Wheatstone bridge circuit

Discussion in 'General Electronics Chat' started by kai88, Jan 26, 2010.

  1. kai88

    Thread Starter New Member

    Jan 26, 2010
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    Newbie here ..I would like to ask whether the dc/ac wheatstone bridge circuit is power source dependent/ power source independent( some sort like ratiometer type).What i mean is will te indicator(replaced the Null-meter) change its value if the power source varies? TQ in advance..
     
    Last edited: Jan 26, 2010
  2. Metalfan1185

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    Sep 12, 2008
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    I know a wheatstone bridge is a very sensitive way to find a resistance value by balancing a meter with voltage dividers.

    in this sense, i would think that no matter what the voltage changes to, the correct resistance would always be needed to "Balance" the meter at the center of the bridge.

    I havent seen too many of these though, maybe someone else's input could benefit you more.
     
  3. kai88

    Thread Starter New Member

    Jan 26, 2010
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    Thanks...Any more inputs?
     
  4. BillB3857

    Senior Member

    Feb 28, 2009
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    My thought is that the higher the excitation voltage, the more sensitive the bridge. Having the excitation voltage 10 times higher would be like making the same meter 10 times more sensitive. The bridge works on a ratio principle. The same percentage output of a higher voltage results in a higher output voltage. Of course some will say the bridge is balancing currents but the same principle applies.
     
  5. MikeML

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    Oct 2, 2009
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    As Bill said: the output of a bridge is directly proportional to excitation voltage, i.e. ratiometric....
     
  6. kai88

    Thread Starter New Member

    Jan 26, 2010
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    I understood that the higher the supply the higher sensitivity.

    Is it true that if the Input voltage varies due to instability or i purposely increase the input voltage by a significant amount ,then my indicator value will oso change?Will it show 4unit instead of 2 unit(initial value) after i double up the voltage input?
     
  7. kai88

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    Jan 26, 2010
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    [​IMG][​IMG]

    Lets read the article..
     
  8. MikeML

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    The cited article is misleading. It should say: "A Wheatstone bridge used as method of measuring an unknown resistance, the null is independant of excitation voltage".

    This should be added to the article: "If a Wheatstone Bridge is used as a means of converting a varying resistance to a differential voltage, then the magnitude of the differential voltage is directly proportional to the excitation voltage, as well as the bridge imbalance".
     
  9. Metalfan1185

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    Sep 12, 2008
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    The way this question is worded is like "If a balanced wheatstone bridge has it's voltage increased, will the meter value change?"

    If its balanced the answer is No, so that would be Power source independant based on that article in previous response





    If the Bridge is unbalanced, then the answer would still be no because as long as all 4 resistances are present and unchanging, the ratio between the left and right side will always be the same

    So if your meter value is at the center and the voltage increases, it will not move.
    If the meter is +25% and you increase the voltage, it will stay a +25%.
     
  10. someonesdad

    Senior Member

    Jul 7, 2009
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    A stumbling block for students in analyzing the Wheatstone bridge is that the circuit looks complicated. I know, cuz I tripped over it when I was a student.

    But it's an easy circuit with one "aha!" insight: remove the galvanometer (OK, maybe I date myself by using that term :p). Now, you're looking at two voltage dividers in parallel. The bridge will be balanced when the voltage dividers have the same ratio. Here, "balanced" simply means that the voltages at the two middle nodes are equal. You can then write down the balance condition as (R1 and R2 are the resistances of the left voltage divider; R3 and R4 are the right resistances):

    R1/(R1 + R2) = R3/(R3 + R4)

    which, after simplification, becomes R1/R2 = R3/R4.
     
  11. MikeML

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    Oct 2, 2009
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    While technically correct, your answer is misleading. Practically, if the bridge is unbalanced, and the excitation voltage is increased, the meter reading will increase proportionally!

    Look at the first simulation (P60).

    The more general case, where both the excitation and one of the bridge resistors varies is in the second simulation (P59). Note that the slope (gain) of the differential output is a function of the excitation voltage...
     
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    Last edited: Jan 28, 2010
  12. kai88

    Thread Starter New Member

    Jan 26, 2010
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    thanks to ans anywer..

    thanks to [​IMG] Metalfan1185 for reminding me ..

    however there is a little conflict between [​IMG] Metalfan1185 ans and MikeML ans ...

    Sorry for causing headache to u all ..;)
     
  13. kai88

    Thread Starter New Member

    Jan 26, 2010
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    Thanks to Mike ML for the simulation ..it helps me alot ..

    However I have some doubts about the simulation Draft 59,the voltage u measured is POTENTIAL DIFFERENT between 2 arms or individual voltage on each arm(left n right)..It seems to me that the potential different is remain constant while the excitation voltage varies, since is the potential different measured by the meter.
     
    Last edited: Jan 29, 2010
  14. MikeML

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    NO! That simulation illustrates EXACTLY the point I am trying to make!

    If you measure the differential output voltage from an unbalanced bridge (i.e. non-zero output), the voltage is proportional to the bridge excitation output.

    The only time the output of a Wheatstone Bridge is independent of excitation voltage is when the bridge is balanced! ( Mathematically: Zero multiplied by anything is still zero)
     
  15. BillB3857

    Senior Member

    Feb 28, 2009
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    X2... exactly my point in Post #4. At Null, it doesn't matter what the excitation. If the excitation is higher, the meter will move further for the same amount of imbalance, therefore, higher sensitivity.
     
  16. Metalfan1185

    Active Member

    Sep 12, 2008
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    This is true...

    I am mistaken in my idea that if the meter is unbalanced (+25%) and the voltage is increased it will stay at the same (+25%). This is false.

    This only holds true if the meter is balanced...If it is not, then the higher the voltage, the higher the sensitivity.

    I apologise again for my misleading statement. I simulated it myself and now i know for the future :cool:
     
  17. jans123

    New Member

    Jan 30, 2010
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    I use bridges frequently to measure stress in metal structures. They are extremely sensitive to variations in excitation voltage as well as to un-balance in the bridge, or to be correct: the signal is direct proportional to the voltage on the bridge and almost proportional to variations in the four branches of the bridge. To get an accurate reading of the deviation from a "balanced bridge" you need a very defined voltage -or at lest know what the excitation voltage is...

    Very useful things those bridges, you find them in many places like: oil pressure meters in cars, level meters in washing machines, accurate temperature meters, altimeters, barometric pressure instruments, instrument detectors for a lot of things... Almost every one has some bridges without knowing it.
     
  18. Metalfan1185

    Active Member

    Sep 12, 2008
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    Ive seen them used where a lot of sensitivity is needed. Like if you were to replace one of the resistors with a Cds Cell type Photoresistor, you would have an extremely sensitive light meter...ect.
     
  19. kai88

    Thread Starter New Member

    Jan 26, 2010
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    Thanks to every1 ..
     
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