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TPS715-Q1: Soft-Start Time Evaluation

Evan Wang
Expert 8550 points
Part Number: TPS715-Q1

Tool/software:

Hi Team,

May I know how to evaluate the Soft-Start Time of our TPS71533QDCKRM3Q1?

Datasheet mentioned that we have internal SS but we're not sure how it works and how to evaluate the SS time based on different COUT.

Looking forward to your reply.

Thank you.

  • 0
    TI__Genius 14245 points

    Hi Evan, 

    For TPS71533QDCKRM3Q1, the internal soft start slows the rise of the internal reference voltage in order to help control the inrush current needed to charge Cout during startup. The internal reference takes approximately 1ms to fully charge, this time is independent of Cout. 

    If Cout were large, such that the LDO goes into current limit during startup that would change the output voltage's startup time. You can approximate at what capacitance the Cout will start to affect the startup time using using the below equations

    • For Vin < 3.5V
      • Cout = (0.09A/Vout)*0.001sec 
    • For Vin >= 3.5
      • Cout = (0.125A/Vout)*0.001sec

    If you are using a capacitor that is significantly larger than the calculated value above, then you can estimate the startup time as if the LDO were a constant current source charging the capacitor with the current limit being the constant current source value. So startup time=(Cout*Vout)/current limit.

  • 0 in reply to Kyle Van Renterghem
    TI__Expert 8550 points

    Hi Kyle,

    Thank you for the comment.

    Kyle Van Renterghem said:
      • Cout = (0.125A/Vout)*0.001sec

    Assume we have 5V -> 3.3V, we will get around COUT = 37.88uF.

    If COUT is larger than 37.88uF then the entire SS timer will be longer than 1ms, is that what you mean?

    We use ICL = 125mA to get min COUT threshold. 

    Summarization:

    If COUT < =  37.88uF, SS time will be 1ms.

    If COUT > 37.88uF(ex: 50uF), SS time will be 50uF*3.3V/(125mA~350mA) = 1.32ms  ~ 0.47ms.

    => Since SS time already take 1ms for internal VREF to build up so 0.47ms doesn't make sense, in that case the SS time will be 1.32ms~1ms.

    Kindly let me know whether my understanding is correct or not.

    Thank you.

  • 0 in reply to Evan Wang
    TI__Genius 14245 points

    Hi Evan, 

    Your understanding is correct. 

    For your example using a 50uF capacitor the SS time could be between 1.32ms and 1ms. The reason the fastest it would startup is limited to 1ms is that if the current limit for a specific unit comes in sufficiently high, then the Vout ramp will be limited by the reference ramp rate since current limit didn't engage until a higher value.