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Isolation amplifier applications; optically coupled circuit

Short answer

Isolation amplifiers are used in medical instruments (ECG, EEG), high-voltage measurement, breaking ground loops, industrial process control, and protecting ADCs. In an optically coupled amplifier, an input op-amp drives an LED; light falls on matched photodiodes PD1 (feedback to linearize) and PD2 (output side), with output amp converting PD2 current to voltage, providing kV isolation.

Write down the application of isolation amplifier. Draw a circuit diagram for an optically coupled isolation amplifier.

An isolation amplifier is an amplifier whose input and output circuits have no ohmic (galvanic) connection. The signal crosses the barrier by optical, transformer (magnetic) or capacitive coupling, and each side has its own ground and power supply.

Applications of isolation amplifiers

  • Medical instruments (ECG, EEG, patient monitors): they protect the patient from leakage currents and mains faults.
  • High-voltage measurement: sensing current and voltage in motor drives, power converters and power lines without exposing low-voltage circuits.
  • Breaking ground loops in industrial instrumentation and data acquisition.
  • Industrial process control (4–20 mA loops, thermocouples on high common-mode voltage).
  • Protecting computers and ADCs from high-voltage transients. They are also used in nuclear and chemical plant monitoring.

Optically coupled isolation amplifier

       INPUT SIDE (GND1)    |   OUTPUT SIDE (GND2)
                            |
 Vin -[R1]-> A1 --> LED ~~~~|~~~~> PD2 --> A2 --> Vo
             ^       ~      |      (I2)   (I to V,
             |       ~      |              Rf2)
             +----- PD1     |
           (feedback, I1)   |
                    isolation barrier
  • A1 drives an LED. The light falls on two matched photodiodes.
  • PD1 (input side) feeds current back to A1's inverting input. This servo loop forces the LED light to be exactly proportional to Vin, which removes the LED's non-linearity and temperature drift.
  • PD2 (output side) receives the same light, so its current is also proportional to Vin. A2 converts this current to the output voltage.
  • With matched photodiodes: Vo = (Rf2/R1)·Vin.

There is no electrical path between GND1 and GND2, so isolation of several kV is obtained.

✓ Answer checked by Claude Opus 5.5 (high effort)

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