---
title: "Ideal IA characteristics; optically coupled isolation"
subject: "Advanced Electronics"
chapter: 4
key: "semester_4/advanced_electronics:ia-ideal-optical-isolation"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/ia-ideal-optical-isolation
site: BCT Engineering
---

# Ideal IA characteristics; optically coupled isolation

Advanced Electronics · Chapter 4: Instrumentation and Isolation Amplifiers

- Asked in: 2072 Chaitra, 2+5 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/ia-ideal-optical-isolation
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-4

## Question

List out the ideal characteristics of Instrumentation Amplifier. Explain the operation of optically coupled isolation amplifier.

## Short answer

Ideal IA: infinite input impedance, zero output impedance, infinite CMRR, finite accurate gain set by one resistor, zero offset/drift, infinite bandwidth, no noise. Optically coupled isolation amplifier: input amp drives an LED; light hits matched PD1 (feedback to linearize) and PD2; output amp converts PD2 current to voltage, giving Vo = (Rf2/R1)Vin, with kV isolation.

## Answer

#### Ideal characteristics of an instrumentation amplifier
- Infinite input impedance on both inputs (no loading of the source)
- Zero output impedance
- Infinite CMRR (zero common-mode gain)
- Finite, accurate and stable differential gain, set by one resistor
- Zero input offset voltage, offset current and drift
- Infinite bandwidth and slew rate, and no noise
- Gain independent of temperature and supply voltage

#### Optically coupled isolation amplifier
An isolation amplifier passes the signal from input to output with **no electrical (galvanic) connection** between them. In the optical type, the signal crosses the barrier as light, from an LED to photodiodes. The input and output sides have separate grounds and power supplies.

```text
       INPUT SIDE (GND1)    |   OUTPUT SIDE (GND2)
                            |
 Vin -[R1]-> A1 --> LED ~~~~|~~~~> PD2 --> A2 --> Vo
             ^       ~      |      (I2)   (I to V,
             |       ~      |              Rf2)
             +----- PD1     |
           (feedback, I1)   |
                    isolation barrier
```

**Operation:**
1. Vin drives a current Vin/R1 into the inverting input of A1. A1 drives the LED.
2. The LED light falls equally on two **matched photodiodes**, PD1 on the input side and PD2 on the output side.
3. **Linearizing feedback:** PD1's current I1 goes back to A1's inverting input. A1 adjusts the LED drive until I1 = Vin/R1. The light output is therefore forced to be proportional to Vin, and the LED's non-linear and temperature-dependent light-current curve is removed from the transfer function.
4. PD2 receives the same light, so I2 = I1 = Vin/R1 (matched diodes).
5. A2 works as a current-to-voltage converter: Vo = I2·Rf2. So:

```text
Vo = (Rf2 / R1) · Vin
```

**Features:**
- Isolation of several kV, with very high isolation resistance and low coupling capacitance
- Good linearity due to the matched-photodiode feedback
- Wide bandwidth (DC to tens of kHz)
- Needs isolated power for the input side

**Uses:** medical (ECG) instruments, high-voltage sensing and breaking ground loops.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the ideal CMRR of an instrumentation amplifier?
   A. Zero
   B. Infinite (zero common-mode gain)
   C. 100 dB
   D. 1
   Answer: B. Ideally common-mode signals are completely rejected. (Key point: Ideal CMRR)

2. Ideal input and output impedances of an IA are:
   A. Zero input, infinite output
   B. Infinite input, zero output
   C. Both infinite
   D. Both zero
   Answer: B. No source loading and ideal drive of the load. (Key point: Ideal impedances)

3. In the optical isolation amplifier, what condition does A1 enforce?
   A. I1 = Vin/R1
   B. I1 = 0
   C. LED current constant
   D. I2 = 2 I1
   Answer: A. A1 adjusts the LED so PD1 current equals Vin/R1, linearizing the light output. (Key point: Linearizing feedback)

4. How is the output Vo produced?
   A. PD2 current I2 = I1 passes through A2 as a current-to-voltage converter: Vo = I2 Rf2
   B. PD2 drives the LED directly
   C. A2 is a comparator
   D. Vo = Vin directly through a wire
   Answer: A. Matched diodes give I2 = I1 and A2 gives Vo = (Rf2/R1)Vin. (Key point: Output side)

## Related in Advanced Electronics

- [Isolation amplifier applications; optically coupled circuit](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aisolation-applications-optical) — Advanced Electronics, chapter 4
- [Practical IA characteristics; optical isolation](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aia-practical-characteristics-optical) — Advanced Electronics, chapter 4
- [3 op-amp IA advantages; electromagnetic isolation](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Athree-opamp-ia-electromagnetic-isolation) — Advanced Electronics, chapter 4
- [Electromagnetic isolation; IA CMRR numerical](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aelectromagnetic-isolation-cmrr-numerical) — Advanced Electronics, chapter 4
- [Short note: Instrumentation amplifier](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Ashort-note-instrumentation-amplifier) — Advanced Electronics, chapter 4

## Related in other subjects

- [Ideal transformer and coupled circuit model](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aideal-transformer-coupled-circuit-model) — Electrical Machines I, chapter 2
- [Instrumentation amplifier and its gain](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_i%3Ainstrumentation-amplifier-gain) — Instrumentation I, chapter 4