Training Signal Conditioning Instrumentation Amplifier
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Instrumentation Amplifier

30 min Signal Conditioning

Op-Amp Signal Conditioning: Instrumentation Amplifier

The instrumentation amplifier (in-amp) is the workhorse front-end of every precision measurement system. Built from three op-amps — two as non-inverting buffers with a shared gain-setting resistor, followed by a unity-gain difference stage — it provides the high input impedance, large common-mode rejection ratio, and programmable gain that sensor signals demand.

Webster and Pallàs-Areny show that the in-amp is the natural partner to the Wheatstone bridge: the bridge produces a small differential voltage riding on a large common-mode level, and the in-amp rejects the common mode while amplifying the difference. This lesson derives the classic gain formula and computes the CMRR benefit of a well-matched resistor network.

By the end you will be able to pick a gain resistor for a specified full-scale output and estimate the offset and noise referred to the sensor input.

Three-Op-Amp In-Amp Gain

$$G = \left(1 + \frac{2 R_1}{R_G}\right) \cdot \frac{R_3}{R_2}$$

For matched resistors $R_1, R_2, R_3$, the output stage is unity gain and the front-end sets the gain via $R_G$. Typical in-amps have $R_1 = 25\,\text{k}\Omega$ and $R_G$ externally programmed for $G = 1$ to 10 000.

Common-Mode Rejection Ratio

$$\text{CMRR} = 20\log_{10}\frac{G_{\text{diff}}}{G_{\text{cm}}}\,\text{dB}$$

A good in-amp achieves 100–120 dB CMRR, rejecting the 60 Hz mains hum that inevitably appears on long sensor leads. This is the real reason bridges + in-amps dominate industrial measurement.

Example 1 — Gain resistor for full-scale output

A 10 mV bridge signal must be amplified to 5 V. Using an AD620-style in-amp with $R_1 = 24.7\,\text{k}\Omega$, find $R_G$.

Target gain: $G = 5\,\text{V}/10\,\text{mV} = 500$.

$500 = 1 + 2\cdot 24700/R_G \;\Rightarrow\; R_G = 49400/499 = 99\,\Omega$.

A standard 100 Ω resistor gives $G = 495$; use a 0.1 % metal-film for accuracy.

Example 2 — CMRR suppresses 60 Hz hum

A sensor lead picks up 3 V of 60 Hz common-mode noise. The in-amp has 100 dB CMRR. How large is the residual differential noise at the output, at gain $G = 500$?

100 dB = $10^5$ rejection. Common-mode referred to differential input: $3\,\text{V}/10^5 = 30\,\mu\text{V}$.

At the output: $30\,\mu\text{V}\cdot 500 = 15\,\text{mV}$ — small compared to a 5 V full-scale signal.

Example 3 — Offset referred to input (RTI)

The in-amp has $V_\text{OS} = 50\,\mu\text{V}$ typical. At $G = 500$, find the output offset.

$V_{\text{OS,out}} = G\cdot V_\text{OS} = 500\cdot 50\,\mu\text{V} = 25\,\text{mV}$.

At a 10 V ADC range, this is 0.25 % — usually acceptable, and completely removable via a single-point calibration.

Interactive Demo: Instrumentation Amplifier Design
Gain G =495V/V
V_out =4.95V
Output offset =24.8mV
CM noise rejected =30µV (RTI)

Practice Problems

1. Find $R_G$ for $G = 100$ with $R_1 = 25\,\text{k}\Omega$.
2. Gain $G = 10$, $V_\text{OS} = 100\,\mu\text{V}$. Find output offset.
3. CMRR = 110 dB. What is the rejection ratio?
4. Why are the three-op-amp gain setting resistors chosen as $R_1 = R_1'$, not independently?
5. What is the typical input impedance of a monolithic in-amp vs. a simple differential amplifier?
6. At $V_s = 10\,\text{V}$ and a full bridge with $\Delta R/R = 1\%$, find the bridge output and the required in-amp gain for 5 V full-scale.
Show Answer Key

1. $100 = 1 + 50000/R_G \Rightarrow R_G = 50000/99 \approx 505\,\Omega$.

2. $G\cdot V_\text{OS} = 10\cdot 100\,\mu\text{V} = 1\,\text{mV}$.

3. $10^{110/20} = 10^{5.5} \approx 316{,}000$.

4. To keep the two gain paths symmetric; otherwise a common-mode input would see different gains, wrecking CMRR.

5. Typical in-amp: $10^9$–$10^{12}\,\Omega$; simple diff amp: dominated by the input resistor, $10^4$–$10^5\,\Omega$.

6. $V_o = 10\cdot 0.01 = 100\,\text{mV}$. Required $G = 5\,\text{V}/100\,\text{mV} = 50$.

The Wheatstone Bridge Filters, Noise, and the ADC