2nd Order low pass filter - Why a topology? - Page 1
Forum section: electronics / projects

author: BartK
author_rank: Newbie
timestamp: May 07, 2019, 07:42:49 am
content: Hi all,Although I am new to this forum, I am familiar with electrical designs and such. I am designing a second order low pass filter to filter a 0-10V signal from a sensor. The theory states that there are two possibilities for an active low-pass filter; the Sallen-Key toplogie and the Multiple Feedback topology.However, I wonder why I should use one of these topologies. Can't I just take a second order passive low pass filter and place an opamp buffer behind it?I used LT spice to compare the two and this is what the simulation shows:Can someone explain to me why this is "not" possible? I can't find anything about this. For many of the articles found, an active filter is always made with one of the two topologies mentioned.Thanks in advance for your great help!

author: iMo
author_rank: Super Contributor
timestamp: May 07, 2019, 08:11:29 am
content: Because they are not identical?

author: AndyC_772
author_rank: Super Contributor
timestamp: May 07, 2019, 08:37:20 am
content: You can use whichever topology you prefer; there's no definitively "right" or "wrong" option.Tip: do always take into account the tolerances of your components, and the parasitic elements that you can't eliminate. You might be surprised just how much of an effect they can have on the behaviour of an analogue filter.

author: rs20
author_rank: Super Contributor
timestamp: May 07, 2019, 08:40:40 am
content: As hinted at by imo's somewhat cryptic post, the two filters have a different response. What is generally considered ideal for a 2nd-order filter is aButterworth filter(often implemented using the Sallen-Key topology), which is sharper and generally considered superior to two standard 1st-order filters in series. (Although I suppose it depends on the application as suggested by AndyC)I think your error was assuming that the ideal output of a 2nd-order filter of cutoff frequency f is the same as two 1st order filters in series. But we can actually do better.Notice, in imo's post, how the Butterworth filter (Sallen Key) filter's response stays nice and flat real close to the cutoff frequency, and then quite sharply transitions into the roll-off region? Whereas your proposed topology takes a much more lazy route. More numerically, the amplitude response of a 1st-order filter at the cutoff frequency is -3dB, which means putting two in series would yield a 6dB attenuation at the cutoff frequency. By contrast, the Butterworth filter only attenuates by 3dB at the cutoff.

author: BartK
author_rank: Newbie
timestamp: May 07, 2019, 08:56:05 am
content: Quote from: imo on May 07, 2019, 08:11:29 amI now see that I have accidentally taken the butterworth response for the Sallen-Key topology. I meant to use one of my other models. This was indeed my mistake.The difference that can be seen comes from the other response, namely the quality factor of 0.7071(butterworth). However, if I take a quality factor of .5 for the Sallen-Key, the graph is almost the same.QuoteSee:https://en.wikipedia.org/wiki/Q_factorStil though, this is debating about the response.My question is more about the topologie. In the case of a second-order active low-pass filter, no mention is made of the option of taking a second-order passive filter with a buffer, why not?

author: rs20
author_rank: Super Contributor
timestamp: May 07, 2019, 09:03:38 am
content: Question that I actually don't know the answer to but I suspect might be relevant: is it actuallypossibleto realise a Q of 0.7071 with your topology?Because if not, that would answer your question: people want the best response (e.g. Q=0.7071), so they go for the topology that can deliver it (Sallen Key).The other thing I'd point out is that your topology uses a 1.5 meg resistor, to prevent loading down the previous stage. This is rather big. The Sallen Key topology doesn't require this hack workaround; avoiding big resistors (or, to take it the other way, big capacitors) removes the need to compromise between the Q of the filter, the size of the capacitor, and the noise/offset introduced by the high-valued resistor.

author: AndyC_772
author_rank: Super Contributor
timestamp: May 07, 2019, 09:04:03 am
content: What's the output impedance of whatever it is you're filtering? It might be worth considering how that affects the response in each case.

author: The Electrician
author_rank: Frequent Contributor
timestamp: May 07, 2019, 09:06:56 am
content: The poles of a filter consisting of a cascade of two RC low-pass stages are on the negative real axis; they cannot be complex, so the rolll-off is not very sharp.To get a decent filter response, complex poles are needed, and this requires an active filter, or a passive filter consisting of a RLC topology.

author: BartK
author_rank: Newbie
timestamp: May 07, 2019, 09:14:47 am
content: Quote from: rs20 on May 07, 2019, 09:03:38 amIf I am right, it is indeed not possible to make a passive butterworth filter with only R and C components.See:https://en.wikipedia.org/wiki/Butterworth_filterHowever, because I only want to filter out noise and want to keep the 0-10V DC component, the reaction at the cornerpoint doesn't really matter to me.With regard to the high impedance of the 1.5Meg resistor you are indeed right, I had not considered this yet. Thank you! Although in the Sallen-Key of my 2nd example with the Q of .5, I had to use a 3Meg resistor.Quote from: AndyC_772 on May 07, 2019, 09:04:03 amThis is not entirely clear, because the circuit must be able to accept universal sensors that operate on the 0-10v principle. My assumption is that the impedance of these sensors will be between 10k and 100k.Behind the opamp is a voltage divider with a factor .3 (10V -> 3.0V), so that I can read the voltage with an ADC.

author: AndyC_772
author_rank: Super Contributor
timestamp: May 07, 2019, 09:27:00 am
content: OK, so try modelling your sensor as a voltage source in series with 100k, then connect that to each of your filter designs and see what happens to the response. Repeat with 10k, then 0R.Also try putting 5pF of capacitance across each of your resistors, because it'll be there whether you like it or not.

author: iMo
author_rank: Super Contributor
timestamp: May 07, 2019, 09:47:56 am
content: Except the freq responses, impedance matching, etc., plz consider your requirement 0-10V DC cannot be maintained by single rail opamp, imho.For "DC accurate" low pass filters choppers are used..

author: BartK
author_rank: Newbie
timestamp: May 07, 2019, 09:53:33 am
content: Quote from: imo on May 07, 2019, 09:47:56 amWhy not? My prototype works very well, which trap do you foresee?

author: iMo
author_rank: Super Contributor
timestamp: May 07, 2019, 10:05:40 am
content: Which opapm do you use in your prototype?

author: BartK
author_rank: Newbie
timestamp: May 07, 2019, 10:07:57 am
content: Quote from: imo on May 07, 2019, 10:05:40 amI use the NCS20074DR2G Rail-to-Rail OpAmp.Farnell:https://uk.farnell.com/on-semiconductor/ncs20074dr2g/op-amp-quad-3mhz-2-8v-us-soic/dp/2774608Datasheet:http://www.farnell.com/datasheets/2337856.pdf?_ga=2.49050722.573137400.1557223578-350974861.1556633241

author: iMo
author_rank: Super Contributor
timestamp: May 07, 2019, 10:22:45 am
content: The output does not go to 0V in single rail Vcc.

author: BartK
author_rank: Newbie
timestamp: May 07, 2019, 10:27:59 am
content: Quote from: imo on May 07, 2019, 10:22:45 amAh well, I was aware of this. This is sufficient for the application, the signals that will mainly be used are 1-10V.I had already devised a workaround for this, namely to raise the input voltage with a volt. Then after the opamp I wanted to remove this by reading with ADC. However, this results in a lower resolution with the ADC.To keep the complexity and price of the circuit low, this has not been done.Thank you for your good advice!

author: emece67
author_rank: Frequent Contributor
timestamp: May 07, 2019, 03:37:40 pm
content: .

author: Benta
author_rank: Super Contributor
timestamp: May 07, 2019, 04:28:39 pm
content: It's quite simple: two first order filters in series do not equal a second order filter. Do the maths.

author: emece67
author_rank: Frequent Contributor
timestamp: May 07, 2019, 04:36:42 pm
content: .

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 07, 2019, 05:05:00 pm
content: The maximum Q factor for an RC network is 0.5 -- this limits how sharp the filter transitions from pass to cutoff. A higher Q is possible with passive LC or active RLC filters, giving sharper cutoffs (maximally flat (Butterworth), peaked (Chebychev), etc.), or more precise profiles (Bessel is very gentle, but just not quite as soft as an RC response, and most significantly so for high-order filters).MFB is generally preferred for lower sensitivity to component variations. There are some other topologies which are preferred for high-Q filters, but this is only important in high-order and narrow band-pass/stop filters.Incidentally, you do get one order almost for free -- the fact that the named filter profiles have poles positioned evenly around an ellipse, means that odd-order filters always have one real pole (and the rest as complex conjugate pairs). RC filters make real poles (this is the abstract mathematical reason why LC or gain is required to make a sharp filter), so you can solve for the pairs with active stages and throw in a single RC to finish it up.The topology for that case can be found here (with calculators!):http://sim.okawa-denshi.jp/en/Fkeisan.htmSimply, it's another RC put in front of the usual RCRC filter. The values affect each other, so a different formula applies.Tim

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: May 07, 2019, 05:18:41 pm
content: Hi,Here is a set of models to illustrate the difference.There is loading in the RCRC filter that modify the response from a true double pole filter.Because the second RC is 10x higher impedance than the first RC, the loading is minimal and the deviation from the true double response is about 0.5dB. This is probably acceptable in most applications.I have attached the modelI used 'E' which is a voltage-controlled voltage source, an ideal buffer.Regards,Jay_Diddy_B

author: BartK
author_rank: Newbie
timestamp: May 08, 2019, 07:52:52 am
content: Because I only want to filter out noise on my DC signal, I have taken a cornerfrequency of 1Hz. The frequency response of the filter is therefore not very important to me, but the step response is. For this reason I probably go for a Bessel response, so I would choose Sallen-Key compared to the passive filter + opamp.Improving the Q is not relevant for the application in which this filter is to be used. However, I now understand why the Sallen-Key has certain advantages, certainly when a different response is required.Due to its capabilities and possible configurations, the Sallen-Key topology is superior.Quote from: T3sl4co1l on May 07, 2019, 05:05:00 pmSince I do not use amplification, an MFB has a noice amplification of A+1=2 instead of A=1.Quote from: Jay_Diddy_B on May 07, 2019, 05:18:41 pmThank you for your clear explanation!