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Latching Relay Circuit Help

Discussion in 'Electronic Basics' started by resrfglc, Dec 25, 2006.

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  1. resrfglc

    resrfglc Guest

    Latching Relay Circuit Help

    Years ago John had one of those cheap alarm switches for his van. It was
    supposed to work by making contact if the van was disturbed. It consisted of
    a springy piece of metal that would make contact if it was shaken.

    Problem was, it would only make a series of momentary contacts while
    whatever energy that was shaking his van continued and stop doing so as
    things quieted down.

    So I devised a circuit that would be triggered by the initial momentary
    contact and "latch" for X seconds providing the 12VDC to the Alarm Bell.

    Over the years I LOST the circuit diagram but kept the circuit. As I failed
    to label anything on it, I now have a circuit with six leads (One Red, One
    Black, two ORANGE and two WHITE with ORANGE strip(s).

    These attach to a small circuit board and two resistors [Red Red Green Gold
    1W(?), one capacitor [22mf 50v Electrolytic) and third device with three
    leads and a heat sink affair drilled for an 1/8" screw or bolt. It is
    labeled Motorola 827 and, above that appears "F 511" and above that what
    looks like an "I" and an "R" on either side of a stamped impression I cannot
    read. It is about 1/8" thick and 3/8" x 3/8" with the heat sink stuck on the
    back and extending 1/4"

    Looking at the black, three-lead device from the "front" I'll call the leads
    1, 2 & 3.

    I'll call the BLACK lead GROUND (-) and the RED lead 12V+ (good guess,
    probably correct - no?)

    That leaves two other "pairs of leads," ORANGE and WHITE/ORANGE.

    I assume one of these must be toe leads that when to the N/O Momentary
    contact switch (the alarm "sensor") and that the other two ran to the
    signaling device (12VDC bell).

    Which, he is asking you all, is which?

    Now, the Positive lead runs to a junction with: One of the ORANGE, one of
    the ORANGE WHITE leads simply tying all three together.

    The BLACK lead connects to PIN3 of the Motorola Device both resistors and
    the (-) end of the capacitor.

    PIN2 of the device is connected to the other ORANGE WHITE lead.

    PIN1 of the device connects the other ORANGE lead, the (+) end of the
    Capacitor and the other ends of the two resistors (probably could not get
    the resistor in the OHMS needed so ran two in parallel as a work around).

    As I recall, shorting across either the two ORANGE leads or the two ORANGE
    WHITE leads would send 12VDC down the ORANGE WHIITE or ORANGE leads (ringing
    the bell) for a period of time determined by the capacitor or resistors or
    both.

    If this circuit (or the Mororola device) is recognizable to you, please help
    me figure out 1) where to connect the trip NO MC Switch and load or (and) 2.
    where to find a schematic.
     
  2. Murph

    Murph Guest

    The Motorola device seems to be hooked up as a simple switch - when pin
    1 receives power, then pin 2 is connected to ground. I'm pretty sure
    it's just a simple NPN transistor, or even a FET, but whichever - it
    doesn't matter.

    What you need to do is connect the Normally Open switch between
    positive (either of the wires connected there will work) and the pin 1
    input. Then you connect the pin 2 output to the alarm.

    What happens when the switch closes is that power applied to the input
    of the transistor (we'll assume it is one), which then "turns on" and
    completes the circuit through the alarm (turning it on). At the same
    time, the capacitor is charged from the power coming through the switch
    (A low resistance path, so it charges quickly).

    Then, when the switch closes, the capacitor supplies the power to keep
    the transistor enabled (and the capacitor drains far slower than it
    charged due to the resistors). Since a transistor doesn't require much
    input current (if any), the capacitor only drains from the current
    going through those resistors - which isn't much, since they have a lot
    of resistance.

    Eventually, the capacitor's voltage drops low enough that the
    transistor turns off, and thus the output of the transistor is
    disconnected from ground - breaking the circuit through the alarm and
    turning it off.

    So yes. Pin 1 to the switch, Pin 2 to the alarm. I could draw you a
    schematic if you'd like. Also, if you have any questions on how to
    modify the circuit (changing the timing = changing the resistors'
    values), just ask (I suppose make sure to CC me in the response, I've
    never read this newsgroup before).

    Good luck / I hope I was helpful.
    --Murph
     
  3. Ross Herbert

    Ross Herbert Guest

    Resistor is 2.2Megohm
    Semiconductor possibly an IRF511 N chan mosfet
    http://www.datasheetarchive.com/datasheet.php?article=1807516

    With so few components you should be able to work out the schematic
    pretty easily.
     
  4. Ross Herbert

    Ross Herbert Guest

    Your schematic would be very similar to this circuit with your 2.2Mohm
    resistor in place of the 4.7Mohm and the 22uF 50V electro cap would be
    across the 2.2M (-ve to ground). The load in your case is the
    bell/alarm and the fet is the IRF511. The trembler sensor replaces the
    finger.

    Looking from the front of the IRF511 with the pins downwards, the
    terminal pins L - R are Gate, Drain, Source.
     
  5. Ross Herbert

    Ross Herbert Guest


    Forgot the schematic link
    http://www.discovercircuits.com/PDF-FILES/fingersw.pdf
     
  6. resrfglc

    resrfglc Guest

    Thank you. Device looks a lot like the diagram you referenced.


     
  7. resrfglc

    resrfglc Guest

    Thank you so much. Could not have asked for a better response. This list
    seems well-served by your interest here. My first day, too!

    I'll touch base off the list for that timing advice!


     
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