phone call detector(voltage0

Discussion in 'The Projects Forum' started by sneharaikar32, Sep 1, 2011.

  1. sneharaikar32

    Thread Starter New Member

    Jul 31, 2010
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    0
    take a look at the circuit diagram.... i want to remove the led...and in place of the led want to insert in two wires and connect them to another breadboard...on this breadboard i want to connect leds in series and give them the voltage input from the original circuit...so in short want to take the voltage output from one breadboard and give a voltage input to the other breadbard which in turn will light the LEDS on the second breadboard....but the problem is the voltage from the first breadboard is very low ...i wanted to know how i can amplify this voltage before supplying it to the LEDS.....PLZZZZ HELP!!!!! P.S. IF ANYBODY HAS DONE THE CIRCUIT PREVIOUSLY....PLZ TRY AND HELP ME OUT...my buzzer keeps ringing without any detection....that is when i supply the voltage..the buzzer directly starts when it is supposed to start only when it detects calls...y is this happening....???/:(:(:(:(:(
     
  2. Audioguru

    New Member

    Dec 20, 2007
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    896
    The extremely simple circuit picks up mains hum radiated from mains wiring in your home. It also picks up many local radio and TV stations. So of course the buzzer is on all the time. It is worse if you built the circuit with a mess of wires on a breadboard.

    The LED can be replaced by the base-emitter junction of an NPN transistor that can drive LEDs in series and in series with a current-limiting resistor.
     
  3. wayneh

    Expert

    Sep 9, 2010
    12,100
    3,036
    In series with the LEDs on your second board, put a transistor configured just like the one in the schematic you provided. Use a current limiting resistor just as R4 is used. You will need to reduce the ohms of that resistor if you have several LEDs in series. But you can probably leave it alone to get started. The LEDs will just be dim.

    On your first board, replace the LED by a 5K ohm resistor, or any value between ~1k and 20K. Take the signal from the end of that resistor opposite ground, and feed it to the base of the transistor on your second board. Now when the original LED would have turned on, the second board will be turned on.
     
  4. Audioguru

    New Member

    Dec 20, 2007
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    No.
    The original transistor for the LED is an emitter-follower. The new transistor must be connected as a common-emitter switch with the LEDs and current-limiting resistor at its collector. R4 will limit its base current.
     
  5. wayneh

    Expert

    Sep 9, 2010
    12,100
    3,036
    Oops, yes that's right. The emitter, of the transistor on the 2nd board, should be grounded and not wired the same as the first transistor. I should have looked more carefully at the diagram - I just assumed...

    But even that won't solve the hyper-sensitivity problem. For that you'll need shielding and maybe some attention to filtering.
     
  6. Audioguru

    New Member

    Dec 20, 2007
    9,411
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    Hyper-sensitivity?
    Didn't the very old circuit come from a magazine in India where the mains electricity rarely works and there are hardly any radio and TV stations?
    Then of course the circuit has hyper-sensitivity in a modern Western city.
     
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