---
title: "Electromagnetic isolation; IA CMRR numerical"
subject: "Advanced Electronics"
chapter: 4
key: "semester_4/advanced_electronics:electromagnetic-isolation-cmrr-numerical"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/electromagnetic-isolation-cmrr-numerical
site: BCT Engineering
---

# Electromagnetic isolation; IA CMRR numerical

Advanced Electronics · Chapter 4: Instrumentation and Isolation Amplifiers

- Asked in: 2070 Chaitra, 4+3 marks
- Video: [Trick to Solve CMRR Based Sums | CMRR OP AMP |  CMRR Problems |  CMRR Calclation For OP AMP | 2020](https://www.youtube.com/watch?v=ZpGp3NFU68Q) — Vinay Yadav
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/electromagnetic-isolation-cmrr-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-4

## Question

Explain the operation of Electromagnetic Coupled Isolation amplifier. Certain Instrumentation amplifier has a gain of 40dB and CMRR of 90dB. It is used in a noisy environment in which the signal has a level of 35 mV and common mode noise level of 150 mV. Determine common mode gain, signal output and noise output.

## Short answer

Electromagnetic isolation amplifier modulates the signal onto a carrier, couples via transformer, demodulates and filters to recover output with kV isolation. For IA: Ad = 10^(40/20) = 100, CMRR ratio = 10^(90/20) ≈ 31623, so common-mode gain Acm = 3.16×10⁻³. Signal output = 100 × 35 mV = 3.5 V, noise output = 3.16×10⁻³ × 150 mV = 0.474 mV.

## Answer

#### Electromagnetically coupled isolation amplifier
In this isolation amplifier the signal crosses the barrier through a **transformer**, so there is no ohmic link between input and output. A transformer cannot pass DC, so the signal is first used to modulate a high-frequency carrier.

```text
 INPUT SIDE (floating)  ||  OUTPUT SIDE (grounded)
                        ||
 Vin->[Input amp]->[MOD]=T1=>[DEMOD]->[LPF]->[Out amp]->Vo
                 ^      ||   ^
 [Rect+Reg]<=====T2=====||==[OSCILLATOR]
 (isolated supply)      ||
```

1. The oscillator (about 25–100 kHz) supplies the carrier. Through transformer T2 it also powers the input side via a rectifier and regulator.
2. The input amplifier amplifies Vin, and the modulator amplitude-modulates the carrier with it.
3. Transformer T1 passes the modulated carrier across the barrier (kV isolation).
4. The demodulator and low-pass filter recover the signal, and the output amplifier gives Vo = G·Vin.

#### Numerical
**Given:** Ad = 40 dB, CMRR = 90 dB, Vd = 35 mV, Vcm = 150 mV.

**Differential gain:**
```text
Ad = 10^(40/20) = 100
```

**CMRR as a ratio:**
```text
CMRR = 10^(90/20) = 31 622.8
```

**Common-mode gain:**
```text
Acm = Ad / CMRR = 100 / 31 622.8 = 3.162 × 10⁻³
(in dB: 40 − 90 = −50 dB)
```

**Signal output:**
```text
Vo(signal) = Ad × Vd = 100 × 35 mV = 3.5 V
```

**Noise output:**
```text
Vo(noise) = Acm × Vcm = 3.162×10⁻³ × 150 mV = 0.474 mV
```

**Answer:**
- Common-mode gain Acm = **3.16 × 10⁻³** (−50 dB)
- Signal output = **3.5 V**
- Noise output = **0.474 mV**

The noise at the input (150 mV) was larger than the signal (35 mV). At the output the signal is about 7400 times larger than the noise.

## Check yourself

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

1. A gain of 40 dB corresponds to a voltage gain of:
   A. 40
   B. 400
   C. 100
   D. 10,000
   Answer: C. Ad = 10^(40/20) = 100. (Key point: dB to ratio)

2. With Ad = 40 dB and CMRR = 90 dB, what is Acm?
   A. -50 dB, about 3.16 x 10^-3
   B. 130 dB
   C. -50 dB, about 0.5
   D. 3.16
   Answer: A. Acm = Ad/CMRR = 100/31623 = 3.162 x 10^-3 (40 - 90 = -50 dB). (Key point: Common-mode gain)

3. With Vd = 35 mV, what is the signal output?
   A. 35 mV
   B. 3.5 V
   C. 0.35 V
   D. 350 V
   Answer: B. 100 x 35 mV = 3.5 V. (Key point: Signal output)

4. With Vcm = 150 mV, what is the noise output?
   A. 0.474 mV
   B. 15 V
   C. 4.74 mV
   D. 150 mV
   Answer: A. 3.162 x 10^-3 x 150 mV = 0.474 mV. (Key point: Noise output)

## Related in Advanced Electronics

- [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
- [Ideal IA characteristics; optically coupled isolation](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aia-ideal-optical-isolation) — Advanced Electronics, chapter 4
- [3 op-amp IA output and common-mode rejection](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Athree-opamp-ia-cm-rejection) — Advanced Electronics, chapter 4
- [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
- [Design three op-amp IA with gain 20](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Adesign-3opamp-ia-gain-20) — Advanced Electronics, chapter 4, asked 3×

## Related in other subjects

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