Wheatstone Bridge

A Wheatstone bridge circuit converts an unknown resistance $R_x$ into a measurable voltage difference $\Delta V$, using two voltage dividers with known resistances $R_1$ through $R_3$.

$$\Delta v=v_b-v_a=v_s\frac{R_x}{R_2+R_x}-v_s\frac{R_3}{R_1+R_3}$$

It’s basically just two voltage dividers that we’re subtracting.

This differential voltage output is preferable to single-ended voltage that one voltage divider would produce. Since all the resistors experience the same variation due to operating conditions, those errors should be self-cancelling.

It is also possible to buy highly-matched on-chip resistors. Often one resistor is a potentiometer for calibration.

There are two strategies for finding the unknown resistance:

1. Vary $R_3$ until $\Delta v=0$, then:

$$R_x=\frac{R_2}{R_1}{R}_3$$

2. $R_3$ is set so that $\Delta v=0$ at a known calibration point, then $\Delta v$ is used to calculate the resulting $R_x$ while in operation:

$$R_x=\frac{R_2\left(\frac{\Delta v}{v_s}+\frac{R_3}{R_1+R_3}\right)}{1-\left(\frac{\Delta v}{v_s}+\frac{R_3}{R_1+R_3}\right)}$$

A software lookup table is often used to directly relate the measured $\Delta v$ with $R_x$ or the sensor’s input physical parameter since $\Delta v$ vs. $R_x$ is nonlinear.

Example