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NEED HELP! Simple DC circuit

Discussion in 'Electronic Basics' started by DaveK, Mar 14, 2007.

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  1. DaveK

    DaveK Guest

    What do I need? 555/556 Timer? Flip-Flop? I have a basic
    understanding of electronics, but most likely would not be able to set
    up this circuit without a schematic. At this point in my education,
    telling me what to do, without a schematic, probably wouldn't help me.
    Believe me I'm trying as hard as I can to learn this stuff. I'm
    really getting hooked on it, but I'm not even close. I've played with
    the timers but that's about it. This circuit doesn't seem that it
    would be very hard for someone who knows what they are doing, but then
    again what do I know? Thank you for your time and any help would be
    greatly appreciated.

    I have a 4.5vdc (battery operated) circuit with three momentary
    switches. The 1st two switches are in series and when closed power a
    buzzer for the amount of time that they are both closed. The third
    switch when closed powers a LED. What I want to happen is when the
    third switch is activated, but not continually closed (momentary
    switch) it will trigger power to the LED and continue to stay on till
    one or both of the 1st and 2cd switches are closed once again. The
    third switch may never need to be on, but if it does it will need to
    stay on till one of the other two is closed.
    TThank you for your time and any help would be greatly appreciated.
     
  2. This is a latching circuit.
     
  3. Guest

    Start thinking about your switches not as driving components directly
    but only as means of providing a 5V (high) or 0V (low) level. So you
    want a High state when switch 1 is closed until switch 2 or 3 is
    closed. In other words: 1 sets the output to high and (2 or 3) resets
    it. Take a look at the following schematic which includes an
    interactive demo :
    http://www.play-hookey.com/digital/rs_nor_latch.html
    The website explains how the individual gates can be built from
    transistors. You might need to invert the input/output and OR the 2 &
    3 inputs together. Again, instructions can be found on the website. If
    your transistor count gets too high you might want to look at the
    74HCxxx or similar ICs that contain gates; I don't know how much
    current they can supply, so you might still need to provide a
    transistor to drive your light.
     
  4. Jamie

    Jamie Guest

    you need an R-S flip flop. (R = Reset), (S= Set). they create a
    latching effect to the Q outputs.
    just about any FF with R&S with work.

    Or, you can use 2 Transistors of NPN types.
    Look here, this is what you can do with transistors which is
    a Monostable set/reset flip flop.

    http://ourworld.compuserve.com/homepages/Bill_Bowden/page9.htm
     
  5. ehsjr

    ehsjr Guest

    Here's a circuit that will do what you described. I
    assumed a relatively low current requirement for
    the buzzer.





    PNP 2N3906
    +5 -+--- ----------[1K]--------+
    | e\ /c |
    | --- [LED]
    | | |
    | | [SCR1] P0118MA2AL3
    | | Pb3 /|
    | | ___ | |
    `----------+---o o---[47K]---+ +-----------+
    | | |
    | | Pb2 |
    | | ___ |
    | +---o o---[Buzzer]-----+ |
    | | | | |
    | | +--->|-----+ | |
    | | D2 | | |
    | | Pb1 | | |
    | | ___ | /c |
    | +---o o---[1K]---|---| NPN |
    | | | \e |
    | +--->|-----+ |2N2222 |
    | D3 | | |
    | D1 | | |
    `--->|---+---[.1uF]---+ | |
    | | | |
    | | | |
    [1K] [10K] | |
    | | | |
    Gnd ----------------+------------+-----+--------+

    Pressing Pb3 turns on the SCR, which stays on, lighting the
    LED, until current through it is interrupted. The PNP is
    biased on through D1 and the 1K resistor. When either PB1
    or PB2 is pressed, a pulse is generated through the .1 uf
    capacitor, which briefly biases D1 and the PNP off,
    interrupting the current through the SCR, so it turns off.
    An SCR (Silicon Controlled Rectifier) is a device that
    once turned on by an external signal at its gate, stays on
    even though the external signal goes away, until current
    through it is interrupted.

    Pb1 turns the NPN on while it is pressed, connecting one
    side of the buzzer to (-). If Pb2 is also pressed, the other
    side of the buzzer is connected to (+), and it sounds.

    The SCR is mouser part # 511-P0118MA2AL3, and the diodes
    are 1N4148 or 1N914 If the buzzer draws more than about 200 mA,
    substitue a relay for the buzzer in the circuit, and use the
    relay contacts to complete the circuit to the buzzer. If this
    is a "hardware store" variety buzzer, you'll need the relay with
    a 1N4148 diode across it backwards, and will need to add a 1N4004
    diode across the buzzer too (also backwards). "Backwards" means
    the cathode end (the end with the stripe) connects to (+) instead
    of (-).

    Ed
     
  6. Bill Bowden

    Bill Bowden Guest


    You might need a smaller resistor in series with the LED. I think the
    SCR holding current is around 5mA and you only have about 1 mA in the
    LED. The SCR will drop 1.5 or more, and the PNP will drop another .5
    and the LED drops 1.8, so you only have 5-(3.8) = 1.2 volts across the
    resistor, or 1.2 mA.

    -Bill
     
  7. ehsjr

    ehsjr Guest

    You're right - thanks for picking up on it.
    That 1K should be 81 ohms. I made 2 mistakes -
    I figured 220 ohms, then wrote 1K. And I didn't
    figure in the drop in the SCR. The SCR would
    hold at 220 ohms, but current would be only a
    little over 5 mA, so that 1K needs to change to
    81 ohms. That'll give ~ 15 mA with a 5v supply.

    Ed
     
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