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Often that the OPAMP has an internal noise of 1uV. If I have an extreme high gain of 10000, the noise will be amplified up to 0.01V.

Then, when I input a very low input signal of 10uV, after amplify 10000 times, it will measure 0.1V at the output.

According to typical SNR equation:

SNR (dB) = 20 * log ( Vsignal / Vnoise) = 20 * log (0.1/0.01) = 20dB

It yields a very low SNR of 20dB.

So, with the above mention, does this mean that my amplifier has a very low SNR characteristic?

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I dont think is duplicated with my previous question. This question is obviously pointing to a different direction. –  jhyap Jul 22 at 6:15
3  
Forget the gain for the moment. You have noise of 1uV and signal of 10uV. SNR is therefore 20dB. Amplify both by the same gain and it's still 20dB. –  EJP Jul 22 at 11:59
    
If you're trying to build a good EEG circuit without using an instrumentation amp as your front end, you'll have a very rough time. –  Scott Seidman Jul 22 at 13:51
    
Agree with @EJP and edit the title to "extreme low input signal" –  jhyap Jul 23 at 0:10

2 Answers 2

up vote 9 down vote accepted

Your amplifier has average noise characteristics, the problem is that your signal is very, very weak. The amplifier is responsible only of the noise part of the SNR, so it has not a "poor SNR", but a "poor input referred noise".

To obtain a better SNR you can either amplify your signal, without adding noise, or reduce the noise.

Since amplifying without adding noise is quite a task I'd say that searching for low noise op amp is what you should do. A quick googling landed me right on the TI low noise amp page. As you can see there are sub \$\frac{nV}{Hz}\$ amplifiers, which may be a good start.

Please note that if your signal is low pass with a narrow band (such as and EEG) you need to consider also flicker noise and offset. If this is an issue you should really switch to an instrumentation amplifier that includes a chopping modulattor or at least an auto-zero or a correlated double sampling input stage. This first amp can help you give a boost to your signal, somewhere around 40dB, you can then add another stage to add the rest of the gain. Please note that in a multi stage amplifier the first stage is the most important noise wise, i.e. your second stage may be much more noisy than the first.

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For those who may have missed your humor, "quite a task" means "impossible". –  scld Jul 22 at 13:26
    
Well you can design a better sensor with higher sensitivity thus more signal and less noise –  Vladimir Cravero Jul 22 at 22:34
    
My point was that there is always a non-zero noise figure. –  scld Jul 24 at 16:38
1  
Well that's true for sure... Unless your signal is noise ;-) –  Vladimir Cravero Jul 24 at 17:05

Your CMRR equation is wrong. CMRR is not the same as signal to noise -- and has little to do with it, in fact (though poor CMRR can result in noise).

When building a differential amplifier, like EEG, there are two gain "modes" (not a formally correct term) -- there is differential gain, Ad, which you want, and there is common mode gain (Ac), which you don't.

CMRR is 20log10(Ad/Ac), and you want this to be a big number. To get the common mode gain, you put the SAME SIGNAL into both inputs of your amplifier. Say you put a 100 mVolt sine wave in to both inputs. Ideally, Ac is zero, and CMRR is infinite. Lets say, though, that you get 1 mVolt sine wave out. Now, Ac is 0.01. For a differential gain of 10000, CMRR is then 20log10(10000/0.01), and you have 120dB CMRR. Before you ask, yes, increasing the Ad usually increases CMRR, almost by definition.

The ONLY way to measure CMRR is to put the same non-zero signal into both differential inputs of your amplifier, and observe the output.

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Original post has been edited to correct the CMRR refs. –  Scott Seidman Jul 22 at 13:49
    
I am indeed referring to SNR in this post. The exact SNR equation should be the ratio of Vrms signal to Vrms noise. –  jhyap Jul 23 at 0:08

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