I want to calculate theoretically my effective resolution from C8051F350 in-built sigma delta ADC with different interface options like ISL28134, ADA4528, ADA4898, OPA211, mcp6v07.
I want to know their noise in 0.1 to 10 Hz range.
Shown below is my ADC interface.(it will be different if I use ad620)
Actually its a old schematic My LPF before ADC will have cutoff of 10Hz.
So when I look at datasheet of ad620 it gives Noise peak to peak at different gains RTI, my question is why did noise reduce at higher gains.?
My second question is, I know my ADCs rms noise at different PGA gains so how should I combine that noise with the noise of interface circuit to find effective resolution?
Also is it possible to get differential o/p from AD620, and will it be beneficial to reduce noise?
Can someone give me math to find ADC ENOB with above interface circuits.
Best Answer
I assume you are referring to this figure: -
And if so then the input noise does appear to decrease with higher gains. However, if you look at the data in the table below: -
You will see that the voltage noise in the graph comprises two noises - real input noise and real output noise - total noise (referred to input) as shown in the graph is a combined version of the two.
So, when you have a gain of 1 the dominant voltage noise is output noise and if referred to the input (gain of 1) remains the same figure. At a certain gain, both noises will contribute equally and that will be approximately at a gain of 8. At a gain of 10 and at 1kHz, the input noise of 9nV / rt(Hz) will be 90 and the output noise will remain the same at 72. These are then vectorially added like so: -
\$\sqrt{90^2 + 72^2}\$ = 115
And, as you can see, the referred input noise at a gain of 10 (at 1kHz) is about 11 or 12 on the graph.
You vectorially add all the noises as per how I did it above (\$\sqrt{A^2 + B^2}\$).
The AD620 is what it is - if you want a differential output then you have to add an op-amp inverter to create that extra output. Whether it will be beneficial depends on the ADC used and I remember going thru the data sheet (of biblical length) and losing the will to live during the process.
Here is some background - you need to understand SNR first and, for an ADC it is asssumed the signal input p-p voltage is at about 95% full scale and a sinewave. If your input ADC range is 2.5 volts then the RMS value of the 95% signal is about 0.84 volts - this is the signal. Why 95%? Because the theoretical full range of the ADC can never be guaranteed for every device (due to offsets and gain errors within the device).
SNR is therefore your signal divided by the total noise into the ADC. This kind of also gives you the ENOB but it doesn't take into account non-linearites in the ADC producing harmonics of the input signal - commonly called distortion.
You then end up with what is called SINAD - signal in noise and distortion as the replacememt "quality figure" for assessing one ADC against another.
ENOB = \$\dfrac{SINAD - 1.76 dB}{6.02}\$
I'd encourage you to read this fine paper by ADI.