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Topic: Futurlec Opto Relay with Arduino (Read 1 time) previous topic - next topic

FishArduino

I need to control 2 Futurlec Opto 4 boards from an arduino. I am using a 12v supply to the board and I have the cable from the board wired as follows:

+5v to Arduino +5v
GND to Arduino GND
Data 1-8 to arduino digital pins

I am using the digitalWrite(x, HIGH) command to turn the pins on.

All of this to say:

IT DOESN'T WORK!

What am I doing wrong?

retrolefty

#1
Mar 30, 2011, 01:34 am Last Edit: Mar 30, 2011, 01:37 am by retrolefty Reason: 1
Did you use the proper mode commands in your setup() section to set each digital pin to be an output pin?

pinMode(Pin#, OUTPUT);


Lefty    

FishArduino

Yep.

An LED hooked up to one of the pins I am using will DIMLY turn on and off. Could I be pulling to much current from the Arduino?
The only other thing I have hooked up is a 20x4 back lit LCD.

Thanks

retrolefty


Yep.

An LED hooked up to one of the pins I am using will DIMLY turn on and off. Could I be pulling to much current from the Arduino?
The only other thing I have hooked up is a 20x4 back lit LCD.

Thanks


Without a schematic drawing of the relay board and posting your actual sketch I can't think of anything else for you to check. I searched for that board on the web, found a page but no schematic link shown.

Lefty

FishArduino

Schematic:



From what I can gather the board will switch for a "transistor logic 0". Now from my knowledge of serial communication, that would be a 0 bit, which is no current applied to the line in question. If this is the case, what keeps the board from switching all the relays on if the arduino is off?

Because this cannot be the case, I must a assume that the communication is more complicated than a simple on-off.

retrolefty

Link to schematic broke?

Quote
If this is the case, what keeps the board from switching all the relays on if the arduino is off?


Powered off would mean the pin is not able to pass any current. Powered on, but outputting zero is capable of sinking current up to 20-30ma, so active low to turn on a relay is fine if the other side of the relay coil (or opto-isolator emitter) is wired to +5vdc.

Lefty


Terry King

OK Got the schematic....

This is simple active-LOW inputs with a 1K resistor so less than -5 ma from Arduino,but you have to set the output LOW to turn these on...

At Arduino startup the outputs are in high impedance and the OPto outputs will be off. 

Have to set an Arduino pin to OUTPUT and LOW to turn an OPto output ON.

Opto Outputs have an LED that will go on when output is active.

You need +12V and 12V common to output section, +5 and Gnd from Arduino to input section...

There is no "Serial Communication" involved...
Regards, Terry King  ..On the Red Sea at KAUST.edu.sa
terry@yourduino.com  LEARN! DO! (Arduino Boards, Sensors, Parts @ http://yourduino.com

FishArduino

So, when I want the relay to be "on" I use the digitalWrite(pinX, LOW) command?

Thanks for the help!

retrolefty


So, when I want the relay to be "on" I use the digitalWrite(pinX, LOW) command?

Thanks for the help!


Yes, and use the digitalWrite(pinX, HIGH) command to turn the relay off.

This of course assumes you have already used a pinMode command to make pinX an output pin.

Lefty

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