31.5MHz Band pass filter design - Page 1
Forum section: electronics / projects

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: December 19, 2015, 09:58:39 am
content: Hi,I have used Filterpro from TI in several designs with excellent results, there is a catch about using the tool, it does not support More than 10MHz designs!So I have stocked, I need a tool or way of designing a band pass filter with these specsPass Frequency 31.5MHz with 2MHz bandwidth and 0dBm on a 50 ohm impedance, the harmonic should be -50dB and Spur under -60dB,where should I start?Q

author: German_EE
author_rank: Super Contributor
timestamp: December 19, 2015, 10:21:46 am
content: http://www.raltron.com/cust/tools/band_pass_filters.asphttp://www.wa4dsy.net/filter/filterdesign.htmlThere is also 'Elsie' but you may need to pay for the professional versionhttp://www.tonnesoftware.com/elsie.html

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: December 19, 2015, 12:44:42 pm
content: Thanks for the links,there is no place for entering Spur also the values of RLC are not standard values,do we have better alternatives?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 19, 2015, 10:22:01 pm
content: 1. Use Filterpro, design the filter for 1/10th of all frequencies, then use very fast op-amps and scale all the capacitors down by 10x (to get the frequency back where you want it).Don't expect it to work (fast op-amps like to oscillate).2. Use a passive filter. Of course the values are nonstandard. You'll have to tune it manually, anyway.3. If you need sharper stop bands (I don't know what "spur" means), use an Elliptic design.Tim

author: ali_asadzadeh
author_rank: Super Contributor
timestamp: December 20, 2015, 06:20:44 am
content: Thanks,Spur means "Spurious signals"

author: cdev
author_rank: Super Contributor
timestamp: December 20, 2015, 06:23:24 am
content: RFSIM99 is perfect, its free, and it runs under wine so it does all three platforms. Google for it, the original site is long gone, its abandonware. But it works great.

author: dmills
author_rank: Super Contributor
timestamp: December 20, 2015, 12:23:52 pm
content: Up there, in a 50 ohm network you are probably looking at passive filters anyway (No RF designer would use an opamp here), so spurs are (for the most part) an non issue (As long as the inductors are physically large enough to remain linear, at 0dBm not a huge deal).Elsie can probably do this for you, or just use the tables in the "Handbook of filter synthesis".Regards, Dan.

author: MagicSmoker
author_rank: Super Contributor
timestamp: December 20, 2015, 12:59:25 pm
content: Quote from: T3sl4co1l on December 19, 2015, 10:22:01 pmOops... Decreasing the value of the capacitors alone by 10x only increases the cutoff frequency by 10^0.5 (~3.16), but it also increases the characteristic impedance by the same amount, neither of which is what the OP wants; the value of both the inductors and capacitors have to be changed by the same amount to change the cutoff frequency by that amount without changing the characteristic impedance. Probably just an oversight on your part, I'm guessing, based on the quality of your other posts.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 21, 2015, 01:51:04 pm
content: Quote from: MagicSmoker on December 20, 2015, 12:59:25 pmNot an oversight, a subtlety.I haven't used Filterpro, but at a glance, it appears to be anactivefilter designer. In other words, RC filters plus op-amps. Here, C is the only time dependent component, so frequency is inverse to it.You are absolutely correct, for a passive LC filter: inductance must be reduced by the same amount as well, to keep impedance constant and to achieve the desired frequency. Impedance and frequency are related to inductance and capacitance by:Z = sqrt(L/C)F = 1 / (2*pi*sqrt(L*C))The actual values used in a given filter are chosen proportionally above or below the ideal L, C values, to achieve a desired filter characteristic, with larger ratios (and requiring higher Q factors) for sharper filters.Tim

author: MagicSmoker
author_rank: Super Contributor
timestamp: December 21, 2015, 04:14:55 pm
content: Quote from: T3sl4co1l on December 21, 2015, 01:51:04 pmD'oh; I see what you did there... That said, you have to admit that an op-amp active filter design would not be the first choice for a 31.5MHz bandpass filter with a 2MHz BW and a -50dB/oct rolloff - that's going to be at least an 8-pole filter, depending on desired response.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 22, 2015, 03:10:19 am
content: And you'll be spending the money on the high speed (GBW > 300MHz?) amps, like I said. And like you said, with a high order like that... have fun...(Actually GBW is probably way higher than that, because of the Q factor required to achieve the narrow bandwidth. Plus more to account for the higher coefficients of whatever filter characteristic is desired. Such amps might only be $5 each, but needing 8 of them...)A coupled resonator design would be quite reasonable here. Prepare a set of 31.5MHz resonant tanks (each can be tuned independently), with component Q factor >30. The impedance (Zo = sqrt(L/C)) doesn't matter much, since you'll probably have to match to it anyway. Go for maximum Q instead (which is usually a modest impedance in the 100 to 1000 ohm range). At a system Q of 31.5MHz / 2MHz ~= 15 or more, the signal resistance will be the Q factor above or below Zo, or 30-300 ohms in series, or 1.5k-15k in parallel).Finally, the filter characteristic is implemented by coupling the resonators together. The typical k factor will be 1/Q, or about 0.06. The couplings are symmetric end to end, i.e., the first and last resonators have the same couplings (to the in/out ports, and to their neighbors), and the second-to-(first/last) have equal couplings to their respective neighbors, and all that. All the factors will average to the common factor (by geometric average, i.e., (k12 * k23 * k34 * ...) ^(1/N) = 1/Q), and the amounts by which couplings are above and below that average defines the characteristic. You'd need a filter table, or a suitable transformation equation, to figure out exactly what coefficients are needed.There used to be a Java calculator for this, but sadly, it's been taken down (and, after all, Java died pretty hard in the last few years). And I don't see anything like it online anymore. Shame.You can also couple the resonators by using any means of passing voltage, current or both, between pairs of resonators. If the coils are solenoids, then simple proximity can provide enough coupling (though at this rate, I think they'd have to be too close). They can be coupled by using tapped inductors. The resonant capacitors can be turned into capacitive voltage dividers, and those taps connected. Very large inductors or very small capacitors can be connected in turn, from the top of each resonator to the next. (Note that using any reactive (L or C) method, besides pure electromagnetic coupling, shifts the resonant frequency and changes the slope in the cutoff region (i.e., a filter with all capacitors has poor attenuation at high frequencies).To get an ideal filter, you will need a spectrum analyzer, hopefully with a tracking generator (otherwise, a wideband noise source), and a lot of trimmer adjustments.Those $5/ea op-amps are starting to look kind of tempting, really...Oh, one more thing: if this is for a radio application, an active filter is probably straight out, because each op-amp (and all those resistors) add noise. That can be manageable with a preamp, but then your dynamic range goes down, too. LC filters are worthwhile.Tim

author: rfbroadband
author_rank: Unknown
timestamp: December 24, 2015, 05:08:58 am
content: maybe this helps as a starting point5th Order Chebyshev Type 1Center Freq. :31.5 MHzBW: 2MHzout of band attenuation 60dB at 7 MHz offsetZin, ZOut: 50 OhmRL < -15 dBPlease note the real painful part will be to replace ideal components with realistic components. You will need to find (or create yourself) accurate models that model resonance frequencies, losses, Q, ESR etc of the passive elements. But if you do that you pretty much have a 1st pass design success guaranteed.

author: dmills
author_rank: Super Contributor
timestamp: December 24, 2015, 01:02:37 pm
content: :palm:Those shunt inductor values are ah, optimistic if you want any Q at all.At a minimum you need to apply the standard transforms to get the shunt C down and the shunt L way up, think hundreds of nH not single digits.Yes, the math works, but you try to build that thing!Regards, Dan.

author: G0HZU
author_rank: Super Contributor
timestamp: December 24, 2015, 03:12:06 pm
content: One of the best transforms to use for a filter with a narrow %BW is the SC coupled.This gives control over the choice of inductance value and it gives sensible values for all L and C. it does give a slightly assymetric passband response but the OP didn't give a spec for rejection on the low side...I knocked together a few simulations to show this with a 2, 3 and 4 section design as in the images below. The 560pF caps would ideally need to be decent SMD types with fairly short connections to minimise lead inductance. Otherwise, the bandwidth will go down and the insertion loss will go up.I chose inductors with an unloaded Q of 100 but you can find inductors with higher Qu than this and (plenty of) inductors with Qu lower than this. But the inductors would need to be tunable. If you ended up using tunable inductors with a Qu of 80 then the insertion loss would get slightly worse.This filter topology yields sensible values for L and C and as long as you lay it out sensibly on a proper RF PCB it will behave itself and give results very similar to the simulation. But if you lay it out badly (don't use a breadboard) then it will all turn to mush.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 24, 2015, 03:24:39 pm
content: Basically, for good HF rejection, the ESL of the parallel caps must be less than the parallel L of the standard prototype filter (which was 17nH or whatever). Not hard to achieve with chip caps, microstrip and ground planes, but you can't be careless with it, and definitely can't breadboard, as you say.It's kind of fun because a crystal filter looks much the same way: main difference being the crystal has terminal-to-terminal capacitance (which leads to a poor HF skirt), and terminals-to-case (usually GND'd) capacitance (which must be accounted for in the values of the capacitors around it).Tim

author: G0HZU
author_rank: Super Contributor
timestamp: December 24, 2015, 03:29:11 pm
content: If you want a minimal parts design but with symmetrical stopbands then this transform below works well although your choice of suitable inductors reduces right down... i.e. you would be winding custom coils by hand.I definitely would not recommend this transform for mass production because of the need to wind high performance inductors and also this filter would require several attempts at finding inductors to meet the requirements (probably powdered iron toroids?)

author: G0HZU
author_rank: Super Contributor
timestamp: December 24, 2015, 04:07:24 pm
content: Quote from: T3sl4co1l on December 24, 2015, 03:24:39 pmI've used this shunt C bandpass filter topology numerous times in the past and it works wellThe hairiest version I ever designed was for an alias BPF for a high performance two channel downconverter that used 14 (fourteen) sections in this filter and each filter in each signal path had to be phase matched as a pair. So that meant adjusting all 14 inductors in each filter to align the phase across the passband.

author: G0HZU
author_rank: Super Contributor
timestamp: December 24, 2015, 06:29:18 pm
content: I wound up some toroids and dug out some old school leaded ceramic caps and had a go at building the minimal parts count filter. I also noticed that my simulation was centered on 31MHz and not 31.5MHz so see below for a comparison of the 31.5MHz simulation and the measured filter on a VNA.You can see that the traces overlap really closely but above about 40MHz the stopband performance isn't quite as good on the real filter (green trace) compared to the simulation (red trace). This is because the model I used for the inductor in the simulation was a very crude model. In order to predict the behaviour of the toroid above 40MHz I would need to create a better model that could model it as it approached self resonance.But the model was reasonably goodIt helped to predict the insertion loss correctly. I used an old set of Q curves for powdered iron toroids and I guessed the self capacitance of each inductor to be <=1pF to help produce the model.

author: G0HZU
author_rank: Super Contributor
timestamp: December 24, 2015, 11:09:08 pm
content: I tried entering a 2 port model of a decent Coilcraft SMD inductor for each of the three inductors in this filter and simulated the response.These inductors are non tunable but if you fitted some trimmer caps for C1, C3 and C5 that have a decent Qu then the simulation predicts a reasonable response with a slightly higher insertion loss (red trace). Trimmer caps are usually avoided in my designs but the Coilcraft 1812CS SMD inductors perform reasonably well here when used with trimmer caps.The green trace is the measured response of the filter I built using toroids and this has a lower insertion loss. I left it in the simulation to use as a benchmark.Coilcraft provide fairly accurate two port models of their SMD inductors on their website and I downloaded them here:http://www.coilcraft.com/1812cs.cfmAlthough it looks like I'm using inductors in the schematic below, the inductor symbol actually represents a 2 port data file (eg 18CS222.s2p) that is provided by Coilcraft. This is a 2 port s parameter file that I use as a model for their 2.2uH 1812CS-222 SMD inductor.The design uses a couple of 1812CS-222 (2.2uH) inductors and an 1812CS-272 (2.7uH) in the centre.