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Electrical Circuits and Sensors · Module 9 of 10

Signal Conditioning and the Wheatstone Bridge

A raw sensor signal is tiny, noisy, and riding on an offset. Signal conditioning fixes all three: the bridge converts a small resistance change to a voltage, an amplifier scales it, and a filter cleans it.

01

Readiness check

This module assembles the conditioning chain. Tick only what you can do closed-notes.

  • Use the voltage divider rule.
  • Recall the strain-gauge change ΔR/R = GF · ε.
  • Find the gain of a difference amplifier.
  • Recall capacitive reactance 1/(ωC).
  • Read a ratio as a decibel value, roughly.
0 or 1 weak itemsContinue with this module.
2 weak itemsRevisit the difference amplifier in Module 7.
3 or more weak itemsReview sensor characteristics in Module 8.
02

The core idea

A Wheatstone bridge turns a small resistance change into a voltage difference; an amplifier raises it to a usable level; a low-pass filter removes the noise above the signal band. Together they condition a sensor for measurement.

balanced bridge: Vo = 0quarter bridge: Vo ≈ (Vs/4)(ΔR/R)low-pass: fc = 1/(2πRC)

Signal conditioning sits between the sensor and the data system. The Wheatstone bridge is four resistors in a diamond; when all are equal it is balanced and its output is zero, so a small change in one arm produces a voltage proportional to that change, free of the large common offset. That difference is then amplified, usually by an instrumentation amplifier with high gain and high input resistance. Finally a filter shapes the frequency content: a first-order RC low-pass passes signals below its corner frequency fc = 1/(2πRC) and attenuates faster noise above it, rolling off at 20 decibels per decade. The corner is where the gain has fallen to 1/√2 of its low-frequency value, the half-power or −3 dB point.

The skill works when: you start from a balanced bridge, read the small unbalance, amplify it, and filter above the signal band.
The skill breaks down when: the bridge offset is amplified along with the signal, or the corner frequency cuts into the signal.
The concept. The conditioning chain: a bridge converts a tiny resistance change into a difference voltage, an amplifier raises it, and a low-pass filter strips noise above the signal band.
03

The skills, taught in order

Five skills build the bridge, its sensitivity, the amplifier, and the filter that cleans the result.

9.1 The Wheatstone bridge

Four resistors form two voltage dividers across one supply; the output is the difference between their midpoints. When the ratios match, the bridge is balanced and the output is zero, so it measures a change rather than an absolute, rejecting the large common voltage.

9.2 Bridge sensitivity

For a small change ΔR in one arm of an equal-arm bridge, the output is Vo ≈ (Vs/4)(ΔR/R). Using more active arms (half or full bridge) raises the sensitivity and can cancel temperature effects, multiplying the output by two or four.

Bridge typeActive armsOutput factorBenefit
Quarter1(Vs/4)(ΔR/R)simplest
Half2(Vs/2)(ΔR/R)double output, temp cancel
Full4Vs(ΔR/R)maximum output, linear

More active arms give a larger, more linear output and can cancel common effects such as temperature drift.

9.3 Instrumentation amplification

The bridge output is a small differential voltage on a large common-mode level. An instrumentation amplifier provides high differential gain, very high input resistance, and strong common-mode rejection, so it amplifies the signal without disturbing the bridge or passing the offset.

9.4 The first-order low-pass filter

An RC low-pass filter passes low frequencies and attenuates high ones. Its gain is 1/√(1 + (f/fc)2), flat below the corner and falling at 20 dB per decade above it. It smooths noise while preserving the slower signal.

9.5 The corner frequency

The corner (cut-off) frequency is fc = 1/(2πRC), where the output has dropped to 1/√2 (−3 dB) of the passband. Choosing R and C places fc just above the highest signal frequency, rejecting noise without distorting the measurement.

Engineering connection: this whole chain, bridge, amplifier, filter, is what produces the clean analogue voltage the data acquisition system samples in Module 10.

04

Worked example 1: a Wheatstone bridge output

An equal-arm Wheatstone bridge has R = 1000 Ω in each arm and a 10 V excitation. One arm changes by ΔR = 4 Ω. Find the bridge output voltage, by the linear approximation and exactly.

Figure 1. A 4 ohm change in a 1 kilohm arm unbalances the bridge by about 10 millivolts. The exact value is 9.98 millivolts, the small difference being the bridge's slight nonlinearity.
  1. ProblemFind the bridge output for the unbalance in Figure 1, approximately and exactly.
  2. Given / findR = 1000 Ω, Vs = 10 V, ΔR = 4 Ω, quarter bridge. Find Vo.
  3. AssumptionsEqual arms; one active arm; ideal, unloaded output.
  4. ModelLinear: Vo ≈ (Vs/4)(ΔR/R). Exact: difference of two divider voltages.
  5. EquationsVo ≈ (Vs/4)(ΔR/R)Vo = Vs[(R+ΔR)/(2R+ΔR) − 1/2]
  6. SolveApproximate: Vo = (10/4)(4/1000) = 2.5 × 0.004 = 10 mV. Exact: Vo = 10[1004/2004 − 0.5] = 10 × 0.000998 = 9.98 mV.
  7. CheckThe two agree to within 0.2 percent, confirming the linear approximation is excellent for a 0.4 percent resistance change. The output is millivolt-scale, so it needs amplification.
  8. ConclusionThe bridge converts a 4 ohm change into about 10 mV. The tiny gap between 10 and 9.98 mV is the bridge nonlinearity, negligible here.
Result. Vo ≈ 10 mV (exactly 9.98 mV).
05

Worked example 2: a first-order low-pass filter

A signal is filtered by an RC low-pass with R = 10 kΩ and C = 1 µF. Find the corner frequency and the gain in decibels at that frequency.

Figure 2. Below the corner the filter passes the signal unchanged; above it, the gain rolls off at 20 dB per decade. At the corner of 15.9 Hz the gain is down 3 dB, to 1/√2 of the passband.
  1. ProblemFind fc and the decibel gain at fc for the filter in Figure 2.
  2. Given / findR = 10 kΩ, C = 1 µF. Find fc and the gain in dB at f = fc.
  3. AssumptionsIdeal first-order RC low-pass; unloaded output.
  4. Modelfc = 1/(2πRC); at the corner |H| = 1/√2, and gain in dB is 20 log10|H|.
  5. Equationsfc = 1/(2πRC)|H(fc)| = 1/√2dB = 20 log10|H|
  6. Solvefc = 1/(2π × 10000 × 1×10−6) = 1/(2π × 0.01) = 15.9 Hz. At fc: |H| = 0.707, gain = 20 log10(0.707) = −3.0 dB.
  7. CheckThe −3 dB point is the defining feature of the corner, where output power is halved (1/√2 in amplitude). A decade above, at 159 Hz, the gain would be near −20 dB.
  8. ConclusionThis filter passes signals well below 15.9 Hz and attenuates faster noise. To pass a faster signal, reduce R or C to raise fc.
Result. fc = 15.9 Hz, gain at the corner = −3.0 dB.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Amplifying the offsetAmplifier saturates on the common level"Did I take the bridge difference first?"Use a difference or instrumentation amplifier on the bridge output.
Corner inside the signal bandReal signal attenuated"Is fc above the signal frequency?"Place fc just above the highest signal frequency.
Forgetting the 2πfc off by a factor of 6.28"Is this ω or f?"fc = 1/(2πRC), not 1/(RC).
Ignoring bridge nonlinearityLarge ΔR read with the linear formula"Is ΔR/R still small?"For large changes, use the exact divider expression.
07

Practice ladder

Level 1 · Direct skill

Find the corner frequency of an RC low-pass with R = 1 kΩ and C = 100 nF.

Show answer

fc = 1/(2πRC) = 1/(2π × 1000 × 100×10−9) = 1/(2π × 10−4) = 1592 Hz.

Level 2 · Mixed concept

A quarter bridge with R = 350 Ω and 5 V excitation has ΔR = 0.7 Ω. Find the output.

Show answer

Vo = (Vs/4)(ΔR/R) = (5/4)(0.7/350) = 1.25 × 0.002 = 2.5 mV.

Level 3 · Independent problem

The 10 mV bridge output of Example 1 is amplified by 100 and filtered at fc = 20 Hz. What is the conditioned signal level, and what does the filter remove?

Show answer

Amplified: 10 mV × 100 = 1.0 V. The 20 Hz low-pass passes the slow strain signal and removes higher-frequency noise such as mains pickup and vibration above 20 Hz.

Transfer task | Real engineering

Design a conditioning chain for a load cell whose strain signal is below 10 Hz but sits in an electrically noisy plant. State a bridge, a gain, and a corner frequency.

What good work looks like

A full bridge for sensitivity and temperature cancellation, an instrumentation amplifier with a gain to bring full scale near the data-system range, and a low-pass at perhaps 15 to 20 Hz to pass the 10 Hz signal while rejecting mains and high-frequency noise.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that my corner frequency uses 2πRC, not RC."
"Give me three bridges; I will find the output of each."
"Design the whole conditioning chain for me." Choosing the bridge, gain, and corner is the skill.
"What gain do I need?" Working it from the signal and range is the point.

Portfolio task

Specify a conditioning chain for a real sensor: a bridge output, an amplifier gain, and a filter corner, then justify each value against the signal and noise.

Must include: a bridge output, a gain, a corner frequency, and a reason for each.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What does a balanced bridge output?

Zero: the two dividers match, so it measures change, not absolute value.

2. Give the quarter-bridge sensitivity.

Vo ≈ (Vs/4)(ΔR/R).

3. Why an instrumentation amplifier?

High differential gain, high input resistance, and strong common-mode rejection.

4. Write the corner frequency.

fc = 1/(2πRC), the −3 dB point.

5. What is the gain at the corner?

1/√2 of the passband, or −3 dB.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the bridge output and corner frequency from a blank page.
+3 daysSpecify a conditioning chain for a new sensor.
+7 daysCarry the conditioned signal into data acquisition, Module 10.
+30 daysReuse the bridge and filter when instrumenting a real test.
10

Textbook mapping

This module draws on Alexander and Sadiku, Fundamentals of Electric Circuits, 4th edition for the bridge and filter, with conditioning practice from standard measurement texts.

Topic in this moduleWhere to read more
The Wheatstone bridgeAlexander & Sadiku, Chapter 2
First-order filters and frequency responseAlexander & Sadiku, Chapter 14
Instrumentation amplifiers and conditioningFigliola & Beasley, measurement systems

The bridge and filter follow the 4th edition of Alexander and Sadiku; conditioning practice draws on measurement texts such as Figliola and Beasley.