• State Resolved
  • Locked Locked
  • Replies 5 replies
  • Answers 1 answer
  • Subscribers 19 subscribers
  • Views 492 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

Original question:

SN74LVC16245A: SN74LVC16245A with a floating input pin

SN74LVCH16245A: HOLD option

amnon pomerants
Intellectual 780 points
Part Number: SN74LVCH16245A

Hello

Looking at SN74LVCH16245A (Buffer with HOLD option) datasheet, at Table 3 it states that for OEn pin at "1" the A, B buses are Isolated.

Does it mean that if I select DIR="0"  (B --> A) and OEn = "1" than A bus pins will float at HighZ, or that A and B buses are isolated from each other and which values ("0" or "1")  that were already  "captured" over the A bus will keep their values regardless of changes at the buffer inputs?

I am asking from the point of view of the device which receives the A bus signals, will it fill during OEn="0" a valid value ("0" or "1") or the inputs of this device will start floating?

My motive for the above questions is that I was asked to check my card status when part of my blocks are powered down in standby mode.

I saw some of the inputs to the buffer will start floating (no pull up/ down) so I changed the buffer to the Hold option.

But on the output point of view, if the buffer is powered up and its receiver is powered down, a "1" at the buffer output will feed an input of a powered down device, an action which violates its Absolute Maximum Values. I order to be on the safe side I need the OEn pin to really float the A bus output pins towards the receiver. Is it so?

Thanks

Amnon

  • 0
    TI__Guru* 96711 points

    Hi Amnon,

    The bus-hold circuits are always active -- the OE pin controls the device output drivers (all ON or all OFF). The bus-hold circuits are weak latches, and these will drive the bus to the previous value. if it is left floating by all other drivers.

  • 0 in reply to Emrys Maier
    Intellectual 780 points

    Hi Emrys

    I am not sure I fully understood your answer.

    I want to narrow my question to my "hurting" issue:

    Will the SN74LVCH16245A drive current into the  device (powered down) its outputs are connected to, although its OEn pin is at "1" ?

    ( I think that your answer is "yes", but I must to be sure...)

    Many Thanks

    Amnon

  • 0 in reply to amnon pomerants
    Guru 270760 points

    The OE pin disables the output drivers, so there will not be a high current.

    The input bus-hold circuits are always active, so they will be able to weakly drive current into the powered-down device. That current cannot be larger than II(hold).

    (In the unlikely case that you do not need the A→B direction, you could use unidirectional buffers.)

  • 0 in reply to Clemens Ladisch
    Intellectual 780 points
    Hi Emrys
    Many thanks for your prompt and elaborated answers.

    To complicate this issue a little bit, assume I also have a powered on device (assume an AND gate) connected to one of the HOLD buffer outputs.
    Can I say that no matter what is the level of the buffer OEn pin ("0" or "1") at power up, the AND gate input will never float as the buffer will always issue a steady "1" or "0" at its outputs (even when OEn="1") ?? True ??

    Thanks again
    Amnon
  • 0 in reply to amnon pomerants
    TI__Guru* 96711 points
    Hi Amnon,
    I can't take credit for the last response, that was Clemens, who is a community member that answers many questions on the forums.

    The SN74LVCH16245A is a transceiver, so every signal pin is both an input and an output -- and that means that every signal pin has a bus-hold circuit.

    Yes - as long as there is power to the SN74LVCH16245A, then the bus-hold circuits will hold all of the pins at a valid state, including outputs that have been disabled through OEn.