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TPS92520-Q1: about the tps92520 Start-up waveform

qinlei
Intellectual 430 points
Part Number: TPS92520-Q1

Tool/software:

hello

i have one question about the TPS92520 ， Why is the frequency of the switching node very fast at the moment of startup?

What causes the switch node voltage like this？ please help to explain this.

Operating condition:TPS92682+tps9250 

TPS92680 output voltage:52v

tps92520 ouput voltage:2.6V  LED current:450mA. PWM duty:50%

please  refer to the following pciture.

  • 0
    TI__Mastermind 23226 points

    Hello Qinlei,

    Sorry for the delay I was out of town.  This is not a fixed frequency part and what we called psuedo-fixed frequency.  During ramp up it will hit min on and min-off times (which looks like really high frequency switching) because that is the fastest way to get to the setpoint.  This is expected behavior.  

  • 0 in reply to fhoude
    Intellectual 430 points

    hello 

    thank you for your reply , and i would like to ask if The converter reaches the minimum off-time of 57 ns (typ) when operating in dropout (low input voltage and high output voltage).Is the chip still working properly at this time? Is the LED current still at the desired value? Is there any relevant waveform for reference?

    thank you

    best regards 

  • 0 in reply to qinlei
    TI__Mastermind 23226 points

    The output current will fold back, meaning it will go lower current then setpoint.   For example if you have an open circuit the part will continue to switch but very little current will be output.  You really need the input voltage to be several volts above the output voltage to ensure it continues to regulate the current proproperly and this is also dependence on the switching frequency (CHxTON register setting) because that affect the max duty cycle the converter can achieve based on min off time.  

    -fhoude