Wheatstone Bridge
Table of Contents
1. Balanced Circuits
Consider the following circuit:
We have two voltage dividers, so we know that the voltages at point A and B are:
\begin{align} V_A &= \frac{R_3}{R_1+R_3}V_0 \notag \\ V_B &= \frac{R_4}{R_2+R_4}V_0 \notag \end{align}When \(V_A = V_B\), we call this a balanced circuit:
\begin{align} \frac{R_3}{R_1+R_3} &= \frac{R_4}{R_2+R_4} \notag \\ \frac{1}{1 + \frac{R_1}{R_3}} &= \frac{1}{1+\frac{R_2}{R_4}} \notag \end{align}So, the balanced condition would be:
\begin{align} \boxed{\frac{R_1}{R_3} = \frac{R_2}{R_4}} \end{align}When it is balanced, the current across \(A\) and \(B\) as measured by an ammeter would be zero.
2. Wheatstone Bridge
The circuit above is often called a Wheatstone bridge circuit:
The Wheatstone bridge can be used to determine a component of unknown resistance. We keep measuring the current with our ammeter and change the variable resistor \(R_4\) until the current measured is zero. This means the circuit is now balanced, so:
\begin{align} \frac{R_1}{R_3} &= \frac{R_2}{R_4} \notag \\ R_3 &= \frac{R_1R_4}{R_2} \notag \end{align}Particularly, if \(R_1 = R_2\), then \(R_3 = R_4\).
The resolution of this result is much higher than just directly measuring the current, as measuring zero current is much easier than detecting small changes in current.
2.1. Strain Sensor
Resistance has the following relationship:
\begin{align} \boxed{R = \rho\frac{L}{A}} \end{align}where \(\rho\) is resistivity, \(L\) is length, and \(A\) is cross-sectional area. This formula says that resistance is proportional to the length and inversely proportional to the cross-sectional area of a wire.
When we put strain on a material (e.g. by stretching or bending), we change the length of the material by a small amount, which changes the resistance.
If we put this in a Wheatstone bridge, we should be able to figure out the change in resistance, and from that, determine the strain. Instead of measuring current, though, we measure the voltage:
We know that the voltages at points A and B are:
\begin{align} V_A &= \frac{R+\Delta R}{2R+\Delta R}V_0 \notag \\ V_B &= \frac{R}{R+R}V_0 = \frac{V_0}{2} \notag \end{align}Then, the voltmeter would read:
\begin{align} V_{AB}&= \left(\frac{R+\Delta R}{2R+\Delta R}-\frac{1}{2}\right) V_0 \notag \\ &= \frac{V_0}{2}\left(\frac{2R+2\Delta R}{2R+\Delta R} - 1\right) \notag \\ &= \frac{V_0}{2} \cdot \frac{\Delta R}{2R+\Delta R} \notag \\ &= \frac{V_0}{2} \cdot \frac{\frac{\Delta R}{2R}}{1+\frac{\Delta R}{2R}} \notag \\ &= \frac{V_0}{4} \cdot \frac{\Delta R}{R}\left(\frac{1}{1+\frac{\Delta R}{2R}}\right) \notag \end{align}Since \(\frac{\Delta R}{R} \ll 1\), we can say that this is approximately equivalent to:
\begin{align} \boxed{\frac{V_0}{4}\cdot\frac{\Delta R}{R}} \end{align}