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Circuit on state activates a second higher voltage circuit

Discussion in 'Electronic Basics' started by [email protected], Apr 14, 2007.

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  1. Guest

    Hey. I am sure this is simple, but I dont know the proper electronics
    jargon to search for an answer. Basically, I want the completion of
    one circuit to activate a second circuit but not allow feedback to the
    first. I have an alarm clock running on a AAA battery, and when it
    sends power to the buzzer (detached) I want it to activate a second
    circuit running off a 9V battery. The problem is, I dont want the 9V
    circuit to get back to the alarm clock, otherwise it would overload it
    (Please correct me if I am wrong). Is there any way to do this
    without causing major problems? Thanks.
  2. JeffM

    JeffM Guest

  3. Randy Day

    Randy Day Guest

    The circuit below should work, depending
    on the current needed by the 9v circuit.
    The NPN transistor acts as a diode to block
    9v from the signal input.

    .-. +
    | | your 9v circuit
    | |
    '-' -
    +v |
    signal ___ |/
    from -|___|-|
    clock 1k |>
    neg term.------+----- 9v battery negative terminal
    clock cct

    (created by AACircuit v1.28 beta 10/06/04
  4. Guest

    Thanks. I'll give this one a try.
  5. Chris

    Chris Guest

    Sure -- it's doable. But a simple transistor alone won't cut it, I'm

    Let's start our with your big question -- you're concerned that the 9V
    will get in to the alarm clock circuit and let the smpke out. That's
    a concern, but you can take care of it.

    To begin with, your alarm is probably a piezo beeper which is driven
    by applying the voltage to one side of the beeper (1.5V), and 0V to
    the other side. It then switches the voltage back and forth at the
    sound frequency (probably a few thousand times a second).

    This output is floating (separate battery). So from the outside, it
    looks like a 3V peak-to-peak square wave. You can feed that to a
    voltage doubler with schottky diodes, and that should provide a high
    enough voltage (about 4.5V or so) to drive a logic level MOSFET, which
    can turn on your circuit. Take a look at this (view in fixed font or
    copy&paste to Notepad):

    | .------------.
    | | |
    | .------o------. |
    | | + | |
    | | 9 Volt | |
    | | Circuit | |
    | | | |
    | C = 0.1uF | | |
    | | - | |
    | D=1N5819 '------o------' |
    | | |
    | | +|
    | .-------------. |D ---
    | | | C IRL2703||-+ 9V -
    | | | || D ||<- |
    | | o----||---o-->|---o------o---||-+ |
    | | | || | | | G |S |
    | | Alarm | D - C --- | | |
    | | Clock | ^ --- .-. | |
    | | | | | | |1 Meg| |
    | | o---------o-------o | | | |
    | | | | '-' | |
    | | | | | | |
    | '-------------' | | | |
    | '------o------o------------'
    (created by AACircuit v1.28.6 beta 04/19/05

    Tack solder two wires to the beeper element in the alarm clock. Build
    the circuit with 0.1uF caps and 1N5819 schottky diodes. Then snip the
    negative (black) wire going from your 9V circuit to the battery, and
    attach the MOSFET in series as shown.

    This will reliably turn on when the alarm is on. It may get a little
    iffy at very low AAA battery voltage, though. The MOSFET selected is
    beefy enough to drive any load a 9V transistor battery can supply.

    If this is too complex, or if you need a quick solution and aren't
    able to come up with the parts on short notice, you can use an
    external 12V wall wart, use standard 1N4001-type diodes, and use a
    darlington transistor to drive a relay. The relay contacts can then
    be used to turn on and off your 9V circuit.

    Please post again if this isn't clear, or you need more advice.

    Good luck
  6. Randy Day

    Randy Day Guest

    One caveat, as another poster noted.
    If the buzzer is fed by a square wave signal
    from the alarm, this won't work.

    If the alarm handles its own beeping, and is
    simply switched on by the clock, it will
    run for as long as the beeper does.
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