Relay don't work

Discussion in 'General Electronics Chat' started by hassan maysara, Feb 24, 2013.

  1. hassan maysara

    Thread Starter New Member

    Oct 23, 2012
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    I have made a circuit using Ic 4017 and it worked but when i connect the output (about 4.2v) to 3v relay the relay didn't work so what is the solution????
     
  2. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    You can´t feed an output of a logic IC directly to a relay, you need to use a transistor switch.
     
  3. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    The 4017 will not be strong enough to drive a relays, as it can only have an output current of about 6 mA.

    You will need a driver to drive the relays.

    Bertus
     
  4. hassan maysara

    Thread Starter New Member

    Oct 23, 2012
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    sorry for this question because i am beginner :) any kind of transistor will be useful or a specific one ??? please write the number
     
  5. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    first you need to specify the relay, mainly what current does it need to operate.
     
  6. hassan maysara

    Thread Starter New Member

    Oct 23, 2012
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  7. sheldons

    Active Member

    Oct 26, 2011
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    This gives you an idea of the sort of relay driver you can use
     
  8. hassan maysara

    Thread Starter New Member

    Oct 23, 2012
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    Thanks sheldons... I really appreciate your circuit
     
  9. panic mode

    Senior Member

    Oct 10, 2011
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    note that relays require certain power to work. in general low coil voltage means large current. depending which of the three 3V relays you have, coil current can be up to 150mA. also note that since transistor is used as switch, gain is lower. in general this means only gain of 10 or so. therefore you will want to reduce R1 value from 3.3K to something smaller because:

    if your 4017 is also powered from 3V and it's output is close to 3V, you will have at best base current of
    Ib= (3V-0.7V)/R1 = 0.8mA

    to get 150mA, you would need gain 150/0.8 = 187.5 which is huge for switching application.

    assuming we get 6mA from 4017 that resistor would need to be

    R=V/I = (3-0.7)/0.006 = 390 Ohm (rounded to common resistor value)

    in this case you still need gain of 25 if your relay coil is 0.45W.
    if the relay coil is 0.2W, you only need gain of 11.

    if you are using standard LED (poor efficiency), you will need R3 of about 33 Ohm.
     
  10. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    Be careful, you only have a 3V relay coil. If you apply 12V to it as the example shows, the relay will not last more than a portion of one operation.
     
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