response we have b1=1 (as mentioned before). \$\endgroup\$ – LvW Jan 6 '18 at 16:10
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    \$\begingroup\$ Okay sure might only apply to the "equal-componenet"-case but that was the one he had used. Besides the book only talkes about that and the "unit-gain"-version. But you are probably right that you can get gain of 2 with a different configuration \$\endgroup\$ – Vinzent Jan 6 '18 at 17:25

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