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The example, worked out
Given
I want
The output
Excitation
5 V
R1 (left, top)
1 kΩ
R2 (left, bottom)
1 kΩ
R3 (right, top)
1 kΩ
R4 (the sensor)
1.1 kΩ
Working
two dividers
the difference between the middles
the ratios equal, the output is zero
Results
Output B − A119 mV
A (left middle)2.5 V
B (right middle)2.62 V
R4 that balances1 kΩ
Sensitivity1.13 V / kΩ
Current drawn4.88 mA
The formulas
Vout = Vex × (R4 / (R3 + R4) − R2 / (R1 + R2))the difference between the two middles
balance: R1 / R2 = R3 / R4then the output is zero whatever the excitation
The usual traps
The output is not linear in the sensor resistance: compute the resistance back for wide swings.
The excitation warms the sensor arm: lower it, or pulse it.
Questions
How do I calculate the output of a Wheatstone bridge?
Vout = Vex × (R4 / (R3 + R4) − R2 / (R1 + R2)): the difference between the two middles. balance: R1 / R2 = R3 / R4: then the output is zero whatever the excitation.
What does it give for typical values?
119 mV out of 5 V with 1 kΩ, 1 kΩ, 1 kΩ, 1.1 kΩ. Output B − A: 119 mV.
What are the usual mistakes?
The output is not linear in the sensor resistance: compute the resistance back for wide swings. The excitation warms the sensor arm: lower it, or pulse it.
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Results rounded to standard values (E12, E24, E96) and power ratings.
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Open the calculatorOpens in the app on 1 December, with Circuit Notebook 1.0