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Original question:

LM3478: Total gate charge specifics and minimum Vin in SEPIC converter

LM3478: Total gate charge specifics and minimum Vin in SEPIC converter

Pietro Di Buono 88
Intellectual 360 points
Part Number: LM3478

Hi,
thanks for the answers.

About the inductor, do you think that WE 74485540101 or SRF1280A-8R2Y could be a right choice? From my calculations, for uncoupled inductance, the min value is 9.3 uH, ΔIL = 2.4 A and IL(peak) 7.5 A. If not, can you suggest me a suitable inductance?

Equation 10 refers only to MOSFET power dissipation, so any choice of MOSFET is right if it can dissipate this power; but, I suppose that, increasing the Total Gate Gharge, at a specific switching frequency, it is required more power to drive the mosfet. LM3478 has to provide this power, that can't be limitless, and none information are in the datasheet or in the application note about this; unless you're telling me that, under any capacitive load and frequency conditions, LM3478 automatically limits the power supplied via the DR pin to avoid overheating.

About trouble 4, I've further investigate it. The problem doesn't depend by the input voltage, but by the input current. Limiting the input current, under a specific value at a specific input voltage (for example 20 V, 2 A), the regulator fails to reach the proper output voltage, also without load, because, at the power on, the regulator drains all input current, the input voltage falls (at 4 V in the example) and the output voltage is instable (between 6 and 10 V). In this condition, the gate voltage is almost constant and equal to the input voltage, so all the input current flows in the inductor and in the mosfet.
Could not this be a problem?

I've solved trouble 5 adjusting compensation filter.

Thanks for your help

  • 0
    TI__Guru* 78270 points

    If you calculation is right, the Wurth inductor should be fine.

    The driver losses is calculated by the following equation:

    Qg x VDR x Fsw, where Qg is the total gate charge, VDR is the driver voltage (7V if Vin is higher), and Fsw the switching frequency. The internal LDO losses is calculated by:

    (Vin-7V) x Qg x Fsw.

    The rest of circuit will consume about Vin x 3.3mA (quiescent current).

    The above three items are basically the total losses inside the IC.

    Regarding trouble 4, it is understandable that the output becomes instable because your input does not support that power level, and the control loop becomes open. This is the natural response of the circuit. Not sure what problem do you really talk about. To avoid this, you need to make sure your source does support the power level.