Applying the Wheatstone Bridge Circuit

Applying the Wheatstone Bridge Circuit 

Applying the 

Wheatstone Bridge 

Circuit 

Dirk Eberlein 

dirk.eberlein@hbm.com 

www.hbm.com 

19.03.2008, Folie 1 Hottinger Baldwin Messtechnik GmbH Dirk Eberlein


The relationship between the applied strain ε 

and the relative change of the resistance of a 

strain gauge 

A strain gauge converts a mechanical strain 

into a change of an electrical resistance. 

“gauge factor”: 

characteristic of the strain gauge and has to been checked 

experimentally: 

ΔR 

R 

= 

k 

⋅ε 



Test of the gauge factor 

Thomas Kleckers, HBM 



Value of the relative change of the resistance 

• 

ΔR/R 0 = k · ε k ~ 2 R (typ.) = 120Ω or 350Ω 

ε for example – (strain level for transducers) 

1000 µm/m = 1mm/m (0,1%) 

E.g.:120Ω-strain gauge at at 1mm/m = 0,24Ω change of of 

the the resistance 

Measurements with an ohmmeter are impossible 



Basics 

U 

U 

1 

0 

= 

R 

1 

R 

+ 

1 

R 

2 

U 1 = 5V 

R 1 = 5 Ω R 2 = 5 Ω 

U 0 =10V 

U 2 = 5V 

19.03.2008, Folie 5 Hottinger Baldwin Messtechnik GmbH Dirk Eberlein 

U 

1 

= U 

0 

R 

1 

R 

+ 

1 

R 

2 

Spannungsabfall = 

voltage drop 

potential 

divider


Basics 

U 

U 

1 

0 

= 

U 1 = 4V 

R 1 = 4 Ω R 2 = 6 Ω 

R 

1 

R 

+ 

1 

R 

2 

U 0 =10V 

U 

U 2 = 6V 


1 

= 

U 

0 

R 

1 

R 

+ 

1 

R 

2 

potential 

divider


Basics 

U 1 = 5V 

R 1 = 5 Ω R 2 = 5 Ω 

U a =0V 

U 0 =10V 

U 2 = 5V 

R 4 = 5 Ω R 3 = 5 Ω 

U 4 = 5V U 3 = 5V 

potential 

divider 



Basics 

increasing of R 1 

decreasing of R 1 

U 1 = 5,45V 

R1 = 6 Ω R2 = 5 Ω 

+ 

U a =0,45V 

U 0 =10V 

U 2 = 4,54V 

- 

R4 = 5 Ω R3 = 5 Ω 

U 4 = 5V U 3 = 5V 

output voltage U a will be positive 

output voltage U a will be negative 



Basics 

increasing of R 2 

decreasing of R 2 

U1 = 4,54V U2 = 5,45V 

R1 = 5 Ω 

- 

R2 = 6 Ω 

Ua = - 0,45V 

R4 = 5 Ω 

+ 

R3 = 5 Ω 

U 4 = 5V U 3 = 5V 

U 0 =10V 

output voltage U a will be negative 

output voltage U a will be positive 



Basics 

U a 

U a 

= 

= 

U 

U 

0 

0 

⎛ 

⎜ 

⎝ 

⎛ 

⎜ 

⎝ 

R 

R 

1 

1 

R 

+ 

1 

R 

2 

− 

R 

U 1 

U 4 

3 

R 

+ 

4 

U a 

R 


U 0 

4 

⎞ 

⎟ 

⎠ 

R1 

+ ΔR1 

+ ΔR 

+ R + ΔR 

1 

2 

2 

U 2 

U 3 

− 

R 

3 

U 

1 

U 

4 

= U 

0 

= U 

0 

R 

1 

R 

R1 

+ R 

3 

2 

R4 

+ R 

R4 

+ ΔR4 

+ ΔR 

+ R + ΔR 

3 

4 

4 

4 

⎞ 

⎟ 

⎠


Basics 

U a 

= 

U 

0 

⎛ 

⎜ 

⎝ 

R 

ΔR 


Basics 

Ua U 

0 

ΔR 

R 

= 

with: 

U a 

U 

0 

= 

ΔR 

R 

1 

1 

k ⋅ ε 

= k 

4 

− 

⋅ 

ΔR 

R 

2 

2 

+ 

ΔR 

R 

3 

3 

ΔR 

R 


− 

that implies: 

( ε − ε + ε − ε ) 

1 

2 

4 

4 

3 

4


Basics 

• we use every time the Wheatstone Bridge 

Circuit with 4 resistors 

• these 4 resistors could be 1, 2 or 4 strain 

gauges 

• modern amplifiers completing “missing“ 

resistors in the Wheatstone Bridge Circuit 

• with 4 active strain gauges we get the largest 

output value 



Basics 

R 1 

R 2 

U A 

R 4 

R 3 

( ε − ε + ε − ε ) 


U 

U 

U B 

A 

B 

= 

k 

4 

1 

2 

3 

4


Elementary circuits with strain gauges 

ε1( + ) 

(R1) 

R2 

R4 

R3 

Half bridge 2 active strain gauges 

ε1 ( + ) 

(R1) 

ε 2 ( −) 

(R2) 

Quarter bridge 1 active strain gauge 

R4 

R3 

ε 2 ( −) 

(R2) 

Full bridge 4 active strain gauges 

ε 1 ( + ) 

(R1) 

ε 2 ( −) 

(R2) 


R1 

R4 

R3 

ε 4 ( −) 

(R4) 

ε 3 ( + ) 

(R3)


Measurements on a bending beam (Quarter Bridge) 

U 

U 

A 

B 

= 

k 

4 

( ε ) 

1 

strain gauge 1 


F 

F 

tensile bar / 

bending load 

strain gauge 1: + 

Temperature compensation: No! 

R 1 

R 2 

U A 

R 4 

R 3 

U B


Calibrating measurement equipment (Quarter Bridge) 

U 

U 

k = 2 gauge factor (written on the strain gauge package) 

estimated strain level ε = 1000 µm/m (estimation!) 

U 

U 

U 

U 

U 

U 

A 

B 

A 

B 

A 

B 

A 

B 

= 

= 

= 

= 

k 

4 

k 

4 

k 

4 

( ε − ε + ε − ε ) 

1 

( ε ) 

1 

0, 

5mV 

( ε − ε + ε − ε ) 

1 

= 

2 

2 

( 1000µm 

/ m) 

4 

/ V 

2 

3 

3 

4 

4 

= 

0, 

5 

⋅1000 

⋅10 

⋅10 


−6 

= 

0, 

5 

−3


Measurements on a bending beam (Half Bridge) 

U 

U 

A 

B 

= 

k 

4 

( ε − ε ) 

1 

2 




F 

bending load 


strain gauge 2: - 

Temperature compensation: Yes! 

R 1 

R 2 

U A 

R 4 

R 3 

U B


Measurements on a bending beam (Half Bridge) 

U 

U 

A 

B 

= 

k 

4 

( ε − ε ) 

1 

2 




F 

R 1 

R 2 

U A 

R 4 

R 3 

Superimposed normal 

strains are compensated 




U B


Calibrating measurement equipment (Half Bridge) 



U 

U 

U 

U 

U 

U 

U 

U 

A 

B 

A 

B 

A 

B 

A 

B 

= 

= 

= 

= 

k 

4 

k 

4 

0, 

5 

( ε − ε + ε − ε ) 

1 

( ε − ε ) = ( 1000µm 

/ m − [ − 1000µm 

/ m] 

) 

1 

1mV 

⋅ 2000 ⋅10 

/ V 

2 

2 

3 

2 

4 

−6 

4 

= 1⋅10 

−3 

2 times the output 

value of a 1/4-bridge 



Optimisation of of a wish mop 

mop 



Measurements on a bending beam (Full Bridge) 

U 

U 

A 

B 

= 

k 

4 

( ε − ε + ε − ε ) 

1 

2 

3 





4 






F 

R 1 

R 2 

bending load 


U A 

R 4 

R 3 

U B


Measurements on a bending beam (Full Bridge) 

U 

U 

A 

B 

= 

k 

4 

( ε − ε + ε − ε ) 

1 

2 

3 





4 

F U A 


strain gauge 2: + - 


strain gauge 4: + - 


R 1 

R 2 

R 4 

R 3 




U B


Calibrating measurement equipment (Full Bridge) 



U 

U 

U 

U 

U 

U 

U 

U 

A 

B 

A 

B 

A 

B 

A 

B 

= 

= 

= 

k 

4 

2 

4 

0, 

5 

( ε − ε + ε − ε ) 

1 

2 

3 

( 1000µm 

/ m − [ − 1000µm 

/ m] 

+ 1000µm 

/ m − [ −1000µm 

/ m] 

) 

⋅ 4000 ⋅10 

= 2 mV / V 

−6 

4 

= 2 ⋅10 

−3 





... and the right strain gauge for this application 

DY11... 



Measurements on a tension/compression bar 

(Half Bridge) 

εq 

− ν = 

ε 

U 

U 

A 

B 


F 

l 

νKPoisson' 

s ratio ( steel ≈ 

= 

k 

4 

( ε − ε ) = [ ε − ( − νε ) ] 

1 

⇒ 

2 

ε 

q 

= 

k 

4 

−νε 

1 

l 

⇒ 

0, 

3) 

1 

ε 

2 

= −νε 

1 




F 

strain gauge 2 U A 

tensile bar 


R 1 

R 2 

R 4 

R 3 

U B



(Half Bridge) 

U 

U 

A 

B 

= 

k 

4 

strain 

gauge 1 

( ε − ε ) = [ ε − ( − νε ) ] 

1 

2 

k 

4 

1 


1 

Superimposed bending 

strains occur in the result 


F 




R 1 

R 2 

U A 

R 4 

R 3 

U B


Calibrating measurement equipment (Half Bridge) 

k = 2; ν = 0,3 


U 

U 

U 

U 

U 

U 

U 

U 

A 

B 

A 

B 

A 

B 

A 

B 

= 

= 

= 

= 

k 

4 

k 

4 

0, 

5 

( ε − ε + ε − ε ) 

1 

( ε − ε ) = [ ε − ( − νε ) ] = ( 1000µm 

/ m − [ − 0, 

3 ⋅1000µm 

/ m] 

) 

⋅ 

1 

0, 

65 

2 

2 

( 1000µm 

/ m − [ − 300µm 

/ m] 

) 

mV / V 

3 

k 

4 

1 

4 

1 

2 

4 

⋅1300 

⋅10 

0, 

65 

⋅10 


= 

0, 

5 

−6 

= 

−3 

1.3 times the output 

value of a 1/4-bridge



(Full Bridge) 

U 

U 

U 

U 

A 

B 

A 

B 

F strain gauge 2 F 

UA = 

= 

k 

4 

k 

4 



( ε − ε + ε − ε ) 

1 

[ ε − ( − νε ) + ε − ( − νε ) ] 

1 

2 

1 

3 

3 

4 

3 


tensile bar 







R 1 

R 2 

R 4 

R 3 

U B



(Full Bridge) 

U 

U 

A 

B 

= 

k 

4 



( ε − ε + ε − ε ) 

1 

2 

3 

4 








F 

R 1 

R 2 

U A 

R 4 

R 3 

Superimposed bending 


U B



k = 2; ν = 0,3 


U 

U 

U 

U 

U 

U 

U 

U 

A 

B 

A 

B 

A 

B 

A 

B 

= 

= 

= 

= 

k 

4 

k 

4 

2 

4 

0, 

5 

( ε − ε + ε − ε ) 

1 

[ ε − ( − νε ) + ε − ( − νε ) ] 

1 

2 

1 

3 

( 1000µm 

/ m − [ − 0, 

3 ⋅1000µm 

/ m] 

+ 1000µm 

/ m − [ − 0, 

3 ⋅1000µm 

/ m] 

) 

⋅ 2600 ⋅10 

−6 

3 

= 

4 

1, 

3 

⋅10 

3 

−3 

= 

1, 

3 

mV / V 

2.6 times the output 





XY11... XY31... 



Measurements on a shaft under torsion (Full Bridge) 

U 

U 

A 

B 




UA strain gauge 2 

Torsion shaft 

= 

k 

4 

( ε − ε + ε − ε ) 

1 

2 

3 

4 







R 1 

R 2 

R 4 

R 3 

U B


Measurements on a shaft under torsion (Full Bridge) 

U 

U 

A 

B 

F 

= 



k 

4 

( ε − ε + ε − ε ) 

1 

2 

3 



4 






F 

R 1 

R 2 

U A 

R 4 

R 3 


and bending strains are 

compensated 


U B



k = 2 


U 

U 

U 

U 

U 

U 

U 

U 

A 

B 

A 

B 

A 

B 

A 

B 

= 

= 

= 

k 

4 

2 

4 

0, 

5 

( ε − ε + ε − ε ) 

1 

2 

3 

( 1000µm 

/ m − [ − 1000µm 

/ m] 

+ 1000µm 

/ m − [ −1000µm 

/ m] 

) 

⋅ 4000 ⋅10 

= 2 mV / V 

−6 

4 

= 2 ⋅10 

−3 






XY11... XY31... 



strain gauge rosettes for stress analysis 

two basic shapes (geometries): 

0°/45°/90°-rosettes 

0°/60°/120°-rosettes 

c 

a 

b 

strain measurement in 

3 directions (a, b, c) 



strain gauge rosettes for stress analysis 

Connection at the amplifier always as three 

different quarter bridges! 

measuring grid a, b, c 

e.g. Spider8-30 



Calibrating measurement equipment 

(3 different quarter bridges) 

c 

1000µm/m correspond to 2mV/V (at a gauge factor of 2), 

valid for amplifier channel (measuring grid) a, b, c 

a 

b 



M b -bending moment; M d -torque; T-temperature; F-normal force 

active strain 

gauge 

strain gauge for 

temperature 

compensation 


ε 

passive strain 

gauge or resistor 

ε ϑ








Reduction and elimination of measurement 

errors, especially temperature effects 

• temperature error can be corrected mathematically 

• temperature compensation using Wheatstone 

bridge circuit 

-quarter bridge with compensating strain gauge 

-half and full bridges 



Basics 

Every strain gauge gives out a value when the 

temperature varies – without any mechanical load. 

Cause: 

• Work piece expansion with temperature 

• Change in specific resistance of strain gauge 

• Strain gauge grid expansion with temperature 

HBM, Thomas Kleckers (Januar 1999) 



Basics 

Thermal output of a quarter bridge 

ΔR 

R 

0 

= 

k 

⋅ 

( ε + ε ) 


M 

ϑ 

Strain gauge 1 

F


Temperature compensated strain gauges (series Y) 

1 for ferritic steel α = 10,8 ·10-6 / K 

3 for aluminium α = 23 · 10-6 / K 

5 for austenitic steel α = 16 · 10-6 / K 

6* for quartz α = 0,5 ·10-6 / K 

7* for titanium /grey cast iron α = 9 ·10-6 / K 

8* for plastic material α = 65 ·10-6 / K 

9* for molybdenum α = 5,4 ·10-6 1 for ferritic steel α = 10,8 ·10 

/ K 

-6 / K 

3 for aluminium α = 23 · 10-6 / K 

5 for austenitic steel α = 16 · 10-6 / K 

6* for quartz α = 0,5 ·10-6 / K 

7* for titanium /grey cast iron α = 9 ·10-6 / K 

8* for plastic material α = 65 ·10-6 / K 

9* for molybdenum α = 5,4 ·10-6 / K 

* not for all measuring grid length available 



Temperature Compensation 

(Material, thermal expansion coefficient 

remaining temperature variation 

polynomial (of (of the remaining 

temperature variation) 

so so that that the the arising arising measurement fault fault 

can can be be corrected mathematically 



temperature compensation using Wheatstone 


( + ε ) 

ε M 

ϑ 

( + ε ) 

ϑ 



F=0! 


F 

R 1 

R 2 

U A 

R 4 

R 3 

quarter bridge with 

compensating strain gauge 

U B




strain 

gauge 1 

strain 

gauge 2 

U a 

R 4 

R 3 

U e 

quarter bridge with 

compensating strain gauge 

strain 

strain 

gauge 

gauge 

( εM 

+ εϑ 

) 

( + ε ) 


U 

U 

a 

e 

= 

k 

4 

1 

2 

⇒ 

⇒ 

( ε − ε ) 

1 

2 

ϑ


quarter bridge with compensating strain gauge 

The compensating strain gauge: 

- must be applied in a spot where it will be subjected to 

the same interference effect as the active strain gauge 

- must have the same physical properties as the active 

strain gauge (same package or batch) 

- must only be subjected to the interference effect and 

never to the quantity to be measured ε M or its side effects 

- must be applied at the same material (e.g. steel) as the 

active strain gauge 





Advantages: 

- it is not necessary to measure the temperature during the 

strain measurement 

- it is not absolutely necessary that the strain gauges are 

adjusted to the material 







Changes of the strain gauge resistance of the 

same sign appearing in neigh boring arms 

will be subtracted with respect to the bridge 

output signal 

ε ϑ(+) 



half bridge 


R 1 

R 2 

U A 

R 4 

R 3 

U B






Changes of the strain gauge resistance of the 

same sign appearing in neigh boring arms 

will be subtracted with respect to the bridge 

output signal 



ε ϑ(+) 

ε ϑ(+) 

s. g. 1: + 

s. g. 2: + 

s. g. 3: + 

s. g. 4: + 


R 1 

R 2 

U A 

R 4 

R 3 

full bridge 

U B


thank you... 

... for your attention 

Hottinger Baldwin Messtechnik GmbH 

Im Tiefen See 45 

D-64293 Darmstadt 

www.hbm.com 

Dirk Eberlein 

Tel. 06151/803-763 

dirk.eberlein@hbm.com

Applying the Wheatstone Bridge Circuit

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