resistor value in wheatstone bridge

I have a resistance based sensor which gives the resistance change when the strain/force is applied. How to measure the unknown resistance change using the Wheatstone bridge circuit.? I mean what are the important things we have to keep in mind while selecting the resistor value. The initial resistance of the sensor is not the same. it always give the creep resistance value while no force is applied or under the constant load/Force/strain.

• You know the schematic and the formula of the Wheatstone bridge – so what exactly are you asking? Dec 21 '16 at 14:24
• I have a force sensing resistor which gives the resistance change from 1.8 Mohm to 0.7 Mohm when the 0.45 N is applied and i want to measure the resistance change using the Wheatstone bridge and fed it to the instrumentation amplifier. so how to select the value of the bridge circuit.? Dec 21 '16 at 14:29
• Is this a single ended sensor or do you have two with one on each side to make a more acurate differential measurement? Then you decide to use a CC source or CV source for wide spaning R of 50% vs 1%. I would use a CC source instead of a Wheatstone bridge with Vref set to null force and I= Icc. Then V=IccR eg using CC=1uA Dec 21 '16 at 14:36
• Is this a single ended sensor or do you have two with one on each side to make a more acurate differential measurement? Then you decide to use a CC source or CV source for wide spaning R of 50% vs 1%. I would use a CC source instead of a Wheatstone bridge with Vref set to null force and I= Icc. Then V=IccR eg using CC=1uA then INA gives excellent CMRR with Zin >> Zsource like 100M with FET INA Dec 21 '16 at 14:42
• the sensor is like a rubber strip with the copper electrode at the end, where i get to measure the resistance change when force input is applied. Dec 21 '16 at 14:45

A Wheatstone bridge is used to measure tiny changes in R using a fixed V which gives a fairly constant I. When R changes 50%, I is no longer constant. So this bridge is no longer linear or balanced with a wide spread.

This "Wheatstone bridge" does not pass the criteria of a small change so it is not linear as Wheatstone intended when balanced. Normally Rg changes are small in such a bridge which makes this configuration useful. A more complex ratiometric bridge is needed to make it more linear, if that is what is desired.

F(x)= Vout = aRg + b This is probably what you want for Force or Displacement

• for a = gain which will be -ve and b = offset

But since Force is inverse to R what is the actual transfer function you want? positive V for Inverse R or something else?

I would suggest something else like 1/R gain with null offset adjust around 1.5 to 2Mohms. try to get a 10:1 linear range to 100:1 but null adjust will be a problem for an "autozero design solution."

1st Define your specs for Forces vs R and then tolerance or error budget for component variance, temperature, aging etc.

Below is a "Wheatstone Bridge"

• Note the formula when Rg changes is not linear. Change 2M or use INA gain control iwth Pot refereence for 0 force R. to get 0 to 5V for linear 0 to Smax. for Strain in mm or cm.

Note the hysteresis on R is much larger than you think. The datasheet shows a mean of 2M not a peak. So I expect there is large variance from sample to sample for gain and offset of mean.

• The graph data indicates a mean of 2M but no units for Strain, however it shows only 25% excursion, so 100% excursion must be even more non-linear, The Hystersis may look like a large capacitor for short term AC actuation as shown on my schematic, the C value can be increased to match this. There is an equation I have computed in a spreadsheet for Strain vs R ( not stress in Nt) Dec 22 '16 at 20:12
• +1 as you don't deserve a -ve score after all that work. Your last diagram appears to show an arrangement of resistors on the Mohm scale indistinguishable from a Wheatstone bridge, followed by a diff-amp. Dec 22 '16 at 21:32
• Yes there are many reasons for the other resistors to make this wide excursion. Once we get a true transfer function for force, it can be made non-linear to get Force vs V. Those who voted -1 may be just too lazy to comment. the major difference here is a wide range but limited range Vnull adjust for zero stress. which I believe can vary from 2 to 2.7M so further refinements can be made. Since there are NO Specs from OP, I shouldn't have even posted a design as I was presuming he had no idea how to define specs, while my motus operandi is good f(x) Specs 1st , Design later. Dec 22 '16 at 21:45
• @TonyStewart.EEsince'75 I want to build a sensor which can record the force that has been applied by bending the sensing element. Jan 12 '17 at 15:02
• Do you have a spring or a scale to measure force? vs R, if not get one then calibrate spring f=kx or use known weights Jan 12 '17 at 15:06

Ideally a Wheatstone bridge is balanced. Where you have one sense resistor, this means choosing the other three resistors to be the same as its mid range value.

Under certain circumstances, you may want to use lower value in the reference arm for slightly lower noise, or lower values to ground to offset the common mode from mid scale, or higher values to reduce current consumption and/or heating. But usually, all balanced is the place to start.

In your case with a sensor that ranges from 0.7 to 1.8Mohms, the geometric mean of those at 1.1 or 1.2Momhs would be ideal, but 1M would work quite well with little increase in the required common mode range.

• I think you know a wheatstone bridge unlike this application. This is not a linear Resistance Amplifier nor an inverse Resistance amplifier when the deviation is so large > 50% as in this case. So it is a misnomer. It amplifes, may look like a Wheatstone bridge but not in performance since there is no balance at any time. The null force just drifts near/past the null operating point somewhere between 1.5M and 2M depending on retention, temp etc. Do you agree it is a misnomer? Dec 22 '16 at 0:04
• The graph data indicates a mean of 2M but no units for Strain, however it shows only 25% excursion, so 100% excursion must be even more non-linear, and not return to max R of 2.7M due to hysteresis. Dec 22 '16 at 20:16