Activate a rele with 0.5v DC, question.

Discussion in 'The Projects Forum' started by seraser, Jan 8, 2014.

  1. seraser

    Thread Starter New Member

    Jan 7, 2014
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    Hi people, great forum, thanks.

    I am join to this topic because my I have a very similar source signal, I need activate a relay with a signal of 0.6v, DC, measure with an oscilloscope.

    Is the option by tom66 of this topic valid for my situation?

    http://forum.allaboutcircuits.com/showthread.php?t=43203

    Thank you very much.
     
    Last edited: Jan 8, 2014
  2. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    Can you explain the nature of this .6V source?
    What is it..What are you trying to do?
    details get proper answers..
     
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  3. wayneh

    Expert

    Sep 9, 2010
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    Triggering a relay with a small change of input voltage is a good job for a comparator. The output can control a transistor which in turn can operate the relay.

    Is this signal DC, so that the relay triggers at any voltage above, say 0.4V, or is it an AC signal that you want to trigger when the AC amplitude exceeds some voltage?
     
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  4. seraser

    Thread Starter New Member

    Jan 7, 2014
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    Hi, thank you for the responses.

    This signal comes from an optimus device with an output of alarm, a residential and hospital alarm system for residents.

    The signal is 0.6v DC pure measured with an oscilloscope, one second 0.6v DC and one second 0v, cyclic, this signal ony appears if the alarm is powered for one resident using one switch.

    I want activate one relay with this signal and use 220v AC on outputs of the relay to turn on lights or a clausor.

    Please, one schematic?

    Tom66 schematic is not clear for me.

    Thank you very much.
     
  5. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Are you taking that output across an LED or at the base-emitter junction of a transistor? If so, try probing around and you will find a different node on the circuit board where a similar signal will swing a few volts...
     
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  6. seraser

    Thread Starter New Member

    Jan 7, 2014
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    Hello, thank you for your reply.

    This output is not based on led or sound signal, is pure output with connector posibiities.

    But yes, the board have to much relays and maybe one of these is for this signal, I need take a look.

    Thank you.
     
  7. wayneh

    Expert

    Sep 9, 2010
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    Maybe the output you are looking at is an open collector for sinking current, not sourcing it. It may be designed to ground one side of an external relay coil.
     
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  8. MikeML

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    Oct 2, 2009
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    Good suggestion. Try this external test circuit connection: The LED should light when your output is at 0V, and should go off when your output goes high...
     
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  9. seraser

    Thread Starter New Member

    Jan 7, 2014
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    Hi all, thank you very much for your replys.

    Finaly I found a 1.9v DC voltage provided to one led when the alarm is turn on, I think this is best option, and maybe I can use an operational amplifier, any little help please?

    Thank you.
     
  10. MikeML

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    Oct 2, 2009
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    Did you try the 9V battery and LED test?

    What did you measure the 1.9V delta with respect to? Red meter probe on LED anode; Black meter probe on LED cathode?

    Do you have any reason to think that the LED cathode is connected to the Optimus pcb ground? Does that pcb ground connect to the Power LineCord building safety ground on the Optimus?

    I am trying to figure out if we can share a common ground between the Optimus an your external circuitry?

    It would be cleanest if you could make the connection as you said: "This output is not based on led or sound signal, is pure output with connector posibiities."
     
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  11. seraser

    Thread Starter New Member

    Jan 7, 2014
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    I am not in house, I need some components, for sure on monday I will do it, thanks.
    Black meter probe on LED cathode, Red meter probe on LED anode.

    LED cathode, ground, is not pure ground, is provided by one IC, if you want I can share his nomenclature, on monday.
    The optimus have one conector for these outputs, and on these outputs are very clear the signal - and ext 0.6v DC.
    Ok, Ill go try the outputs signals, no led.
     
  12. seraser

    Thread Starter New Member

    Jan 7, 2014
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  13. MikeML

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    And I predict that it will not work. The coil current to operate the AGQ2001H relay will be about 90mA. There is no way that much current can be diverted from the LED circuit.
     
    Last edited: Jan 14, 2014
  14. seraser

    Thread Starter New Member

    Jan 7, 2014
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  15. MikeML

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    An opto-coupler will work, but you will have to put it in series with the existing LED (assuming you want it to keep functioning normally), and possibly reduce the existing current limiting resistor to compensate for the forward drop of the existing LED plus the forward drop of the newly-added photo-emitter side of the opto-coupler.

    Have you ever tried to see if the existing connector output is an open-collector type output? The 9V battery/LED/Resistor circuit I posted will discover if it is. If it is, your solution is at hand.
     
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  16. seraser

    Thread Starter New Member

    Jan 7, 2014
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    Thank you MikeML.

    Ill go try the two variants, first ext output, but one question please, I am employeer of one enterprise and I need request the components, not with my money.

    If this led circuit work what about the materials to run in this condition?, I want request all components and one MOC 3010 and work with 2 posibilities, first your solution, for sure.

    Thank you very much.
     
  17. MikeML

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    If the external LED test circuit I posted lights the LED during the alarm condition, here is how to proceed: If the original equipment maker provided an open-collector type of output, it is likely that it will drive a small relay directly.

    Try a relay with a 12Vdc coil. Get one that has a coil resistance of 240Ω or higher. You can wire the contacts to run your external alarm ciruit.

    The diode is used as a transient suppressor (snubber). Almost any small Silicon rectifier diode will work for this.

    The 12V source is any small plug-in nominal 12Vdc power supply (here we call them a wall-wart) capable of supplying 100mA or more. (Looks like this:click here) Voltage regulation is not critical; no load voltage could be as high as 15V which can drop as low as 11V when the relay coil is connected to the supply.

    You will use the same pins on the output connector as with the LED test. Almost certainly, one of the pins in that connector is intended to be used as the Common between the internal and added external relay circuit.
     
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