Tricky Circuit

A matched difference amplifier

Keep the difference, reject the shared part

A difference amplifier subtracts one input from another. Op amp means Operational Amplifier, a powered difference-amplifying circuit. Common-mode voltage is the part shared by both inputs. CMRR means Common Mode Rejection Ratio: how well an amplifier rejects that shared part compared with the desired difference.

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EXPERIMENT BENCH

Change the circuit settings, predict the readings, then check a challenge. These are learning models, with the assumptions explained in each lesson.

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How it works

Ideal four-resistor amplifier with exactly matched resistor ratios and ±12 V output limits. Common-mode rejection is perfect here; actual resistor mismatch, input range, noise and bandwidth are omitted.

Input difference is V2 − V1. Matching both resistor ratios makes output Vout = g × (V2 − V1). Increasing both inputs equally leaves their difference unchanged.

Difficulty

  1. Build · Observe

    Power on at V1 = 1 V, V2 = 3 V and g = 1. Obtain 2 V.

    Need a hint?

    V1=1 V; V2=3 V; g=1 → 2 V

  2. Explore · Explore

    Set V1 = V2 = 2 V and g = 2. Obtain 0 V.

    Need a hint?

    V1=V2=2 V; g=2 → 0 V

  3. Solve it · Solve

    Set V1 = 3 V, V2 = 1 V and g = 2. Obtain −4 V.

    Need a hint?

    V1=3 V; V2=1 V; g=2 → −4 V

Extra reading and circuit recipes

A matched difference amplifier

Two sensor wires may share an unwanted voltage. A difference amplifier asks what differs between them, making subtraction useful for measurements rather than merely arithmetic.

Vout = g × (V2 − V1)

A common shift and a difference are independent ideas. This model preserves the difference while eliminating the shared shift.

A useful connection

Equal inputs ideally cancel even when neither is at ground.

Beyond the model

Real resistor-ratio mismatch can turn a shared voltage into output error.

A useful trap to avoid

Matched ratios matter. Four unrelated resistors do not guarantee perfect common-mode rejection.

Both inputs rise by 1 V while gain stays fixed. The ideal output…

  1. stays unchanged
  2. rises by 1 V
  3. doubles
Read more

Both subtraction terms rise equally, so V2 − V1 is unchanged.

Measure a separation

Use inputs 1 V and 3 V, gain 1.

The 2 V difference becomes a 2 V output.

Reverse the measurement

Use inputs 3 V and 1 V, gain 2.

The negative difference becomes −4 V.

Try another change

Compare 1 V/3 V with 2 V/4 V at gain 1. Both outputs are 2 V although the shared level changed.

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