Hey guys. I have recently started living with a bunch of people, which means that there is a laundry room that we all share. THe problem is that it takes a long time to go from my door all the way to the laundry room door, open it, and check whether a machine is available.
There is an intelligent controller on the side where you pretty much swipe your card and enter the machine number to enable the washer. Somehow, this controller knows which are "available" and which are not. I looked inside the control panel and found out that there are two screw terminals for each washer/dryer. One GND and another AVAIL (available) wire. They are 3.31 V on high and .10V on low. I am thinking about making a simple LED row of lights that I can stick on the OUTSIDE of the room. Since the only really long wire I have is a telephone cord, I plan to multiplex 4 LEDs and maybe a speaker on that line.
I was just wondering if I should be aware of anything before I start. Like whether I'm doing the input monitoring correctly: common all the grounds, attach each AVAIL pin to an INPUT on the Arduino with a 10K to the common ground on each input pin.
Since you're modifying someone elses stuff you want to be careful.
The 3.3v can drive a photo couple that can then drive a transistor and relay. You'll need you're own power supply on the output side of the photocouple.
The arduino is over kill for that application.
Heh, this would actually be pretty simple. It would just be 4 LEDs just outside the laundry room so that people don't have to open the door.
I also believe that the wires used to tell whether the machine is available and the wires used to turn it on after you pay are different.
This will all be powered off a 9V wall wart.
This is how I was thinking of hooking it up, but since hooking it up to input pins are like hooking it to a 100M-ohm resistor, is it necessary to use R1 and R2?
(and is there better software to make these kinds of drawings? this was painful. Fritzing seems to crash a lot so I didn't use it.)
Yes, we understand that the inside part will be powered from the 9V wall wart. My question was how you are going to power the FAR end.
AH. So I mentioned I am going to use a telephone wire. The telephone wire has 4 conductors inside, which I think are all the same. I was planning on using Charlieplexing.
Eagle! Of course...how could I forget.