Q factor for a low pass filter - Page 1
Forum section: electronics / projects

author: Gibson486
author_rank: Frequent Contributor
timestamp: June 25, 2018, 08:38:19 pm
content: I am designing a low pass filter using a sallen key topology to be used to filter out oscillations from a load cell. I picked my values by setting c1 = c2 and r1 = r2. However, I was told that it is naive to do that. I asked why and he said because it will make a q factor of 0.5. I asked why that was bad, and he could not explain further and just said it is understood. So, I started doing my own research and now I am just confused. It looks like Q factor is more for band pass filters. Also, why would I want a Q value so high? Wouldn't I want energy loss at that break frequency, so a .5 value would suffice? My design works, but if I could improve it, I would like to know how, not just "make it better". Any chance that there is an EEvblog video on this subject?

author: CopperCone
author_rank: Super Contributor
timestamp: June 25, 2018, 08:43:46 pm
content: q does not make sense, order makes sense

author: ahbushnell
author_rank: Frequent Contributor
timestamp: June 25, 2018, 08:52:06 pm
content: I set the r1=r2 and c1=c2 all the time. If you need a special response you can use other values.Sent from my SM-G930V using Tapatalk

author: Benta
author_rank: Super Contributor
timestamp: June 25, 2018, 10:01:12 pm
content: Q is relevant for any filter that's second order and higher.For a low pass filter with low Q you'll have a slow rolloff, with high Q you'll have peaking at the cutoff frequency. The graph shows damping, which is the inverse of Q.BTW, don't use Sallen-Key, MFB filters work much better.

author: ahbushnell
author_rank: Frequent Contributor
timestamp: June 25, 2018, 10:09:14 pm
content: Quote from: Benta on June 25, 2018, 10:01:12 pmWhy are MFB better? What does MFB stand for.Sent from my SM-G930V using Tapatalk

author: Benta
author_rank: Super Contributor
timestamp: June 25, 2018, 10:36:42 pm
content: MFB is the abbreviation for "multiple feedback filter". It's almost as simple as the Sallen-Key, but somewhat harder to calculate.The advantage of the MFB filter is, that with proper design, component sensitivities can be kept at 1 or lower.Component sensitivity means: what happens when a 5% resistor or capacitor is actually 5% off? With a sensitivity of 1, the filter parameters shift 5%. With a sensitivity of 2 it shifts 10%.The Sallen-Key filter is atrocious in this regard with sensitivities higher than 1, it only really works at very low Q and very low gain.Link:http://www.ti.com/lit/an/sloa049b/sloa049b.pdf

author: ahbushnell
author_rank: Frequent Contributor
timestamp: June 25, 2018, 11:03:02 pm
content: ThanksSent from my SM-G930V using Tapatalk

author: Audioguru
author_rank: Super Contributor
timestamp: June 26, 2018, 03:38:03 am
content: Your Sallen-Key filter has equal resistor and capacitor values so the opamp gain determines the Q. With a gain of 1 the Q is low and the response is droopy. With a gain of about 1.6 then the response is Butterworth with a flat level then a sharp cutoff. With a gain higher than 1.6 then there is a peak at the cutoff frequency and if the gain is 3 of more then the circuit oscillates at the cutoff frequency.The ratio of the resistors or capacitors also affects the Q.

author: bson
author_rank: Unknown
timestamp: June 26, 2018, 04:28:05 am
content: More importantly, in a Sallen-Key filter the sum of the resistors is a factor.

author: Gibson486
author_rank: Frequent Contributor
timestamp: June 26, 2018, 12:30:20 pm
content: Quote from: bson on June 26, 2018, 04:28:05 amCan you elaborate?

author: Gibson486
author_rank: Frequent Contributor
timestamp: June 26, 2018, 12:31:23 pm
content: Quote from: Audioguru on June 26, 2018, 03:38:03 amWhat do you mean by droopy?

author: Gibson486
author_rank: Frequent Contributor
timestamp: June 26, 2018, 12:51:27 pm
content: Quote from: Benta on June 25, 2018, 10:36:42 pmOn the MFB, we go into the negative terminal. Does this mean the signal will come out inverted (positive terminal would go to you reference voltage)? This could be a good thing in my case....Also, because of the feedback resistor, I have to be careful with my values to ensure I do not apply too much gain?

author: CopperCone
author_rank: Super Contributor
timestamp: June 26, 2018, 12:53:24 pm
content: how do you describe q for a high order butterworth LPF?i would think applying Q to a LPF is focusing on something usually irrelevant.

author: Benta
author_rank: Super Contributor
timestamp: June 26, 2018, 03:29:52 pm
content: Quote from: Gibson486 on June 26, 2018, 12:51:27 pmYes, the MFB filter inverts.Concerning feedback resistor: if you use the correct formulas, gain, cutoff frequency and Q can be chosen freely.

author: Benta
author_rank: Super Contributor
timestamp: June 26, 2018, 03:36:00 pm
content: Quote from: CopperCone on June 26, 2018, 12:53:24 pmYou are partially right. For most filters, you'd normally target one of the classic transfer functions, Butterworth, Bessel or Chebyshev. For Butterworth and Bessel, the Q is given for the individual second-order sections.For Chebyshev it's a bit different, as you have the option of playing around with passband ripple vs. initial cutoff. This is where you can play with the Q of the sections.And if we move to Cauer/elliptical filters there's alotto play with.

author: Lee Leduc
author_rank: Regular Contributor
timestamp: June 26, 2018, 08:01:10 pm
content: TI application note on Sallen Key filters. Good intro on designing Sallen Key Filters.http://www.ti.com/lit/an/sloa024b/sloa024b.pdfLook at page 10, section 4.4 "Simplification 4: Set Filter Components as equal (R1=R2=R, and C1=C2=C)".

author: Benta
author_rank: Super Contributor
timestamp: June 26, 2018, 08:09:29 pm
content: Quote from: Lee Leduc on June 26, 2018, 08:01:10 pmNoooo!

author: ahbushnell
author_rank: Frequent Contributor
timestamp: June 26, 2018, 10:29:36 pm
content: Quote from: Benta on June 26, 2018, 08:09:29 pmthis makes it clear

author: bson
author_rank: Unknown
timestamp: June 27, 2018, 05:06:17 am
content: Quote from: CopperCone on June 26, 2018, 12:53:24 pmQ=1/sqrt(2) for any 2nd order active Butterworth filter. That's what makes it Butterworth and gives it a flat passband frequency response.QuoteQ will vary for each stage in a cascade. For a Butterworth response in a cascade \$Q=\frac{1}{2\cos(|\theta_n|)}\$ wherethetanis the angle(s) against the real axis of the poles of thenth stage, in the S plane. (For even order filters.) Calculating Q and f for a cascade is a very workable design strategy for simple filters that don't require much else in terms of pole-zero matching or stage ordering. You can log-divide the gain across the stages, or put it all in the first stage (cf. Friis, a good rule of thumb) and rearrange the cascade so the first stage has the lowest Q. In most cases this is simply how the cookie crumbles - for example if you solve for the poles in increasing order of angle.In the simple 2nd order single stage above where the pole angles are +45 and -45 degrees, Q=1/(2*cos(45)) = 0.7071For a 4th order however for a Butterworth response the poles are at +/-22.5 and +/-67.5 degrees. So the stages will have Q's of 0.5412 and 1.3066.I think analog filter design iswaybeyond what's explainable on an online forum however... and there are plenty of textbooks out there.So the answer to your question is the Qs of the stages correspond to roots of a polynomial and form a sequence. ThisQ vectoris what governs the behavior of the filter. (There also afrequency vector, but for a Butterworth series all stages have the same frequency.)

author: b_force
author_rank: Super Contributor
timestamp: June 27, 2018, 01:54:52 pm
content: Q-factor in these kind of cases isn't extremely relevant.Maybe the guy who advised you has an audio background?There it's traditionally common to use these kind of Q-factors (but often wrongly)In general I wouldn't use an high Q-factor (bigger than 0.707), since you will introduce extra energy in your circuit (so to speak).It's a trade-off between bandwidth, stability and how "fast" your system settles.You don't want to much overshoot, but also not a very slow system.But to be very honest, a Q of 0.707 is in most cases just perfectly fine and easy on the component values.Sometimes I choice a little lower because of the tolerances in your passive parts.