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IO light stuck ON

JOHNBONZJOHNBONZ Junior MemberPosts: 99
On my NI-3100, there are 8- IO ports, with amber led. My first light, channel 1, is stuck in an ON state . Does anyone know of a way to reset the led's to off?
I set up a screen to toggle the lights ON/OFF and all that works fine as the LED's toggle ON/OFF

But I ran into issue when I set up a doorbell push button ---one wire to ground and the other connected to channel 1 on device 17 on back of controller.
Then I connected 12 volt to other end of the door bell push button, and the other to the +12 volt on the port 17 of the IO in back of controller

So when I press door bell button the led lit up,then when I released door bell button, it turned off -- but after playing around the led on channel one stayed lit.
After I removed all the wiring to the IO port in back of the controller and rebooted, the led for channel 1 stayed lit.

so:
on door bell push button
- screw 1 - I have black and green wire
- screw 2 - I have red wire feeding 12 volt

on IO port in back of controller 17
- Gnd - black wire
- channel 1 - green wire
+12 volt - red wire

So is there a command , a setting to reset the LED's on the IO port to be reset to OFF?

Comments

  • MLaletasMLaletas Junior Member Posts: 226
    Just try to switch it off in control a device. Device 5001 Port 17 System 1, Channel 1, then select Off. If it turns off without the wire connected your good to go. If not that doorbell may have had some voltage on it and may have caused some harm in which case youll have to step it down.
  • JOHNBONZJOHNBONZ Junior Member Posts: 99
    yes I tried the control a device and it doesn't do anything to that LED channel 1. I removed all wiring to the IO port (doorbell button) and the light is still stuck ON. I was wondering if there is a command or setting somewhere to clear out the LED's
  • MLaletasMLaletas Junior Member Posts: 226
    You could try sending a command to master, LED-DIS which disabled all LEDs and LED-EN which renables them. Not sure if that will clear out your issue.
  • richardhermanrichardherman not-so-junior member Posts: 242
    Does that I/O channel actually still works?

    I do not understand exactly what you did by reading your post, but if somewhere in the setup you connected 12V (through the doorbell button) to that I/O input while it was pulled low, there is a real possibility that it is now broken. In that case the transistor where you applied 12V to it's collector is now effectively a short circuit because there ran too much current through it and the LED is actually tellling you the correct state.

    Happened to me once when a colleague accidentally connected a diode accross an external relay with the wrong polarity. The maximum current of the power supply ran through the I/O pin and it was forever short circuited....

    Richard
  • JOHNBONZJOHNBONZ Junior Member Posts: 99
    I could have shorted it but the manual states:

    Note: This IO port uses 5V logic but can handle up to 12V without harm. It
    can handle up to 12V on the input. At higher voltages you run a higher risk of
    surge damage.

    I used a 12 volt power supply when I hooked up door bell button to channel 1 of IO port


  • richardhermanrichardherman not-so-junior member Posts: 242
    Sure, it will handle 12V and probably higher than that, but what it can't handle is too much current. From memory the max stated current for an IO port is 200mA. If you connect a power supply without any current limiting to the pin and then pull it low (turn the channel on), something close to the maximum current the power supply can deliver will flow through the port en will damage it.

    But why not find out? connect a resistor (10K) from the 12V on the controller to the IO pin and measure the voltage on the pin with a multimeter. It should be a few 100mV when the channel is on and around 10V or so when it is off.
  • GregGGregG Just some guy... Posts: 249
    Last time I connected a push button to the NI it was as a dry contact closure. One leg of the button to ground on the IO ports and the other leg right to the IO port in question. No +voltage was needed.

    More info from the manuals:
    A contact closure between the GND and an I/O port is detected as a PUSH.
    When used for outputs, the I/O port acts as a switch to GND and is rated for 200mA @ 12 VDC.
    The PWR pin provides +12 VDC @ 200 mA and is designed as a power output for the PCS Power Current Sensors, VSS2 Video Sync Sensors (or equivalent).
    The GND connector is a common ground and is shared by all I/O ports.
    The input impedance on the I/O port is 22k.

    -That last item about the input impedence means that if you hit it with 12v directly, there is at least a 22k resistance in line already, so the current should be limited.
  • JOHNBONZJOHNBONZ Junior Member Posts: 99
    well if you put a volt meter black probe on the IO port ground and red probe on the channel 1 which is always lit, I get a reading of 1.7 volts. If I then take red probe from volt meter and test channels 2-8 I get a reading of 5 volts. If I test 12V slot I get 13.2 volts.
    so there is a 5 volts to each channel, except the one I shorted apparently - or since it is in ON state I get a lower voltage reading. so I wonder if I can get that channel to shut off and maybe return to 5 volts
  • richardhermanrichardherman not-so-junior member Posts: 242
    GregG wrote: »
    Last time I connected a push button to the NI it was as a dry contact closure. One leg of the button to ground on the IO ports and the other leg right to the IO port in question. No +voltage was needed..

    Correct, no voltage is needed, but it seems that's what the OP did. If you want to control a relay, a light bulb or a LED (with a series resistor), you would connect one end to, say, 12V and the other end to the port. That's completely valid.
    GregG wrote: »
    More info from the manuals:


    -That last item about the input impedence means that if you hit it with 12v directly, there is at least a 22k resistance in line already, so the current should be limited.

    No, I don't think that's what it means.If there was a 22K resistor in series with the channel when it is 'ON', you could get about 500uA (0.5mA) before the voltage on the channel would reach 12V and it would be impossible to switch something like a relay. Although not exactly clear, it probably is the input impedance (that's AC 'resistance') when the channel is off, that is when it is used for voltage input. The OP seems to have used it as an output AND applied 12V to it. That can go wrong...

    @JOHNBONZ: If you shorted it (and it looks like it) most likely you can't get it to shut off any more. The transistor that 'acts like a switch to ground' is likely shorted and you either live with that (there are 7 other channels...) or send it in for repair.


  • GregGGregG Just some guy... Posts: 249
    "When a circuit is driven with direct current (DC), there is no distinction between impedance and resistance"

    From:
    https://en.wikipedia.org/wiki/Electrical_impedance

    My guess is that the 22k impedance is when the IO is being used as a passive input. There would have to be an impedance difference between the Off and On state when being used as an output.

  • richardhermanrichardherman not-so-junior member Posts: 242
    We could discuss the impedance case in depth, but let's not :). It's not related to the question.

    When the port is used as an output, it acts like a current sink. It is most likely configured as an open collector (or drain) with some sort of weak pull-up to 5 volt. So when used as an output, when actvated, the port effectively shorts the pin to ground. I don't know if it's internally current limited, but would not be surprised if it isn't, that's more difficult than it seems, especially if you want to sink a substantial amount of current (which 200mA is, in this context).

    So my best guess, without knowing anything specific of the electrical internals of the I/O ports, is that the OP connected the power supply to the pin when it was 'on' (pulled low) and destroyed the internal current sink (transistor). So the stuck LED is actually correct, the output is low and wil stay that way, short of repairs.
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