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For the txs0102 there is a requirement for VCCA<=VCCB. What happens to the device if this condition is not met?

This may seem like a strange question but I am inheriting a design where VCCB = 2.5v and VCCA = 3.3v and need to understand the criticality of this specification. I most likely will have to recommend a re-layout of the circuit board but need to be able to explain what will happen if the design is not changed.

  • Hello Scott,
    You will draw more current for Vcca because you are greater than 0.4V below Vcca. You are basically biasing a diode, which is not recommended. Power dissipation is increased but only very slightly and you may have some other issues as it was not designed to operate this way.
    Sincerely,
    -Francis Houde
    p.s. If you need a greater level of detail for your management then let me know.
  • Hello Francis,

    This is helpful, thank you. I am recommending that the board layout be corrected in the next spin but mostly needed to know if it would be a catastrophic failure or as you indicate, additional power draw with the possibility of a malfunction. In the meantime, we will work with the board "as is" and be on the lookout for other issues that might arise.

    Thanks again!

    Scott

  • Hi Francis,

    On a very similar note:

    I am planning on having VCCA and VCCB powered by *independent* 3V3 supplies.
    Due to noise, ripple etc it is possible that at any one time, VCCA or VCCB could be higher. (Despite them both being 3V3 nominal)
    Are you saying that there is only a problem if VCCA goes greater than 0.4V above VCCB?

    Additionally, under Absolute Maximum Ratings it mentions "Voltage range applied to any output in the high-impedance or power-off state".
    Does "Power off state" mean BOTH VCCA and VCCB are off, or does it mean that if VCCA was off and VCCB was on, that I could still apply up to 4.6V to VCCA with no damage?

    Many thanks for your help,
    James