Wheatstone bridge sensitivity

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  • #1
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I'm struggling to understand what makes the wheatstone bridge such a good circuit for measuring small resistances. Essentially the bridge is just two voltage dividers and the output voltage is the difference between the two dividers.

Ok so the only advantage I can see over a conventional single voltage divider is that when the instrument being measured is zero, the output voltage will be zero. Which if I am correct is not achievable with a single voltage divider?

So in terms of the accuracy and precision of the measurement at Vout the bridge has virtually no effect it is simply there to provide zero voltage when the instrument is zero?

Thanks
 

Answers and Replies

  • #2
828
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I'm struggling to understand what makes the wheatstone bridge such a good circuit for measuring small resistances. Essentially the bridge is just two voltage dividers and the output voltage is the difference between the two dividers.

Ok so the only advantage I can see over a conventional single voltage divider is that when the instrument being measured is zero, the output voltage will be zero. Which if I am correct is not achievable with a single voltage divider?

So in terms of the accuracy and precision of the measurement at Vout the bridge has virtually no effect it is simply there to provide zero voltage when the instrument is zero?

Thanks

One way to understand is to derive the transfer function of a wheatstone bridge and see how the resistors play a part in the sensitivity. Can you solve for Vout/Vin of a wheatstone bridge? The unknown in the bridge will be proportional to a ratio of two voltages rather than an absolute voltage, and so you can adjust this ratio for sensitivity.

Also, have you seen the term ratiometric before? This is known as a ratiometric measurement. One advantage of ratiometric measurements is that they reduce multiplicative errors.
 
Last edited:
  • #3
146
5
One way to understand is to derive the transfer function of a wheatstone bridge and see how the resistors play a part in the sensitivity. Can you solve for Vout/Vin of a wheatstone bridge?

I have seen the equation derived, it makes little sense to me. What I don't understand is that fundamentally you are interested in Vout for a given change in resistance. Why is the equation solved for Vout/Vin? I don't really understand the concept of transfer function to be honest. Is this something that can be put into simple terms?

Thanks
 
  • #4
828
1
I have seen the equation derived, it makes little sense to me. What I don't understand is that fundamentally you are interested in Vout for a given change in resistance. Why is the equation solved for Vout/Vin? I don't really understand the concept of transfer function to be honest. Is this something that can be put into simple terms?

Thanks

Ya, it can be simplified in terms. Vout and Vin are kind of arbitrary terms, but it means you want to find the output of a circuit when an input is applied, and so you choose the appropriate Vin and Vout.

All you need to do is pick the node on the circuit that has the applied voltage, and then find the node where the voltage is output (the one you are using as a measurement, for example). Solve for the relationship (Vout/Vin) between these two voltages with circuit analysis/algebra, and that gives you your transfer function.

For example in a resistor divider, the Vin is the initial voltage, and the Vout is the voltage after being halved. If you solve for Vout/Vin, you get the transfer function as R1/(R1+R2). The use of this form is that multiplying a transfer function by a Vin voltage will tell you what the output voltage, Vout, is.

Honestly, the best way for you to learn why/how a wheatstone bridge works is to perform the circuit analysis. Substitute your measurement resistor [itex]R[/itex] with "[itex]R + \Delta R[/itex]", and see how the output voltage is affected by this [itex]\Delta R[/itex] term, which represents incremental differences in R. Often times circuit analysis is confusing and the results don't always make sense, but when I had to do this for a wheatstone bridge when I studied it, the equations actually made it intuitive of how it works.
 
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  • #6
jim hardy
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Think old days when meters were analog and power supplies were not stable.
The wheatstone bridge is immune to power supply variations
and it doesn't need an accurate voltmeter, just 3 good resistors.

Zero is very easy to resolve even with primitive instruments.
But no other number is.
 

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