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Indicator LED

Discussion in 'High Voltage' started by jack0987, Feb 14, 2010.

  1. jack0987

    jack0987 Member

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    In my house, I have some GFCI breaker plugs with what appear to be indicator LEDs that are on when the circuit is hot.

    How could I build a circuit to do the same thing and also give me an isolated TTL level output?
    A quick diagram would be most helpful.

    Later edit: I should say that I am only interested in the indicator LED here and not in the function of the GFCI.
     
    Last edited: Feb 14, 2010
  2. Hero999

    Hero999 Banned

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    The most important thing is whatever you do, do not open the GFCI up.

    The circuit should respond to the light emitted from the LED: I'd recommend using a photo-transistor/diode
     
    Last edited: Feb 14, 2010
  3. jack0987

    jack0987 Member

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    Thanks for the reply. I will not open the GFCI. I only used it as an example.
    I kindof need a circuit diagram because I do not know much at this point about non-mains circuits.

    Later edit:
    Prehaps this illustration of what I am thinking will help.

    [​IMG]
     
    Last edited: Feb 14, 2010
  4. dave

    Dave New Member

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  5. Hero999

    Hero999 Banned

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    There is no circuit.

    Providing the LED is bright enough, a photo-transistor will act like a normally open switch which is closed when the LED turns on. For a 5V TTL signal connect the photo-transistor's collector to +5V and the emitter to 0V via a pull down resistor, try 10k and the emitter goes to the logic input.

    If you like, you can draw the circuit and I'll check it.
     
  6. jack0987

    jack0987 Member

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    Thanks for the reply.
    The reason for this circuit is to remotely tell if the line is live and later add a latching relay to control its on-off state. I tried X10 modules but they are not reliable.
    I will try and draw it for you to check.
    Could you please suggest a photo-transistor to use.
     
  7. jack0987

    jack0987 Member

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    Please do not laugh. I really do not know what I am doing in designing this circuit.
    Here is my guess as to about how it should be:

    [​IMG]
     
  8. Hero999

    Hero999 Banned

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    No, that's not right.

    You don't use 110VAC, it will damage the logic input.

    Remove both the diodes, the top resistor and replace the 110VAC supply with 5VDC and it'll work.
     
  9. jack0987

    jack0987 Member

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    Hi:

    The 110VAC is the supply. The purpose of the indicator LED (the second diode) is to tell me there is power.
     
  10. Hero999

    Hero999 Banned

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    I think I can see what you're trying to do. The resistors act as a potential divider and the rectifier converts it to DC.

    This will not work becuase an opto-transistor will not have a high enough voltage rating to work from the mains and even if it did, it will produce 60Hz pulses not a steady high logic signal when the photo-transistor turns on.

    You need to use a separate 5V power supply to power the circuit and connect it as I described in my previous post. If you want an indicator you can still connect an LED in series with the pull-down resistor.

    I recommend using a wall plug 5V power supply but you can build your own using a transformer, rectifier, smoothing capacitor and an LM7805.
     
    Last edited: Feb 14, 2010
  11. jack0987

    jack0987 Member

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    Thank you so much.

    My reason for the first resistor is to drop the voltage to 5 or twelve volts.

    I am not sure I care if there are 60hz pulses as long as the led lights up and I can trigger a relay coil.
    I could add a small bridge rect. and a cap. but did not think it would be needed.

    Of course the purpose of the transistor is to isolate the output.

    What do you think? Maybe?
     
  12. Hero999

    Hero999 Banned

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    Here's how to use it to activate a relay.

    The LED and series resistor are part of the GFCI.

    EDIT:
    The 12V battery should be replaced with a 12V mains powered power supply.
     

    Attached Files:

    Last edited: Feb 14, 2010
  13. jack0987

    jack0987 Member

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    Thanks for the illustration. It enabled me to overcome some of the things I did not understand.

    Using that information, I have come up with a revised version as shown in the following picture:

    [​IMG]

    I will need a power on indicator LED as I am not using a GFCI (it was only an example of a device with one) and
    have added it to the circuit.
    Also I have added a second photo-transistor which is intended for isolated remote power on sensing.

    Please review what I have done and make suggestions. For example, I was thinking I may be able to use a cmos package which will contain 4 opto isolators. Please suggest a package and I do not know what to use for R values in some cases.
     
  14. Hero999

    Hero999 Banned

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    That's far more complicated than it needs to be.

    Where's the LED on the GFCI?

    You can't use an opto coupler because the IR transistor is encapsulated in black resin.

    I think my schematic might have confused you because it contained an opto-isolator.

    This simple circuit will do what you want.
     

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  15. jack0987

    jack0987 Member

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    Your first schematic was exactly what I was looking for and I only needed to add the power on LED to it which you have done in your second schematic.
    Perfect.
    At this point, I need to select a part number for the transistor and the photo-transistor.
    (I had said opto before but I was wrong.)
     
  16. Hero999

    Hero999 Banned

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    Well the BJT needs to be able to carry enough current to power the relay coil, assuming it's pretty small I'd use a generic low power transistor such as the BC547, BC337, 2N2369A etc.

    The photo transistor needs to be able to supply enough current when illuminated by the LED. Obviously it'll need to be shielded from ambient light. You also need to choose one in a clear package, not one with an IR filter.
     
  17. jack0987

    jack0987 Member

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    The relay will turn on a computer. The computer power supply is most likely 450 watts. At 110VAC that's ( I think) about 4.5 Amps.
    So, the relay will not be huge, but not tiny either.

    I was thinking the photo-transistor and led were all in one package?
     
  18. Hero999

    Hero999 Banned

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    No, that's an opto-coupler, forget the opto-coupler.

    Place the photo-transistor next to the indicator LED on the GFCI and cover with black tape so it doesn't get activated by ambient light. The relay will trigger when the LED on the GFCI turns on.

    The current through the relay contacts doesn't matter, it's the coil current you need to worry about.
     
  19. jack0987

    jack0987 Member

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    You have been most patient with me. Thanks.
    Due to my lack of understanding, I am now thinking I may have started this thread out on the wrong foot.

    First, I am not using a GFCI at all.

    Second, I am basicly looking to build a remotely low voltage controlled latched relay switch where the control line and sensor lines are isolated to protect the controlling equipment.

    In view of this, maybe we could use a latching relay with a AC coil and remotely trigger it through prehaps a SRC. This would keep the component count down and eliminate the need for a power supply.

    What do you think.
     
  20. Hero999

    Hero999 Banned

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    Why did you say you were using a GFCI when you're not.

    This is why it's important to give the whole picture in the first place.

    A relay provides isolation so there is no need for an opto-isolator.

    Build the normal latching relay circuit but connect the trigger to another relay with normally open contacts.
     
  21. jack0987

    jack0987 Member

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    I would still like to add the indicator LED. Could that be done as simply as this:

    [​IMG]

    Please suggest a value for R4.
     

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