Instrumentation amplifier

Why it matters

A strain-gauge bridge, an ECG electrode pair or a thermocouple on a grounded pipe gives a millivolt difference signal sitting on volts of common-mode voltage, from a source that may have kilohms of resistance. A plain difference amplifier loads such a source and its CMRR collapses with source-resistance imbalance. The instrumentation amplifier (in-amp) solves both problems and is the standard front end of industrial and biomedical measurement channels.

Key ideas

What an in-amp must provide.

• Very high, equal input impedance on both inputs (no loading, no imbalance).
• High CMRR (typically 80–120 dB), ideally rising with gain.
• Gain set accurately by one resistor, from 1 to about 1000.
• Low offset, low drift and low noise; a single-ended output referenced to a REF pin.

The three-op-amp circuit.

• Input stage: two non-inverting amplifiers (A1, A2), each with a feedback resistor R₁, joined at their inverting inputs by a single gain resistor R_G. By the virtual-short rule, the full differential input v_d = v₂ − v₁ appears across R_G, so the current v_d/R_G flows through R₁ + R_G + R₁. The differential output of this stage is v_d·(1 + 2R₁/R_G).
• Common-mode signals: if v₁ = v₂ = v_cm, no current flows in R_G and both outputs simply follow v_cm with gain 1. So the first stage amplifies the difference by (1 + 2R₁/R_G) but the common mode only by 1, raising the CMRR in proportion to gain before the signal reaches the resistors of the next stage.
• Output stage: a difference amplifier (A3) with resistors R₂ and R₃ converts the differential signal to single-ended, rejects the remaining common mode, and adds gain R₃/R₂ (often 1).
• Because the inputs drive op-amp non-inverting terminals directly, input resistance is that of the op-amps (GΩ for FET inputs) and equal on both sides.

Gain set by one resistor. Changing R_G changes the gain without touching the matched resistor network of the output stage, so CMRR is not disturbed. Monolithic in-amps laser-trim R₁ and the output network and specify the gain equation in the datasheet (for example G = 1 + 49.4 kΩ/R_G for one common device); take the constant from the datasheet.

Two-op-amp in-amp. Uses fewer parts but has poorer CMRR at high frequency and cannot reach a gain of 1.

REF pin. The output-stage reference terminal shifts the output; it must be driven from a low impedance (a buffer), because any resistance in series with it unbalances the output-stage resistors and degrades CMRR.

Limitations. Common-mode range is limited: the input op-amp outputs must stay within their swing at v_cm ± (gain × v_d)/2. Bandwidth falls as gain rises. Input bias currents need a DC return path to ground; floating sources (thermocouples, transformers) need a high-value resistor from each input to ground.

Formulas

G = (1 + 2R₁/R_G)·(R₃/R₂) — three-op-amp in-amp; R₁ input-stage feedback resistors (Ω), R_G gain resistor (Ω), R₃/R₂ output-stage gain.

G = 1 + 2R₁/R_G — when the output stage has unity gain (R₃ = R₂).

R_G = 2R₁/(G/(R₃/R₂) − 1) — gain resistor for a required G.

v_o = G·(v₂ − v₁) + V_REF — ideal output.

CMRR(dB) = 20·log₁₀(G/A_cm); output error from common mode = v_cm·G/CMRR.

v_bridge ≈ V_ex·ΔR/(4R) — quarter-bridge output for small ΔR, used as the in-amp's input.

Worked examples

Example 1 (standard: gain). A three-op-amp in-amp has R₁ = 10 kΩ, R_G = 1 kΩ and an output stage with R₃/R₂ = 2. The differential input is 50 mV. Find the gain and output.

1. Input-stage gain: 1 + 2R₁/R_G = 1 + 20/1 = 21.
2. Total gain: G = 21 × R₃/R₂ = 21 × 2 = 42.
3. v_o = G·v_d = 42 × 50 mV = 2.1 V.

Answer: G = 42, v_o = 2.1 V.

Example 2 (GATE level: bridge front end). A quarter-bridge of 350 Ω strain gauges is excited by 10 V. Under load the active gauge rises by 0.7 Ω. An in-amp with R₁ = 25 kΩ and a unity-gain output stage must give 2.5 V out. Its CMRR at that gain is 100 dB. Find the bridge output, the required R_G, and the output error due to the bridge's common-mode voltage.

1. Bridge output: v_d ≈ V_ex·ΔR/(4R) = 10 × 0.7/(4 × 350) = 5.0 mV (the exact divider gives 4.995 mV).
2. Required gain: G = 2.5 V / 5 mV = 500.
3. R_G = 2R₁/(G − 1) = 50 kΩ/499 = 100.2 Ω (use 100 Ω, 0.1 %).
4. The bridge's common-mode voltage is V_ex/2 = 5 V. CMRR = 100 dB = 10⁵, so A_cm = G/CMRR = 500/10⁵ = 0.005.
5. Common-mode output error = 5 V × 0.005 = 25 mV.

Answer: v_d ≈ 5 mV, R_G ≈ 100 Ω, common-mode error ≈ 25 mV (1 % of full scale). A plain difference amplifier with 1 % resistors would have a common-mode error tens of times larger; this is why in-amps are used.

Common mistakes

• Writing the gain as 2R₁/R_G (forgetting the 1) or as 1 + R₁/R_G (forgetting the factor 2).
• Forgetting to multiply by the output-stage gain R₃/R₂ when it is not unity.
• Assuming the input stage alone rejects common mode; it passes common mode with gain 1 and the output stage removes it.
• Leaving a floating source without a bias-current return path; the inputs drift to a rail.
• Driving the REF pin through a resistor divider, which ruins CMRR.
• Ignoring the common-mode range: a large v_cm plus a large amplified signal can saturate the input op-amps even when the output looks reasonable.

For GATE IN

Expect gain and R_G calculations for the three-op-amp circuit, output voltage for a bridge input, CMRR and common-mode error, and conceptual questions on why the in-amp beats a single difference amplifier. Bridge-plus-in-amp combinations are a favourite instrumentation question. Practise deriving the gain using the current through R_G.

Quick check

1. R₁ = 20 kΩ, R_G = 2 kΩ, unity output stage. Gain?
2. For G = 101 with R₁ = 25 kΩ (unity output stage), what is R_G?
3. With v₁ = v₂ = 3 V, what are the input-stage op-amp outputs?
4. Why is the gain set by a single resistor an advantage?

Answers: 1. 21. 2. 500 Ω. 3. Both 3 V (no current in R_G). 4. Gain can be changed without disturbing the matched resistors that set CMRR.