Another Latching Circuit - For Car Ignition

Discussion in 'The Projects Forum' started by BurninBri, Aug 18, 2011.

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  1. BurninBri

    Thread Starter New Member

    Aug 16, 2011
    I've looked quite a bit for the correct latching circuit, but I haven't found exactly what I'm looking for.

    I'm using a TPS 56121 as well as a TPS 51315 power regulator to power some automobile electronics. It takes 12V in and provides 3.3V out to a CPU and more. This regulator has a "logic input" level to enable or disable power output. The power circuit is all done. Currently, the enable pins are not used, though, and they are just left floating to a capacitor (in the case of the 56121) or tied high (in the case of the 51315). Note that to shut off the power, these would need to be brought to ground.

    So here's what I'm looking for. I would like to use the "ignition sense" wire (12V when on, 0V off, but not much current) in the car to control turning on power. Otherwise, the electronics could easily kill the car battery if left on for more than 3 days without driving. Also, I would like to be able to keep the electronics on for a while after the ignition sense wire goes low.

    Fortunately, I am using a Max 7300 for GPIOs and I have a few extra GPIO pins to sense when the ignition goes low and also to reset a latch to finally kill power. (So I was thinking I would just use these instead of a 555 timer circuit.)

    So I would like a latch that would:
    a) not have a lot of quiescent current when the car is off (ideally a self-latch or something?)
    b) take in a 12V signal and output a logic level that can be tied into the power supply. The output would need to be pulled to ground (pull-down I'm assuming) to turn off the power converters when the latch finally goes low.
    c) Can be reset easily with a GPIO pin to "unlatch" and shut down power.

    I was thinking of just using an S-R flip-flop for this. I would run the ignition sense wire through a resistor divider circuit to get 5V (or 3.3V) logic levels. When this is off, it will be ground and no current would flow. But there is one problem with this: I need the power for the flip-flop to come from the power supplied by the regulator (which is only being turned on by the flip-flop). So I guess I would need to run the output of the 12V ignition sense into the output of the flip-flop as well... would this go through another resistor? Or maybe a diode here? This would make sure the power regulators would come on (even without the flip flop starting), then the flip flop would start in the correct state. Then when the ignition sense was put back to zero, the flip-flop's output would hold it high until a GPIO reset was given to the flip-flop. This would be based on the CPU sensing it with a sense GPIP. (But then this would cut the power to convert, flip-flop, and GPIO reset output. Hope this is fine since the pin would be held down to ground through the resistor wire.)

    I haven't done engineering for a long time. Thoughts? Will this work? Any thoughts on the resistor values?

    It would look like this (sorry if there are rules on ascii circuits - didn't see them):

    12V Ign. --------------*----------*------5Vzener(?)------------\
    | | |
    58.3kΩ | 1kΩ(?)
    | \-------- S-Pin of FF |
    GPIO Sense---* FF Output --*--> OP
    | GPIO Reset-- R-pin of FF |
    41.6kΩ 100kΩ
    | |

    So basically, the circuit's output (OP) would be tied to the voltage regulators as an enable/disable. When we get 12V on the ignition sense, it goes through a 5V zener which basically should put almost 5V (maybe 1% less with the 1K/100K divider) onto to the output. This should enable the power regulators and it would supply power to the flip-flop. ***Note: I need the flip flop to come up and not make the OP output goes low!! If it goes low upon flip-flop power-up, then the circuit will basically come on, then off, then on, then off, etc.!!! So I wanted an S/R flip-flop, and the 12V ignition sense gets voltage-divided down to 5V to the "Set" of the flip-flop so it's outputting 5V at power-up. We sense the ignition sense with another GPIO, and if it goes low then the power is still on (because of the output of the flip-flop). We wait a certain amount of time (programmed into the CPU) and then we send a GPIO reset to the flip-flop which should make the output go low and everything shuts down (since the OP is controlling the voltage regulator).

    Please feel free to comment before I try to hook this up. It seems this could have many issues (like the output going low upon powering up the flip-flop). There could also be a much easier way?!? (I was even thinking a capacitor and a diode to hold the enable high, but I would have needed a resistor to ground to keep the circuit off as well.) If there are any other ideas, let me know.

    Thank you so much for any suggestions or pointers to better reference material.
  2. bertus


    Apr 5, 2008

    I am closing this thread as it violates AAC policy and/or safety issues.

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