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How does the core size affect an Inductor's efficiency?

  1. Oct 10, 2016 #1
    Hello,

    I am wondering if anyone can help me with the following question. I want to find out how the size of the core in an inductor influences the current induced in the secondary coil, and whether a smaller core is more efficient if efficiency is defined as inducing the highest current in the secondary coil relative to the power required by the primary coil.

    Let’s say there are two O core inductors, operating at the same frequency. The core in Inductor A is twice the size as Inductor B. The primary coil in Inductor A has twice the number of turns as B, for example Inductor A’s primary coil is 100 mm long and has 100 turns, while Inductor B’s primary coil is 50mm long with 50 turns.

    They both have the same turn density, and the FEMM simulations I have done show each Inductor producing the same Magnetic Field strength. Does this mean that the current induced in the secondary coil would be the same in both inductors?

    Both Inductors receive the same level of current in the primary coil, however there is less resistance in Inductor B’s coil because the wire is half the length. Does this mean that Inductor A will need more power in the primary coil to produce the same magnetic filed strength and induce the same level of current in the secondary coil?
    I know that the inductance is greater in Inductor A, but I am not sure how that fits into the equation. If someone could point out the errors in my thinking, it would be appreciated. Thanks!
     
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  3. Oct 10, 2016 #2

    anorlunda

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  4. Oct 10, 2016 #3

    Hesch

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    With your example, you are right about the field strength ( double length of the magnetic path through core A, and double number of turns ). But it's not the field strength ( H-field ) that induces EMF in the secondary windings, but magnetic flux. The flux is proportional to the cross section area of the core, and since A = 2*B, the flux in A will be 4 times greater than in B.

    You must consider the magnitude of the magnetic induction ( B-field ), since a too large B-field results in huge hysterisis losses in the core. Having chosen a B-field, ( say 1 Tesla ) you can calculate the number of turns in the windings from

    EMFrms = 4.44 * f * N * A * B

    f = frequency ( Hz)
    N = number of turns
    A = cross section area ( m2 )
    B = induction ( Tesla ).
     
  5. Oct 16, 2016 #4
    Hello Hesch,

    Thanks very much for your answer - it was a great help.

    I used the formula EMFrms = 4.44 * f * N * A * B to compare two transformers, one that was 80% the size of the other - 80% of the core cross section, and 80% the number of turns in the coil (same wire cross section, same current). The result was that the smaller transformer produced an EMF that was 57% of the larger one. However, given that the smaller transformer had a shorter wire in the primary coil, and therefore less resistance in the wire, it drew less power to produce the required current - 63% of the larger one to be precise. Does this mean that the smaller transformer is more efficient?
    Thanks again, sorry if I have misunderstood and made some basic error somewhere, but I did think long and hard, checked and rechecked.
     
  6. Oct 16, 2016 #5

    Hesch

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    How come that you have two transformers ( same shape? ), A and B, size is scaled 1:0.8, and then their cross section area is also scaled 1:0.8 ?

    The cross section area must be scaled 1:0.64 ( 0.8 square ). Thus the number of turns cannot be 1:0.8, using the formula.

    ( To be continued later ).
     
  7. Oct 16, 2016 #6

    Hesch

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    To make an "optimal" transformer, do the following steps:

    1) Choose a core. Often the documentation states how many VA the core is made for.

    2) Use the formula to calculate the number of turns in the windings ( big core . . small core → B = 0.9 . . 1.2 T ).

    3) Calculate the maximum cross section area of the wire that can be wound on the core. Remember space for insulation.

    For an optimal transformer, Iron losses ≈ copper losses.
     
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