Difference Amplifier

A difference amplifier outputs a voltage proportional to the difference between two input signals while rejecting any voltage that appears equally on both inputs. Bridge sensor amplifiers in weighing scales and pressure transmitters use this circuit to extract small differential signals from a large common-mode background.

Core Concept

The difference amplifier combines an inverting and a non-inverting signal path in one circuit. V1 drives the inverting input through R1, with feedback resistor Rf. V2 drives the non-inverting input through a voltage divider formed by R3 and R4. When resistors are matched in pairs, the common-mode rejection ratio becomes very large and the output reflects only the difference V2 - V1.

For the standard case where R1 = R3 and Rf = R4, the transfer function simplifies to Vout = (Rf/R1) × (V2 - V1). If all four resistors are equal, gain is unity and Vout = V2 - V1 exactly. This is proved by superposition: each input contributes its scaled value, and common-mode voltages cancel.

In practice, resistor mismatch is the dominant limitation. A 1% tolerance mismatch in any one resistor degrades the theoretical infinite CMRR to a finite value. Precision resistor networks like the INA105 combine matched resistors on a single substrate to achieve CMRR above 86 dB with a single IC.

Key Equations

General output (R1=R3, Rf=R4): Vout = (Rf/R1) × (V2 - V1)

Unity gain case (all resistors equal): Vout = V2 - V1

CMRR limited by resistor mismatch: CMRR ≈ (1 + Rf/R1) / (4 × tolerance). For 1% tolerance and gain 1: CMRR ≈ 2/0.04 = 50 (34 dB).

Input resistance (V2 side): Rin2 = R3 + R4. Input resistance (V1 side): Rin1 = R1

Given:
  V1 = 3.5 V, V2 = 5.0 V
  R1 = R3 = 22 kΩ
  Rf = R4 = 100 kΩ
  Find: Vout

Why this formula:
  Matched pair difference amplifier: Vout = (Rf/R1) × (V2 - V1)

Formula:
  Vout = (Rf / R1) × (V2 - V1)

Substitution:
  Vout = (100,000 / 22,000) × (5.0 - 3.5)

Calculation:
  Rf/R1 = 4.545
  V2 - V1 = 1.5 V
  Vout = 4.545 × 1.5

Final Answer:
  Vout = 6.82 V

Exam Tip: GATE frequently tests the condition for a pure difference amplifier: R1/Rf = R3/R4 must hold. If this ratio is not met, the circuit has both differential and common-mode gain. Also, input resistance is not equal on both sides: Rin at V1 is R1, while Rin at V2 is R3+R4. This asymmetry is why the instrumentation amplifier is preferred for sensor applications.

Key Properties

• Output is proportional to V2 - V1 when R1/Rf = R3/R4; any mismatch introduces common-mode error.
• Input resistance is unequal: Rin1 = R1 (typically 10-100 kΩ), Rin2 = R3+R4 (twice that).
• Differential gain is set by Rf/R1; using matched resistor networks improves CMRR substantially.
• Resistor tolerance directly limits CMRR; 0.1% resistors give approximately 20 dB better CMRR than 1% resistors.
• The INA105 monolithic difference amplifier achieves 86 dB CMRR minimum with laser-trimmed on-chip resistors.
• For bridge sensor outputs in the millivolt range, even 0.1% mismatch can introduce errors larger than the signal itself.

Quick Revision

• Vout = (Rf/R1) × (V2 - V1) when R1=R3 and Rf=R4.
• All resistors equal: Vout = V2 - V1 (unity differential gain).
• CMRR degrades with resistor mismatch; matched networks are essential.
• Rin at V1 = R1; Rin at V2 = R3 + R4 (not equal).
• Condition for pure differential operation: R1/Rf = R3/R4.
• INA105 implements this with on-chip matched resistors.
• Single op-amp difference amplifier loads the source; instrumentation amp solves this.
• Exam trap: students apply the simple Vout = V2 - V1 formula without checking whether the resistor matching condition R1/Rf = R3/R4 is satisfied in the problem, leading to incorrect results when resistors are not matched pairs.