PI LC filtering and ferrite bead

[EDA_hg81]

I am designing power supply for a 16bit ADC.

I used PI LC filtering to reduce power supply ripple at the switching regulator side.

My question is: should I still need to connect ferrite beads to analog power pins at the ADC side?

If I use both of them, is there any potential interference?

Do you think the ferrite bead connected with caps make another PI LC filter?

Thanks.

 

[FvM]

Do you think the ferrite bead connected with caps make another PI LC filter?

Yes. A ferrite bead is a lossy inductor, but below a few 10 MHz, it's purely inductive. You can calculate an inductance from the impedance characteristic.

should I still need to connect ferrite beads to analog power pins at the ADC side?

Possibly yes. It's purpose is different, mainly to block the on board generated power plane noise.

If I use both of them, is there any potential interference?

I won't expect it. LC supply filters have generally a certain risk to create ringing with current steps. An ADC can be expected to have allmost constant current consumption, also a ferrite bead filter is typically low Q because of the low inductance and respective low filter impedance. LC filter ringing may be rather a problem at the power regulator side.

 

[EDA_hg81]

Thank you FVM.

I will use both of them.


I used on line calculator to find out the attenuation dB.

http://www.pentodepress.com/calculator/LC-ripple-filter.html

inductor is 1uH and cap is 10uF and ripple frequency 1.2M.

why I can get -55.0792attenuation dB result?

 

[flatulent]

The problem with all filter calculations is knowing the source and load impedance. If you have the wrong value, your calculations are off. For instance, if you have a good power supply, the source impedance is near zero. If you have an additional bypass capacitor on the load, the load impedance is near zero.

Most simple attenuation calculators assume 50 ohm source and load.

 

[FvM]

why I can get -55.0792attenuation dB result?

It's the attenuation of an ideal 2nd order low pass. It's easily calculated with a pocket calculator, but you don't have an ideal filter. The capacitor has a series inductance and the inductor has a parallel capacitance. Furthermore you have no ideal ground connection on your circuit.

It's not generally a problem to achieve > 40 dB attenuation at 1 MHz for a particular power supply node, at a particular point in the circuit. But this won't block other existing interference pathes in your circuit.