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LM5176: full load issue.

henry-liu
Intellectual 280 points
Part Number: LM5176

The attachment is the SCH, the full load of the design is 20V 4.5A, input voltage range is 9~32VDC.

Test condition: input 24V, output 20V 2A, this PCBA works normally. when turn the load to 20V 4A, the output voltage will drop to 4V, after a minute, the IC (LM5176) damaged.

Could you please check the SCH, thanks!

SAWA-81-45020A -SCH.PDF

  • 0
    TI__Guru 68495 points
    Sounds like hitting the current limit. Can you double check if the two paralleled resistors for current senses are both well soldered? Please check both the sense resistor below the switching legs, and the one for the average current limit. Please also show the waveforms of SW1 and SW2 at 2A and 4A load, respectively.
  • 0 in reply to Youhao Xi
    Intellectual 280 points

    Hi YouHao:

    Thanks for your reply. after check, the current sense resistors are soldered well. 

    Enclosed the Q2 driver waveform.  The PCBA can load 3A, (input 31V，output 20V), 

    When the PCBA load 4.5A, (less than 30 seconds), the LM5176 will damage，remove the load, the PCBA cannot work again, change a new IC, the PCBA will works.

    LM5176 driver waveform.doc

  • 0 in reply to Youhao Xi
    Intellectual 280 points
    Hi YouHao:

    solder the damaged IC off and test, pin 27 (HDRV1) is short to pin 28 (SW1).
  • 0 in reply to henry-liu
    TI__Guru 68495 points
    Reviewed your waveform and you can see the driver of Q2 showing negative spike, and the higher load the larger spike. Did you measure with the short ground lead of the probe? If using the regular probe with long ground lead, such negative spike may not be real. However, if it is real, it may overcharge the BOOT1 capacitor and damage the high side driver (BOOT1 as well as HDRV1).

    This ringing is mostly related to the layout parasitic inductance along the ac current loop running through the input capacitor, Q1, Q2, the current sense resistor, and back to the input capacitor. There is an internal clamp circuit to protect the BOOT pins, and it will work for most cases. However, if the parasitic inductance is too strong to induce larger energy in the negative spike, damage can occur.

    Improving the layout by following the datasheet layout guide lines will help resolve the issue. In additions, replacing the 0 ohm gate resistor with a a few Ohm resistor for Q1, Q2, Q3 and Q4, should slow down the switching transient and reduce the said negative spike.