IA features; 3 op-amp IA output voltage

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.