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TXS0104E: One side VCC power down

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Replies: 5

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Part Number: TXS0104E

Hi sir,

I have some inquiries on this device.

If the VCCA only power supply, and VCCB power down.

1) the state of the I/O pin on the B side is Hi-Z?

2) When the OE is set to L, does the B port side without power supply become high impedance?

Thank you.

  • Frank,

    During operation, Vcca must less than or equal to Vccb to function properly without damaging the device. During power sequencing this doesn't matter but during operation that equality (Vcca <= Vccb) must hold true. Doing this in operation can cause higher currents and can lead to damaging the device. Is switching side A to B not an option?


  • In reply to Rami Mooti1:

    Hi Rami,

    Thank you for your response.

    Yes, I knew the voltage should be Vcca <= Vccb in the normal operation,

    We consider the use of txs0104pwr in the buffer for the connection between the FPGA and the sensor.
    The connection is:

    A Port: FPGA
    B port: sensor

    Power supply
    A Port: 1.8 V
    B port:2.5V or 3.3V from sensor.

    There is a potential situation that a voltage is supplied only by the VCCA when the power is turned on.

    So I confirm the inquiries above, wanna know the I/O state, ensure that there is no hidden danger of IC damage. 

    I can find a description in the datasheet.
    8.3.3 Power Up
    During operation, ensure that VCCA ≤ VCCB at all times. During power-up sequencing, VCCA ≥ VCCB does not
    damage the device, so any power supply can be ramped up first.

    Is it also apply to the voltage be only supply to VCCA?

    Thank you for your clarification.


  • In reply to FRANK1:


    I'm not fully certain i'm following your explanation but what i'm getting is that there may be a scenario where VCCA has power before VCCB can ramp up. This is fine as long as when you're in that state there isn't a signal being passed. I'm understanding that you're aiming to hold the OE pin low to have the outputs in an high-impedance state until VCCB is powered on. That is fine and maybe not fully necessary. But if that's what they're looking to do, OE is referenced to VCCA and pulling it low will do just that.

    If you're aiming to have the device outputs be High-Z while the Vccb supply is off without the OE pin, this is something that the TXB family has. It has the VCC Isolation Feature. So when either VCCA or VCCB are at GND, all outputs are in the high-impedance state. TXB is more suited for push-pull architecture however.

    Let me know if i've answered your questions properly here


  • In reply to Rami Mooti1:

    Hi Rami,

    I'm sorry for my unclear expression.

    It is not required that I/O is Hi-Z when VCCB is power down,

    I wanna know how I / O work of TXS series.

    Please confirm the following points again.

    VCCA power supply, VCCB power down

    1) the state of the I/O pin on the B side is H, L, Hi-Z? which one?

    2) the state of the I/O pin on the B side H, L, Hi-Z? which one? when OE is set to L


  • In reply to FRANK1:


    Okay I follow. For a general answer here for both cases, when the OE pin is low the device is in HI-Z when it's not low, it isn't. For the first case, if you're trying to operate it in this conditions then I can't guarantee anything as this is outside of the recommended use case in the datasheet but I would warn against this, as it could damage the device. 
    Now if you're worried about a moment where Vccb isn't powered up yet but it will ramp up soon after Vcca but before the device becomes operational, it's recommended to hold OE low to remain in HI-Z until both supplies are ramped up.

    See in datasheet :

    To ensure the high-impedance state during power up or power down, OE should be tied to GND through a pulldown resistor; the minimum value of the resistor is determined by the current-sourcing capability of the driver.