Wheatstone bridge broken resistor LED monitor

Discussion in 'The Projects Forum' started by TylerVB, May 6, 2015.

  1. TylerVB

    Thread Starter New Member

    May 6, 2015
    2
    0
    I'm currently designing an anemometer that utilizes a Wheatstone bridge within a feedback loop. My attached schematic is not a Wheatstone bridge, but it is the most simplified way to ask my question.

    Essentially I have a voltage input to the top of the circuit which can range from 0V to -15V (as a result of closed loop feedback not shown in the diagram). I would like to have an LED light up if R2 is broken. It is imperative that whatever is added to this circuit does not impact the circuit function or it will be detrimental to the feedback (whatever is attached should have very high impedance and be powered separately). I have separate references of +15V, -15V, and Ground available.

    Logically, this means that the voltage at point 1 (dashed circle) will be zero and at point 2 will be not zero (I don't need the LED to go off when there is 0V coming in). I thought I could use two comparators feeding into an AND logic gate (which would power an LED), but the negative voltage is throwing me off and there are many properties of both comparators and AND chips that are not easy to a novice (open collector, TTL, CMOS).

    Any suggestions on strategy, or at least point me in the right direction of part names to look at or solutions to similar problems? I feel like I'm making this more difficult than it seems.
     
  2. Søren

    Senior Member

    Sep 2, 2006
    472
    28
    Although making it harder to give you an answer. We're not oracles, whatever it seems ;)


    Logically, whether you break R2 or not, the "point 1" will follow "point 2" as drawn, a they have a theoretical zero impedance between them - breaking R2 will make both equal to Vin as loaded by R3.

    If you want a solution to a particular problem, you really have to give us the actual circuit (with component values), or at least a substantial fraction of same, to enable us to tell you what you need, which you don't seem to be too sure about ;)

    As is, we can't even begin to guess what will constitute a high (enough) impedance.

    Would R2 be a platinum wire?
     
  3. TylerVB

    Thread Starter New Member

    May 6, 2015
    2
    0
    Thanks for the reply Soeren!

    Firstly, on the order of megaohms would suffice as high enough impedance, the resistors are all on the order of 100s of ohms. The wires are not necessarily platinum, they can vary in size and resistance.

    I can see how my explanation was confusing, you are most definitely right about point 1 and 2 and it was silly placement on the diagram, making my post convoluted. I do believe that the diagram the simplest way to express my problem, as this is essentially what I have.

    If R2 was to break, there would be no voltage drop across R1 (which could be detected with a comparator I think?). Assuming voltage was still entering the circuit, there would be a voltage drop across R3 (which could be detected with a different comparator). A logic AND gate could drive an LED with this information from the comparators.

    My question was if this was really the easiest way to do this. If so, I was hoping to be pointed in a direction of components that would be useful (ideally using the least amount of board space) or maybe information sources that explain how to start using comparators and logic circuits because datasheets are convoluted and unapproachable.

    Thanks again!
     
  4. Søren

    Senior Member

    Sep 2, 2006
    472
    28
    If the impedance of your resistors are less than 1k, a 100k impedance will mean less than 1% error and from what I'm reading, this will not be of any real consequence to your circuit.


    I'd split R3 into two, placing a R4 under R3. Then I'd make R3 = R1+5..10% and R4 approximately equal to the max wire resistance (if that doesn't wreck something else in your circuit).

    As is, I see no point in R3 as a single resistor, as it won't influence your readings and you cannot use it as a measuring bridge without splitting it up anyway.

    Then an op-amp used as a normal Wheatstone bridge amplifier, will work as a comparator from the two middle nodes and with a mild hysteresis (positive feedback), it will provide a very clean "wire broken" signal - I can draw you a schematic if you need it.

    Normally, I'll recommend not to use op-amps as comparators, but in this case it would be fine, as there's no critical speed requirement and you get a higher input impedance from an op-amp than a comparator, at least when we're talking common stuff like LM393 vs. LM358.

    If you used the wire in a real bridge configuration, it would be very obvious if the wire broke, as the op-amp used would saturate to one of the rails (which depends on inv/non-inv configuration of the op-amp) and that could be detected on the output side of said op-amp so no interference at all.

    You're using the feedback to keep the voltage drop over the wire constant with changing winds I'd guess and that can be done just as well with a Wheatstone bridge.
     
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