ADC: μV measurement and noise

Question

I am trying to measure current thru load from 1 mA to 10 A at 1 mA accuracy. I am using a 10 mΩ current shunt. I was also thinking of doing a low-side measurement (since input voltages can get to 100 V). I basically want to measure DC currents or pulsed currents as high as I can get. I've read thru numerous documents and became out more baffled than I was before, ADC really is a complex subject (or maybe its just me :-P).

So basically I need 14-bit ADC but since those are rather rare I decided to look for a 16-bit and found ADS1148 which is a 16-bit 2 kSPS \$\Delta\Sigma\$ ADC.

With 10 mΩ and 1 mA going thru it I get 10 μV across the current shunt resistor. Or 10 μV/mA.

With an external reference of 2.048 V, I get \$ \frac{2\times2.048}{2^{16}}= \small 62.5 \mu V\$ of resolution, I could further use a x16 amplifier and thus get 3.9 μV of resolution or 390 μA.

1. Analog input voltage range is +-Vref (thats why 2*Vref), so if I have Vref+ connected to 2.048 V and Vref- to 0 V I can read voltages in +-2.048 V range? Or will those values below Vref- read as 0x00?

2. Since this particular ADC has internal PGA can I connect its inputs directly across current shunt and set gain to something like 16 V/V? Or is it better practice to connect differential amplifier across current shunt amplify the signal and then measure its voltage via ADC (in this case I only need single-ended ADC?)? (the diff amp will still be present tho since I already need it for something else)

3. 62.5*16=160 μV seems rather low to me, is this in the range of "normal" noise floor? Will I really get 16-bits of accuracy from this? (I am highly skeptical of this) But if I connect a differential amplifier across the current shunt I am also amplifying this noise (or am I not because of high CMRR)? If not how can I calculate what CMRR I need to achieve 16-bit accuracy?

4. At page 15 in the datasheet there are noise and ENOB tables. But refeering to MAXIM definitions ENOB is in this case really Noise-Free Resolution? Following from this, at PGA gain of 1 (using external differential amplifier) and 640SAS I can expect 15.4bits of noise-free resolution or \$\frac{2\times2.048}{2^{15.4}}= \small 94.7 \mu V\$ or 9.47 mA or accounting for x16 amplification about 592 μA of resolution?

I apologize in advance if this is too long of a question, but I've just not been able to understand these concepts on my own.

Answer 1

The ADS1148 is not the right ADC for this purpose.

It has a differential input, which is useful if you want to measure positive and negative voltages. In your application the polarity is (most likely) fixed and therefore you are wasting half of the measurement range.

The differential input has a certain "input common range", which is the average of the two input voltages. It has to be within a certain range given by equation (3) in the datasheet. Therefore you can't measure small voltages or small currents in your case.

The amplification of the PGA should be such that you make use of the full range of the ADC. For example 2.048V / 10mV ~ 20. So 16 is a good choice. Without looking at the datasheet I would assume that the built-in PGA shouldn't degrade the performance too much.

The ENOB of 15 is OK and using averaging you should be able to get even more accuracy out of this ADC.