Noise evaluation with AD8692

Hi Topspin,

Good day. It was correct to neglect the effect of the 1/f noise since you are using a broadband value. I agree with you that the RMS value of the output noise for your voltage follower circuit would be 6500 nV.

Just to add up, in a typical noise analysis, one must consider the stages of the amplifier, the equivalent input resistances, noise coming from external circuitry, and/or shot noise. These considerations would most likely affect the output noise of your amplifier. An elaborate analysis of how noise performance is calculated for an amplifier circuit was discussed in AN-940 (https://www.analog.com/media/en/technical-documentation/app-notes/an-940.pdf). 

Regards,
Gilbeys

Hi GLADION ,

thanks for the reply and for sharing the application note.

What happens if there is a passband amplifier stage after the buffer? The noise at the Vbuffer node is always the same ( 6500nV ) or I have to consider the bandwidth (1-15Hz) of the second stage?  I am a bit confused about this point because if I consider 6500nV as the RMS noise at Vbuffer, the noise contribution of the buffer at Vout should be 6500nV*NoiseGainU2.

I would like to assess the main noise contributors in a circuit made like the one shown above.

Regards

Hi Topspin, 

Good day. That is correct, the noise at the buffer node is always the same. However, please note that the buffer's noise contribution would be uncorrelated to the other noise of the second stage. 

To analyze the noise in your 2nd stage, one must consider the other noises coming from the different sources of the 2nd stage. Please see a sample circuit shown in MT-050 (https://www.analog.com/media/en/training-seminars/tutorials/MT-050.pdf).

Because of C1 and C2, the noise gain is a function of frequency and has to peak at the higher frequencies (assuming C2 is selected to make the second-order system critically damped). A flat noise gain can be achieved if one makes R1×C1 = R2×C2. You can refer to this table to determine all the noise present in the circuit:

You must include the noise contribution of the buffer at Vout (which is 6500nV*NoiseGainU2). This means that the total RMS output noise would be the root-sum-square of all the output contributions including the noise coming from the buffer stage. You can refer back to the application notes on the formulas. Hope this helps.

Regards,

Gilbeys

Hi,

I believe that the question has already been answered so I am closing this thread. Thank you!

Regards,
Paul