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LMR36015-Q1: Peak current capability of integrated MOSFETs body diode

Part Number: LMR36015-Q1
Other Parts Discussed in Thread: DRV8872-Q1,

Hi there,

Is there a specified value for the maximum current capability of the integrated MOSFETs body diode?

Best Regards,

Christian

  • Hi Christian,


    There is no specified value for this one but the device should be capable of delivering 1.5A of current and we also have a current limit to protect the MOSFET from being damaged. 

    Thanks

    -Arief

  • Hi Arief,

    Thanks for your quick response.

    There are some tests we have to perform with our ECUs. One of them simulates a loss of supply voltage together with line faults.

    Loss of supply voltage means in that case, that there is a broken connection of the battery voltage to the ECU but GND connection is still there or vice versa.

    The LMR36015-Q1 supplies a motor driver (DRV8872-Q1 from TI). If there is a broken connection of the battery voltage and at the same time a short circuit to battery voltage at one of the motor driver's output pins, the ECU can get supplied via this output path. In that case, a current flows through this output pin via the motor driver's body diode of the High Side MOSFET to VM (motor driver's supply pin) which is connected to the voltage output pin SW of the LMR36015-Q1. That means, after the motor driver the current flows through the SW pin via the converter's body diode of the High Side MOSFET to VIN and then to other circuits of the ECU.

    The other case would be, a broken connection of GND and at the same time a short circuit of GND. Then the Low Side MOSFET would be affected.

    In both cases, it is important that the device gets not damaged by this "feedback current". And if this current exceeds the specification of the body diode, we need external protection elements.

    So, back to your answer:

    • Am I understand it right, that if one of these cases occur, the body diode is capable of handling 1.5 A?
    • How is the body diode related to the thermal performance of the device?
    • Can you explain this current limit mechanism which protect the MOSFETs? Does this apply to these cases?

    Thank you very much for your time.

    Best regards,

    Christian Edhofer

  • Hello

    Here are the answers for your questions:

    1. Usually it is not a good idea to allow voltage on the output of a regulator when the input is zero volts (or even floating in some cases). This condition could be caused by a very large output capacitor that is charged when the input is removed, or some kind of active load suppling a voltage. There is a diode between the drain(VIN) and source(SW) of the power MOSFET inside the converter. With a voltage on Vout, and the input held at zero volts, there will be a path from Vout to Vin through the inductor and this diode. Even if the input is floating, a current will flow and try to charge the input caps, possibly to a high voltage due to resonance with the caps and the inductor.  This phenomenon has been seen in the past under some conditions. Allowing current to flow through this diode can cause a parasitic to turn on or even damage the diode.  Since it is not a "true" diode but a parasitic formed when the MOSFET is constructed on the die, it is not designed to take current for more than a few 10s-100s nsec. One way to avoid this issue is to use a diode from input to output to provide a path for the output capacitors to discharge around the device.  A Schottky diode with cathode connected to Vin and anode connected to Vout can be used. A blocking diode in series with the load can also be used. Please also see app note SLYT689 for more information.

    2. The body diode does not relate to the thermal performance of the device, thermal performance is more related to the layout of the device. 

    3. I don't think the current limit will be able to protect the device in this manner. The current limit only protect the device when the current is in the forward normal condition. If you have a short at the output then yes the current limit is meant to protect the excessive current through the inductor. 

    Thanks

    -Arief

  • Thank you very much for that answer.

    Best regards

    Christian