Switch Relay from just 2v ?

Discussion in 'General Electronics Chat' started by swifty, Aug 28, 2006.

  1. swifty

    Thread Starter New Member

    Aug 15, 2006
    6
    0
    Hi all,

    I have a small problem - I am working on a little project that requires me to switch a 12v supply on using a 2v signal.
    I have been searching Farnell and Maplin (UK) but can't find any relays that will activate on my 1.8 ~ 2v supply
    Any ideas how I can do this or if such relays even exist ?

    Cheers
     
  2. n9352527

    AAC Fanatic!

    Oct 14, 2005
    1,198
    4
    You can use the signal to switch a transistor and then a relay which is powered from the 12V line.
     
  3. swifty

    Thread Starter New Member

    Aug 15, 2006
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  4. Gadget

    Distinguished Member

    Jan 10, 2006
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    Well, the transistor will turn on at approx 0.7 volts. A resistor is series from the switch signal to the base to limit the base current should be all thats required. Try something like 1 or 2 k ohms. Fairly non critical.
     
  5. swifty

    Thread Starter New Member

    Aug 15, 2006
    6
    0
    Thanks for your reply gadget, would you mind clarifying one more thing for me as I'm a little confused when it comes to selecting the correct transistor.
    Looking at this page there are values VCEO,VCBO,VEBO - what are these as I can't figure out which one I need.
    http://www.maplin.co.uk/module.aspx?TabID=1&ModuleNo=19061&doy=29m8

    Sorry for my ignorance! :p

    Thanks again
     
  6. nomurphy

    AAC Fanatic!

    Aug 8, 2005
    567
    12
    Note that the relay you suggest has rated contacts of 2 Amps Max, therefore your supply load and the current going through those contacts should always be less than that 2 Amps.

    The relay coil for the 12V version is rated at 960 ohms (see spec sheet), therefore:

    0.0125A = 12V / 960 ohms

    This means the xstr you choose should have an Ic of at least twice that or 25mA (this math indicates that the Ic should not be much of an issue in your application).

    Now, technically, what you would do is divide 12.5mA by the xstr's hFE to find the maximum base resistor value (e.g., 12.5mA / 100 = 125uA, then 1.8V - 0.7V / 125uA = 8.8K).

    The minimum base resistor value is mainly determined by the amount of loading the input signal can tolerate, or the extent to which you want to load it. Figure that 2.0V - 0.7V = 1.3V and 1.3V / 2.2K = 590uA, which indicates that a 2.2K resistor shouldn't be a problem (assuming your input signal can drive at least 1mA).

    The upshot is, just about any NPN xstr with more than about twice the operating voltage (2 x 12V = 24V CEO) will work with the circuit you describe, and a 2.2K base resistor will be more than adequate. Therefore, the listed 2N3904 is quite common and would be a good choice.
     
  7. swifty

    Thread Starter New Member

    Aug 15, 2006
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    0
    Thanks for the great explanation - its made things a whole lot clearer :)
     
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