In a three-op-amp instrumentation amplifier, derive the overall gain expression in terms of RG, R1, and the differential-stage resistors R3 and R4 (assuming ideal matching).

• A. G ≈ (R4/R3) × (1 + 2R1/Rg); overall Vout = G × (V2 − V1) with proper resistor matching.
• B. G = RG/R1 and Vout = G (V2 + V1)
• C. G = (R3/R4) × (1 + RG/R1)
• D. G = 1 and Vout = V2 − V1

Answer: (A)

The gain of a three-op-amp instrumentation amplifier comes from stacking two stages: a first stage that amplifies the difference at its inputs with a gain set by RG, and a second stage that is a differential amplifier with gain set by the ratio of its output-stage resistors. In the first stage, each input is buffered and fed back through a resistor R1 to the corresponding inverting input, while RG ties the two inverting inputs together. This configuration makes the combined gain of the first stage equal to 1 + 2R1/RG. The second stage then converts the buffered outputs into a true differential output, with gain determined by the ratio of the differential-stage resistors, R4/R3. When the resistors are ideally matched, these two gains multiply, giving an overall gain G = (R4/R3) × (1 + 2R1/RG). Therefore the output voltage is Vout = G × (V2 − V1). This matches the provided expression, since the output scales with the difference between the inputs and does so by the product of the two stage gains.