Boost-Flyback converter

Discussion in 'General Electronics Chat' started by anhnha, May 7, 2014.

  1. anhnha

    Thread Starter Well-Known Member

    Apr 19, 2012
    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?

  2. AnalogKid

    AAC Fanatic!

    Aug 1, 2013
    You have a model of a transformer, not a schematic of one. The model shows the turns ratio, leakage inductance, and magnetizing inductance as separate parts. They can be thought of in that way for analysis, but they are not three physically separate components.

    I have not seen this topology before. I suppose that for very high output voltages it would be more efficient than a straight non-isolated boost circuit, but it seems like a lot of work to get there. RB probably has more to say on this since he has done work in this area.

    My guess is that the idea here is to combine isolated and non-isolated boost circuits to get a little higher output voltage than either would create individually. When S is on, current through the "transformer primary" charges up the inductance. D1 and D2 both are reverse biased so no energy is transferred to the output.

    When S turns off the magnetic field collapses generating the standard flyback inductive kick and voltage spike. The collapsing field is of the opposite polarity from the charging field, so D2 conducts. Also, the voltage spike is positive so D1 conducts. Because of the voltage isolation provided by the transformer, the two diode outputs can be stacked on top of each other and sum together for the total output voltage.

    Note that above I put transformer primary in quotes. In a flyback circuit that big iron thing does not function as a transformer in the traditional sense because continuous current is not flowing through the primary and secondary at the same time. That's why both dots are not at the same end of the core. The part functions as two coupled inductors, and energy transfer happens only during the primary side turn-off transient.

    anhnha likes this.
  3. anhnha

    Thread Starter Well-Known Member

    Apr 19, 2012
    Thank you for the detailed answer. I was confused about Lm.
    I just read that Lm is magnetizing inductance that exists when core reluctance is nonzero.
    Could you tell me how to calculate ig(t2) - ig(t1) as in the picture?