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car battery auto shutoff problem.......

Discussion in 'Automotive Electronics' started by hcchuar, Feb 3, 2009.

  1. hcchuar

    hcchuar New Member

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    below is overall of my project..
    i'm using solar panel to charge up my car battery and use the car battery to power my system

    As a example
    when the car battery is without loaded, LED1 will light on to show the voltage level of the battery, when i connected to my system, the car battery voltage will drop and LED2 will light on, when i shutoff my system then the LED1 will turn on again

    my planing is when the car battery voltage is going low same as LED3 and it will automatically switch off my system, right now i'm using single relay switch (SRD-9VDC-SL-C) in this application, but when it reach LED3, it does'nt switch off my system, because when it reach to LED3 the voltage of the system will off and the voltage of car battery will increase, when the voltage is increase, then my system will on, after my system is on, voltage will drop to LED3 and it will shutoff again, so my system will on off on off and LED4,5,6,7 will blind

    anyone know how to solve this problem?
     

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  2. crutschow

    crutschow Well-Known Member Most Helpful Member

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    You need to add some type of hysteresis to the relay control. For example LED3 could set a flip-flop which turns off the relay. LED2 (or any appropriate higher voltage LED) could reset the flip-flop and turn the relay back on. Thus there would be a selectable voltage difference between the relay opening and the relay closing.

    A latching relay with two coils could also be used. One coil operated by LED3 and one coil by LED2.
     
  3. hcchuar

    hcchuar New Member

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    the flip-flop u means is SR flip flop?
    what flip flop can find from market?



    i found that there are many latching relay with two coil in internet......and the price is very expensive


    may i know which 1 is suitable? for 9Vdc 7A or 10A
    below is the connection for single coil......is it i just make a same connection at the second coil?
     

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  4. dave

    Dave New Member

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  5. crutschow

    crutschow Well-Known Member Most Helpful Member

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    Yes, you coud use an SR flip flop such as a CD4013.

    The relay current rating should be greater than the current you expect it to carry. You want to energize the relay "load off" coil when LED3 goes off. You want to energize the relay "load on" coil when LED2 goes on.
     
  6. hcchuar

    hcchuar New Member

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    so i need use 2 coil latching relay with SR flipflop?
    there are many types of 2 coil relay
    do u know which 1 is suitable?for 9 or 12 VDc 7A or 10A

    CD4013 have 2 set of RESET,SET,DATA, should i only connect 1 set?
     
  7. crutschow

    crutschow Well-Known Member Most Helpful Member

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    You can use either a 2 coil latching relay or an SR flip-flop driving a single coil relay. You don't need both.

    The voltage and current rating of the relay are determined by the voltage you have to operate the coil and how much current you want the relay contacts to carry.

    The CD4013 had two flip-flops in one package. You only need to use one of the two. But connect all unused inputs to ground.
     
  8. hcchuar

    hcchuar New Member

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    may i know where the DATA and the CLOCK pin connect to?
    is CLOCK connect to 555 timer?
    then how about DATA?
     
    Last edited: Feb 4, 2009
  9. crutschow

    crutschow Well-Known Member Most Helpful Member

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    To use the FF as an RS (Reset-Set) type you don't use the DATA (D) and CLOCK inputs. Connect them to common. You use just the SET and RESET inputs. A high level on the SET line sets the Q output high and /Q low. A high level on the RESET sets the Q output low and the /Q output high. Of course the FF can only supply a few mA (see data sheet) at it's output so you will need an amp to drive the relay from the FF.
     
  10. hcchuar

    hcchuar New Member

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    i'm facing failure in this problem
    is my connection wrong?
    thanks for ur help
     

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  11. crutschow

    crutschow Well-Known Member Most Helpful Member

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    The problem is that both the SET and RESET inputs are high when LED 2 and 7 are both off. Assuming you want the relay off when LED 2 goes off, then you want the RESET input to be low and the SET input high. Thus connect the SET input (pin 6) to LED 2. Then connect LED 7 through an inverter to the RESET input (pin 4). That way the relay will turn off when LED 2 turns off and will turn back on when LED 7 lights.

    If you do a truth table you'll see how it works. Remember that the logic level is low from the LEDs when they are on and the relay is on when the Q1 output is low.

    The inverter function can be performed by a CMOS circuit such as a CD4069.
     
  12. shokjok

    shokjok Member

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    An 2N3904 or 2N2222 NPN transistor with a pair of resistors works as well as an inverter. Pins 7 through 11 of the 4013 should be grounded. A Darlington transistor could eliminate the relay requirement and drive the LEDs directly.
     
  13. crutschow

    crutschow Well-Known Member Most Helpful Member

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    I believe he's using the relay to control a load, not drive other LEDs.
     
  14. hcchuar

    hcchuar New Member

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    ya........i using relay to control the load

    but i still facing some problem

    i'm testing the relay by using CD4013 flip-flop and i use 12V supply as "1" and ground as "0", for the SET and RESET, and it's working well

    but unfortunately when i replace it by connect RESET and SET to LED, all LED will show "1", even which LED is ON (only 1 LED will on at a time at 1.8V,the rest of LED are showing 0V)


    if let say i connect LED2 to RESET pin of CD4013 flip flop through CD4069 inverter, the result i get is the RESET pin is always at "0"

    may i know how to solve this problem?

    thanks for your help again
     
  15. hcchuar

    hcchuar New Member

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    any can help me?
     
  16. crutschow

    crutschow Well-Known Member Most Helpful Member

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    Yes, I forgot that the voltage only changes a couple volts (the 1.8V LED forward voltage you measured) at the LM3914 output since it is current regulated. The easiest way to convert this signal to give a 0 to 12V logic signal is to use an PNP transistor such as 2N2907A, the emitter connected to 12V, the base connected in series with the LED anode (the LED current will go through the transistor base-emitter junction) and a 10k to 20k load resistor from the collector to common. You will need one transistor for each LED you monitor (two in this case).

    This transistor output will be near 12V when the LED is ON and 0V when the LED is OFF, thus the logic is inverted from our previous setup. Therefore the transistor collector output connected to LED 2 should be connected through the inverter to the FF SET input, and the transistor output connected to LED 7 should be connected the the FF RESET input.

    Hope that does it for you.
     
  17. hcchuar

    hcchuar New Member

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    Hi crutchow,

    what are u talking is like this?
     

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  18. crutschow

    crutschow Well-Known Member Most Helpful Member

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    Not quite. It should be like this

    Auto shutoff.JPG
     
  19. hcchuar

    hcchuar New Member

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    ok.......thanks crutchow
    i will try on that
    hopefully i will get it
     

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