mantisboy1 Posted April 2, 2008 Posted April 2, 2008 Can anyone explain how ferrite beads work without mentioning "impedance?" How exactly does the noise get filtered out? Thanks in advance.
mantisboy1 Posted April 2, 2008 Author Posted April 2, 2008 Ferrite beads as in "a passive electric component used to suppress high frequency noise in electronic circuits." Someone wanted me to explain in terms of Maxwell's equations???
swansont Posted April 2, 2008 Posted April 2, 2008 Ferrite beads as in "a passive electric component used to suppress high frequency noise in electronic circuits." Someone wanted me to explain in terms of Maxwell's equations??? Which of Maxwell's equations would apply here?
SH3RL0CK Posted April 3, 2008 Posted April 3, 2008 Why can I not use "Impedance" as part of the explaination? But I like a challenge. Rather than using equations (which will eventually lead to this verbotten word), think of it intuitively. Think of the ferrite as an inductor. An inductor will build a magnetic field in response to the current flowing through. When the current changes, the magnetic field must also change in this inductor. Now if the inductor were designed to be inefficient (as a ferrite is) there will be energy lost when this magnetic field changes, which will results in less current flow. At low frequencies, the field does not change much and there is little energy lost. At high frequencies, there is more energy lost because the field changes often. Therefore with much energy lost, the current is suppressed. Its quite a bit more complicated than this, but this is the best I can do without going into a great deal of mathematical equations.
miknsk Posted April 4, 2008 Posted April 4, 2008 Think of the ferrite as an inductor. An inductor will build a magnetic field in response to the current flowing through. When the current changes, the magnetic field must also change in this inductor. Now if the inductor were designed to be inefficient (as a ferrite is) there will be energy lost when this magnetic field changes, which will results in less current flow. Energy losses? But if you take usual inductance coil without any ferrite core in it, it would cut-off high-frequency too!
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