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TUSB321: Maximum drive current for Open-Drain I/O

Don Schelle
Expert 4150 points
Part Number: TUSB321


Hello,

I'd like to connect an LED to the VCON_FAULT# pin.    The LED requires about 10mA of current.   I plan to drive it from 5V with a series connected 300 Ohm resistor.

Will this configuration in any way damage the TUSB321?   

Don

  • 0
    TI__Guru*** 147390 points

    max current is 20ua, so 300ohm is too low.

  • 0 in reply to Brian Zhou
    TI__Expert 4150 points

    Hi Brian,

    I'm unsure of what you mean.   

    I see in the EC table that the maximum input low-level current is 10uA (not 20uA); however, I suspect that spec is targeted toward the devices input pins (PORT, CURRENT_MODE, OUT1, OUT2).    Regarding the open-drain output pins on the TUSB321 (VCONN_FAULT#, OUT1, OUT2, and DIR), we spec 0.4V with a 1.6mA sink current which implies that drawing 1.6mA is not only OK, but also guaranteed to produce no more than 0.4V at the output.

    My question is with regards to sinking 10mA.   While I understand that the output voltage won't be guaranteed to produce 0.4V, I'm not sure if this will adversely impact the long-term performance of the part.   There is no specified maximum current in the abs-max table, hence my question.

    What are your thoughts on this?

    Don

  • 0 in reply to Don Schelle
    TI__Guru*** 147390 points

    sorry, for output , it's 1.6ma

  • 0 in reply to Brian Zhou
    TI__Expert 4150 points

    Hi Brian,

    Will the part be damaged if this current is exceeded?

    Don

  • 0 in reply to Don Schelle
    TI__Guru*** 147390 points

    Those open drain pins are active low. When open drain pin is high, main current is leakage and pretty low.

    But when open drain pin is low, current is (5v-Vol )/Pu=(5-0.4)/300=15.3ma. So it's way too high than 1.6ma in datasheet. Pullup resistor is used to limit the current. Even if it does not damage the unit, it will still hurt its liability.