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UCC21520-Q1: It is not recommended to leave DT floating.

Part Number: UCC21520-Q1

Hi TI-team

My customers want to know exactly what happens when they float the DT pin.

The description of the DT pin in the past data sheet stated the following.

”Leaving DT open sets the dead time to <15 ns. ”

(Rev.B)

Could you please tell me why the DT pin shouldn't be floating and what exactly happens ?

Best Regards,

Koji Hayashi

  • Hello Koji,

    The reason we don't recommend leaving the DT pin floating because high voltage noise coupling in can disturb the deadtime duration which can cause unintended behavior from the driver.

    It is best to tie DT to VCCI if not needed (allowing overlap of outputs).

    Or place a 500Ω - 500kΩ resistor (RDT) between DT and GND which will adjust DT(in ns) = 10 x RDT(in kΩ). Please place a 2.2nF capacitor in parallel with this resistor as close to the pins as possible to achieve better noise immunity. 

    Thanks,

    Krystian

  • Hi Krystian-san

    Thank you for response.

    >can cause unintended behavior from the driver.

    Does this mean that the dead time will be 15ns or more, or that the dead time will not be specified ?

    Also, it does not mean that the output (OUTX) fluctuates by H or L regardless of the input (INX).

    Do you agree with this recognition ?
     
    Best Regards,
    Koji Hayashi
  • Hi Koji-san,

    Leaving the DT pin floating will cause the device to set to the minimum DT, about 10ns.

    This is not recommended because the DT pin in the floating state can cause the DT circuit to trigger unintentionally causing outputs to glitch low for periods of time. The actual dead time window won't vary that much.  

    Regards,

    Krystian Plaskota

  • Hi Krystian-san

    Thank you for response.

    I understand.

    I have answered my customers.

     

    Best Regards,

    Koji Hayashi

  • Hi Koji-san,

    You're absolutely welcome! Let us know if you have questions in the future!

    Regards,

    Krystian