I am from IT background so I don't have much knowledge on electrical. This is maybe a very simple question. I want to connect a switch to act as a master "on/off" to the IND.IO. If the switch is off, the program in loop will not be executed. According to arduino samples, I should use pull down resistor to get rid of the "floating" state in the digital input pin. Is it the same in IND.IO-1286? The switch has two NO ports and both ports will be connected when the switch is on. I am thinking to connect +V from power supply to first NO port and second NO port to one of the IND.IO digital pin as INPUT. Will it work? I want it to send signal HIGH (24Vdc from psu) to the input pin when the switch is on but worry I would damage the IND.IO if I wire it wrongly. The same power supply is used to power up the IND.IO. Please kindly advise. Thank you =)
Hi Tom. So there is no need to connect any resistor in between the psu and digital input? I can connect V+ from psu straight to digital input when switch is on?
Hi Albert, that's right, you don't need a resistor between PSU and digital I/O. But as Stefano mentions, it's probably a good idea to use GND for your switch instead of 24V; define the channel as digital output, write HIGH (to enable the pull-up resistor), change the channel to digital input and read LOW when the switch is on.
I usually do the opposite for the very reason you state.
I write HIGH on the input to enable pull-up resistor and then connect the switch between the I/O and GND.
This way I feel more safe against mistakes.
But either way is just fine.
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|Asked: 10/9/16, 5:39 AM|
|Seen: 2229 times|
|Last updated: 10/11/16, 1:04 AM|
Hi Albert, the wiring you describe is correct, you can connect the 24V of your PSU to a digital I/O channel of the IND.I/O via your switch. As far as i know, the best way would be to first declare the channel as digital output, write LOW to it, then change it to digital input, e.g. in the setup of your sketch.