Author Topic: Solved: How to convert an LC filter to FDNR form?  (Read 894 times)

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Offline DunckxTopic starter

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Solved: How to convert an LC filter to FDNR form?
« on: March 24, 2021, 08:20:10 pm »
Hi Folks,

I am trying to remember how to convert a 1Hz 1 ohm 3rd order Chebychev T LPF to 10Hz 1kohm in FDNR form.  Three years ago I knew how to do this, and succeeded with an 11th order elliptic, but most irritatingly it appears I have totally lost the plot!

I have used the late AADE's filter design program to get the 1Hz 1 ohm version (see 1Hz 1ohm 3rd order Chebychev LPF AADE.png) and simulating it in MicroCap yields the appropriate response (see 1Hz 1ohm 3rd order Chebychev LPF analysis.png).

The problems begin when I start the conversion process. I am trying to follow the method in the Analog Devices Design book, chapter 8, Analog Filters.  The relevant section is on pages 8.134 - 8.136, attached below as pngs.

In the LC version the two inductors are both 0.195H and the capacitor is 0.183F.

If I want to convert to 10Hz, I calculate the scaling factor as 2*pi*10 = 62.8
I now choose C arbitrarily to be 1uF.  The impedance scaling factor will be (C[old] / C[new]) / 62.8 = (0.183 / 1E-6) / 62.8 = 2921

The two inductors become resistors with the value 0.195*2921 = 571 ohms

Then comes the "gotcha": there's no explanation regarding the conversion from 1 ohm to 1kohm, and I have no idea how to calculate R6,
See 10Hz 1ohm FDNR 3rd order Chebychev LPF circuit.png.  I also suspect that I am missing something else of significance, but after several hours of beating my head against the wall, I am out of ideas and appear to be suffering a severe old git moment. |O


I won't trouble you with simulation results at this point, it doesn't work and at a characteristic impedance of 1 ohm, it would be near-miraculous if it did.

Any ideas?
« Last Edit: March 26, 2021, 03:00:33 pm by Dunckx »
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Offline mawyatt

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Re: How to convert an LC filter to FDNR form?
« Reply #1 on: March 24, 2021, 08:49:59 pm »
The general method for lumped element filters is as impedance increases the capacitors decrease and inductors increase by the impedance ratio. As the frequency goes up then both the inductors and capacitors decrease with the normalized ratio. Note some filter networks in older books are referenced to 1radian/s not 1Hz.

Best,
« Last Edit: March 24, 2021, 08:51:46 pm by mawyatt »
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Offline DunckxTopic starter

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Re: How to convert an LC filter to FDNR form?
« Reply #2 on: March 24, 2021, 09:14:01 pm »
Thanks for that.  I am aware of the trend of the relative sizes of these components as they scale with frequency, and that the literature sometimes uses 2*pi*f instead of f.  What I don't understand is exactly how the transformation is calculated in this case.  I've tried stepping R6 in MicroCap but it doesn't get close to the correct response.  Something more is involved and that's the piece of the puzzle which I am missing.

It's something like R6 = (R4 * unknown?) / (R5 * C1 * C2) but I'm still missing how AD did it in the book.  They talk about scaling from 1 ohm to 1 kohm but then the magic number 1000 is never mentioned again.  It's like something got left out by mistake. ???

Any more for any more?
"God help us, we're in the hands of engineers." - Dr. Ian Malcolm, Jurassic Park
 

Offline mawyatt

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Re: How to convert an LC filter to FDNR form?
« Reply #3 on: March 24, 2021, 10:21:54 pm »
You can solve the network function to see how the values scale. The 3rd Order Butterworth I know because I've used it so much, it's S^3 + 2S^2 + 2S +1 normalized. An old text by Zverev (The Handbook of Filter Synthesis) is a good reference for popular filter transfer functions.

Best,
Curiosity killed the cat, also depleted my wallet!
~Wyatt Labs by Mike~
 
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Offline DunckxTopic starter

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Re: Solved: How to convert an LC filter to FDNR form?
« Reply #4 on: March 26, 2021, 03:42:31 pm »
Hi Folks,

After an hiatus of a few days, I have succeeded in finding a remarkably simple solution to my problem and will post it here for reference, as there isn't much on FDNR filters for the amateur on the web.  Lots for professional engineers but not a lot for hobbyists.  My mistake was trying to be too clever and the equation I was struggling to find was for the D element of an FDNR filter, which it turns out you don't really need!

I will use a simple 3rd order Chebychev low pass filter as an example.  In these circuits, R1 (10Meg) in the simulation is the element which the output is being measured across and is there for the analysis only, it should not appear on the pcb.

First, using software or tables of your choice (I used the Filter Design program by the late AADE) create a passive LC filter of the required order and configuration.  Capacitors should have one end grounded.  Design impedance is 1ohm and frequency 1Hz, see attachment 1Hz 1ohm 3rd order Chebychev LPF AADE.png and if you were to run a simulation it will work just fine.  However, the impedance of one ohm, the frequency of one hertz plus the one farad capacitors is a problem and this circuit is not practical.

To convert this to FDNR form, the circuit shown in 3rd order Chebychev low pass 1Hz 1ohm FDNR circuit.png is used and it is necessary to scale the values.  R3 and R4 can both be changed to 1k (the end result of scaling is shown in 3rd order Chebychev low pass 10Hz 10kohm FDNR circuit.png, see later).  To minimise the capacitances required, decide on a scaling factor for the circuit impedance.  Let's choose 10k.  Divide the 1F capacitances in the 1Hz 1ohm FDNR circuit by 10,000 to get 100uF and divide by another ten to increase the frequency to 10Hz from 1Hz, so these caps are now 10uF.  The resistors R6 and R7 represent the inductors L3 and L5 in the AADE screenshot, so an inductance of 0.1954H multiplied by 10,000 becomes a resistance of 1954ohms.  Finally, resistor R2 represents the AADE capacitor C4, and its value of 0.1834F is also multiplied by 10,000 to give R2 = 1834ohms.  This completed circuit is shown in 3rd order Chebychev low pass 10Hz 10kohm FDNR circuit.png  In practice, the 1954ohm resistors R6 and R7 would be 2k and R2 would be 1k8, and there would be no significant difference in frequency response.  The series input capacitor and shunt output capacitor (terminators) are equivalents to the original 1ohm input and output resistors, scaled to 10uF, and are essential for correct operation.

The 1Meg resistors connected across the capacitors are there to allow for the input bias and offset currents of the op amps.  Without these, the outputs would tend to go to one or other of the power rails and the filter would not work.  As their value is 100x the circuit impedance, their effect on performance is negligible.

For MicroCap users, the source file is attached, 3rd order Chebychev FDNR 10Hz 10kohm.txt, you will need to modify the txt extension to .cir to get it to run.

The frequency response is shown in 3rd order Chebychev low pass 10Hz 10kohm FDNR ac analysis.png and is commendably flat in the passband :)

The noise analysis results are shown in 3rd order Chebychev low pass 10Hz 10kohm FDNR noise analysis.png and evidently there is nothing to worry about regarding the noise performance.  It makes very little difference what op amp you use here, a low bias and offset current JFET input op amp like TL081 may even be preferable to a low noise chip like NE5532, but the op amp must be unity gain stable, which the NE5532 is and the NE5534 isn't, so choose wisely.

I hope that this spares someone the frustration and futility I have just been through!
« Last Edit: March 26, 2021, 03:51:32 pm by Dunckx »
"God help us, we're in the hands of engineers." - Dr. Ian Malcolm, Jurassic Park
 


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