is there a way I can get a "sharper" cut-off/slope from a  passive Low Pass fil. - Page 1
Forum section: electronics / beginners

author: Deactivated-1
author_rank: Frequent Contributor
timestamp: January 18, 2021, 04:22:05 pm
content: I want to have a sharper cutoff/slope/rolloff from a low pass filter, but I want it to be passive. is there any trickery or anything that can achieve this or do I need to go active if I want a sharp ... cutoff or whatever it's called?

author: TimFox
author_rank: Super Contributor
timestamp: January 18, 2021, 04:27:37 pm
content: There is a large literature concerning passive filters employing R, L, and C in higher-order filter configurations. “Elliptic” filters have sharp cut-off slopes, but suffer from some responses in the “stop band” above the cut-off (low-pass filter). They are often used after the DAC in DDS synthesizers.

author: emece67
author_rank: Frequent Contributor
timestamp: January 18, 2021, 05:03:32 pm
content: .

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 18, 2021, 05:05:01 pm
content: Quote from: ELS122 on January 18, 2021, 04:22:05 pmThere are no tricks. Building a passive filter with a sharp cutoff would be a challenge. You are better off going active.

author: TimFox
author_rank: Super Contributor
timestamp: January 18, 2021, 05:07:36 pm
content: Note that “Cauer” and “elliptic” filters are synonymous. These complex passive filters are normally used at high frequencies, due to the required inductors.

author: Kleinstein
author_rank: Super Contributor
timestamp: January 18, 2021, 05:59:50 pm
content: A first step is to use LC (with R) filters instead of just RC filtering. You can't get the steeper filters like Butterworth or Tschebyscheff with only passive RC - though it is possible to get a simple (but not a good one) passive elliptic filter with only RC.For higher frequencies ( e.g. > 1 MHz) LC filters are very viable, while active filters get increasingly difficult (need faster OP). For low frequencies LC gets tricky and active filters get more viable.The 100 kHz to 1 MHz range can use both, but neither solution is easy.

author: Terry Bites
author_rank: Super Contributor
timestamp: January 18, 2021, 06:35:05 pm
content: See if you can get hold of "Newhertz Technologies Filter Solutions". Simple fast filter design software.Fun, helpful and educational!

author: TimFox
author_rank: Super Contributor
timestamp: January 18, 2021, 06:49:08 pm
content: Active filters are great for the audio range, and RLC filters for RF range. With low-pass filters, the attenuation in the step band can be compromised by low high-frequency open-loop gain for op-amp active filters, and parasitic capacitance (etc.) in the inductors for passive filters. These are all quantitative questions. If you know the op amp response and all parasitic, Spice simulation can be used to check the design calculations, since the problem is linear and .AC analysis is accurate.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 18, 2021, 06:52:01 pm
content: - An RC filter is very soft, a single (real) pole. The roll-off is gradual, and the asymptote is -20dB/dec.- You can chain RCs, but the roll-off gets even softer. An RCRC filter (equal values) has two poles, but the effect of each RC on the other causes the poles to spread apart, by a factor of 3 or so IIRC. Effectively it's a cutoff at Fc then another cutoff at 3Fc. So the roll-off is very gradual indeed. You eventually get a -40dB/dec asymptote.- You can mitigate the interaction effect, by scaling the values geometrically. You might do an R1-C1-R2-C2 filter with R2 = 10*R1 and C2 = 0.1*C1. This puts the poles within about 10% of each other, almost as sharp as a single RC but twice as effective.- This doesn't take you very far of course, as three stages you need 100x the values, etc.- Aside: you can actually get voltage gain from a passive (RC) filter. The trick is to use phase shift to your advantage (it is a non-minimum phase filter). The maximum gain per stage is very small, a few percent. Just a neat curiosity.- Active filters use gain elements to raise the Q factor of RC networks, or transform them into other impedances (negative resistance, inductance, etc.). This is also an aside, as you've specified passive.- And finally, that leaves inductors. An LC filter can have voltage or current gain (just not power gain), can have complex poles (giving sharper cutoffs, peaked or ringing response, etc.), but requires impedance matching for consistent response -- the network is part of a power transfer system, and that transfer is completely dependent on the source and load impedances.You can also mix them, so you have RLC filters with responses somewhere inbetween; some power is lost in the filter, and the cutoff may be softer, but maybe it's more tolerant of impedance mismatch. Resistance of course is unavoidable in any filter*, but we generally design LC filters with low enough component losses (high Q inductors, capacitors) that the insertion loss is usefully low.*Yes even superconducting filters! There are still loss mechanisms even in otherwise-ideal conductors. The trick is, superconductors are fine at DC, but at AC, all bets are off. In the extreme, consider a puck of YBCO: it's black, and doesn't change color with temperature. It's not superconducting at optical frequencies at least. It actually follows that, somewhere between DC and light, there is a cutoff where resistance takes over, and indeed there's an obvious quantum-mechanical reason for that. They can be quite excellent though: Q factors of 10^7 or so, better than most quartz crystals. Optical resonators too (built from dielectric supermirrors).Tim

author: Benta
author_rank: Super Contributor
timestamp: January 18, 2021, 08:26:58 pm
content: For higher orderpassivefilters (= sharper cut off), you need both inductors and capacitors.There's no way around it, and the maths concerning this subject have been around for centuries.

author: srb1954
author_rank: Super Contributor
timestamp: January 19, 2021, 04:36:26 am
content: Quote from: Wimberleytech on January 18, 2021, 05:05:01 pmThis is incorrect. It is usually easier to design a sharp cut-off passive filter than an active filter.Passive filters can be easily synthesised using classic design techniques or from filter design tables and the resulting filter designs are less sensitive to component tolerances than in active filters.Active filters are more difficult to design, particularly for high performance filters with sharp cut-offs, as there are fewer standard design techniques for these. The resulting designs can be quite sensitive to component tolerances and active component performance. In the case of very high performance active filters trimming of components is often required to meet performance targets.

author: ejeffrey
author_rank: Super Contributor
timestamp: January 19, 2021, 04:51:16 am
content: Quote from: Benta on January 18, 2021, 08:26:58 pmYou can make high order passive RC filters with no inductors. The component values get pretty annoying, but it certainly works. High order != sharp cutoff. High order refers to the number of poles and the ultimate slope of the rolloff. For a sharp transition region you need complex pole pairs which requires either inductors or active filters.You can also make buffered passive filters where you use a passive RC for each stage with a buffer amplifier between them. It's not what I would generally call an active filter, but it lets you make a multi-stage filter without the component values getting silly.

author: RandallMcRee
author_rank: Frequent Contributor
timestamp: January 19, 2021, 04:55:14 am
content: Quote from: ELS122 on January 18, 2021, 04:22:05 pmA sensible answer requires as input the frequency Fc (-3dB point) and the required steepness.Get out yee old filter calculator and do some checking! Lots of tradeoffs.E.g. Marchand electronics sells passive filters for audio purposes you can get 24dB slopes at 80 Hz.https://www.marchandelec.com/xm46.htmlThey are expensive.

author: srb1954
author_rank: Super Contributor
timestamp: January 19, 2021, 05:01:48 am
content: Quote from: ELS122 on January 18, 2021, 04:22:05 pmThe first step is to list the performance requirements and consult a standard filter design textbook to determine the required filter configuration (Butterworth/Chebyshev/Elliptic ...) and the number of poles (basically the number of reactive components in the circuit) required to meet the performance specifications.The next step is to determine whether to go passive or active. High performance passive filters require inductors but are easily designed using standard synthesis techniques or from tables in a filter design text. Passive filters are usually more appropriate for higher frequencies where the inductor and capacitor sizes are more manageable.Where you don't want to use inductors or for lower frequency filters where inductors become prohibitively large active filters are another option. The overall filter response has to be split up into multiple stages for synthesis into standard active filter building blocks, which usually only handle 2 poles per active stage. Standard design techniques such as Sallen & Key and Multiple Feedback exist for these active filter building blocks but these are generally only usable for lower performance filters with Butterworth or Chebyshev response. More elaborate active filter blocks and design techniques are required for producing active elliptic filters.

author: Terry Bites
author_rank: Super Contributor
timestamp: January 19, 2021, 08:27:42 am
content: What is your cut off frequency (-3dB) point?what slope do you need db/decade or db/octave?What is the maximum attenuation you need?Is your signal pulsed or continuous?Is this an AF or RF filter?

author: Benta
author_rank: Super Contributor
timestamp: January 19, 2021, 06:28:59 pm
content: Quote from: ejeffrey on January 19, 2021, 04:51:16 amPlease show us how, with the associated topologies and transfer functions.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 19, 2021, 06:54:44 pm
content: Order is simply the number of poles. Not how sharp they are.RC filters are all real poles (or zeroes). Soft cutoff, same asymptote.Tim

author: Benta
author_rank: Super Contributor
timestamp: January 19, 2021, 07:09:28 pm
content: Quote from: T3sl4co1l on January 19, 2021, 06:54:44 pmThe problem is, that a higher number of real poles does not result in a higher order filter. They can be lumped for a final cutoff slope, which will still be -20 dB/decade = 1st order filter.Like a ski slope that undulates a bit, but the final slop is still the same.@ejeffrey: please show us your n-order RC-only passive filter.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 19, 2021, 08:15:36 pm
content: I'm not sure how long it's been since you solved a transfer function? It's not hard to show, good homework, do recommend.Example sim attached.Tim

author: TimFox
author_rank: Super Contributor
timestamp: January 19, 2021, 08:20:56 pm
content: The "trick" for the example above is that the resistors increase by a large factor from input end to output end, with capacitors scaling in the opposite direction for equal time constants.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 19, 2021, 08:32:30 pm
content: Quote from: TimFox on January 19, 2021, 08:20:56 pmAs I described above -- the geometric series is only used to keep the poles close; the asymptote is identical in any case.Tim

author: TimFox
author_rank: Super Contributor
timestamp: January 19, 2021, 08:46:50 pm
content: Yes. The poster was asking for "trickery" to obtain the "sharper slope".

author: Benta
author_rank: Super Contributor
timestamp: January 19, 2021, 10:36:22 pm
content: Quote from: T3sl4co1l on January 19, 2021, 08:15:36 pmConceeded.You've demonstrated a triple first-order filter. Bravo! I don't want to bicker about naming conventions here.Your professorial tone about homework is not nice, though. I'm 60+ and have solved plenty of transfer functions, but mostly in audio/video, where somewhat more advanced filters than your three resistors and caps are needed.Cheers.

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 19, 2021, 11:11:45 pm
content: Quote from: T3sl4co1l on January 19, 2021, 08:15:36 pmJust for the sake of argument, and I know you are a smart guy (I read your posts), would you truly characterize a three-pole real-axis filter as one having "sharp" cutoff?

author: RoGeorge
author_rank: Super Contributor
timestamp: January 19, 2021, 11:14:13 pm
content: The one posted byT3sl4co1lis a 3rd order filter. Three 1st order filters chained as shown in the above schematic will make a 3rd order filter. Also the slope being 60dB/decade tells that's a 3rd order filter indeed.Never heard of a triple first-order filter before.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 20, 2021, 12:27:11 am
content: The confusion seems to be on terminology:Order refers to the number of poles.It doesn't matter where they are (as long as they are finite). A filter with a pole each at 1kHz and 1GHz, is still a two-pole filter.Sharpness is due to the distribution of poles.In particular, poles should cluster together, typically zero or one real poles and the rest as complex pairs, and typically in some convenient geometric distribution (IIRC, Bessel lie on an ellipse, Butterworth on a circle centered on the origin, Chebyshev on a circle offset right of the origin?), but there is no necessity for this to be the case. Indeed, practical filters might have a multitude of intentional and parasitic poles and zeroes, and perhaps additional poles added to smooth them out (for example, a filter that needs good attenuation over many decades, might have "clean up" stages with poles deep in the stopband to anticipate those zeroes)."Homework" is... perhaps insensitive, granted. But these definitions are readily found in the textbook, and easily seen when working equations (in simplified rational form, the order of the denominator polynomial is the order of the filter, and its roots are the filter's poles). It's an invitation to refresh rusty knowledge and practice -- I'm half tempted to do the homework myself, honestly, I could probably use it. I've done the 2nd order (RC) case before, and that was many years ago. The 3rd order case is sure to be a mess, but in there, shall be found a cubic polynomial all the same.Tim

author: The Electrician
author_rank: Frequent Contributor
timestamp: January 20, 2021, 12:44:29 am
content: [ Specified attachment is not available ]Here is a plot of the frequency response using the R and C values in Reply #19 (blue curve) and with all the R and C values equal to the middle values in Reply #19 (red curve). Scaling the R and C values helps, but the blue curve is about as sharp as you can get with an RC low pass, no matter how many stages are cascaded, because cascading more stages only makes it less sharp.

author: Benta
author_rank: Super Contributor
timestamp: January 20, 2021, 01:07:42 am
content: Teslacoil, you're cheating a bit here.Your filter is three RC stages with impedance stepping. A real circuit would be be three RC dividers with opamp buffers in between.But you're right. It's a terminology question.Is stacking three 1st order filter responses on top of each other a 3rd order response? Put on your professor hat, raise your index finger and chastise me.

author: The Electrician
author_rank: Frequent Contributor
timestamp: January 20, 2021, 02:11:20 am
content: Quote from: Benta on January 20, 2021, 01:07:42 amWhy is the circuit shown in reply #19 not "real"?Terminology is important in the hard sciences. I don't understand what is meant by "stacking" filter responses, but I do understand what is meant by "cascading" filter responses; is that what you mean?Here is the response of 3 RC lowpasses with (ideal) opamp buffers in between (red curve) and the circuit in reply #19 (blue curve). It's hardly worth adding buffers versus scaling impedance levels:

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 20, 2021, 02:43:02 am
content: IIRC, the pole splitting, for a factor of 10 impedance scaling, is around 20%. So, roughly speaking, the cutoff will be 10% softer than for the buffered case (which approaches a repeated pole; ideally equals, but the hedging language allows for considering nonideal opamps). The all-equal case I think is close to 250% worse, at least for two poles (as you can see above, it's... not very good!). Dunno for 3+.Hrm, I'm not sure that you'd ever use the buffered case, actually? As long as you're putting gain into the system, you might as well make some complex poles. And make it odd order, because you get the real pole for free (passive).Might be some special applications where repeated real poles are desirable, in which case of course that'd be the way to go.Tim