INTRODUCTION TO ELECTRONICS: The Unbalanced Wheatstone Bridge and Sensor Technology

A transducer is an electronic device that measures physical parameters such as mechanical strain, pressure, optical density, and/or temperature. If temperature is being measured, an instrument called a thermistor acts as a variable resistor at the R₁ location of a Wheatstone bridge. At a known neutral temperature, a zero reference voltage ( V_out ) value is recorded. As temperature derivations occur, proportional changes in the thermistor’s resistance ( R ) occur as well. As a consequence, changes in the voltage gradient between points A and B are recorded and converted into either analogue or digital outputs that are interpreted by human eyes and/or ears. Consider the diagram below:

Q: A bridge circuit is situated parallel to a 12 V source. Its resistors have the following values at 25 ℃ :

R_therm = 1.0 kΩ

R₂ = 1.0 kΩ

R₃ = 1.0 kΩ

R₄ = 1.0 kΩ

At 50 ℃, the thermistor’s resistance drops to 900 Ω. What voltage reading will register across the A-to-B terminal?

A: Let’s first use the voltage-divider formula to determine the potential difference from node A to B at 25 ℃.

Note: At node A and B, charged particles have lost half of their energy potential; therefore, we use R₃ and R₄ in the voltage divider formula to determine what voltage value remains at each node.

V_A = [ R₃ / ( R₃ + R_therm ) ]( V_s )

V_A = ( 1.0 kΩ / 2.0 kΩ )( 12 V )

V_A = 6.0 V

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V_B = [ R₄ / ( R₂ + R₄ ) ]( V_s )

V_B = ( 1.0 kΩ / 2.0 kΩ )( 12 V )

V_B = 6.0 V

ΔV = V_A – V_B 

ΔV = 0 V

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Let’s now see what differential exists when the thermistor is exposed to a 50 ℃ temperature. Since R_therm at 50 ℃ is 100 Ω lower than its reference resistance value, we can expect to obtain a non-zero value for V_A:

V_A = [ 1.0 kΩ / ( 1.0 kΩ + 900 Ω ) ]( V_s )

V_A = ( 1.0 kΩ / 1.9 kΩ )( 12 V )

V_A = 6.3 V

ΔV = 6.3 V – 6.0 V

ΔV = 0.3 V

Since node A is at a higher energy potential, charged particles will flow from node A to node B.