Short life failure explanation

Discussion in 'General Motoring' started by nothermark, Oct 7, 2005.

  1. nothermark

    nothermark Guest

    Hyundaitech said he did not understand what I was talking about in
    another note so here goes. I think it's something that should be
    understood. ;-)

    When one designs a circuit or mechanical device part of the design
    process is to take into account the power to be transmitted. In a
    mechanical part it would be the material choice, cross section area,
    shape, etc. In electrical components it is mostly rated heat
    dissipation. Electrical waste is generally produced as heat. To keep
    it simple if I use a 12 ohm load in the control module sensor
    connected to it will cause at most a 1 amp current at a nominal 12
    volt system. Amps=volts/ohms. That is with the sensor resistance at
    0 ohms. As the sensor increases reistance the current drops and the
    power decreases. The power also divides between the sensor and the
    control module. Max power is 12x1=12 watts.

    There are 3 failure models.
    1. Open circuit - no current flow.
    2. Short circuit - currnet limited by source resistance - classic arc
    and spark with charred remains.
    3. Lowered impedance. Result is more current, more power
    dissipation. If it's low enough you approach failure 2 but what if it
    drops something less? To pick a number sya a 30% drop. My 12 ohm
    resistor becomes 8 ohms. Max current goes to 12/8=1.5 amps. Max
    power becomes 12x1.5=18 watts.

    OK - what's it mean? Not enough power for arc and spark but the extra
    power becomes extra heat that slowly simmers the parts instead of
    flash burning them. It is hard to predict what will happen as there
    are many failure modes as well as conditions. For example if the
    sensor is the gas gauge and you never let it get below half full you
    probably won't get to max power so nothing happens. On the other hand
    if you consistently run low the gauge fails prematurely when either
    the sensor or the load in the meter circuit cook off. The failure is
    as likely to be a type 1- open circuit as a type 2 short. Either way
    it's broken but a type 1 will not damage the rest of the circuitry
    even thogh the part that fails may be the "good" part that is being
    operated out of it's design range of power dissipation. Most sensor
    circuits work at lower power levels than this but the principle
    remains the same.

    nothermark, Oct 7, 2005
  2. nothermark

    hyundaitech Guest

    I think I understand your point (and understood it when you posted about
    the TCM earlier), but I hadn't thought about the lowered impedance idea
    until I had read your post, which was unfortunately after I made mine.

    What I don't know is whether the TCM provides any significant resistance
    in the pulse generator circuit. They are analog A/C voltage generating
    sensors. If I remember correctly the sensors provide about 220 ohms
    resistance. I have no idea what order of magnitude the TCM might be.

    Furthermore, I definitely agree that if the TCM provides a significant
    amount of resistance to the circuit that it's *possible* it caused the
    sensor to fail. I've only ever seen a failure like this once-- a Kia ECM
    was causing an ignition coil to misfire. It would literally blow up the
    coil after the car was driven awhile. And it's probably exactly the
    problem you're referring to. The ECM's resistance was too low, causing
    the coil to overheat and fail.
    hyundaitech, Oct 7, 2005
  3. nothermark

    hyundaitech Guest

    I should have mentioned this in the above post, but sorry to say I forgot.

    You've given an excellent explanation and I appreciate your efforts very
    much. I hope nothing in the above post led you to believe otherwise. In
    fact, you've made me think about some things I hadn't thought about
    before, and that might help me fix some more cars right the first time.
    hyundaitech, Oct 7, 2005
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