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Original question:

LM2775: short-duration overcurrent

LM2775: Output current protection

user6527039
Prodigy 70 points
Part Number: LM2775


Hello,

In my application, large load current(About 1A) occurs on output side when output is rising to 5V.

After Vout reaches to 5V, that current vanishes and the output seems to be fine. 

According to the datasheet, the output current limit protection is 250 mA (typical).

How does the protection circuit work?

Limit the output current without stopping the output (in other words, nothing is to be damaged)

or stop the output temporally or else? 

looking forward to your reply.

Thank you.

Jie Zheng

  • 0
    TI__Guru 67515 points

    Hello Jie,

    Input current limit limits the current in every cycle. This means that the part will go on switching and most probable not reach the desired output voltage due to the current limit. With 1A load current during startup, it is possible that the part is not able to start at all. As your load seem to be very specific, you need to test this with the real load.

    If the device is able to start because your load drops fast enough to a value low enough for the part to start, then the current limit will not stop the device from switching.

    The thermal shutdown might cause a temporary shutdown when the device gets too hot during this time. But as soon as the temperature reduces, the part will restart, so you might se thermal cycles during the high load.

  • 0 in reply to Brigitte
    Prodigy 70 points

    Hello Brigitte,

    Thank you for prompt relpy.

    I realized my expression was not propriate, which made you misunderstand.

    There is no load during startup.

    I think that large rush current, tested in my real systerm, may be caused by the output capacitors.

    Can this explain why the startup is successful and conclude the part won't fail to start because of this current?

    By the way, the current lasted only 200us during startup. I'm quite sure it won't trigger the thermal shutdown.

    Jie Zheng

  • 0 in reply to user6527039
    TI__Guru 67515 points

    Hello Jie,

    Did you measure the current between the VOUT pin and the output capacitor(s)? If yes, then the current coming out of the IC is for charging the output capacitors if there is no load connected.

    During startup switching power supplies normally draw more current than in normal operation as the output capacitors need to be charged to the desired voltage during startup.

    So as long as your input voltage does not drop too much because of the startup current, this is normal behavior.

  • 0 in reply to Brigitte
    Prodigy 70 points

    Hello Brigitte,

    As you assumed, that is the current between the VOUT pin and the output capacitor.

    >>So as long as your input voltage does not drop too much because of the startup current, this is normal behavior

    Do you mean there is no requirment on the current for charging capacitors during startup，

    as long as input voltage doesn't drop down to 2.7V or less?

     

    Best regards.

    Jie

  • 0 in reply to user6527039
    TI__Guru 67515 points

    Hello Jie,

    I mean that it makes sense to make sure that the input voltage is even stable during the startup. If you do have too little input capacitance, it could happen that during startup the voltage on the input drops again below UVLO and then you get 2 startups directly after each other.

    The device will take care on the charging of the output capacitor and you only need to take care that your load is not too high and your input capacitors are big enough and close to the IC.

  • 0 in reply to Brigitte
    Prodigy 70 points

    Hello Brigitte,

    Thank you.

    your reply just resolved my issue.

    Thank you again for your kindness.

    Best regards.

    Jie