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Electromagnetic isolation; IA CMRR numerical

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.

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.

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.

 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:

Ad = 10^(40/20) = 100

CMRR as a ratio:

CMRR = 10^(90/20) = 31 622.8

Common-mode gain:

Acm = Ad / CMRR = 100 / 31 622.8 = 3.162 × 10⁻³
(in dB: 40 − 90 = −50 dB)

Signal output:

Vo(signal) = Ad × Vd = 100 × 35 mV = 3.5 V

Noise output:

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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