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Original question:

TXS0102: What will happen with I2C if loosing power on VCCA or VCCB

TXS0102: TXS0102

Lin Sen
Intellectual 260 points
Part Number: TXS0102

Hi，

The TXS0102's recommended low-level input voltage, VIL is 0.15V, which is too low even for AP from QUALCOMM, APQ8053, in which VOL max is 0.4V. But per notes in TXS0102, 'The maximum VIL value is provided to ensure a VOL is maintained. The VOL value is VIL plus the voltage drop across the pass-gate transistor'.

My questions are:

1. Does it mean even the actual VIL is higher than 0.15, for example, 0.4 of APQ8053, the VOL is simply 0.4+voltage drop? Because per block diagram of TXS0102, input and output are linked by a pass-gate transistor;

2. what is the value of 'voltage drop' across the pass-gate transistor? I need this value to evaluate the 'digital low' voltage of input and output voltage by using TXS0102.

Thank you very much and look forward to hearing from you soon.

Rgs/Sen. 

  • 0
    Guru 243180 points

    1. Yes, the specified VIL is needed only if you want to get the specified VOL.

    2. The voltage drop is the difference between the voltages, i.e., 0.4 V − 0.15 V = 0.25 V.

    Please note that the pass-gate transistor and the AP's output transistor have a constant resistance (0.25 V / 1 mA = 250 Ω for the TXS; see the IV curve for VOL in the AP datasheet), so if the actual current is less than 1 mA, both voltage drops will be smaller.