Opto interrupter control for relays

Discussion in 'The Projects Forum' started by craiggoff, Nov 10, 2011.

  1. craiggoff

    craiggoff Thread Starter New Member

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    I need to use a couple of transmission type opto interrupters to drive the coils of latching 12vdc relays. One opto interrupter will drive one relay coil and the other interrupter drives two relay coils. Each 12vdc relay coil requires 15mA of current to actuate. I am assuming that when the LED emitter beam is broken, the output of the interrupter's transistor will rise to allow the activation of the relay coil.

    Any suggestions for how to spec out the opto interrupters, assuming a 12vdc supply will be providing the voltage for the interrupters' LED and output transistor? Also, how do I size the dropping resistor for the interrupter's LED voltage and the output collector resistor? Perhaps there is already a simple schematic of this application. Thanks to you all!
  2. praondevou

    praondevou Well-Known Member

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    You could use any high CTR optocoupler with open collector output, e.g. the 4N46.

    However it sinks current (open collector). So when there is no LED current then the output transistor will not be conducting, i.e. your relay is OFF.

    With the 4N46 you could adjust the LED current to 2mA and should be good to go for 30mA output current.

    [​IMG]

    Attached Files:

  3. craiggoff

    craiggoff Thread Starter New Member

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    Thanks for the input and the schematic! I see that you suggest an optocoupler circuit, but in my application I have opto interrupters. I assume that the concepts involved are similar; would setting the resistance values for the opto interrupters follow the same guidelines?
  4. praondevou

    praondevou Well-Known Member

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    Oops, sorry about that, I didn't read correctly your post.

    I had a quick look at digikey.com, the common output current for opto interrupters seem to be much lower, not enough to drive 2 x 15mA relays.

    The ones I saw have an open collector output too. So you could simply put a pull-up resistor of let's say 2.7k at the output and then feed the signal to drive a N-channel MOSFET. 2.7k to charge a gate is ok of it's not a high-speed application.
  5. craiggoff

    craiggoff Thread Starter New Member

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    Thanks for the re-direction, it sounds like a simple enough way to go. I'm off to Digikey!
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