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SN65C1167E: Behavior of DE pin in open #2

Part Number: SN65C1167E

Please add this to Liu Yang's question and let me know.

Looking at the DE pin waveform on the PCB board,
The level is maintained at around 0.4 V,
but it rises to around 1.1 V instantaneously.
(Width of 5us to 15us)

How many microseconds does the input signal need to be H to operate the drive circuit?

Is the signal ignored (the drive circuit does not move) if the H level time is short?

  • Hello,

    Once the DE pin toggles its logic state (low to high or high to low), the output of the device will become enabled within 19 ns (see tPZH/tPZL specifications in Table 6.7 of the datasheet).  So, I would expect a pulse with width ranging from 5 us to 15 us to enable the driver circuit and create a visible pulse on the output.

    Regards,
    Max

  • Hi Max,

    Thank you for your answer.

    I'm sorry, there was an error in the content of the previous question.
    I was "width from 5 us to 15 us" but it was correctly "width from 5 ns to 15 ns".

    If the DE pin signal is "5-15 microsecond pulse width Higher than 0.6V",
    is the drive circuit enabled?

    For reference, the waveform measured with an oscilloscope is attached. 
    Such a waveform sometimes occurs (about every few seconds), 
    but the time width of 0.6V or more is about 5ns.

    Regards,
    kanetomo

  • Hello Kanetomo-san,

    I would not expect these glitches to result in the driver becoming enabled.  It would be difficult to say that it could never become enabled given that the device's response to these kinds of short-duration, low-amplitude pulses was not fully characterized, but I believe the probability of this happening should be low.  If you wanted to remove the concern entirely you could try filtering the DE pin with a small capacitance to ground; this should be effective at suppressing the ~5-ns glitches.

    Regards,
    Max Robertson