I have a 16 bit ADC, I use the +/1 V range, differential channel. If I used the full range of the ADC to provide full resolution (I am aware non-linearity and noise may affect this) this calculates as:
2^16 = 65536 Therefore 2 V / 65536 bits = 30.52 µV
However my signal has a voltage between 0-0.6 V. That covers approx. 1/3 of the ADC range.
I know I lose resolution and signal to noise ratio but what is the calculation for this? I presume it is going to be approx. 90 µV.
Please clarify.
It doesn't change. If your input range is from 0 to 65535 for 0 to 2V, applying an input of 0.6V will simply give an output of (0.6/2) * 65535 = 19661. The resolution has not changed, you are simply using less than the input range of the DAC.
The ideal situation is to do some signal conditioning (i.e. preamp and maybe level shift) to match the signal to use most of the DAC range. Then you use pretty much all the bits that you paid for. In your situation, your 16 bit DAC is actually something like 14 1/2 bits. That might be quite OK and not worth improving, or sometimes you might really care. This depends completely on the application.
The question isn't what is the real resolution it's what the noise floor is. If you have one millivolt of noise then for a 30uV count you'll have ~32 bits that are noisy which is 2^5 or 5 bits that will be unusable. Minimizing the noise floor is the best way to improve SNR.
