timed relay closure on power loss

Discussion in 'The Projects Forum' started by justinlw, Jul 22, 2007.

  1. justinlw

    Thread Starter Member

    Jul 22, 2007
    11
    0
    I need to build two circuits to activate when trigger power is lost. One to run 5-10 seconds, and on at 60-90 seconds. Power to the circuit will be 12vdc in a vehicle.

    Better explanation: When I turn the key off and accessory power is cut by the ignition switch, this should trigger a timer circuit to activate a relay to continue power supply to the ignition power circuit. This will keep the vehicle running for an additional 60-90 seconds after key off.

    The second circuit will be the same, but to provide a 5-10 second output to power another accessory relay.

    I am quite the beginner in circuitry, any help will be very much appreciated.

    Thanks in advance,
    Justin
     
  2. justinlw

    Thread Starter Member

    Jul 22, 2007
    11
    0
    Anybody? I know I can build a 555 or 556 circuit to run this. I read this in other posts. But can the 556 run two different time functions? Also, to trigger I guess I could use a relay that closes on discharge to send the trigger to activate the timer. Does this sound correct?
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Sorry. I started a long-winded reply to you earlier, but got disgusted and gave up when my computer crashed. :p

    To keep the engine running for a period of time after the operator has decided to turn the machine off may not be a wonderful idea. For example, if the throttle cable became stuck wide open, you would want to turn the thing off in a really big hurry. Or, for example, the engine caught on fire (happens all too often with fuel injection) - not being able to turn it off would result in a rather spectacular (and costly) fire.

    I suspect that the reason you wish to keep the engine running for a period of time is to avoid the phenomenon of "heat soak", where the temperature continues to climb after the engine is turned off due to having been run hard prior to shutdown. Or is it for another reason?

    Are you by any chance working on a Jaguar V12?

    A 555 is a versatile timer chip. A 556 is two 555's in one IC.

    Turning the ignition on immediately after it's been switched off may result in a loud "bang" followed by a very loud noise coming from a now-blown-open muffler or catalytic converter, due to the possibility of an unburned charge of fuel/air mix being dumped into the exhaust manifold (the engine's still turning for a revolution or two with no ignition). It could be possible for the relay to be transitioning when the ignition system should've been firing. You'd need a latching-type relay circuit that would always be engaged with the ignition on, and your circuit to break the path to the relay's coil.
     
  4. justinlw

    Thread Starter Member

    Jul 22, 2007
    11
    0
    This is a diesel engine. I'm converting it to vegetable oil. The 5-10 second pulse will open an oil purge solenoid on shutdown. You have to purge the vegetable oil out of the system so that it doesn't harden in the fuel rail. The 60 sec. timer will keep the engine running to purge the rest of the oil out of the injectors. Basically, the engine starts with diesel fuel, and once the engine hits operating temp. it switches to veg. oil. On shutdown it has to purge the oil and replace it with diesel. I'm building the timer to automate this process.
     
  5. justinlw

    Thread Starter Member

    Jul 22, 2007
    11
    0
    Exactly. I need to have a relay hot, and then use my circuit to keep it hot for the set time. I don't understand the build of the 556 circuit completely. I've found a couple schem. and still don't quite understand how to control the timing. I'm guessing I'd need a monostable circuit? I have years of mechanical engineering, and auto mechanic experience. The problem is that I see things mechanically, and don't quite get all the variables in circuits.:confused:
     
  6. John Luciani

    Active Member

    Apr 3, 2007
    477
    0
    A 555 monostable will work.

    There should be application circuit diagrams in the 555 datasheet showing the
    monostable configuration. IIRC you need to be careful with long time periods.
    Another good source that has the 555 circuits are Don Lancaster's TTL Cookbook
    and CMOS Cookbook.

    On the www checkout http://www.uoguelph.ca/~antoon/gadgets/555/555.html

    If you think you may want to add other *features* to your circuit you may want
    to use a uC.

    (* jcl *)
     
  7. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Like John said, long time periods can be tricky.

    Be sure to use a diode across the relay, and another to protect the relay driver circuit - otherwise, when the power is removed from the relay, the coil will try to maintain the current flow, causing a large reverse polarity voltage spike. The diode across the relay will provide a short path for the reverse voltage.

    Here's one for you to work with, using a 555 timer IC:
    1) Connect pins 8 and 4 to 12v
    2) Connect pin 1 to ground.
    3) Connect one end of a 1M potentiometer (VR1) to +12v, the wiper and other end to pins 7 and 6.
    4) Connect a 100uF capacitor (C1) from pin 6 to ground.
    5) Connect pin 5 to ground using a .01uF capacitor (C2, prevents false triggers)
    6) Connect pin 2 to ground using a normally-open momentary SPST switch (S1).
    7) Connect pin 2 to +12v using a 10K resistor (keeps trigger high until S1 is closed)
    8) Connect pin 3 (output) to the + side of the relay's coil, using a 1N914 diode, and the other side of the relay's coil to ground.

    When S1 is depressed momentarily, a timing cycle is begun, and the relay is energized for the entire cycle. C1 being 100uF allows a range of around 10 to 175 seconds, or about 70 seconds at the midpoint of adjustment. If C1 is replaced with a 10uF cap, you'll have a range of about 2 to 15 seconds. Once you have the desired amount of time "dialed in", you should consider replacing the pot with a fixed resistance value for higher reliability.

    I suggest you use industrial or preferably mil-spec components, as the environment under the hood of a vehicle is a harsh one, easily exceeding commercial specs. It would be better yet to enclose your project somewhere in the passenger compartment, to protect it from the elements.
     
  8. JohnnyD

    Well-Known Member

    Aug 29, 2006
    79
    0
    Hi,

    I designed a circuit last year that outputs an adjustable length pulse (from memory I think it is adjustable between 5 seconds to 1 minute) to activate the total closure feature on my car from the remote central locking. if you PM me your email address I can email you the schematic and a few notes if you'd like?
     
  9. JohnnyD

    Well-Known Member

    Aug 29, 2006
    79
    0
    Just a thought....

    If I were designing this circuit, I'd make it so that it will only keep the engine running if you are holding down a pushbutton at the same time as killing the ignition. That way it is much safer, as you can still kill the engine normally in an emergency.
     
  10. justinlw

    Thread Starter Member

    Jul 22, 2007
    11
    0
    I'm including a bypass switch that will be flipped off when instant shutdown is needed without total system purge, like at a drive thru window...

    Thanks again for all the help, you guys are great.
     
  11. n9352527

    AAC Fanatic!

    Oct 14, 2005
    1,198
    4
    This might not be appropriate, but on my wvo system I only use the wvo on longer journeys. So I just make sure that about 5 minutes before reaching the destination or stopping, I switch back to diesel. Wouldn't it be annoying to wait for the engine to shutdown when you've arrived? :D
     
  12. justinlw

    Thread Starter Member

    Jul 22, 2007
    11
    0
    that's the idea behind the timer, and automatic purge cycle. That way, you shut the key off and walk away. The timer does the rest.

    They market a box called a diesel timer for about $200 that runs the engine after key off to prevent coking of the turbo bearings. This is basically the same idea, with an added circuit for WVO purge.

    I have a temp switch for automated start up of the system once the oil reaches temperature, so the system is mostly automatic. Here in Houston, TX, it only takes about 6-10 minutes to activate the WVO system, and I drive an hour or so to work, and a little longer on the way home.

    I drive a 6.0l F250 Crew Cab with upgrades producing about 475-500 rear wheel horsepower, so I want to burn free fuel every chance I can.
     
  13. n9352527

    AAC Fanatic!

    Oct 14, 2005
    1,198
    4
    Interesting, an off topic question if I may, what make of injector pump does that engine have?
     
  14. JohnnyD

    Well-Known Member

    Aug 29, 2006
    79
    0
    Here is a link to my total closure schematic. You can ignore all the circuitry before the 555 trigger input, thats just stuff specific to my application.

    When triggered, it should output a pulse on the OUT pin which is variable between 2.5 and 39 seconds. A pulse applied to the 'Motors Unlock' terminal will stop the OUT pulse at any time.

    I have used the wrong voltage regulator for my circuit. It should be either 9 or 5 volt, not 12.

    http://img.photobucket.com/albums/v330/agua-moose/Electronics/TOTAL_CLOSURE_REV-B.jpg
     
  15. justinlw

    Thread Starter Member

    Jul 22, 2007
    11
    0
    The injectors are driven by a high pressure oil system, in which each injector is electronically controlled. The fuel suppy on the diesel side is the oem pump, external electric. The WVO pump is a FASS brand, 80psi., 90gph.

    When the engine reaches 160 degrees, it triggers the WVO relay. Once the WVO gets up to pressure, it triggers a pressure switch which cuts voltage to the factory pump. The fuel rail has check valves installed on the diesel side to stop WVO from flowing into the petro system.
     
  16. justinlw

    Thread Starter Member

    Jul 22, 2007
    11
    0
    if I'm looking at this correctly, a monostable 555 circuit can be set up so that trigger voltage remains constant, and on low votage (key off) the system will start the timed event? I see that it will activate on downward spike, or upward spike with a transistor in place. If this is true, I have my application figured out. Any comments on this?
     
  17. JohnnyD

    Well-Known Member

    Aug 29, 2006
    79
    0
    Would that depend on if the 555 is triggered by the falling edge or the rising edge?
     
  18. justinlw

    Thread Starter Member

    Jul 22, 2007
    11
    0
    The trigger in the circuit I'm looking at looks for falling edge. The transistor is what would make it look for the rising edge.
     
  19. JohnnyD

    Well-Known Member

    Aug 29, 2006
    79
    0
    I just had a thought...

    If your circuit detects the ignition turning off, and the keeps it on for a number of seconds, would it keep repeating this cycle over and over? Because once the timing circuit has elapsed, the ignition will turn off which would trigger the whole process over again?

    Like I say, just a thought, I may be wrong.
     
  20. justinlw

    Thread Starter Member

    Jul 22, 2007
    11
    0
    Just figured that out this afternoon, so I'm going to use a normally closed relay with a diode on the ignition trigger. The contacts will open when the key is on, and close with key off sending a ground signal to trigger the timer.
     
Loading...