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Unconnected pins on a 26LS33AC Quadruple Differential Line Receiver

Colin Foster
Prodigy 30 points

We recently purchased a product that uses the 26LS33AC chip as an encoder counter.  The manufacturer suggests leaving pins 1, 6, 7, 9, and 10 disconnected if we are only using a single digital input. What does the manufacturer recommend be done with these connections when not in use?

Our end result is that the pin 1 has floated to a higher voltage and we are seeing only 40 kohms to ground but are seeing 100 kohms for the rest of the pins, as the document suggests we should. Is there something that we can do to prevent this from happening in the future?

We were able to fix this problem by adding a 4.3kohm resistor to ground and the voltage on pin 1 seems to be what it should.

Thank you very much


  • 0
    TI__Mastermind 28265 points

    Colin,

    you are a single digital input of the entire chip or each receiver as a single input?

    The reason I'm asking, there are a few pins missing in your list of "unconnected pins"

    Please clarify.

    Thanks, Thomas

  • 0 in reply to Thomas Kugelstadt
    Prodigy 30 points

    Yes, I neglected to list every unconnected pin of the system.

    The design is to allow a maximum of pins 1, 2, 6, 7, 9, and 10. It is suggested that if you don't use the negative signals (1B, 2B, 3B; pins 1, 7, and 9) they should be left unconnected.  Additionally if you don't use signals 2A and 3A, they should be left unconnected.

    Eventually, our 1B signal floated to a high voltage and we had to pull down the voltage in order for the chip to function.

    I also believe that pins 13, 14, and 15 are unused and left unconnected.

     

    Thank you for the help.

  • 0 in reply to Colin Foster
    TI__Mastermind 28265 points

    Because in the last couple of weeks I received similar questions, I currently putting some material together that explains what to do and why. For a kick-off, you CANNOT leave both inputs of a receiver unconnected. The internal pull-up resistors will put both inputs on the same common-mode potetnial. From the truth table you can see that in order to get a deterimned output state, the differential input voltage must be either greater than +200mV or smaller than - 200mV.

    Leaving both inputs floating means zero differential input voltage. if that happens the output state depends on the position of the 60mV hysteresis band within the +/- 200mV input range. If you don't use a receiver unit at all, one input can be floating, but the other input must be connected to either Vcc or ground or any potential that is 200mV larger or lower than the 2.5V common-mode of the floating input. Then, and only then you will have a stable output state for an unused receiver.

    If you use the receiver in a single-ended configuration then you can leave one input open (which is internally pulled to Vcc/2 or 2.5V) while the other input receives the input signal.

    As I said I'm creating a short write up that should clarify this whole thing.(hopefully this afternoon).

    regards,

    Thomas 

       

     

  • 0 in reply to Thomas Kugelstadt
    Prodigy 30 points

    Excellent. Thank you for the help and please let me know when and where the write up is available.