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LM5117: Driver Real Gate Peak Current

Part Number: LM5117

Tool/software:

In my CC/CV DC-DC converter project I need to put 2 MOSFET in parallel to manage more current and I need a very high peak current to charge quickly the two gates, so I choose le LM5117 because it has 2.2Amps Gate peak current.  

Reading the datasheet I learned the VCC regulator can supply maximum 30mA (typ 5.8mA) of current towards the VCC pin.
On the VCC pin there is 7.6Volts with 30mA of maximum current available to supply the HO Driver and LO driver.

The question is: How the Drivers can deliver 2.2Amps or 3.3Amps to the MOSFETs gates if their power supply can deliver only 30mA?  

Do they maybe get peak energy from Cvcc capacitor?

In the above table the peak source current, for the HO driver, is 2.2Amps with the conditions "Vho=0V" and "Vsw=0V", but this condition will depolarize the QH N-MOS (Vgs=0) and the driver current will be in "sink" mode... sorry but I don't understand.
The same for the "sink" condition: sink should be when the current flows from the gate capacitance to the driver port... but this is impossible with "Vho=Vhb=7.6Volts".

Can you explain me what is the logic?

Can you confirm me the peak current in the MOSFETs gate is about 2.2Amps during the gate charging?

thank you,
Michele Faini

  • Hello Michele 

    Yes, your understanding is correct. The peak sourcing currents are supplied by CVCC and CHB capacitors. 

    The test conditions are correct. Please assume there is a 10mΩ resistor between and LO pin and the MOSFET gate, and we measure the peak LO sourcing current in~ 100ns after short the MOSFET gate to ground. During this 100ns, LO is 7.6V but the gate voltage is 0V.    

    I confirm 2.2A is the peak sourcing current. Please keep in mind that the MOSFET driver cannot source the 2.2A peak current for a long time. If 7.6A CVCC voltage drops down, the peak current will drop down. If the MOSFET gate voltage is greater than 0V, then the peak current also will drop down.  

    -EL

  • Ok Eric, now is all clear! The conditions are forced externally during the measure!


    Another question: since there is the possibility to supply the VCC pin from an external power supply, in your opinion, could an external "strong" 7.6Volt power supply help me to better polarize the gates of multiple MOSFETs connected in parallel?


    Could this method speed up the gates charging also when the total gate capacitor is high (10nF or higher)?


    Do you thing the HO & LO driver outputs will be more stressed (and possibly demaged) if I use a strong power supply? Can I adopt some precautions to avoid drivers demage? 

  • Hello Michelle

    The external 7.6V helps the internal VCC regulator to drive multiple MOSFETS, which require more than 30mA(min) VCC current, but doesn't help increasing its peak sourcing /sinking current. 

    No it doesn't help much. You have to supply the external power which is greater than 7.6V in order to increase the peak sourcing/sinking current.

    Don't supply more than 14V at the VCC pin. 

     

    Supply the external VCC power after supplying VIN. The VIN pin voltage should be always greater than the external supply voltage.

    -EL 

     

  • Thank you Eric,

    I think I'll use en external source to help the MOSFET's gate charging.
    In the datasheet, for Vout > 14.5Volts, this circuit is suggested:

    Sorry but I have the last questions:


    1) Do I have to switch off the internal regulator through VCCDIS control or it must be on?

    2) Using this circuit, I think the VCC power supply will be applied automatically after the VIN, won't it?

    2) About the 30Kohms resistor, is it placed inside the LM5117 chip?

    3) R1 is necessary to limit the Zener Diode current. What limit current do you suggest for protecting the chip? I think it depends by the Izener_max... 

  • Hello Michele

    1) No, you don't need to. 

    2) Not exactly, the external VCC will be applied after VIN applied and also after the device is enabled by UVLO. 

    3) 30kOhm is not a real resistor inside the resistor. It represents the amount of current consumption of the IC. 

    4) Yes, you are correct. the worst case Izener will be (14.5V-7.6V)/R1 

    -EL 

  • Thank you Eric for your explanations.
    Have a nice day!

    Michele

  • Thanks for choosing TI. Please feel free to contact us if you have any question.
    -EL