--- title: "Isolation amplifier applications; optically coupled circuit" subject: "Advanced Electronics" chapter: 4 key: "semester_4/advanced_electronics:isolation-applications-optical" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/isolation-applications-optical site: BCT Engineering --- # Isolation amplifier applications; optically coupled circuit Advanced Electronics · Chapter 4: Instrumentation and Isolation Amplifiers - Asked in: 2073 Chaitra, 2+2 marks - Video: [Isolation Amplifier Basics](https://www.youtube.com/watch?v=uIWsTHZBAdA) — Electronics Circuit Hub - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/isolation-applications-optical - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-4 ## Question Write down the application of isolation amplifier. Draw a circuit diagram for an optically coupled isolation amplifier. ## 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. ## Answer 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 ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does an isolation amplifier provide between input and output? A. A shared ground B. No ohmic (galvanic) connection, with separate grounds and supplies C. A common power supply D. Higher gain only Answer: B. The signal crosses a barrier by light, transformer or capacitor coupling. (Key point: Meaning of isolation) 2. Why are isolation amplifiers used in ECG and patient monitors? A. They increase gain B. They protect the patient from leakage currents and mains faults C. They remove the need for electrodes D. They filter 50 Hz only Answer: B. The barrier stops dangerous currents reaching the patient. (Key point: Medical use) 3. In the optical isolation amplifier, what is the role of PD1 feedback to A1? A. It powers the output side B. It forces LED light to be proportional to Vin, removing LED nonlinearity and drift C. It amplifies PD2 current D. It grounds the LED Answer: B. The servo loop linearizes the LED. (Key point: Role of PD1 feedback) 4. With matched photodiodes, what is Vo? A. (R1/Rf2) Vin B. (Rf2/R1) Vin C. Vin Rf2 R1 D. Vin Answer: B. I2 = I1 = Vin/R1 and A2 gives Vo = I2 Rf2. 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