What is needed to achieve 1 nanoVolt RMS noise, in a sampled-capacitor system?
We will use the formula Vnoise_rms = sqrt( K * T / C)
and K = Boltzmann Constant, T is Kelvin Temperature and C is the capacitance.
For 290 degree Kelvin (17 C) and 10 picoFarad, the formula spits out
Vnoise_rms = 20 microVolts RMS
We want to reduce that noise to 1nanoVolt.
But first, how about we compute what is needed for 1 microvolt RMS, which many 24-bit ADCs approach.
For 1 uV, we need to increase the C by 20*20 or 400X, to 4,000 picoFarad. Thus in an oversampled 24-bit ADC, you will notice plenty of charge being demanded from the sensor.
Suppose you have 5 volts from the sensor. How much energy/power is needed?
Energy_Cap = 0.5 * C * V * V, = 0.5 * 4nanoFarad * 5 * 5 = 2nF * 25 = 50 nanoJoule, for 24 bit ADC --------at a minimum, per conversion, for 1uV RMS noise. Ignoring the internal comparators and opamps, and any VREF noise and VDD noise. And ignoring any EFI coupling thru the plastic surrounding the silicon, and ignoring any HFI (magnetic coupling) onto the silicon or onto Vin+ and Vin- and Vref.
OK Now for the original question ---- 1 nanoVolt.
The capacitor must be 1000 * 1,000 larger, at 0.004 Farad.
The energy must be 1,000 * 1,000 larger, at 50 millijoule, per conversion.
What does the math tell us? for 40,000 conversion/second, we need 0.05joule * 40,000
we need 2,000 watts signal power into the 32-bir ADC.
And the ADC, in generating the 32 bits, converts the 2,000 watts into heat.