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TPSM861257: Clarification on Enable pin behavior

Patrick Theisen
Prodigy 10 points
Part Number: TPSM861257

Tool/software:

Hi TI Expert,

I'm having issues with my component. I believe I interpreted the documentation incorrectly and I wanted to double check. My input voltage is 14.4V and I was using a 110K resistor to a 10K resistor for my voltage division. This resulted in a voltage of 1.2V to the Enable pin. I now realize 1.2V is the minimum value for enable high and my enable pin voltage should be within 1.25V to 5.5V. I'm switching my 10K resistor to 30K for a 3.08V input to my Enable pin. I attached the original schematic below for reference.

My question relates to how the 2M ohm internal pulldown resistor functions. When calculating my voltage division should I take into account the 2M? Or does it only play a role when there's no bottom resistor?

I also wanted to know if the low voltage to my Enable pin could cause the component to pull too much current and burn? I'm trying to figure out what caused my component to catch fire and that is my best theory.

Thank you so much,

Patrick

  • +1
    TI__Intellectual 2600 points

    Hi Patrick,

    If two resistors are in parallel (R1//R2), the equivalent resistance value will be nearly equal to lowest resistance value.

    i.e. 10k//2M = 9.9k.

    So, if you take that 2M is in parallel, it won't affect too much in the calculation.

    but if we don't use the voltage divider at the enable pin, then by default 2M (internal resistor) at enable pin pulls down the maximum voltage at that pin. and enable pin has maximum voltage range is 5.5V, so it can cause damage to that pin.

    Regards,

    Manish