that all six interrupters got encoder wheels,
You never mentioned encoder wheels before, this changes everything! What are they like?
What we are trying to do here is second guess the intentions of the engineers who designed it, previously I was only thinking about a blocked / unblocked sensor but if there are encoding wheels then it is likely that these are not just simple opto slot detectors. It could be that they are incremental encoders like you find in computer mice.
I am not sure where you got the C & E markings from on your initial diagram. It is possible that this is wrong and that would make the other lines (the ones going to your connector) from the detector part not the emitter part.
So I think you will have to be a bit more precise in what is actually happening. It would be good to see what sort of signal is actually being produced by this unit when it is in operation. Can you measure them on a scope and see what signals they have.
I mean to say that, this works if and only if the wheel moves.
Is this ALL wheels move? If so over what time period. That is what that chip could be about. It is a retriggerable monostable acting as a missing pulse detector. Movement could reset the monostable, if no movement is detected in a certain time then the monostable times out and produces a falling edge, then that edge triggers the other monostable in the package which fires an interrupt to stop the machine.
These other questions may or may not be relevant now I know about encoder wheels.:-
Yes a series resistor is there,
Where? we are talking about one resistor per opto slot, is it on another board?
And by the way, what do u mean by arduino outputs??
Do u mean to define 6 pins as output(out of 0 to 13 digital pins).
Yes but never use pins 0 or 1 because they are used by the serial port.
I'm not able to understand what is the connector here.
The 12 pin header on your board.
Also, please tell how to provide power to this arrangement
Wire up +5V and ground to the correct pins on the header. You have identified the +5V as being at the end and connecting to the +ve of the IC but you need to trace the ground.
I see you have sent me a PM, at the moment these do not work and so I can't read it or reply to it.