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0E mounted instead of inductor (L) of TPS54319RTER output

Mohammed Azlum
Intellectual 835 points

Hello TI,

We are using TPS54319RTER in our design. As per reference design, inductor (L) need to be connected between PH pin and Vout. During board assembly 0E (zero ohm) resistor has been mounted instead of inductor (L). As a result, when board POWERED-ON, the components (processor and other chips) related to Vout been damaged. Please let us know why the load components (processor and ICs) have been damaged when the 0E mounted instead of inductor (L).

Regards,
Azlum

  • 0
    TI__Guru**** 191445 points
    The inductor is required for proper operation. with 0 ohm in place of the output inductor, the input voltage may be connected directly to the output when the high side FET turns on. that is the probable cause of your failure.
  • 0 in reply to JohnTucker
    Intellectual 835 points
    Hello,

    The output pulse depends only the duty cycle. In buck regulator, the inductor is used to control the sudden change in current. Then how can relate the inductor with voltage pulses ?

    Also I didnt touch the voltage smoothing caps. So hope the voltage pulses will not create the problem. The sudden raise in current may damage load.

    Regards,
    Azlum
  • 0 in reply to Mohammed Azlum
    TI__Guru**** 191445 points

    There cannot be any current in the inductor unless there is a voltage across it.  During normal operation, the output side is at Vout, and the SW node side switches between Vin and GND.  With 0 ohm in place of L, those voltages can be impressed across the output.

  • 0 in reply to JohnTucker
    Intellectual 835 points

    Dear JohnTucker,

    So you meant to say that, the output of switching buck regulator voltage is controlled by the inductor ? Before inductor it pulse between 0V and Vin. After inductor the desired output will be present. So I just little confused how the inductor is controlling the output ? My understanding was duty cycle is adjusting the output voltage amplitude.  

    Regards,

    Azlum

  • 0 in reply to Mohammed Azlum
    TI__Guru**** 191445 points

    It is a closed loop system with feedback.  A portion of the output voltage is fed back to the error amplifier via a resistor divider.  The ratio of that divider sets the output voltage.  The fed back signal is compared to a reference.  The transconductance error amplifier adjust the COMP pin voltage so that the fed back voltage is = VREF.  The COMP voltage is proportional to the peak switch (inductor) current.  when the high side FET turns on, SW is at VIN and the output side of the inductor is at Vout, so the switch (inductor) current is rising at the rate of (Vin - Vout) / L.  When the current rises to a level determined by the COMP pin voltage, the high side switch turns off and the low side switch turns off.  Now the inductor voltage is 0 - Vout and the current starts to decrease.  For current mode control, this ramping action of the switch (inductor) current acts as the PWM ramp signal and is necessary for proper regulation.  The inductor is a critical part of the loop control.