Return to Answer

Consider for simplicity the setup with only one capacitor. The main purpose of the second capacitor in the pi setup is to provide low impedance to power at lower frequencies:

Both setups may work. Which is better is governed by capacitor values, their ESLs and the power delivery network downstream.

In the left-hand setup, the PDN should provide low impedance path at lower frequencies. This is the requirement for this setup to work.

The potential advantage of paralleling two capacitors is lower power impedance in a broader range (assuming 0.1 uF and 10 uF cover different frequency ranges). As for the notorious anti-resonance of the two capacitors - look at impedance frequency curves. The situation when it happens is when one capacitor is still capacitor and another one is an inductor. This should not be the case. So, the answer provided by Spehro makes sense as well.

As for the right setup, it may work also. But note that C1 is the only one to provide power when the bead is closed - so its responsibility is huge. The left larger capacitor may not be needed in close proximity (as assumed by the pic I guess). If the bead closes early (say in units of MHz or tens of MHz), then it should provide low impedance path at kHz (or units of MHz) frequencies where location requirements are relaxed (as light wavelength is on the order of tens of meters at these frequencies). But it depends.

Appendix

Below are some general considerations re ferrite beads that might be interesting.

Consider for simplicity the setup with only one capacitor. The main purpose of the second capacitor in the pi setup is to provide low impedance to power at lower frequencies:

I guess, the way the derived the formula was as follows. They assumed reactance of the inductor and the capacitor equal (Lw=1/cw), calculated frequency, expressed Zt in terms of frequency to get the equation. This is not correct in general. First, impedance of a capacitor in general does not equal to 1/Cw, especially at high frequencies where ESL dominates. Second, the impedance of the capacitor should be much (orders of magnitude) smaller than the impedance of the inductor, not just smaller (2x or 3x times smaller wouldn't work).

So, the correct way is to compare the impedance-frequency curves of the capacitor and inductor (accounting for the DC bias used, ideally) and to make sure the impedance of the capacitor is much smaller then the impedance of the inductor where it needs to be. It is not simply some capacitance value needed. The required value of the capacitor's impedance (at some frequency) may be calculated as deltaV/current, where deltaV is an allowable voltage fluctuation and current is the current amplitude at this frequency.

I guess, the way the derived the formula was as follows. They assumed reactance of the inductor and the capacitor equal (Lw=1/cw), calculated frequency, expressed Zt in terms of frequency to get the equation. This is not correct in general. First, impedance of a capacitor in general does not equal to 1/Cw, especially at high frequencies where ESL dominates. Second, the impedance of the capacitor should be much (orders of magnitude) smaller than the impedance of the inductor, not just smaller (2x or 3x times smaller wouldn't work).

The correct way would be to compare the impedance-frequency curves of the capacitor and inductor (accounting for the DC bias used, ideally) and to make sure the impedance of the capacitor is much smaller then the impedance of the inductor where it needs to be. It is not simply some capacitance value needed. The required value of the capacitor's impedance (at some frequency) may be calculated as deltaV/current, where deltaV is an allowable voltage fluctuation and current is the current amplitude at this frequency.

Ferrite beads for power filtering are usually designed as low-q inductors to prevent parasitic resonance. So, DC resistance of ferrite beads is made intentionally high. Often it is about 500 mOhm or even several Ohms. Make sure you can tolerate IR drop given your DC current (say, 10 mA current at 500 mOhm produce 5 mV drop).

Ferrite beads for power filtering are usually designed as low-q inductors to prevent parasitic resonance. So, DC resistance of ferrite beads is made intentionally high. Often it is about 500 mOhm or even several Ohms. Select a bead with an appropriate DC resistance (there are special series for power lines with relatively low DC resistance). Make sure you can tolerate IR drop given your DC current (say, 10 mA current at 500 mOhm produce 5 mV drop).