# ADC: μV measurement and noise

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.

• At 10 A, your current shunt will be dissipating 1 W. With a desired 1 mA accuracy, have you checked the temperature coefficient of the current shunt? Or at least that it doesn't get significantly warm? – Andrew Morton Sep 5 '16 at 19:10
• @AndrewMorton I havent really looked for shunt yet. I've been to busy with ADCs :-P – Golaž Sep 5 '16 at 19:20