Common-mode rejection ratio and Instrumentation and difference amplifiers

[lavster]

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! :)

 

[CWatters]

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.

 

[lavster]

thats what i thought i was doing 1/0.05 = 20/1

or am i misunderstanding it?

thanks

 

[CWatters]

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.

 

[lavster]

s youd agree that the answer is -66dB?

 

[CWatters]

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.