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.

Why did noise reduce at higher gains?

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.

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?

You vectorially add all the noises as per how I did it above ($\sqrt{A^2 + B^2}$).

Also is it possible to get differential o/p from AD620, and will it be beneficial to reduce noise?

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.

Can someone give me math to find ADC ENOB with above interface circuits?

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 = $\dfrac{SINAD - 1.76 dB}{6.02}$

• thank you very much sir, so I have to, calculate all noise at o/p of ADC interface circuit and then add it vectorially to the input noise of opamp to get my total RMS noise.? Oct 4, 2014 at 11:34
• Thats right. All the individual noises multiplied by the gain downstream to the ADC input are added as sums of squares. Don't forget current noises too. If impedances are low these are usually ignorable. Oct 4, 2014 at 12:04
• Yep. current noise only of opamp right. Btw my sensor sm5652 states o/p resistance of 2.5k so should i take this as resistance to calculate resistor noise and as well as multiply it with my current noise? Oct 5, 2014 at 5:13
• Yes, it's worth checking this way. Oct 5, 2014 at 8:25
• Ok thank you. Ill post another question regarding resistor noise. Oct 5, 2014 at 9:04

Andy's answer gives a fine account of part of the problem.

Another important factor is, that gain is controlled by the resistor Rg in the schematic shown, and Rg is operating on signals at the same voltage level as the input signal.

Therefore to reduce gain in this style of amplifier, you increase Rg, and thus increase its Johnson noise. I haven't checked the numbers so can't say whether Rg or the output noise is the dominant noise problem here, however for low noise audio amplifiers, e.g. migrophone amplifiers where input noise is well below 1 nV/rtHz, values of Rg > 50 ohms can easily be the dominant noise source.

Here, the input voltage noise is quite high at 9 nV/rtHz so unless Rg is several kilohms or higher, it can be discounted as the problem. If it is higher, it is worth calculating its Johnson noise and factoring it into the calculation.