MEGR 3171 — Week 4: Advanced Signal Conditioning

Learning Objectives

Explain the architecture and key specifications of an instrumentation amplifier (INA).
Calculate INA gain from the external gain resistor and interpret CMRR in dB.
Derive the output voltage of quarter-, half-, and full-bridge Wheatstone configurations.
Select bridge excitation method and apply lead-wire compensation for remote gauges.
Design a bridge circuit for a target output voltage given a strain gauge resistance and gauge factor.

1. Instrumentation Amplifier (INA)

A standard op-amp rejects common-mode signals imperfectly when built with discrete resistors due to resistor mismatch. The instrumentation amplifier solves this with a three-op-amp architecture where a single external resistor R_G sets the gain precisely without affecting CMRR.

INA Gain and CMRR

Gain (A_d) = 1 + (2R / R_G)

where R is an internal precision resistor (e.g., 24.7 kΩ for INA128)

CMRR (dB) = 20 · log_10(A_d / A_cm)

A good INA: CMRR > 100 dB at DC, > 80 dB at 60 Hz

Common-Mode Rejection Ratio (CMRR) quantifies how well the amplifier rejects a voltage that appears equally on both inputs (e.g., 60 Hz interference from power lines or ground loops). Higher is better. CMRR degrades at higher frequencies.

Differential Input

The INA amplifies only the difference between the two input terminals (V+ − V−). This is the desired sensor signal.

Common-Mode Input

Any voltage present equally on both inputs (noise, ground offset). The INA suppresses this. CMRR = 20·log(A_d / A_cm).

Input Protection

INAs include clamping diodes and series resistors to protect against overvoltage on the input pins from sensor cable transients.

Input Offset Voltage

A small DC error voltage at the input. Multiplied by gain, it appears at the output. Reduced by trimming or choosing a low-offset INA.

2. Wheatstone Bridge Theory

A Wheatstone bridge converts a small resistance change (from a strain gauge, RTD, or thermistor) into a measurable voltage. A balanced bridge (all arms equal) produces zero output. A resistance change in one or more arms produces a differential output proportional to that change.

Bridge Output Voltage (general)

V_out = V_ex · [ R_4/(R_3+R_4) − R_2/(R_1+R_2) ]

For a quarter-bridge (one active arm, gauge factor G_f, strain ε):
  ΔR/R = G_f · ε

V_out ≈ V_ex · (G_f · ε) / 4     (for small ΔR/R)

Bridge Configurations

Quarter-bridge: One active arm (one strain gauge). Smallest output. Temperature compensation requires a dummy gauge in an adjacent arm.
Half-bridge: Two active arms on adjacent or opposite sides. Output doubles for adjacent, cancels for opposite if both see same strain.
Full-bridge: All four arms active. Maximum output (4× quarter-bridge), best noise immunity, and inherent temperature compensation.

Temperature Compensation Placing a dummy (unstrained) gauge in an adjacent bridge arm ensures that temperature changes affect both arms equally, producing no net output change. Both the active gauge and dummy must be the same type at the same temperature.

Lead-Wire Compensation

For remote installations, the resistance of lead wires adds to the gauge resistance and changes with temperature. A 3-wire connection moves one lead wire into each of two adjacent bridge arms, so their resistances cancel in the bridge output. A 4-wire Kelvin connection completely separates force-current and sense-voltage paths.

3. Bridge Excitation

Constant-voltage excitation is simpler and most common. Self-heating error exists because power dissipated in the gauge = V_ex² / (4R). Constant-current excitation eliminates self-heating variation with resistance change but requires a precision current source. The choice depends on gauge resistance, thermal conductivity of the substrate, and required accuracy.

Practice Problems

Problem 1 — INA Gain An INA128 has an internal resistor R = 24.7 kΩ. What value of R_G is needed to achieve a gain of 100?

A_d = 1 + 2R/R_G → R_G = 2R / (A_d − 1) = 2(24700) / (100 − 1) = 49400 / 99 ≈ 499 Ω

R_G ≈ 499 Ω (use standard 499 Ω 1% resistor)

Problem 2 — Quarter-Bridge Output A 350 Ω strain gauge with G_f = 2.05 is configured in a quarter-bridge with V_ex = 5.0 V. The applied strain is 800 µε. Calculate the bridge output voltage.

V_out = V_ex · (G_f · ε) / 4 = 5.0 × (2.05 × 800×10&sup-⁶) / 4

= 5.0 × 0.001640 / 4 = 5.0 × 0.000410 = 0.00205 V

V_out ≈ 2.05 mV

Problem 3 — CMRR An INA has A_d = 200 and A_cm = 0.001. Calculate the CMRR in dB.

CMRR = 20 · log₁₀(200 / 0.001) = 20 · log₁₀(200,000) = 20 × 5.301 = 106 dB

CMRR = 106 dB (excellent)