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LM25011: startup problem with (precharged) load capacitor

Arndt Kritzner
Intellectual 720 points
Part Number: LM25011

Hi,

we have a circuit with a large capacitor (2.5F) as short-duration power source. The intended operation after applying VIN to the circuitry is to charge this capacitor with constant current to 5V and afterwards start the cpu module. Under most conditions this works as intended:

yellow: VIN, blue: SS, magenta: SW, green: VOUT

Under some circumstances precharging the load-buffering capacitor fails. This might (not sure about it) be caused by remained charge in the load-buffering capacitor:

In this cases the SW voltage does neither reach VIN nor GND and the load is not charged at all.

Can this be caused by load-side voltage in an inappropriate window at circuitry startup?

Kind regards, thanks for any help

Arndt

  • 0
    TI__Guru 72625 points

    It would be helpful to see a little more zoom-in on the switch waveform.

    How much residual voltage is on Vout when the start-up fails ??

  • 0 in reply to Frank De Stasi
    Intellectual 720 points
    Residual voltage in the failing case was about 3.8V. Eventually my colleague can specify the "window of failure" a bit more precise tomorrow.

    Kind regards
    Arndt
  • 0 in reply to Arndt Kritzner
    TI__Mastermind 23845 points

    Hi Arndt,

    Could you also share the schematic? In particular the soft-start capacitor value that you use. On the first picture, i can see the SS pin rise faster than VOUT.

    Thanks

    -Arief

  • 0 in reply to Arief Hernadi
    Intellectual 720 points

    Hi Arief,

    attached the schematic clip. The soft-start capacitor is 100n, but the load capacitor is 3F. The idea is to charge the load capacitor over about 15 seconds with 1A (limited by the resistor between CS and CSGND).dimm-cpu-cb17.pdf

    Kind regards

    Arndt

  • 0 in reply to Frank De Stasi
    Intellectual 720 points

    Hi Frank,

    here some scope measurements with more zoom:

    "normal" (good) startup with residual voltage near 0V looks as this:

    With a load-side residual voltage of 2.9V the startup fails:

    Similar situation (fail) with further more zoom:

    yellow: VIN, blue: SS, magenta: SW, green: VOUT

    Kind regards

    Arndt

  • 0 in reply to Arndt Kritzner
    TI__Mastermind 23845 points

    Hi Arndt,

    There is a possibility that the boot capacitor is not properly charged during this condition. Is the failure to startup only happens at higher VOUT residual voltage?
    What happens when the VOUT dropped to a lower voltage? Will it start ok?
    Also when this happen, could you try to short the SS pin and release it and see if the part startup ok? This will engage another soft-start and hopefully it will charged the boot capacitor. 

    Thanks

    -Arief

  • 0 in reply to Arief Hernadi
    Intellectual 720 points

    Hi Arief,

    yes, if residual voltage at VOUT goes below about 1V a normal startup happens (value depends on the actual chip sample).

    Unfortunately an additional soft-start event does not force the device to a successful startup.

    Regards

    Arndt

  • 0 in reply to Arndt Kritzner
    TI__Mastermind 23845 points
    Hi Arndt,

    It seems that the boot capacitor is not properly charged. I talked to Frank too about this and you can try this idea.
    Isolate the 3F super capacitor with a diode. So basically you put a diode in series between the output of the inductor and the 3F capacitor.
    Make sure that the FB resistor network is connected to the anode of this diode. Also put enough capacitor to GND from the anode of this diode just like you design the normal converter. Therefor you are only isolating the 3F capacitor with a diode.

    Thanks
    -Arief
  • 0 in reply to Arief Hernadi
    Intellectual 720 points
    Hi Arief and Frank,

    yes, I will try this. But the diode solution will induce major drawbacks to the supplied circuitry. Because of the forward voltage drop in the diode, regulation precision on this rail also drops. Efficiency and amount of in the capacitor stored energy too. The load conditions in my circuitry are variable and the diode's forward voltage also will be.

    Thanks and regards
    Arndt
  • 0 in reply to Arndt Kritzner
    TI__Mastermind 23845 points

    Hi Arndt,

    Yes, go ahead and try the diode method. 
    Other things to try without the diode:

    1) Increase the value of boot capacitor maybe to 0.22uF or 0.47uF and see if that improves. 

    2) Put a schottky diode from output to boot pin (Anode at the output and cathode at the boot pin). I assume your output maximum voltage is 5V correct? 

    Thanks

    -Arief