Where does noise filtered from ferrite core go?

Discussion in 'General Electronics Chat' started by Dong-gyu Jang, Jul 29, 2015.

  1. Dong-gyu Jang

    Thread Starter Member

    Jun 26, 2015
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    Hello.

    Let's take a picture that live and neutral wires wrap around ferrite core several times for common-mode noise filtering.

    As fas as I know high frequency noise is filtered by ferrite core in inductive way, which means noise is stored in form of magnetic field such as normal inductor.

    But ultimate aim is not to store noise but eliminate it.

    Eventually, where does the noise trapped within ferrite core go? Is it reflected back to original noise source or released to wires for a relatively long time thus noise "amplitude" falls acceptable range?
     
  2. Brevor

    Active Member

    Apr 9, 2011
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    It is lost within the ferrite as heat.
     
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  3. Dong-gyu Jang

    Thread Starter Member

    Jun 26, 2015
    100
    4
    Hello.

    Do you mean ferrite core has resistance thus eddy current induced by high frequency noise on the core will be dissipated as a heat? Can you tell me typical resistance of ferrite core? Altough I guess it is frequency dependent..
     
  4. crutschow

    Expert

    Mar 14, 2008
    13,028
    3,238
    The inductor has a high impedance to the high frequency noise, so the noise is simply reduced by this impedance along with the usual capacitor to ground at its output, which has a low impedance to ground for the noise.
    The noise doesn't "go" anywhere except for the small amount that goes through the inductive impedance and capacitor to ground.
     
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  5. RichardO

    Well-Known Member

    May 4, 2013
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    It can be both what Bevor and crutschow say. Some "EMI" core material is conductive and both acts as an inductor and converts the EMI to heat.
     
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  6. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    I can add a little here.

    For a signal that is somewhat high in frequency, lets say short DC pulse, it will cause a small increase in inductor current which will be stored temporarily as a higher magnetic field, and if there is a capacitor to ground then the current will also flow through the cap which will cause the cap voltage to rise slightly, and so the cap stores some of the energy of the pulse also. This would be the case for lossless components. But real noise is considered random which means that the area above zero is equal to the area below zero, so simply put it is like we have a positive pulse followed by a negative pulse of the same amplitude and duration. Since energy is stored when the positive pulse occurs, the energy is taken away when the negative pulse occurs, and thus the output only changes slightly up and then slightly back down. Since the input source caused energy storage during hte positive pulse it must have taken back the energy during the negative pulse. If we had a load however, some of the energy would have ended up in the load as random noise but of lower amplitude.

    For an inductor with loss, which is what most real inductors have, some of the energy could get dissipated as heat in the core or in the wire. The process of flipping domains will eat up some energy, and the eddy currents will eat some energy, and the wire will eat some energy, and not all of the energy that went into making the field will go into making the reverse current, so some energy will radiate away from the inductor. How much of all these things actually happens depends on the quality of the inductor. Manufacturers will publish core specs which tell you about the kind of material that is being used for the core per part number. That can help you determine what kind of material the core should be.
    Because of the non ideal input source, the input source resistance can also eat up some of the energy involved in the pulse. For an ideal set of components except for the input source (with its resistance) the only energy dissipator in the circuit is the source resistance, and some radiation, so the source resistance would eat up some energy.
     
    Last edited: Jul 29, 2015
  7. Dong-gyu Jang

    Thread Starter Member

    Jun 26, 2015
    100
    4
    All right.

    Everyone has good points and the most dominating factor to eliminate noise energy looks resistance of real material.

    Then, Let's talk about ideal case (no resistance) in which there is a 'live' line with series connection of an inductor. All system is based on the battery and no ground is connected. Neutral line (return current path) is well insulated and far away enough such that parasitic capacitance between two lines is ignorable. All wires are superconductor (no resistance) and the system is in vacuum. Then I would somehow manage to shine high frequency EMI to only small wire section before inductor.

    The aim of the inductor is block this noise from being existed on the load and the inductor is simply made by superconductor solenoid thus it also has no resistance.

    Then will the noise energy be stored in form of magnetic field within the inductor during EMI on then eventually released to the load after EMI off? If this is true, I would say this is not EMI-filtering anymore but simply delay of the presence of EMI on the load as long as EMI frequency is preserved. (I'm even not sure the frequency is preserved when stored EMI is released to the load.)
     
    Last edited: Jul 30, 2015
  8. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    What you are talking about sounds like a lossless system except for the load and battery.
    For this kind of system, the load, assuming it has a resistive component, will absorb some of the energy and since the battery can be charged it could absorb some of the energy in the noise also.

    If the filter itself was ideal, then it would resonate, unless the load caused the entire system to be over damped or close to it (critically damped). To understand this kind of situation, you should probably look at the response of a filter to an impulse. Depending on the makeup of the filter and load, it can either resonate or damp out, or resonate a little and then damp out. This depends highly on the component values. Perhaps you should try to do a simulation.
    For a perfectly ideal filter with no load and no other resistive components, the filter would resonate or simply store the energy as a DC value forever, or for an AC excitation it may simply store energy for a little while and then give it back to the source. If the source was removed at some point, then the last state before removal would hold in the system forever. That's because the only thing that eats energy is resistance (not considering radiation).

    It might be better though if you could describe what it is you are really after here. Sometimes these kinds of question border on the philosophical and so they get very multi conditional where it is very hard to determine without applying some constraints. Even a simple RLC filter has three possible conditions of operation, depending on the component values and of course the topology could change things too.
     
  9. jjw

    Member

    Dec 24, 2013
    173
    31
    That is not true. The ideal lossless inductor is perfectly good in filter.

    You talk about noise energy.
    Think about noise voltage source.There is no noise energy unless there is a load connected.
    The inductance makes an impedance X ( reactance ) which is 2pi×f×L
    The voltage at load R approaches U×R/X when X is much higher than R, so less voltage and power at load with higher frequency.
    No energy is consumed in the inductance.
     
    Last edited: Jul 30, 2015
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