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Boost-Flyback converter

  1. May 7, 2014 #1
    I am reading about Boost-Flyback converter and this is a bit confusing.
    Could you explain the role of D1 and C1 here?
    The lecture says that "D1 and C1deliver to the output the energy stored in the transformer leakage inductance Ld." I can't understand what is meant here.
    And could you explain why there is Lm there?


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  2. jcsd
  3. May 7, 2014 #2


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    Lm is the (non-ideal) leakage inductance of the transformer. You need to understand that first.

    The leakage inductance stores energy the same as Ld does, and that energy needs to go somewhere as the field collapses. It doesn't go through the transformer because that's what Lm is representing, the inductance of the transformer as seen from the primary.
  4. May 7, 2014 #3

    jim hardy

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    I don't know your source for that circuit so take this as speculation.

    I'd wager leakage inductance includes not only transformer leakage as pointed out by mBGuy, but inductance of the loop (of wires or pc traces) connecting transistor S to the power supply and transformer as well.
    That inductance can be quite troublesome if the layout is not very good as in the first switcher i ever built.
    That's clever, capturing its stored energy with D1-C1 and adding it to output. The dissipative snubbers in my amateur project got quite hot.
  5. May 8, 2014 #4


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    this isn't my field of expertise
    so just wanting to learn....

    mebigguy .... You are saying that Ld and Lm are not physical inductors but are the "product" of the primary of the 1:n transformer ?
    At least I think that's what you are saying ?

  6. May 8, 2014 #5


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    I am saying Lm is inside the transformer. I think Ld is a real inductor.
    http://www.ijrte.org/attachments/File/v2i6/F0948012614.pdf [Broken] uses a series inductor for boost.
    Last edited by a moderator: May 6, 2017
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