A circuit to temporarily maintain 12V for 5 second window

Discussion in 'Power Electronics' started by meld2020, Jun 21, 2016.

  1. meld2020

    Thread Starter New Member

    Mar 14, 2016
    21
    0
    This is for a project that utilizes a v-twin 25hp motor, but I feel the nature of the question is more electronics oriented, thus I have posted here.

    We have constructed a device that drives itself with a small 25hp gas engine. The transmission is hydraulically actuated and we use a series of remotely-controlled valves to spin one side of the drivetrain (left front and back, or right front and back,) so effectively a skid loader in a sense. Anyhow, the remote control for this unit has e-stop functionality, which is provided in the form of a 12VDC normally-closed line available off the hydraulic transceiver's harness. The hydraulic transceiver is powered by the 12V battery, which is the primary cold cranking source for the engine. Simply put, everything is sourced from the battery.

    I want to integrate this function to the engine to shut down the fuel pump. On the engine it's simple: ground a wire and it does just that. My original thought was to run the 12V (NC) from the hydraulic harness into the control line/coil of a normally-closed relay so that if the e-stop is engaged, it open circuits the e-stop line and the relay is closed, grounding the ignition shutoff circuit and disabling the fuel pump. This circuit should work fine, but I need something that won't be disrupted when the engine cranks. While cranking, the 12V is temporarily unavailable, thus the primary relay channel will close and ground the engine shutoff circuit, presumably not allowing it to start.

    Is a timer or a capacitor the best fundamental approach to this issue? My thought is to either create a timed latch or create a temporary capacitance, though admittedly the capacitance might not be the safest approach if an emergency stop situation is desired.

    Thoughts? References to example circuits?

    Thank you,
    Mel
     
  2. FrancescoC

    Member

    Nov 22, 2014
    30
    5
    Hi,
    For precise timing I would use a small Microcontroller (Try PIC micro). You can then control a relay, or a MOSET directly, to control your voltage.

    This would mean you have to program the Micro. The tool to do that are freely available.

    Regards

    Francesco C
     
  3. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
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  4. BobaMosfet

    Active Member

    Jul 1, 2009
    109
    11
    Nothing complex is necessary-- All you need is a 12VDC source, and an RC circuit that will keep a transistor holding the 12V feed open for 5 seconds.
     
  5. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
    1,144
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    That too, if the currents are not excessive. A diode isolated capacitor can keep deal with droop too.
     
  6. ian field

    Distinguished Member

    Oct 27, 2012
    4,415
    783
    If the current is high - you can get one farad supercapacitors from places that sell high power car audio systems.

    But they're not cheap - there are cheaper supercapacitors, but they may have to be stacked in series to handle 12V.
     
  7. meld2020

    Thread Starter New Member

    Mar 14, 2016
    21
    0
    What about addressing it at the battery? Would a "higher rated," presumably referring only to cold cranking amps capability, significantly prevent voltage drop, or is this usually inevitable during starting?
     
  8. KeepItSimpleStupid

    Well-Known Member

    Mar 4, 2014
    1,144
    203
    Theoretically, it could lower the voltage drop.

    With a "traditional" automotive ignition switch, the accessories turn completely off during starting. Note the radio doesn't play. A higher CCA means a lower effective battery series resistance.
     
  9. BobaMosfet

    Active Member

    Jul 1, 2009
    109
    11
    Inevitable. When starter not turning, it's a short circuit. You avoid this to everything else by creating your own 'mini-battery' with an RC circuit that will meter out the cap's charge at 12V until the starter circuit is no longer in a shorting condition.
     
  10. hellifino

    New Member

    Jul 2, 2015
    19
    1
    As everyone has said, an RC circuit.

    However I want to point out that you have to use the RC circuit to engage the starter after the caps charge up. You'd also want your shutoff driven on a p channel so it's only blocked while the 12v is saturated.

    And as far as your question on the battery, only if the starter is limited. Otherwise current follows least resistance. So even though caps act as a short-circuit on the current, you'll need a diode to keep the starter from pulling from the caps... so you have more resistance going to the caps than the starter and hence the caps won't charge.
     
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