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Simple question about this "trip wire" circuit.

Discussion in 'General Electronics Discussion' started by PizzaCombo, Dec 24, 2013.

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  1. PizzaCombo

    PizzaCombo

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    Dec 24, 2013
    Hi all, first post here and still getting the hang of electronics :).

    So I am looking at making the circuit found on this page http://www.instructables.com/id/Extremely-Easy-and-Safe-Trip-Wire/ and I think I understand it all, except for one part.

    So when the "trip wire" has been broken, the piezo buzzer makes noise as the current is forced to go through the base of the transistor, and lets the circuit flow.

    What I'm not understanding is when the trip wire is still connected, why is there not any current being applied to the base of the transistor? Why does the current just go right past the base and straight to ground? I have a feeling it's something to do with "Electricity wants to take the path with least resistance" but I'm not entirely sure what's going on.

    I hope this is clear enough, thanks.
     

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    Last edited: Dec 24, 2013
  2. davenn

    davenn Moderator

    13,812
    1,945
    Sep 5, 2009
    hi there
    welcome to the forums :)

    yes that's basically the answer. to take it further, the transistor needs ~ 0.7 of a volt on the base to turn on to make a conduction path between the collector and emitter to operate the buzzer

    With the loop intact the base is held at 0V ( the negative of the battery) so the transistor cannot turn on
    When the loop breaks, the base voltage will rise to ~ 0.7V or so and the transistor will turn on and allow the buzzer to operate

    cheers
    Dave
     
  3. PizzaCombo

    PizzaCombo

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    Dec 24, 2013
    Thanks Dave, good explanation. So essentially by being connected straight to the negative terminal of the battery, it makes the base of the transistor grounded?
     
  4. davenn

    davenn Moderator

    13,812
    1,945
    Sep 5, 2009
    yup, so a voltage cannot develop on the base pin in that condition

    when the loop is broken, a voltage divider is produced by the 2 resistances
    1) the 10k resistor from base to +V and
    2) the base - emitter junction resistance of the transistor

    knowing that it takes ~ 0.7 - 1V drop across the base - emitter junction to turn the transistor on
    we then know about 5V will be dropped across the 10 k resistor
    from that we can work out the current flowing through the 10k resistor and base - emitter jnct
    if we wanted to..... it will be very low !

    Dave
     
    Last edited: Dec 24, 2013
  5. PizzaCombo

    PizzaCombo

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    Dec 24, 2013
    Cheers Dave, helped me understand it.
     
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