Instrumentation and Isolation Amplifiers

An instrumentation amplifier (IA) is a precision differential amplifier with very high input impedance, high CMRR and an accurately set gain. It is used to amplify small differential signals from transducers in the presence of large common-mode noise.

Features of an instrumentation amplifier

• Very high input impedance (both inputs buffered), so it does not load the source
• Very high CMRR (typically 100 dB or more)
• Gain set by a single resistor RG, adjustable and stable
• Low offset voltage, low drift and low noise
• Low output impedance and high slew rate

Three op-amp IA

 V1 --->(+)\
           A1>---+--Vo1--[R2]--+--[R3]--+
       +->(-)/   |             |        |
       |         R1          (-)\       |
       +---------+              A3>-----+--> Vo
       RG                    (+)/
       +---------+             |
       +->(-)\   R1   Vo2--[R2]+--[R3]--GND
       |   A2>---+--Vo2
 V2 --->(+)/

A1 and A2 form the buffered input stage. A3 is a difference amplifier.

Vo = (R3/R2)(1 + 2R1/RG)(V2 − V1)

Design for gain A = 20

Make the difference stage a unity-gain stage so that all the gain is in the first stage (only RG needs to be adjusted).

1. Choose R2 = R3 = 10 kΩ, so R3/R2 = 1.
2. Then 1 + 2R1/RG = 20, so 2R1/RG = 19.
3. Choose RG = 1 kΩ:

R1 = 19 × RG / 2 = 19 × 1 kΩ / 2 = 9.5 kΩ

Alternatively, fix R1 = 10 kΩ:

RG = 2R1/(A − 1) = 20 kΩ/19 = 1.053 kΩ

Use 1 kΩ fixed plus a 100 Ω trimmer.

Check: A = (10k/10k)(1 + 2 × 9.5k/1k) = 1 × (1 + 19) = 20.

Design values: R1 = 9.5 kΩ (two, matched), RG = 1 kΩ, R2 = R3 = 10 kΩ (four, 0.1 % matched for high CMRR), op-amps for example 741, OP07 or TL084.

The four resistors of A3 must be closely matched (R3/R2 equal on both sides). Mismatch directly reduces CMRR.

An instrumentation amplifier (IA) is a precision difference amplifier with buffered inputs. It has very high input impedance, high CMRR and a gain set by one resistor. It is used to amplify small transducer signals in noisy, high common-mode environments.

Features of an instrumentation amplifier

• Very high input impedance on both inputs
• High CMRR (100 dB or more), which rejects common-mode noise
• Gain set by a single external resistor RG, accurate and stable
• Low offset voltage and drift, low noise
• Low output impedance

Three op-amp IA and gain

 V1 --->(+)\
           A1>---+--Vo1--[R2]--+--[R3]--+
       +->(-)/   |             |        |
       |         R1          (-)\       |
       +---------+              A3>-----+--> Vo
       RG                    (+)/
       +---------+             |
       +->(-)\   R1   Vo2--[R2]+--[R3]--GND
       |   A2>---+--Vo2
 V2 --->(+)/

The current through RG is (V1 − V2)/RG. The same current flows through both R1 resistors, so:

Vo2 − Vo1 = (V2 − V1)(1 + 2R1/RG)
Vo = (R3/R2)(Vo2 − Vo1) = (R3/R2)(1 + 2R1/RG)(V2 − V1)

Design for gain A = 50

Option 1, all gain in the first stage:

1. R2 = R3 = 10 kΩ, so the difference stage gain = 1.
2. 1 + 2R1/RG = 50, so 2R1/RG = 49.
3. Choose RG = 1 kΩ:

R1 = 49 × 1 kΩ / 2 = 24.5 kΩ

(Or fix R1 = 10 kΩ, then RG = 2R1/(A − 1) = 20k/49 = 408 Ω; use a 390 Ω resistor plus a 50 Ω trimmer.)

Option 2, split the gain (10 × 5):

• First stage: 1 + 2R1/RG = 10, with RG = 2 kΩ and R1 = 9 kΩ
• Difference stage: R3/R2 = 5, with R2 = 10 kΩ and R3 = 50 kΩ

A = 5 × (1 + 2 × 9k/2k) = 5 × 10 = 50

Check (option 1): A = 1 × (1 + 2 × 24.5k/1k) = 1 + 49 = 50.

Design values (option 1): R1 = 24.5 kΩ (×2, matched), RG = 1 kΩ, R2 = R3 = 10 kΩ (×4, 0.1 % matched), op-amps such as OP07 or LM324.

Gain can be changed by adjusting RG only, without disturbing CMRR.

An instrumentation amplifier (IA) is a closed-loop differential amplifier with buffered inputs. It has very high input impedance, high CMRR and a precisely set gain, and it is used to amplify low-level transducer signals.

Features of an instrumentation amplifier

• Very high input impedance (hundreds of MΩ), so it does not load the transducer or bridge
• High CMRR (more than 100 dB), which rejects common-mode noise and hum
• Gain set accurately by one resistor RG, typically 1 to 1000
• Low offset voltage, low drift and low noise
• Low output impedance and good gain stability with temperature

Three op-amp instrumentation amplifier

 V1 --->(+)\
           A1>---+--Vo1--[R2]--+--[R3]--+
       +->(-)/   |  (a)        |        |
       |         R1          (-)\       |
       +---------+              A3>-----+--> Vo
       RG                    (+)/
       +---------+             |
       +->(-)\   R1   Vo2--[R2]+--[R3]--GND
       |   A2>---+--Vo2 (b)
 V2 --->(+)/

Derivation of output voltage

Step 1, input stage (A1, A2). By the virtual short, the inverting inputs of A1 and A2 are at V1 and V2. So the voltage across RG is V1 − V2, and the current through it is:

I = (V1 − V2) / RG

The op-amp inputs draw no current, so the same I flows through both R1 resistors:

Vo1 = V1 + I·R1 = V1 + (R1/RG)(V1 − V2)
Vo2 = V2 − I·R1 = V2 − (R1/RG)(V1 − V2)

Subtracting:

Vo2 − Vo1 = (V2 − V1) + (2R1/RG)(V2 − V1)
          = (V2 − V1)(1 + 2R1/RG)

Step 2, difference amplifier (A3). For a difference amplifier with matched resistors:

Vo = (R3/R2)(Vo2 − Vo1)

Result:

Vo = (R3/R2)(1 + 2R1/RG)(V2 − V1)
Ad = (R3/R2)(1 + 2R1/RG)

Example: R1 = 10 kΩ, RG = 1 kΩ, R2 = R3 = 10 kΩ gives Ad = 1 × (1 + 20) = 21.

A common-mode input (V1 = V2) gives no current in RG, so Vo1 = V1 and Vo2 = V2 with gain 1, and A3 cancels it. The gain can be changed with RG alone.

An isolation amplifier is an amplifier whose input and output circuits have no ohmic (galvanic) connection. The signal crosses the barrier by optical, transformer (magnetic) or capacitive coupling, and each side has its own ground and power supply.

Applications of isolation amplifiers

• Medical instruments (ECG, EEG, patient monitors): they protect the patient from leakage currents and mains faults.
• High-voltage measurement: sensing current and voltage in motor drives, power converters and power lines without exposing low-voltage circuits.
• Breaking ground loops in industrial instrumentation and data acquisition.
• Industrial process control (4–20 mA loops, thermocouples on high common-mode voltage).
• Protecting computers and ADCs from high-voltage transients. They are also used in nuclear and chemical plant monitoring.

Optically coupled isolation amplifier

       INPUT SIDE (GND1)    |   OUTPUT SIDE (GND2)
                            |
 Vin -[R1]-> A1 --> LED ~~~~|~~~~> PD2 --> A2 --> Vo
             ^       ~      |      (I2)   (I to V,
             |       ~      |              Rf2)
             +----- PD1     |
           (feedback, I1)   |
                    isolation barrier

• A1 drives an LED. The light falls on two matched photodiodes.
• PD1 (input side) feeds current back to A1's inverting input. This servo loop forces the LED light to be exactly proportional to Vin, which removes the LED's non-linearity and temperature drift.
• PD2 (output side) receives the same light, so its current is also proportional to Vin. A2 converts this current to the output voltage.
• With matched photodiodes: Vo = (Rf2/R1)·Vin.

There is no electrical path between Gnd1 and Gnd2, so the isolation is several kV.

This is the two op-amp instrumentation amplifier. Both inputs go to non-inverting terminals, so both inputs have very high input impedance. The gain resistor R between the two inverting inputs sets the gain.

Circuit (as given)

   GND--[R2]--+--[R1]--+
              |        |
           A (-)\      |
                 A1>---+-Vo1-[R1]-+--[R2]--+
        V1--(+)/                  |        |
              |                B (-)\      |
              |                      A2>---+-> Vo
              |          V2-------(+)/
              |                   |
              +-------[ R ]-------+
 (node A = inverting input of A1,
  node B = inverting input of A2)

Analysis

By the virtual short, the inverting input of A1 (node A) is at V1, and the inverting input of A2 (node B) is at V2. The op-amp inputs draw no current.

KCL at node A (currents leaving through R2 to ground, R1 to Vo1, and R to node B):

V1/R2 + (V1 − Vo1)/R1 + (V1 − V2)/R = 0
Vo1 = V1(1 + R1/R2) + (R1/R)(V1 − V2)

KCL at node B (currents leaving through R1 to Vo1, R2 to Vo, and R to node A):

(V2 − Vo1)/R1 + (V2 − Vo)/R2 + (V2 − V1)/R = 0
Vo = V2(1 + R2/R1) − (R2/R1)Vo1 + (R2/R)(V2 − V1)

Substitute Vo1:

(R2/R1)Vo1 = V1(1 + R2/R1) + (R2/R)(V1 − V2)

Vo = V2(1 + R2/R1) − V1(1 + R2/R1)
     − (R2/R)(V1 − V2) + (R2/R)(V2 − V1)
   = (V2 − V1)(1 + R2/R1) + (2R2/R)(V2 − V1)

Result

Vo = (1 + R2/R1 + 2R2/R)(V2 − V1)
Ad = 1 + R2/R1 + 2R2/R

Example: R1 = R2 = 10 kΩ, R = 2 kΩ gives Ad = 1 + 1 + 10 = 12.

For V1 = V2 the output is zero, so common-mode signals are rejected (with matched resistors). The gain can be adjusted with R alone. Without R (R → ∞), Ad = 1 + R2/R1.

An instrumentation amplifier is a precision differential amplifier with high input impedance, high CMRR and an accurately set gain. It is used to amplify small differential signals from transducers in the presence of large common-mode noise.

Characteristics

• High and balanced input impedance
• High CMRR (more than 100 dB)
• Accurate gain set by one resistor, typically 1 to 1000
• Low offset, drift and noise; low output impedance

Single op-amp IA with mismatched resistors

            R1           R2
 V1 ----/\/\/----+----/\/\/----+
                 |             |
                (-)\           |
                    >----------+--> Vo
                (+)/
            R3   |
 V2 ----/\/\/----+
                 |
                 R4
                 |
                GND

Output voltage (superposition):

V1 alone (V2 = 0) is an inverting amplifier: −(R2/R1)V1. V2 alone is a divider followed by a non-inverting amplifier: V+ = V2·R4/(R3 + R4), so the output is (1 + R2/R1)·R4/(R3 + R4)·V2.

Vo = (1 + R2/R1)·R4/(R3 + R4)·V2 − (R2/R1)·V1

Differential and common-mode gains. Put V1 = Vc − Vd/2 and V2 = Vc + Vd/2, where Vd = V2 − V1 and Vc = (V1 + V2)/2:

Vo = Ad·Vd + Acm·Vc

Ad  = (R1R4 + R2R3 + 2R2R4) / [2R1(R3 + R4)]
Acm = (R1R4 − R2R3) / [R1(R3 + R4)]
CMRR = |Ad/Acm|

With matching R4/R3 = R2/R1: Acm = 0, Ad = R2/R1, and Vo = (R2/R1)(V2 − V1). Any mismatch makes Acm ≠ 0 and limits the CMRR.

Input impedance:

• At the V2 input: the current flows through R3 and R4 to ground, so Rin2 = R3 + R4.
• At the V1 input: the inverting terminal sits at V+ (virtual short), so I1 = (V1 − V+)/R1. With V2 = 0, Rin1 = R1.
• Differential input resistance (matched, R3 = R1): Rid = 2R1.
• Common-mode input resistance (matched, both inputs tied): Ricm = (R1 + R2)/2.

The input impedance is only a few kΩ and the two inputs are unequal. This is the main weakness of the one op-amp IA, and the reason for the three op-amp IA with buffered inputs.

Ideal characteristics of an instrumentation amplifier

• Infinite input impedance on both inputs (no loading of the source)
• Zero output impedance
• Infinite CMRR (zero common-mode gain)
• Finite, accurate and stable differential gain, set by one resistor
• Zero input offset voltage, offset current and drift
• Infinite bandwidth and slew rate, and no noise
• Gain independent of temperature and supply voltage

Optically coupled isolation amplifier

An isolation amplifier passes the signal from input to output with no electrical (galvanic) connection between them. In the optical type, the signal crosses the barrier as light, from an LED to photodiodes. The input and output sides have separate grounds and power supplies.

       INPUT SIDE (GND1)    |   OUTPUT SIDE (GND2)
                            |
 Vin -[R1]-> A1 --> LED ~~~~|~~~~> PD2 --> A2 --> Vo
             ^       ~      |      (I2)   (I to V,
             |       ~      |              Rf2)
             +----- PD1     |
           (feedback, I1)   |
                    isolation barrier

Operation:

1. Vin drives a current Vin/R1 into the inverting input of A1. A1 drives the LED.
2. The LED light falls equally on two matched photodiodes, PD1 on the input side and PD2 on the output side.
3. Linearizing feedback: PD1's current I1 goes back to A1's inverting input. A1 adjusts the LED drive until I1 = Vin/R1. The light output is therefore forced to be proportional to Vin, and the LED's non-linear and temperature-dependent light-current curve is removed from the transfer function.
4. PD2 receives the same light, so I2 = I1 = Vin/R1 (matched diodes).
5. A2 works as a current-to-voltage converter: Vo = I2·Rf2. So:

Vo = (Rf2 / R1) · Vin

Features:

• Isolation of several kV, with very high isolation resistance and low coupling capacitance
• Good linearity due to the matched-photodiode feedback
• Wide bandwidth (DC to tens of kHz)
• Needs isolated power for the input side

Uses: medical (ECG) instruments, high-voltage sensing and breaking ground loops.

Condition to reject common-mode signal (single op-amp IA)

            R1           R2
 V1 ----/\/\/----+----/\/\/----+
                 |             |
                (-)\           |
                    >----------+--> Vo
                (+)/
            R3   |
 V2 ----/\/\/----+---[R4]---GND

Step 1, output by superposition:

V+ = V2 · R4/(R3 + R4)
Vo = (1 + R2/R1)·V+ − (R2/R1)·V1
   = [R4(R1 + R2) / (R1(R3 + R4))]·V2 − (R2/R1)·V1

Step 2, apply a pure common-mode input, V1 = V2 = Vcm:

Vo = Vcm [ R4(R1 + R2)/(R1(R3 + R4)) − R2/R1 ]
   = Vcm (R1R4 + R2R4 − R2R3 − R2R4) / (R1(R3 + R4))
   = Vcm (R1R4 − R2R3) / (R1(R3 + R4))

So the common-mode gain is:

Acm = (R1R4 − R2R3) / [R1(R3 + R4)]

Step 3, condition for rejection. Acm = 0 requires R1R4 = R2R3, that is:

R4/R3 = R2/R1   (resistor ratios matched)

Usually R3 = R1 and R4 = R2. Then:

Vo = (R2/R1)(V2 − V1),   Ad = R2/R1,   CMRR → ∞

With mismatched resistors, Acm ≠ 0 and CMRR = Ad/Acm is finite. Even a 1 % mismatch limits CMRR to about 40 dB. For this reason the resistors are trimmed (for example R4 made adjustable) or laser-trimmed in IC form.

Applications of isolation amplifiers

• Medical electronics (ECG, EEG, patient monitoring): protect the patient from leakage and fault currents
• High-voltage sensing in power electronics, motor drives and power systems
• Breaking ground loops between remote sensors and the data acquisition system
• Industrial process control: thermocouples and 4–20 mA loops at high common-mode voltage
• Protecting low-voltage control or computer circuits from high-voltage transients. They are also used in nuclear and chemical plant instruments.

Advantages of the 3 op-amp IA over the 1 and 2 op-amp IA

• Very high input impedance on both inputs. Both inputs go to non-inverting terminals of buffer op-amps. The 1 op-amp IA has an input impedance of only R1 or R3 + R4.
• Gain set by a single resistor RG without changing CMRR. In the 1 op-amp IA, changing the gain requires changing two matched resistors.
• Higher CMRR. The first stage has a common-mode gain of 1 but a differential gain of (1 + 2R1/RG). This raises the CMRR before the difference stage.
• Symmetric signal paths (unlike the 2 op-amp IA, where V1 passes through two op-amps and V2 through one). This gives better CMRR at high frequency.
• Low output impedance and wide gain range.

Applications of isolation amplifiers

• Medical equipment such as ECG and EEG, for patient safety
• Measuring high voltages and currents in power electronics and motor drives
• Breaking ground loops in industrial instrumentation
• Industrial process control in noisy, high common-mode environments
• Protecting computers and ADCs from transients and high voltage

Electromagnetically (transformer) coupled isolation amplifier

 INPUT SIDE (floating)  ||  OUTPUT SIDE (grounded)
                        ||
 Vin->[Input amp]->[MOD]=T1=>[DEMOD]->[LPF]->[Out amp]->Vo
                 ^      ||   ^
                 |      ||   |  carrier
 [Rect+Reg]<=====T2=====||==[OSCILLATOR ~ 100 kHz]
 (isolated +/-V)        ||        ^
                        ||    +/-Vs (output side)

Operation:

1. An oscillator on the output side (for example about 100 kHz) does two jobs:
   • Through transformer T2 it sends power across the barrier. A rectifier and regulator then make the isolated ±V supply for the input side.
   • It also provides the carrier for the modulator and demodulator.
2. The input amplifier amplifies Vin (it is often an IA).
3. The modulator varies the carrier amplitude (or pulse width) with the signal. This turns DC or low-frequency signals into AC, which a transformer can pass.
4. Transformer T1 carries the modulated signal across the isolation barrier, with no ohmic contact (isolation of 1–5 kV).
5. A synchronous demodulator and low-pass filter recover the original signal. The output amplifier buffers it:

Vo = G · Vin

It gives high isolation voltage, high CMRR and good accuracy. Example ICs: AD202 and AD210.

Electromagnetically coupled isolation amplifier

In this isolation amplifier the signal crosses the barrier through a transformer, so there is no ohmic link between input and output. A transformer cannot pass DC, so the signal is first used to modulate a high-frequency carrier.

 INPUT SIDE (floating)  ||  OUTPUT SIDE (grounded)
                        ||
 Vin->[Input amp]->[MOD]=T1=>[DEMOD]->[LPF]->[Out amp]->Vo
                 ^      ||   ^
 [Rect+Reg]<=====T2=====||==[OSCILLATOR]
 (isolated supply)      ||

1. The oscillator (about 25–100 kHz) supplies the carrier. Through transformer T2 it also powers the input side via a rectifier and regulator.
2. The input amplifier amplifies Vin, and the modulator amplitude-modulates the carrier with it.
3. Transformer T1 passes the modulated carrier across the barrier (kV isolation).
4. The demodulator and low-pass filter recover the signal, and the output amplifier gives Vo = G·Vin.

Numerical

Given: Ad = 40 dB, CMRR = 90 dB, Vd = 35 mV, Vcm = 150 mV.

Differential gain:

Ad = 10^(40/20) = 100

CMRR as a ratio:

CMRR = 10^(90/20) = 31 622.8

Common-mode gain:

Acm = Ad / CMRR = 100 / 31 622.8 = 3.162 × 10⁻³
(in dB: 40 − 90 = −50 dB)

Signal output:

Vo(signal) = Ad × Vd = 100 × 35 mV = 3.5 V

Noise output:

Vo(noise) = Acm × Vcm = 3.162×10⁻³ × 150 mV = 0.474 mV

Answer:

• Common-mode gain Acm = 3.16 × 10⁻³ (−50 dB)
• Signal output = 3.5 V
• Noise output = 0.474 mV

The noise at the input (150 mV) was larger than the signal (35 mV). At the output the signal is about 7400 times larger than the noise.

Practical characteristics of an instrumentation amplifier

• Differential gain from 1 to 1000, set accurately (about 0.1 %) by one external resistor RG
• Very high input impedance, about 10⁹ Ω or more, on both inputs
• Low output impedance, below 1 Ω
• High CMRR of 100–130 dB, which falls at higher frequency
• Low input offset voltage (µV range) and low drift (µV/°C)
• Low input bias current and low noise
• Moderate bandwidth (tens to hundreds of kHz, falling at high gain) and a good slew rate
• Good gain linearity and stable gain with temperature

Advantages of the 3 op-amp IA over 1 and 2 op-amp IAs

• Both inputs are buffered, so both have very high and equal input impedance. The 1 op-amp IA has only a few kΩ.
• The gain is adjusted by a single resistor RG without affecting CMRR.
• The input stage has unity common-mode gain but high differential gain, so CMRR is higher.
• The signal paths are symmetric, which gives good CMRR at high frequency. In the 2 op-amp IA one input passes through two op-amps.

Optically coupled isolation amplifier

       INPUT SIDE (GND1)    |   OUTPUT SIDE (GND2)
                            |
 Vin -[R1]-> A1 --> LED ~~~~|~~~~> PD2 --> A2 --> Vo
             ^       ~      |      (I2)   (I to V,
             |       ~      |              Rf2)
             +----- PD1     |
           (feedback, I1)   |
                    isolation barrier

1. A1 drives an LED with a current that depends on Vin.
2. The light falls on two matched photodiodes, PD1 (input side) and PD2 (output side).
3. PD1's current I1 is fed back to A1's inverting input. A1 sets the LED drive so that I1 = Vin/R1. This feedback cancels the LED's non-linearity and temperature drift.
4. PD2 gets the same light, so I2 = I1. A2 converts it into voltage:

Vo = I2 · Rf2 = (Rf2/R1) · Vin

There is no conductive path across the barrier, so isolation of several kV is obtained. It is used in medical and high-voltage measurement.

The three op-amp instrumentation amplifier has two buffer op-amps (A1, A2) forming an input gain stage, followed by a difference amplifier (A3).

Circuit

 V1 --->(+)\
           A1>---+--Vo1--[R2]--+--[R3]--+
       +->(-)/   |             |        |
       |         R1          (-)\       |
       +---------+              A3>-----+--> Vo
       RG                    (+)/
       +---------+             |
       +->(-)\   R1   Vo2--[R2]+--[R3]--GND
       |   A2>---+--Vo2
 V2 --->(+)/

Expression for output voltage

Input stage. By the virtual short, the two ends of RG are at V1 and V2. The current through RG is:

I = (V1 − V2)/RG

No current enters the op-amp inputs, so the same I flows through both R1 resistors:

Vo1 = V1 + I·R1 = V1 + (R1/RG)(V1 − V2)
Vo2 = V2 − I·R1 = V2 − (R1/RG)(V1 − V2)

Vo2 − Vo1 = (V2 − V1)(1 + 2R1/RG)

Difference stage (A3). By superposition:

V+ = Vo2 · R3/(R2 + R3)
Vo = (1 + R3/R2)·V+ − (R3/R2)·Vo1
   = (R3/R2)(Vo2 − Vo1)

Result:

Vo = (R3/R2)(1 + 2R1/RG)(V2 − V1)

How it rejects common-mode signal

Let V1 = V2 = Vcm (pure common-mode input):

1. The voltage across RG is zero, so I = 0 and no current flows in R1.
2. So Vo1 = Vcm and Vo2 = Vcm. The first stage passes the common-mode signal with gain 1 only, while the differential signal gets a gain of (1 + 2R1/RG).
3. A3 with matched resistors subtracts its two inputs: Vo = (R3/R2)(Vcm − Vcm) = 0.

Rejection happens in two steps. The first stage raises the differential signal relative to the common-mode one by (1 + 2R1/RG), which multiplies the CMRR by this factor. The difference amplifier then cancels what remains. Any residual mismatch in the R2/R3 network is trimmed with a potentiometer in one R3 to maximise CMRR.

An instrumentation amplifier (IA) is a precision differential amplifier with buffered inputs. It amplifies small differential signals (mV or µV) from transducers such as strain gauges, thermocouples and ECG electrodes, while rejecting large common-mode noise.

Features:

• Very high input impedance, so it does not load the source
• High CMRR (100 dB or more)
• Gain set accurately by one resistor RG, typically 1 to 1000
• Low offset, drift and noise; low output impedance

Three op-amp IA:

 V1->(+)A1-Vo1-[R2]-+-[R3]-+
       |R1         (-)A3---+-> Vo
       RG          (+)
       |R1          |
 V2->(+)A2-Vo2-[R2]-+-[R3]-GND

A1 and A2 are input buffers with gain. A3 is a difference amplifier.

Vo = (R3/R2)(1 + 2R1/RG)(V2 − V1)

A common-mode input causes no current in RG, so it passes the first stage with gain 1 and is cancelled by A3.

Applications: bridge and strain-gauge amplifiers, temperature measurement, medical instruments (ECG, EEG), and data acquisition. Example ICs: AD620, INA128.