App description

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.

<img src="http://drr.ikcest.org/static/upload/c7/c7129530-ecc4-11e9-80df-00163e0618d6_m.jpg" />

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:

<!-- <img src="http://mathtex.yunsuan.org/cgi-bin/mathtex/yunsuan? V_B = V_{in} \times [ Rx/(R3+Rx)- R2/(R1+R2)] ">

-->

If R3=R1, and Rx=R2+delta, then

<!-- <img src="http://mathtex.yunsuan.org/cgi-bin/mathtex/yunsuan? V_B= V_{in}*[ (R2+delta)/(R1+R2+delta)-R2/(R1+R2)] ">

-->

Now, if we assume that delta is smaller than R1 + R2, then

<!-- <img src="http://mathtex.yunsuan.org/cgi-bin/mathtex/yunsuan? V_B= ~ V_{in}*[delta/(R1+R2)] ">

-->

Therefore, we can see that the bridge voltage is approximately proportional to the error delta, divided by the sum of the resistors.

Due to the bridge voltage, we can calculate an unknown resistance value.

<!-- <img src="http://mathtex.yunsuan.org/cgi-bin/mathtex/yunsuan? (R1+R2)(R3+Rx)V_B/V_{in}= Rx(R1+R2)+ R2(R3+Rx) ">

<img src="http://mathtex.yunsuan.org/cgi-bin/mathtex/yunsuan? Rx(R1+R2)V_B/ V_{in} + R3 (R1+R2)V_B/V_{in}= RxR1+RxR2 - R2R3- Rx*R2 ">

<img src="http://mathtex.yunsuan.org/cgi-bin/mathtex/yunsuan? RxR1 - Rx(R1+R2)V_B/ V_{in} = R2R3 + R3* (R1+R2)V_B/V_{in} ">

<img src="http://mathtex.yunsuan.org/cgi-bin/mathtex/yunsuan? Rx = (R2R3 + R3 (R1+R2)V_B/V_{in} )/ (R1- (R1+R2)*V_B/ V_{in}) ">

-->