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Original question:

[FAQ] Can I leave "...." pin floating on my I2C I/O expander or I2C switch? (INT, Reset, Port I/O, ect.)

TCA6424A: Port input without pull-up

Gaurav Gothoskar
Prodigy 110 points
Other Parts Discussed in Thread: TCA6424A, TMAG5231

Tool/software:

Hello,

I have a circuit in which I have 2 TMAG5231 sensors connected in a diode-or configuration and the output is going to a pin of the TCA6424A. The or circuit has an 10k external pullup such that when the TMAG sensor is active, the TCA6424A pin pulls low and when the sensor is inactive the 10k pullup resistor pulls it high (please refer to the diagram shown below). both VCCI and VCCP are tied to the same 3.3V source. I have multiple of such sensors connected in the same configuration and it works as expected. 

on some sensors the PCBA house did not install the 10k resistor but even then I can see VCC on the TCA6424A pin when the sensor is not active and is pulled to gnd when the sensor is active indicating that it has some sort of an internal weak pullup. I measured the resistance between the VCC and the pin of the TCA6424A and it shows around 345K. is this expected? 

I am going to add the pullup going forward but I am curious to know where this leakage path is coming from. also what is the max value of this resistor pullup I can use and still get reliable operation? I have 24 of such circuits and when all 24 paths are active, the effective resistance is around 416ohms (24 10K resistors in parallel) increasing the current draw to around 8mA. I need to minimize this current draw by increasing the value of the pullup resistor.

Thank you.

  • 0
    TI__Mastermind 47328 points

    Gaurav,

    The input leakage for the Pxx pins is +/- 1uA. The pin can either be leaking a small amount of current to VCC or GND when the CMOS input is left floating (in your case, if the 10k PU resistor is omitted). 

    If operating at a VCC = 3.3V, then theoretically the PU equivalent resistance could be ~ 3.3V / 1 uA = 3.3 Mohm. To measure this accurately, you would need to remove the IC from the PCB. If you measured from the Pxx pin to VCC while the IC was still soldered to the PCB, I expect the resistance measurement to be less since there are other devices connected to that pin. 

    The Pxx pin should not be switching fast enough for a large PU resistance to matter. You could try 100k PU resistors instead of 10k to reduce current 10x. 

    Regards,

    Tyler