Do inductors have power ratings?

Discussion in 'General Electronics Chat' started by iinself, Feb 17, 2015.

  1. iinself

    Thread Starter Member

    Jan 18, 2013
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    Hi,
    I have some 4700uH inductors with me, they seem very small. Do they have power ratings, if so how can I find it, the site where I bought them from did not mention any power ratings like they do for resistors?

    Also when I tried to find out how much wire I need for a 5mH inductor from this site - http://www.colomar.com/Shavano/inductor_info.html , it comes out as about 200 feet of wire !! while the one I have is hardly a centimeter long?

    Thanks
     
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    They should have a rated Ohmic resistance. If wound on Ferrite, they may have a maximum current (core saturation) limit.
     
  3. iinself

    Thread Starter Member

    Jan 18, 2013
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    Can I measure the ohmic resistance using a multimeter?
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    If you Google "inductor power ratings," you will find a number of interesting links. Look those over and then let's discuss some more.
     
  5. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    That is the traditional method.
     
  6. JMac3108

    Active Member

    Aug 16, 2010
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    Inductors generally have a max current rating that is specified at a certain temperature rise - this is related to the resistance. In other words, they spec the current that will cause it to heat up by a specified amount. They also have a saturation current which is the point where the core saturates.
     
  7. alfacliff

    Well-Known Member

    Dec 13, 2013
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    gthe reason that the inductor seems small for the amount of wire required by the webside formula is that the inductor is probablyh wound on ferite, and the formula is based on an aircore inductor. the wire size determines the dc current capacity of the inductor as well as the power handling capacity. and yes the core material has a rating where it saturates and becomes less effective as a core.
     
  8. iinself

    Thread Starter Member

    Jan 18, 2013
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    Ok will do.

    So for removing ripple/noise from the walwart, I need to only consider the changing current component through the inductor right? The DC component will just pass through and the inductor will present impedance only to the ripple component? I am guessing the ripple/noise component is only going to be about max of 2 volts in worst case. So in the formula Ipk= V*Ton/L, using 2V, 0.0067ms (assuming switching frequency of 150KHz) and L = 4700uH, I get Ipk at 2.85 mA which seems pretty small. So the inductor should be alright in the filter?
     
    Last edited: Feb 18, 2015
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