Dimensioning variable cutoff 2nd order filters for audio signals

Thread Starter

Hamalz

Joined Oct 14, 2018
13
Hi everyone. I'm creating this thread to discuss another issue I encountered on the synth project I was talking about here.
I want to design a low-pass and a high-pass filter that I will ad in series to my oscillator. Both filters will have variable cutoff frequency but I'm not sure that using potentiometers will be enough. This project is my bachelor degree thesis, so I cannot use relatively complex designs, and since my knowledge is limited to 2nd order filters, they will be my choice.
Since I don't want any resonance peak and I want to limit ripple as much as I can I decided to go with Butterworth filters using a Sallen-Key design. The design I came up with for the low-pass filter is attached to this... What do you think? This lowpass filter design tool seems to prove me right, but I trust experienced people more than I trust calculators.

Regarding the high-pass filter, I thought I could just reverse the components of the low-pass filter: this works in theory, but I'm not sure why not one design tool suggests that configuration, that I attached down here anyways. Any suggestion?

Texas Instrument filter design tool suggests the OPA2345... Isn't it a bit of an overkill for audio?

LOW-PASS.jpg HIGH-PASS.jpg
 
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danadak

Joined Mar 10, 2018
4,057
For audio work the OPA2345 seems overkill, its GBW >> what you need. And
wideband amps take very careful layout and design to keep from oscillating.
An opamp in range of 1 - 5 Mhz GBW probably more than adequate.

I am making an assumption this is audio range work ?

Not assuming what is the range of freqs you want ?

Regards, Dana.
 
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Thread Starter

Hamalz

Joined Oct 14, 2018
13
Thank you for the reply.
Yes, I'm working with audio signals (20 to 20k Hz) and cutoff frequencies range from about 150 Hz to 15 kHz.
What do you think is better between TLV2462 and TLC072? The TLC072 seems to be cheaper and have better specs, is it possible?
Also, what type of capacitor do you think I should use?
 

Veracohr

Joined Jan 3, 2011
752
Regarding the high-pass filter, I thought I could just reverse the components of the low-pass filter: this works in theory, but I'm not sure why not one design tool suggests that configuration, that I attached down here anyways. Any suggestion?
It's because of the relative impedances that are required to get the Butterworth response. They're not equal at the cutoff frequency. With a simple RC filter you can flip the components around and get the same cutoff in high pass or lowpass, but it doesn't give a Butterworth response.

By the way, are you aware that you can use equal value capacitors if you have some gain in the opamp?
 

Thread Starter

Hamalz

Joined Oct 14, 2018
13
Thanks everyone,
It's because of the relative impedances that are required to get the Butterworth response. They're not equal at the cutoff frequency.
So I need equal value caps? I think I can solve the problem by using digipots instead of a dual potentiometer, which forces me to keep two equal resistor values. I'm not sure I understand why I can't have equal resistor values on the highpass filter however...
By the way, are you aware that you can use equal value capacitors if you have some gain in the opamp?
I didn't know that... I thought the gain only depended on the negative feedback, could you be more specific please?
 

Thread Starter

Hamalz

Joined Oct 14, 2018
13
This explains it much more thoroughly than I can:

http://www.ti.com/lit/an/sloa024b/sloa024b.pdf
I see what you mean, however by reading this paper and the paper you linked I didn't find any reason why swapping resisors and capacitors wouldn't work. I see the only reason to choose two equal capacitors in a high pass filter is that it's hard to find strange capacity values... Please tell me if I missed something.

I think I came up with a solution: using a 2 channel digipot in place of R1 and R2. This way I can change the cutoff frequency mantaining R1/R2 constant and therefore mantaining Q constant even if C1≠C2. Choosing two equal caps would significantly increase the difference between R1 and R2, reducing the variation range of fc (e. g. reaching low cutoff frequencies would require resistance values that are too high). I can increase C1/C2 by choosing easily available capacitance values until the range of R2 and R2 is between 1-100 kOhm and then use a digipot to change the cutoff.
Please feel free to correct any mistake I may have done.
 

AnalogKid

Joined Aug 1, 2013
9,261
Working from memory, I think a SK Butterworth filter topology with a forward gain of 2 has equal value resistors *and capacitors*. This means the corner frequency will track with a tight-tolerance dual-gang pot for the variable resistors. This is not the case with unity gain circuits in post #1. Beware, the gain-of-two circuit might have slight peaking at the corner freq, but it is my fav circuit for this kind of application because it is so easy to tune.

ak
 

Audioguru

Joined Dec 20, 2007
11,249
The gain for an equal R and C Sallen-Key second order Butterworth filter is 1.6 times not 2 times. Editted

Why are you making a variable lowpass and highpass audio filter? A 2nd-order does not have a not sharp cutoff and will simply make sounds muffled or tinny.
 
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Veracohr

Joined Jan 3, 2011
752
I see what you mean, however by reading this paper and the paper you linked I didn't find any reason why swapping resisors and capacitors wouldn't work. I see the only reason to choose two equal capacitors in a high pass filter is that it's hard to find strange capacity values... Please tell me if I missed something.
Think of it in terms of impedance instead of part value. Your original low pass circuit, with 100k resistors, has a cutoff of 164.7Hz. To transform it to a high pass, replace the resistors with capacitors that have 100k impedance at that frequency, and replace the capacitors with resistors that have their values of impedance at the cutoff frequency. Like this:

Screen Shot 2018-10-25 at 8.24.39 PM.png


The two circuits have the same cutoff:

Screen Shot 2018-10-25 at 8.32.38 PM.png


The problem, of course, is that the values are unreasonable. But that's why you can't just swap the parts to transform it. The impedances aren't the same at the cutoff frequency.
 

Audioguru

Joined Dec 20, 2007
11,249
The filter slope is still a second order Butterworth but the RC values are equal if the gain is 1.6 times:

EDIT: I notice that the slope DID change a little maybe because my gain is not exactly what is needed for a Butterworth slope.
 

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Thread Starter

Hamalz

Joined Oct 14, 2018
13
Provided that I introduce a gain of 1.6, I will have to attenuate the output signal afterwards: is a voltage divider enough?
Is it ok if I connect the two filter in series as they are?
 

Audioguru

Joined Dec 20, 2007
11,249
A voltage divider reduces a signal voltage if the load resistance is much higher than the resistances.
Instead of having a gain of 1.6 then attenuating it, use a gain of 1 and use two parallel capacitors for a 2nd-order Butterworth lowpass and use two parallel resistors for a 2nd-order Butterworth highpass, as I showed in my simulation.

Two Butterworth filters in series do not produce a Butterworth output, the output is droopy above and below the cutoff frequency and the cutoff frequency is -6dB instead of -3dB. You must lookup the design for the parts of a 4th-order Butterworth filter.
 

Thread Starter

Hamalz

Joined Oct 14, 2018
13
Sorry if I wasn't specific: I want to put a low pass and a high pass filter in series, I don't want to obtain a 4th-order low pass but a band pass. Is it possible? It's just curiosity, thinking about it, I don't think it will be useful to implement such a circuit since I'll implement adjustable cutoff frequencies and if they exceed or overlap each other I'll end up with nothing good...

use a gain of 1 and use two parallel capacitors for a 2nd-order Butterworth lowpass and use two parallel resistors for a 2nd-order Butterworth highpass, as I showed in my simulation.
Thank you, you really made things much simpler!
 
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