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Timed circuit for operating an automotive relay

Discussion in 'Electronic Basics' started by lon, Jun 17, 2005.

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  1. lon

    lon Guest

    Hi there, wondering if anyone could help me out with a circuit.

    I would like to make a circuit which basically continues to output 12 volts
    for 5-10 secs after an input signal (12 volts) is removed. If, during the
    5-10 seconds the input voltage is restored there would be no loss of output

    Hoping someone could lend me a hand with this. I have no idea if this is
    terribly difficult or not.


  2. It depends on how much current you need the circuit to supply. If it's
    small, then simply use a large capacitor. I've had a red LED flasher
    operate from a charged up half farad supercapacitor for hours.
  3. lon

    lon Guest

    I believe the circuit must supply approx 0.13A @ 12V to drive the relay.
    (Bosch 12v 30 amp relay)
  4. Bob Monsen

    Bob Monsen Guest

    You need to tell us what it's for. Otherwise, we can't even begin to
    tell you if it's possible, or easy. The problem is that you are going to
    power something with that 12V, and how much power it uses influences how
    to do this.

    If there is 12V available elsewhere that can just be switched in, like
    in a car, that makes it easier. Actually storing the energy, and feeding
    it back is possible with a battery, or with a capacitor, depending again
    on what you are powering.
  5. Tom Biasi

    Tom Biasi Guest

    I'm guessing a headlight delay? There may be a setup already for your
  6. This does what you describe:

    A breadboarded version worked OK. Let me know if the schematic and
    notes need further explanation, or if you want to see the output
  7. lon

    lon Guest

    Hi there, wondering if anyone could help me out with a circuit.
    This circuit is to activate an automotive relay. The current needed to
    drive the relay is approx 0.13A @ 12V.
    So this would be the requirements for signal out of the circuit.

    Signal in is actually 2 12 volt lines but i figured i could combine them
    with diodes and a filter into one input line.

    This input line is usually 12 volts, but it may go low for brief periods of
    time. During these transitions I want the circuit to output a continuous
    12v signal. If however the input line goes low for an extended period (lets
    say 5 or 10s), I would like the circuit to produce 0 volts.

    Sorry for any confusion and thank you to the replys.

  8. ehsjr

    ehsjr Guest

    You can do it with a diode, relay 2 capacitors & a resistor.
    Here's a diagram - the parts list is below it.

    | |
    +12 signal in ---------+--relay coil--+--Gnd
    | |
    C2 R1
    | |
    +12---+---relaypoint--+--Bosch relay coil---Gnd
    Existing wiring------+

    Relay - RLY-407 from
    D1 - 1N4001 diode
    C1 - 1000 uF to 2200Uf (~5 sec to ~10 sec) 50v
    C2 - .47 uF
    R1 - 47 ohms 1/2 watt

  9. ehsjr

    ehsjr Guest

    After my previous post, it occurred to me that you might
    want an adjustable time period, so here's a new circuit.
    The existing Bosch relay circuit is assumed to be the
    top line of the diagram - everything below is what you add.

    +12---o/ o--------------------+---BoschRelayCoil-----Gnd
    | |
    ----- |
    / \ D1 |
    --- |
    | |
    | | /
    e \ R1 | \
    Q1 |--/\/\/\---+---------->/ R3
    c / | \
    | --- C1 /
    | --- |
    | | R2 |
    +12-------------+ +---/\/\/---+

    C1 - 470 uF 63v
    D1 - 1N4001
    Q1 - 2N2222
    R1 - 4.7 K 1/4w
    R2 - 10 K 1/4w
    R3 - 50K pot

    The length of time that the Bosch relay stays energized
    after the switch is turned off depends upon the specific
    relay drop out voltage and the specific transistor gain.
    R3 allows for varying the time from a minimum of about
    2 seconds to a maximum of about 10 seconds. If you increase
    the value of C1, minimum on time will increase.

    The circuit was tested with a relay that draws about 100 mA.
    Because your relay draws more, it is likley you will need to
    increase the value of C1.

  10. Jamie

    Jamie Guest

    you could simply drive a heavy current type transistor using
    current source mode (the emitter feeding the 12 volts to the relay coil)
    drive the base from the 12 REF through a silicone diode for isolation
    into a cap that will be used to hold a charge, the - side of the cap
    will go to ground, from the + side of the cap which is also connected to
    the Cathode side of the isolation diode goes to the base of the transistor.
    the collector of the transistor comes from your main 12 supply.
    the emitter will drive one side of the relay coil, the other side of the
    relay coil will go to ground.
    basically the transistor is used to allow the use of a smaller cap which
    will hold a charge ..
    the relay it self will produce its own latching effects.
    its cheap and dirty but it works.

    you should also use a silicone diode across the coil to suppress the
    fly back effects in the relay coil to prevent high voltage damage..
    simply put the cathode of the diode (line side) on the same connection
    that is being connected to the emitter of the transistor, the anode side
    of the diode connected to the other leg of the relay coil.
    not using a transistor will require you to use a much larger cap to
    hold the relay in but with experimentation you may find that 1 diode
    and a cap could be sufficient..!, in this case the cap would be
    connected across the relay coil and thus will suppress the Flyback
    effects when the ref voltage is removed ..
  11. ehsjr

    ehsjr Guest

    Ooops - missing diode D2. Corrected in E-mail to op, figured
    I'd better correct it here, too:

    +12---o/ o---------------------+---BoschRelayCoil-----Gnd
    | |
    ----- ---
    / \ D1 \ / D2
    --- -----
    | |
    | | /
    e \ R1 | \
    Q1 |--/\/\/\---+---------->/ R3
    c / | \
    | --- C1 /
    | --- |
    | | R2 |
    +12-------------+ +---/\/\/---+
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