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TPS7A49: the start up charge current of Cout？

liang ren
Intellectual 420 points
Part Number: TPS7A49

Hi

    This LDO‘ internal current is 309mA， and if I place a 2200uf  Cout， the charge current when start up will crash the device ？  (the output votage is 5V)

  • 0
    TI__Genius 10200 points
    Hi,

    During startup, the output capacitor charging current will be limited to the rated current limit of the TPS7A49 (309 mA typical). This will prevent the device from crashing or being damaged by large inrush currents. Keep in mind that operating the device in current limit conditions for extended periods of time will degrade performance and reliability.

    Thanks,
    Gerard Copeland
  • 0 in reply to Gerard Copeland
    Intellectual 420 points
    Hi
    Thanks . So that means the device could work on limit level 309mA during start up charging. And the safe period is t (ms) = 1.4 C (nF) ? the datasheet gives a 10nf C example, and 14ms start up period. And the Cout need charge 5v*2200uf*10e-6=0.011， and 0.309A*0.014s=0.004326 during the 14ms. So the 2200uf Cout only charged (0.004326/0.011)=39%, the rest 61% need will charge after the soft start period. am I right ?
  • 0 in reply to liang ren
    TI__Genius 10200 points
    Hi,

    On startup, the soft-start feature slowly ramps up the internal voltage reference. Since the error amplifier reference is lower, less current is needed to charge the output capacitor, and the overall inrush current is reduced. However, with a large output capacitor, the internal reference will ramp up faster than the LDO can charge the output. As such, the device will charge the output as much as possible by operating in current limit.

    Your calculations are correct. At the end of soft-start, the output voltage will be:

    dV = Ilim / C * dt = 309 mA / 2200 uF * 14 ms = 1.97 V

    This is 39% of the 5 V output. The total startup time will be approximately:

    dt = C * dV / Ilim = 2200 uF * 5 / 309 mA = 36 ms

    Keep in mind that applying a load during startup will increase the startup time.

    Thanks,
    Gerard
  • 0 in reply to Gerard Copeland
    Intellectual 420 points
    Hi
    Thank you very much. I will place a large capacitor for test. One point : "keep in mind that operating the device in current limit conditions for extended periods of time will degrade performance and reliability. " What Is the "extended periods of time" definitons ?
  • 0 in reply to liang ren
    TI__Genius 10200 points
    Hi,

    Performance metrics are measured with the assumption that the device is being operated within the recommended operating conditions. While current limit is intended to protect the device from permanent damage during fault conditions, the device is not designed to source output currents larger than the recommended maximum. The heat dissipated when sourcing larger currents will damage the device over time. Because of this, it is recommended to minimize the time in which the device operates in current limit. If a large output capacitor and high performance are required for the full lifetime of your application, you may need to externally limit the startup current to ensure that the TPS7A49 does not operate in current limit.

    Thanks,
    Gerard
  • 0 in reply to Gerard Copeland
    Intellectual 420 points
    Hi
    I have add a current limit resistance ctrled by relay after ldo.