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Analog9 min read

Differential Amplifiers

Understand how a circuit amplifies the difference between two signals while rejecting what they share.

Mental model

Imagine two microphones hearing the same background noise, but only one hears a nearby speaker. Subtracting the microphone signals removes much of the shared noise.

−+U1R1R1R2R2V1V2VoutMatched R2/R1 ratios
V1 enters the − input through R1; V2 enters the + input through R1. With the two R2/R1 ratios matched, Vout = (R2/R1)(V2 − V1). Dots mark connected branches; op-amp supply pins are omitted.Scroll sideways to view the complete circuit.
UNDERSTAND THE BEHAVIOR
01

A differential amplifier treats the difference between two inputs as useful information. Any voltage that moves both inputs together is common-mode content and should ideally disappear from the output.

02

In a four-resistor difference amplifier, resistor-ratio matching matters more than the absolute resistor values. If the two ratios differ, some common-mode voltage leaks into the output.

03

Instrumentation amplifiers extend this idea with very high input impedance and precisely matched internal resistors, making them better for small signals from bridges and biological sensors.

Why it matters

Many sensors produce millivolt-level differential signals while sitting on top of volts of common-mode voltage and electrical noise.

Core ideas

  • Differential signal: Vd = V2 − V1.
  • Common-mode signal is present on both inputs.
  • CMRR tells you how well the amplifier rejects common-mode signals.
Key equationsMatched-resistor difference amp: Vout = (R2 / R1)(V2 − V1)CMRR(dB) = 20 log₁₀(Ad / Acm)
EXAMPLE

Worked example — sensor signal

1A bridge sensor gives V2 = 2.53 V and V1 = 2.50 V.

2The difference is only 0.03 V.

3With R2/R1 = 50, Vout = 50 × 0.03 = 1.5 V.

4Both 2.5 V common-mode levels are largely rejected.

Common mistake

A high differential gain does not guarantee good noise rejection. Check CMRR and the allowed input common-mode range.

RememberDifferential signal: Vd = V2 − V1.
DEEPER EXPLANATION

How does a difference amplifier ignore a shared voltage?

Two sensor wires can each carry a large shared voltage while the information is a small difference between them. Write Vd = V2 − V1 first. If V2 = 2.53 V and V1 = 2.50 V, the information is 0.03 V, even though both inputs sit near 2.5 V. That shared level is called common-mode voltage.

With well-matched resistor ratios, a basic difference amplifier gives Vout = (R2/R1)(V2 − V1). If R2/R1 = 50, the 0.03 V difference becomes 1.5 V. Ideally, adding the same 0.1 V to both inputs leaves the output unchanged because the difference stays 0.03 V. Matching the ratios matters: small resistor mismatch lets some common-mode voltage appear at the output.

In practice, check the amplifier input common-mode range, output swing and supply voltage before trusting the answer. A large shared voltage can put the input outside its permitted range even if Vd is tiny. For precision sensor work, an instrumentation amplifier can offer better input impedance and common-mode rejection than a simple four-resistor circuit.

Walk through another example

  1. A bridge has V1 = 1.990 V and V2 = 2.010 V. Subtract: Vd = 0.020 V.
  2. With matched gain 40, Vout = 40 × 0.020 = 0.80 V if the device stays in its linear range.
  3. If both wires rise by 0.100 V, the ideal Vd remains 0.020 V and Vout remains 0.80 V. Any real change points to finite common-mode rejection or other errors.
Check your intuition

Subtract the two inputs before applying the gain. If both inputs move equally, the ideal difference does not change.

Where this model can mislead you

Do not assume perfect rejection with unmatched resistors or assume the inputs are valid just because their difference is small.

TEST YOUR UNDERSTANDING

Three quick checks

Choose one answer per question, then check your work. Each answer includes an explanation.

1. V2 = 2.53 V and V1 = 2.50 V. What is the differential voltage V2 − V1?
2. What does an ideal difference amplifier reject?
3. A gain of 50 acts on 0.03 V. What is ideal output?

Practice and check your understanding

Try each question before opening the worked answer.

1. With matched resistor ratios of 20, V2 = 1.52 V and V1 = 1.50 V, find Vout.

Show answer and reasoning

First subtract: V2 − V1 = 0.02 V. Then multiply by the differential gain: Vout = 20 × 0.02 = 0.4 V. This assumes the input common-mode level and output are within the amplifier's limits.

2. Both inputs rise by 0.1 V. Should an ideal difference amplifier output change?

Show answer and reasoning

No. The difference remains unchanged. Real resistor mismatch and finite common-mode rejection can introduce an output error.

Try the calculation

Open Voltage gain in decibels calculator →

A related calculation tool, not a complete simulation of the circuit. Check its model and units.

Continue learning

  • Operational Amplifiers — Learn why an op-amp tries to make its two input voltages equal and how that creates useful amplifiers.
  • Voltage Buffers — Learn why a unity-gain circuit is useful even though its output voltage is almost equal to its input voltage.

Technical references and further reading

References are for further study, not endorsements. Report an error or suggest an improvement.

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