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UCC28951-Q1: significance of primary side diodes D4 and D5

Arun Thawait
Prodigy 155 points
Part Number: UCC28951-Q1

Dear sir ,

In the document tida01407 what is the role of diode D4 and D5 ?

Regards

  • 0
    TI__Guru* 85416 points

    Hello,

    Those diode provide a path for the difference in primary transformer current and the shim inductor when the design comes out of freewheeling.

    If you do not have these diodes this extra current will cause excessive voltage across the primary of the transformer; as well as, your secondary FETs. 

    These clamping diodes are necessary for the phase shifted full bridge controller to operate correctly. 

    Regards,

  • 0 in reply to Mike O'
    Prodigy 155 points

    Thank you   for your reply sir .

    While selecting these diodes as I can see its voltage rating must be similar to that of primary side MOSFETS. what about the current ratings of these diodes . I cant be able get a clear picture of average current rating of these diodes ??

  • 0 in reply to Arun Thawait
    TI__Guru* 85416 points

    Hello,

    To select these diodes I would select them so they can handle the primary peak current of the Hbridge.  You could set the RMS current to be similar to the primary H Bridge FETs.  Most design

    I gave this some thought.  The current going through the diode when the transformer is first energized should be the difference in reflected output current ripple.

    Iclamp_diode = (ILout peak)*Ns/Np - ((1-D)/fs)*Vout/Lout  

    The energy that needs to dissipated through these diodes is the shim inductance (Ls) and leakage inductance (Llk)

    W = (1/2)*(Llk+Ls)*Iclamp_diode^2 

    You could estimate that the power dissipated by this diode is energy times switching frequency (fsw) of the primary

    Pclamp_diode = W*fsw

    Do be safe I would add a factor of 1.5.

    You could also get a better estimate realizing the shim inductor is clamped to 1V.

    dt = (Llk+Ls)*Iclamp_diode/1V

    I would estimate that the current is triangular in shape so you estimate the RMS current based on a triangle.

    Idiode_clamp_rms = Idiode_clamp* (dt/(3*fsw))^0.5

    Regards,