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Output voltage of two op-amp IA

Short answer

By virtual short and KCL at the inverting nodes, the output voltage is Vo = (1 + R2/R1 + 2R2/R)(V2 − V1). The differential gain is Ad = 1 + R2/R1 + 2R2/R.

Calculate the output voltage for the 2 op-amp Instrumentation amplifier circuit shown in figure below. [Figure: Op-amp A1: V1 to its non-inverting input; its inverting input is connected to ground through R2 and to its own output through R1. Op-amp A2: V2 to its non-inverting input; the A1 output connects through R1 to the A2 inverting input; feedback resistor R2 from the A2 output V0 to its inverting input. A gain resistor R connects the inverting input of A1 to the inverting input of A2. Output V0 is taken from A2.]

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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