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Fly-buck transformer maximum primary inductance

Keith Weller
Prodigy 100 points
Other Parts Discussed in Thread: TPS55010

Does anyone know how the expression in equation (14) of the TPS55010 data sheet (SLVSAV0A - June 2011) was derived? I've done some simulations which seem OK when this value is (even greatly) exceeded. Problems do arise when the leakage inductance becomes too large however, so that sets a practical upper limit, but unrelated to the value given by the expression.

I'm wondering if it is derived from minimising the transformer resistance and core losses? Maybe in due course there will be a definitive analysis of the fly-buck topology, but I've not yet found one...

Keith

  • 0
    TI__Mastermind 35165 points

    The Lomax equation calculates an inductance to give best efficiency. By following this equation it makes sure there is always some negative current flowing in the primary winding of the transformer when the low-side FET turns off. When the low-side FET turns off, since there is still energy in the transformer it casues the PH voltage to rise until it is clamped by the body diode of the high-side FET. This allows the converter to have zero voltage switching (ZVS) when the high-side FET turns on to reduce switching losses. A larger inductance certainly won't cause any issues in the parts actual operation. Like you mention it will increase the conduction losses due to higher resistance but the switching losses can dominate at the lighter loads.

  • 0 in reply to Anthony F
    Prodigy 100 points

    Thanks Anthony, that makes sense, and also explains the downward kick I was seeing in the PH waveform at low input voltage due to residual positive current in the magnetising inductance.

    Do you perchance also know of an expression for the optimum output capacitances to resonate with a given leakage inductance? When I was using a large primary inductance transformer (and hence large leakage inductance) I observe the output dropping off at low input voltage (i.e. as the low-side switch off time reduces), presumably because there's insufficient time for the primary side output capacitor to charge the secondary side capacitor via the leakage inductance. Some texts mention this is effectively a resonant converter, so I guess one has to tune it in some way?