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Question about the ferrite beads we use as RF chokes at DC power lines
I want to ask a question about the ferrite beads we use as rf chokes at dc power lines. These chokes should be open circuit for ac signals. However, I coulnd't get this feeling when i looked through the datasheet for chip ferrite beads of muRata.
At resonances the impedance of the choke is in the range of 100-1000 ohm. How can such a small impedance be open circuit to ac signals, I couldn't understand?
Thanks and best regards,
I think this is a misunderstanding. "Open Circuit" means infinite attenuation/suppression of AC/RF signals. This only exists in theory. In practice you will always have a finite amount of AC signal attenuation/suppression.
The higher impedance the ferrite bead has, the larger AC signal suppression.
I also believe you have a neophyte's view of the purpose of a ferite bead.
You think that a ferrite bead is some magical element that single handedly performs like a 11 pole lowpass filter!
It does not.
The purpose of a ferrite bead is two fold:
1) it presents a moderate AC resistance to dampen out any transmission line resonances.
2) it presents a near zero ohm DC resistance to allow the power supply current to get to an active device
The last thing you would want would be for the ferrite to look like an ~open circuit. Lets say you placed two of them space 2" apart along a bias line. All that means is that you will have nearly NO loss for any frequency traveling along the bias line that had a λ/2 = 2 inches! Same with the harmonics of that resonant frequency.
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