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TPS4H160-Q1: question about drive inductive load

Part Number: TPS4H160-Q1

hi experts

when i read this APP note:How To Drive Resistive, Inductive, Capacitive, and Lighting Loads (Rev. E) . i have some questions about "stored energy during the on time from the inductance value"  and "demagnetization energy"

from my narrow knowledge, i think "demagnetization energy" equal to "stored energy during the on time from the inductance value" , because high-side need to consume this energy(demagnetization energy) which is stored during the on time from the inductance value. what is wrong ?

  • Hi Jay,

    The inductive energy stored in the inductance is not the same as the energy that is dissipated across the FET during turn off.

    During inductive turn off, the current is decaying but it is still flowing through the FET and now the Vds across the FET is larger. In addition, the inductive energy that is stored must also be dissipated.

    Therefore, the energy dissipated during turn off of the HSS is greater than the energy that is stored in inductor itself.

    Thanks,

    Shreyas

  • hi expert

    i draw the following picture. pls read by ①②③. i am not sure whether i have got your full comments.

  • Hi Jay,

    In your picture, 1 is correct but 2 and 3 is incorrect.

    During inductive turn off, D voltage will remain VBAT. S voltage will go negative. Therefore the body diode cannot conduct in this scenario.

    Since S goes negative, the drain to source voltage increases. When this increases to a certain level, the internal gate drive structure will start to drive the gate to turn on the FET into a dissipative state hereby clamping the voltage at the Vdsclamp.

    The dissipative path is through the FET.

    Thanks,

    Shreyas

  • yes,I make a mistake in body diode.

    i got it. thanks.