Training Signal Conditioning The Wheatstone Bridge
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The Wheatstone Bridge

30 min Signal Conditioning

The Wheatstone Bridge

The Wheatstone bridge is the most important analog circuit in all of sensor instrumentation. First described in 1833 and popularized by Charles Wheatstone a decade later, the bridge converts a tiny fractional resistance change into a differential voltage that can be amplified and digitized with negligible loss of accuracy.

Pallàs-Areny and Webster devote an entire chapter to bridges because every resistive sensor — strain gauge, RTD, thermistor — is invariably read through one. Understanding the balance condition, the quarter-, half-, and full-bridge configurations, and the effect of lead resistance is the single biggest skill separating a careful measurement from a sloppy one.

This lesson derives the classic bridge output equation, explores its linearization for large $\Delta R$, and previews the instrumentation amplifier that almost always follows.

Bridge Output (quarter-bridge)

For a single active gauge $R + \Delta R$ and three matched $R$:

$$V_o = V_s \cdot \frac{\Delta R}{4R + 2\Delta R} \approx \frac{V_s}{4}\cdot \frac{\Delta R}{R}$$

for small $\Delta R / R$. Sensitivity is $V_s/4$ volts out per fractional ohm change.

Half- and Full-Bridge
  • Half-bridge (two active arms, opposite strain): $V_o \approx (V_s/2)\cdot \Delta R/R$
  • Full-bridge (four active arms): $V_o = V_s\cdot \Delta R/R$

Each doubling improves sensitivity by 2× and — crucially — cancels the nominal resistance and its temperature drift.

Example 1 — Quarter bridge, small signal

A 350 Ω strain gauge with $GF = 2.0$ sees $\epsilon = 500\,\mu\epsilon$ in a quarter-bridge at $V_s = 10\,\text{V}$. Find $V_o$.

$\Delta R/R = GF\cdot \epsilon = 2.0\cdot 5\times10^{-4} = 10^{-3}$.

$V_o \approx (V_s/4)\cdot \Delta R/R = (10/4)\cdot 10^{-3} = 2.5\,\text{mV}$.

Example 2 — Full bridge

Four active gauges on a bending beam. Same $\epsilon$, $V_s = 10\,\text{V}$. Output?

$V_o = V_s\cdot \Delta R/R = 10\cdot 10^{-3} = 10\,\text{mV}$ — 4× larger than quarter-bridge.

Example 3 — Bridge nonlinearity

A quarter-bridge reads $V_o$ for $\Delta R/R = 0.10$. Compare exact vs. linearized formula.

Exact: $V_o/V_s = 0.10/(4 + 0.20) = 0.0238$.

Linearized: $V_o/V_s = 0.10/4 = 0.0250$.

Error: $(0.0250-0.0238)/0.0238 = 5.0\%$. For $\Delta R/R > 1\%$, use the exact formula or active linearization.

Interactive Demo: Wheatstone Bridge
V_o (linearized) =2.50mV
V_o (exact) =2.49mV
Linearization error =0.25%
Sensitivity =2.50mV per mV/V

Practice Problems

1. Quarter bridge: $V_s = 5\,\text{V}$, $\Delta R/R = 2\times10^{-3}$. Find $V_o$.
2. Half bridge, same $\Delta R/R$ and $V_s$. Find $V_o$.
3. Full bridge likewise. Find $V_o$.
4. Compute the exact $V_o/V_s$ for $\Delta R/R = 0.05$ in a quarter bridge.
5. Why does a full bridge automatically compensate for temperature drift?
6. What role does the instrumentation amplifier play after the bridge?
Show Answer Key

1. $V_o = (5/4)\cdot 2\times10^{-3} = 2.5\,\text{mV}$.

2. $V_o = (5/2)\cdot 2\times10^{-3} = 5.0\,\text{mV}$.

3. $V_o = 5\cdot 2\times10^{-3} = 10\,\text{mV}$.

4. $V_o/V_s = 0.05/(4 + 0.10) = 0.0122$.

5. All four resistors share the same temperature, so common-mode changes in $R$ cancel in the difference.

6. It provides high input impedance, high common-mode rejection, and gain — turning a few mV into a few V suitable for the ADC.

Instrumentation Amplifier