please help a NOVICE understand a circuit diagram that says "load (relay)'

Discussion in 'The Projects Forum' started by helpmeunderstand, Jun 4, 2010.

  1. helpmeunderstand

    Thread Starter New Member

    Apr 23, 2010
    17
    0
    hello.
    I'm setting up beam-break detector. not from scratch mind you. I have a omron E3T-ST22 photoelectric sensor with a built in amplifier. there is a positive voltage output when the beam is interrupted.
    http://www.ia.omron.com/product/item/e3t_1135g/index.html
    [​IMG]
    I have set it up most of the way (that is, attaching blue wires to ground and brown ones to power. in this case 2 9V batteries (in series?) making about 17V of power. I'm using the same batteries to power both the emitter and the receiver. I don't know if that's ok or not.

    I have tested it, and basically it works fine. When I align the emitter bean and the receiver, and measure the voltage across the brown and black wires, I read about 0V, as expected (and the green light is on as described in the specs). when I block the beam, I read 17V (and the orange light also comes on, again, as it should). This seems fine, but high. I was expecting the output to be around 1 or 2 volts. I don't know enough about these things to know whether something is wrong or not. On the ratings/performance section on the website it says this
    Control output (Load current) 0 to 50 mA

    Control output (Residual voltage)
    (DC) 1 V Max. (Load current: Less than 10mA)
    (DC) 2 V Max. (Load current: 10 to 50mA)


    but i don't know what that means (residual voltage etc)

    Also, I'm not sure what the "load (relay)" mentioned in the output circuit diagram on the website is. I don't know what a "load" is. is that just connecting something to the outputs from the power? I suspect it has something to do with making sure the current is less than 50 mA. I do have 2 330 Ohm resistors because I thought maybe that with about 18V, that would keep my current at about 27mA. I did connect the brown wire to the black wire through the 2 resistors. nothing seemed different. My level on these matters is VERY basic. please talk to me like I'm in 3rd grade. I'm intelligent but have no real background in these matters. It would be helpful if someone said "1) connect this to that, then 2) this to that" etc.

    thanks,

    a
     
    Last edited: Jun 4, 2010
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Sounds like you have connected the circuit correctly and it is functioning as per the specs.

    With the beam broken ["dark" condition] the open collector output black wire will be pulled down via the switching transistor. The residual voltage is probably what you would measure between the black and blue terminals when the load is switched on. It seems to be current dependent. Going by the schematic, this voltage will be predominantly determined by the diode in series with the switching transistor. The on voltage of the transistor would normally be quite low - say the order of 200 mV.

    With the beam "on" or unbroken, the output switching transistor will be off and the black-to-blue terminal voltage would be virtually equal to the supply voltage [2x9V in your case]. There is a zener or surge suppressor diode shown on the receiver schematic which probably protects the transistor from damage by any high peak transient voltage generated during switching. It's normally considered good practice to place an anti-parallel diode across inductive loads - such as a relay but the spec doesn't seem to suggest this is required.

    The load can be whatever you want, but a relay is a typical device which would be used to control or operate something else. The advantage of a relay is that you can have electrical isolation of the sensor from other interconnected circuitry - say where you want to control a mains operated device such as an alarm bell.

    The 2x330Ω in series would give you an on state load current of around the 24 mA mark [(18V-2V)/660Ω] - assuming a 2V residual voltage at >10mA.

    All looks OK from my perspective.
     
    Last edited: Jun 4, 2010
  3. helpmeunderstand

    Thread Starter New Member

    Apr 23, 2010
    17
    0
    i have 2 more questions.
    when i tested the setup, I saw a change in voltage between the brown and black wires depending on whether the beam was broken. you mentioned blue and back?

    also, do you know of a way to de-amplify the voltage down to 4-5V after it comes out the black wire. I want to use the output signal as a digital input to a I/O board in a PC.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    The change in voltage will occur on the black wire.

    The blue wire will still have 0v on it.
    The brown wire will still have the supply voltage on it.

    See the attached. I've added a 1k Ohm resistor between the brown and black wires, and a 5.1 Ohm Zener diode between the black and blue wire, cathode (the end with the band) connected to the black wire. The 1k Ohm resistor supplies current to the Zener diode. When the sensor activates, the voltage between the black and blue wire will be around 0.7v. When the sensor is not activated, the Zener diode will limit the output to 5.1v.

    Your 9v batteries won't last very long powering these circuits; maybe 5 hours or so.
     
  5. helpmeunderstand

    Thread Starter New Member

    Apr 23, 2010
    17
    0

    thank you, i'll try it. I never knew about diodes. after looking at the disgram again (not yours), it looks like it has one inside already doing the same thing (ZD connecting black and blue). does this mean mine is broken in some way?
     
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