The Wheatstone Bridge is a very useful circuit. When the bridge is fully balanced, the right resistor is the same, the left resistor（R1=R3,and R2=Rx），The voltage across the bridge is zero. However, due to a small change in resistance, the bridge becomes unbalanced and a voltage difference occurs. Wheatstone bridge applications, such as strain gauges, pressure gauges, sensors and other equipment.
A differential amplifier can be used to extract the common mode signal while rejecting all common mode noise. As a very small signal change can be extracted from the bridge because common mode noise is easily rejected.
The bridge voltage is calculated as follows：
VB= Vin*[Rx/(R3+Rx)- R2/(R1+R2)]
If R3=R1, 和 Rx= R2+delta, then
VB= Vin *[ (R2+delta)/(R1+R2+delta)-R2/(R1+R2)]
Now, if we assume delta is better than R1 + R2 is small, then
VB = ~ V in*[delta/(R1+R2)]
Therefore, we can see that the bridge voltage is about proportional errordelta，Divided by the sum of the resistors.
Due to the bridge voltage, we can calculate an unknown resistance value.
(R1+R2)*(R3+Rx)*VB/Vin= Rx*(R1+R2)+ R2*(R3+Rx)
Rx*(R1+R2)*VB/ Vin + R3* (R1+R2)VB/Vin= Rx*R1+Rx*R2 - R2*R3- Rx*R2
Rx*R1 - Rx*(R1+R2)*VB/ Vin = R2*R3 + R3* (R1+R2)VB/Vin
Rx = (R2*R3 + R3* (R1+R2)VB/Vin )/ (R1- (R1+R2)*VB/ Vin)