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Pull pin input to ground

Sean Reid
Prodigy 50 points

I'm using the G2253 and have to signal something that uses active low logic. The device keeps all the lines high and I need to pull it low to get it to operate. How can I sink the 3.3v to ground with this device? I know that the device in input mode will be high impedance which is fine so I keep it high when I do not intend to signal it, but how can I take this current in and sink it to ground when I do intend to signal it?

Thanks

  • 0
    Prodigy 30 points

    How about add a big value resistor in between and output low from G2253 when you want to signal the other device?

  • 0 in reply to Tao
    Prodigy 50 points

    I'm wonder if I could use the internal pull down resistor when I want to signal a low and then set it to high impedance again when I want it to stay high. Does that sound reasonable?

  • 0 in reply to Sean Reid
    Guru 12050 points

    > signal something that uses active low logic

    "Something" is not a allowed word on this forum, post the device name so we can look at the datasheet.

    Weak pull-down on your side will not work, if the line is pulled-up by 10k for example your weak ~50k will just act as
    a voltage divider and only reduce the voltage by 20%, what you need is complete connection to ground.

    If you are trying to emulate open drain, use DIR as the toggle of state.

    Set OUT=0 and keep it there

    To toggle pin.
    High: DIR=0
    Low: DIR= 1

    Do you have external pull-up resistor or does the something-device have one built-in?

  • 0 in reply to Tony Philipsson
    Prodigy 50 points

    Can't post the device name as it's for a company that I'm working on a school design project for. The schematic is not available publicly. But the device does have pull up resistors (10k). It pulls the lines to 3.3V which I need to bring to ground so that the device on the other side sees the line as a low.

    By setting the direction of the pin to out with a low going out, it will bring the current to ground? Having the DIR as an input I understand because it's a high impedance. Setting the pin to output allows current to come in?

  • 0 in reply to Sean Reid
    Guru 12050 points

    The copper trace (line) is like bath tub, the pull-up resistor slowly (some ns) fills it up until it's full at 3.3v.

    If you want to show a low state, drain the tub by  open a huge hole in the bottom of it
    = connection the trace to ground by: out=0 dir=1

    Sure the 10k resistor is still trying to fill the tub up, but it's trickle compared to the 0.5 Ohm drain hole
    But it's safe as the 10k limits how much water it can waste (current) while the hole is wide open.

    As either side can not (or at least should never) pull the line hard to Vcc,
    a electrical short can not happen in case of miscommunication.

    That your secrete device is active-low, does not mean it can output anything and then regular OUT= 0/1 will be safe.
    And then a pull-up resistor is not needed unless the 5uS boot uptime of mcu to set the pin to 1 is not OK.

  • 0 in reply to Tony Philipsson
    Prodigy 50 points

    Thanks. I appreciate the help. It makes sense.

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