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Ohms to trigger a switch

Discussion in 'General Electronics Discussion' started by Alan P, Mar 12, 2015.

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  1. Alan P

    Alan P

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    Mar 12, 2015
    The sender has its own power supply as it is connected to a gauge. If I'm reading this right, it's going to put 12v into the sender which I don't need. I'm slightly confused by 135R. Is that a resister? Assuming that 12v represents +12v and the bottom right line end is ground, the sender wires connect between full and empty? If so, can I just cut the +12v at the sender since it has its own ps?
     
  2. ADRT

    ADRT

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    Nov 25, 2014
    Ok, so the sender unit actually sources the output? This means that there is a voltage not a resistance reading on the output? If this sources and is not a sink output then that's a whole different ball of wax.
     
    Last edited by a moderator: Mar 14, 2015
    davenn likes this.
  3. davenn

    davenn Moderator

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    indeed
     
  4. ADRT

    ADRT

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    Nov 25, 2014
    Do you have the model number for the sending unit? Mybe a manual or schematic?
     
  5. Alan P

    Alan P

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    Mar 12, 2015
    I probably should have been clearer. There is a voltage reading coming off the sending unit. At empty I read 5.0v dc. My trigger voltage is about .8v dc. .8v is almost full but still leaves me a safe buffer in case there is any delay. The gauge has a delay of about 20 seconds to read. The sender on its own is pure resistance, however There is a voltage which I did not initially expect powering the gauge.
     
  6. ADRT

    ADRT

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    Nov 25, 2014
    Ok so can you separate the unit from the power supply? Use only the resistance part of the unit?
     
  7. ADRT

    ADRT

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    Nov 25, 2014
    The other option would be to use the power supply from the gauge to also power the comparitor circuit.
     
  8. Alan P

    Alan P

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    Mar 12, 2015
    The only option would be to use the existing voltage to power the circuit. I was trying to figure out how to modify the transistor circuit to use my existing power since I already have the parts to make it, but I've since ordered the comparator parts so I might have to go that direction unless I can figure out how to separate the 12v relay power from the sender power.
     
  9. ADRT

    ADRT

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    Nov 25, 2014
    A schematic or part # would be great.:)
     
  10. ADRT

    ADRT

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    Also what comparitor did you order?
     
  11. Alan P

    Alan P

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    Mar 12, 2015
    I'm working with the two schematics you guys posted here. The link you posted and the transistor circuit from Colin. I ordered a few different comparators,

    IC COMP R-RINOUT QUAD 16-SOIC LT1721CS#PBF-ND
    IC COMP PUSH-PULL 1.8V SOT23-5 MCP6561T-E/OTCT-ND
    IC COMPARATOR R-R 1.1V SOT-23-5 497-14339-1-ND
    IC COMPARATOR 2.4V REF SOT-23-6 MCP65R41T-2402E/CHYCT-ND
     
  12. Alan P

    Alan P

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    Mar 12, 2015
    Hey Guys,

    I've put together a test circuit, Unfortunately, it's not working. I have a constant +12v output, the relay never shuts off. The schematic is posted below. For simplification, the input trigger voltage is shown as simply .9v dc. When working properly, the goal is to be able to adjust the pot to the target voltage, once the input trigger matches that target voltage, the relay is triggered. I this case, .9v.

    [​IMG]
     
  13. ADRT

    ADRT

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    Nov 25, 2014
  14. duke37

    duke37

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    You should fit a pull-up resistor, try 4k7.

    To find out where the +12V is coming from, disconnect R2 and R3.
    Connect the input to 1.5V.
    Does it switch when you twiddle?

    The correct operation depends on the trigger input, a high impedance at the input will let R2 override the input.

    Does the comparator use inputs down to 0V?
     
  15. Alan P

    Alan P

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    Mar 12, 2015
    This is a single comparator. The Comparitor has 5 pins. I've used 3 of them, skipping vcc+ and -. Not sure if that was correct or not. The base schematic that I used only showed those 3 connected. The comparator is linked here:
    http://www.digikey.com/product-detail/en/TS881ILT/497-14339-1-ND/4576432

    No, it's .85 to 5.5v.
     
  16. duke37

    duke37

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    Jan 9, 2011
    If all else fails, read the instructions !

    The comparator must have a power supply, not exceeding 5.5V not less than 0.85V.
    The output is push/pull so does not need a pull up resistor.
    The inputs can go between ground and the power supply.

    If you have run the comparator without a power supply, then any input will be above the power supply and you may have popped it off.

    Still do not know where the 12V was measured or where it came from.
     
  17. Alan P

    Alan P

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    Mar 12, 2015
    The 12v is secondary voltage and is used only to power the relay which is switched at ground. If I'm misreading something, please let me know.

    Can you elaborate on that, not sure what you mean. For the sake of this experiment, I have a 12v power supply which I've reduced to 5v to power the comparator and further reduced to .9v via a voltage divider to act as my trigger. I'm using a solid state relay. 12v to power the relay, 5v to power the comparator and .9v as my trigger. Is the only fault in my circuit the missing power supply to the comparator, or am I still incorrect somewhere?
     
  18. duke37

    duke37

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    Jan 9, 2011
    The comparator must have a power supply, 5V is good, positive to P5, ground to P2.
    The inputs should not be outside the power supply range i.e. between 0V and 5V. If the inputs are outside this range, you may damage the device, depending how the input is designed. Sometimes exceeding the supply will feed current into the device which then may have its limit voltage of 5V exceeded.

    Whether this is the only fault I know not.

    Post details of the SS relay, It may be possible to drive it directly from the LT497.
     
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