Common-mode rejection ratio and Instrumentation and difference amplifiers

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Discussion Overview

The discussion revolves around the common-mode rejection ratio (CMRR) required for an instrumentation amplifier to effectively record an ECG signal in the presence of common-mode voltage interference. It includes aspects of theoretical calculations, comparisons between instrumentation and differential amplifiers, and considerations for signal-to-noise ratios (S/N).

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the required CMRR for an ECG signal and suggests that a CMRR of -66 dB is needed based on their calculations.
  • Another participant points out that the starting S/N ratio is 1mV/100mV, which needs to be improved to 20/1, implying a significant increase in the required CMRR.
  • There is a discussion about the advantages of instrumentation amplifiers over simple differential amplifiers, including high input impedance and better accuracy.
  • One participant expresses uncertainty about their understanding of the calculations and whether they are approaching the problem correctly.
  • Another participant confirms the CMRR calculation but notes that it is typically expressed as a positive number.

Areas of Agreement / Disagreement

Participants express differing views on the calculations related to the CMRR and the S/N ratio, indicating that there is no consensus on the correct approach or final answer.

Contextual Notes

There are unresolved aspects regarding the assumptions made in the calculations, particularly concerning the initial S/N ratio and how it relates to the required CMRR. The discussion also highlights potential misunderstandings in the interpretation of the results.

Who May Find This Useful

Students and practitioners interested in electronics, particularly in the context of biomedical signal processing and amplifier design.

lavster
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Homework Statement



1. a) If we wish to record an ECG of amplitude 1mV in the presence of interference that causes a common-mode voltage of 100mV to appear on the inputs of the amplifier, then what common-mode rejection ratio (CMRR) do we need require if we need a signal to noise ratio better than 20/1 on our recording?

b) give reasons why an instrumentation amplifier is better than a simple differential amplifier for biomedical signals

Homework Equations



dB = 20 log(Vout/Vin)

The Attempt at a Solution



Im doing a crash course in electronics - never have done it before so there is some concepts I am struggling with!
im not sure if I am approaching this correctly, and I am not sure what the answer is. But this is my attempt

a) ECG amplitude = 1mV, common mode voltage = 100mV. Common mode needs to reduce by 1/20 which is equal to 0.05mV

dB=20log(0.05/100) = -66dB

can someone tell me if this is correct?

b) High input impedances of the buffers of the instrumentation amplifiers eliminate the need of input impednace matching. Precise resistor matching of differential amplifiers cause problems and reduce the CMRR. I am assuming instrumentation amplifiers have better accuracy and stability?

Thanks! :)
 
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The S/N ratio is currently 1mV/100mV and you need it to be 20/1.

Instrumentation amps have good CMRR but I think your answer on matching is good as well.
 
Last edited:
thats what i thought i was doing 1/0.05 = 20/1

or am i misunderstanding it?

thanks
 
If the starting S/N was 1/1 then multiplying by 20 gets you to 20/1 however the starting S/N is 1/100.

To get from 1/100 to 20/1 you need to multiply by 2000.
 
s youd agree that the answer is -66dB?
 
Sorry yes. I'm half asleep here. I should have read your initial post more carefully. The CMRR is normally expressed as a +ve number.
 

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