To makes things easier I will ignore RL and combine the R1's:
Considering just one opamp, the noise due to the input resistance is
R1=2k and R2=10k so
The feedback resistor R2 (10k) has a noise that is
The opamp (ADA4807) has an input referred voltage noise of 3.1nV/rtHz and its noise gain is (5+1):
The opamp current noise is 10pA/rtHz but splits between R2 in one opamp and 2R1+R2 in the other opamp. This means that most of the current noise is common-moded at the outputs. The difference is
Now we take the squareroot-sum-of-squares to get the total noise. However, this is a differential amp so each vnx term shows up at each output. For an uncorrelated term we can just multiply by sqrt(2) but in the case of vn1 the noises are correlated between the two opamps so it needs to be doubled. So the total differential noise is
This is applied to a difference amp with a gain of 5 so its output noise is 284.2nV/rtHz.
Now we need the noise bandwidth. If we assume we want to respond to a 0.25us target then we need a bandwidth of
Assuming a 1-pole roll-off gives us a NBW multiplier of π/2, or NBW = 1MHz.
The total integrated noise is now
Peak-to-peak noise is ~6x this value, or 1.71mV.
At the ADC input this is 586.5 codes per volt, so a 5V full-scale input would need 2932 codes, or 12 bits. Or, to put another way, for a 16 bit converter @ 5V we have about 22 LSBs of noise. This is not good. I'll stop for now to review these numbers to see if I did something wrong.
WOUW!!!


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