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LM51561H: LM51561H no work issue of boost to 24V

Herbert Wang
Prodigy 90 points
Part Number: LM51561H
  • function role：LAN POE power supply
  • function design：LM5161H chip，input power 12V boost output 24V
  • issue description：

1.power on  supply input 12V power, the function can output 24V.

 as below: picture 1 (out:24V, Gate PWM:11.13%)

2.alter 30s, the inductor L1 and N-MOS VQ1 is burning hot. the output 24V will drop to 12V.

as below: picture 2(out:24V, Gate PWM:64.47%) and pictue 3(out:12V, Gate PWM:0%)

TI LM51561H boost to 24V and 48V power SCH.pdf

  • 0
    TI__Genius 16000 points

    Hi Herbert,

    Thanks for using the e2e forum and sharing the waveforms and schematic.

    If I understand correctly, the device starts up correctly but then fails operating after 30s and shuts down.
    The first two waveforms look okay to me. The device starts with a smaller duty cycle during softstart and then increases the duty cycle until the desired output voltage is reached.
    (For boosting from 12V to 24V, duty cycle of 50-60% seems realistic, so this looks okay)

    Can you give me the maximum load conditions of the application?

    It is important that the inductor current ratings are high enough to supply the peak inductor ripple at max load. Otherwise it may enter saturation, which leads to even higher currents that eventually trigger overcurrent protection or lead to overheat and overtemperature trigger.

    Best regards,
    Niklas

  • 0 in reply to Niklas Schwarz
    Prodigy 90 points

    output of maximun is 24V/4A  and  48V/2A

  • 0 in reply to Herbert Wang
    TI__Genius 16000 points

    Hi Herbert,

    Thanks for the update.
    At Vin of 12V and max load, there can be current peaks up to 9.75A due to inductor current ripple.
    The current rating of the 6.8uH inductor is stated to be 10A, so we might have found the culprit for the heat-up here.

    Could you replace the inductor with part higher rated part? E.g. a current rating of 13A or 15A would leave higher margin and avoid saturation of the inductor.

    Thanks and best regards,
    Niklas