Hi,
I went to the internet and used pi filter calculator
Why a Pi Filter at all?
I guess you did not investigate what a Pi filter is and waht it is used for.
So I do your job in this regard:
* a Pi filter is a symmetric filter. It works in both directions equally.
* a PI filter often is used in the power supplies, since it helps to reduce Noise in direction power_supply -> external world, but also the other way round, it hepls to reduce external noise to enter the power supply.
You don´t need the "bidirectional feature". No need to use a Pi filter.
Nor do you want to use the filter in a "power supply path".
--> I don´t see why you decided to use a Pi filter at all.
(The same way I don´t see why you are fixed on a ferrite bead now)
More information on "power supply path".
You may want to install a Pi filter in the mains power input of an SMPS. This filter is used to suppess RF noise in both directions. RF noise is in the Megahertz region (stop band) and mains frequency is in the low Hertz (50Hz, 60Hz) region (pass band). RF may be induced in the power cord by a cellular phone nearby( example). You don´t want to enter the cell phone´s noise into the SMPS and possibly manipulation it´s regulation. AND the PI filter is used to suppress the swithing noise of the SMPS to leave the SMPS and cause EMI noise .. and possible affect electronics devices neraby.
--> So stop band is far from pass band. You have an unknown mains side impedance ... and you have a rather unknown (because variable by load, by rectifier..) impedance at the SMPS side. Sine you can not rely on these impedances you need to install the two capacitors to ensure low impedance at high frequencies .. to make the filter reliable work (independent of the used power cord for example)
****
Now you are fixed (no one knows why. There may be a valid reason or not) on ferrite beads.
Your job now: Find out the function and the use cases for ferrite beads.
****
For your application: I can only recommend to
NOT look for a
solution first.
But
first find out / decide the
requirements and
after that look for a suitable solution.
And this "finding the requirements" starts with defining the "pass band". What frequencies do you need to pass the filter?
****
Example for "finding a solution first":
Let´s imagine your car has a flat tyre.
* YOU go to the garage and buy a new tyre. You come back to the car and see the tyre does not fit the car. You need help in mounting the wrong tyre to your car, with a lot of effort and time consuming.
* OTHER PEOPLE: .. investigate
why the tyre is flat. They see that the valve of the tyre is loose. They fix the problem by tighten the valve and pump up the tyre. done. focussed, cheap, fast.
***
You have cosen Pi filter, now you choose ferrite bead.
Why not do an internet search for "low pass filter OPAMP".
Klaus