[SI-LIST] Re: Indcutance role in EMI
- From: "Zabinski, Patrick" <zabinski.patrick@xxxxxxxx>
- To: <si-list@xxxxxxxxxxxxx>
- Date: Wed, 30 Jul 2008 07:19:12 -0500
The key difference is related to odd versus even mode.
Regardless if the component is a ferrite bead or a wound inductor,
the component supresses only the even-mode signal passing through it.
All balanced/odd-mode signal is passed without any degradation,
regardless
of frequency.
Another premise is that EMI is generally related to even-mode signals
and
not odd-mode.
In Case #1 where you're trying to supress EMI, the ferrite bead is
placed around both the signal and its return. Ideally, the signal and
its return are identical in strength and opposite in polarity, under
which case the signal is odd-mode and the even-mode current is zero.
As such, the ferrite bead placed around both signal and ground have no
effect on the odd-mode signal, which is good.
Life being non-ideal, there generally is some amount of even-mode signal
on such cables, and the ferrite beads suppresses it (while passing
through the odd-mode signal), which helps reduce EMI.
In Case #2, the inductor is placed just on the ground and not the signal
path, which presents an unbalanced situation. In this situation, the
the inductor is affecting both odd- and even-mode portions of the
signal,
and will generally increase EMI.
I believe the difference you're getting at is not dependent on the
component being a choke or wound inductor, but it is instead dependent
on how the component is used with respect to odd- and even-mode signals.
Pat Zabinski
Mayo Clinic
> Hi all,
>
> I have doubt regarding inductor functionality in power supplies and
> reducing EMI.
>
>
>
> 1. We use common mode choke or ferrite bead to reduce the high
> frequency noise in the circuits
> 2. But the when inductance present in the ground or at the IC pins
> ( lead inductance) it will cause ground bounce or induced
> voltage due to
> the flow of noise current.
>
>
>
> In the first case we can assume that inductor will act as open circuit
> or very high impendance for the high frequecy noise , so it is nopt
> allowing the high frequency noise.
>
>
>
> In the second case the noise current will flow through the inductance,
> due to the change in the current, V=di/dt voltage get induced, this
> will result in ground bounce.
>
>
>
> My questions are: is my view correct in the both cases or
> not. If it is
> correct, why the inductor is behaving differently for the
> high frequency
> noise in the two cases.
>
>
>
>
>
>
>
>
>
> Thanks& Regards
>
> U.K.UMAMAHESWAR
>
>
>
>
>
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