Digital signal, transistor switch, ring the bell

Discussion in 'General Electronics Chat' started by kgstewar, May 10, 2012.

  1. kgstewar

    Thread Starter Member

    Apr 5, 2012
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    6
    Hi all,

    I have a circuit that sends out a 12V pulse from a 40106 inverter. I would like this pulse to ring a mechanical bell. I believe the bell draws too much current to plug directly into the 40106 (and it's rated at 3-6V DC) so I was wondering if the attached circuit would work. If so, I'm not sure how to calculate the value of R1. Many thanks!

    Kevin

    bell is this one: http://www.homedepot.com/h_d1/N-25e...0053&langId=-1&keyword=doorbell&storeId=10051
     
  2. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    Any value from 1K to 10K would suffice
     
  3. kgstewar

    Thread Starter Member

    Apr 5, 2012
    151
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    Thanks, and the rest of the circuit looks ok? 2n2222 should suffice?

    Kevin
     
  4. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    The transistor is chosen according to the load it is driving.
    {ed}
    I believe 2N2222 is OK as long as u don't use a load more than 800mA
     
  5. #12

    Expert

    Nov 30, 2010
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    There is still a limitation not discussed. The transistor base needs a tenth of the current required for the bell. If the 40106 can't do that, you'll need 2 transistors or a Darlington transistor.

    (12V -.6V)/10k ohms = 1.14milliamps. Not enough drive current!
    11.4V/.03 amps = 380 ohms. Try 390 ohms. If that doesn't work, add more current amplification.
     
  6. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    or OP could use a buffer in at the 40106 output
     
  7. kgstewar

    Thread Starter Member

    Apr 5, 2012
    151
    6
    Thanks for all of the good info! I will try the 390 ohm resistor and I have a few ULN2003s lying around so may just go with one of those. I'm not exactly sure which pins to connect to where because I see the data sheet for the ULN2003 contains an inverter. So perhaps I need to bypass the 40106....


    Kevin
     
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