# Can I provide a negative input voltage to this (or any) ADC?

I have a pH probe which is a high impedance voltage source of which I want to measure in increments of 0.5mV (500uV) with an ADC. This voltage can swing between +/- 1V.

I have chosen this ADC as something that looks suitable for the job: http://docs-europe.electrocomponents.com/webdocs/0d2a/0900766b80d2a9fd.pdf

When placed in 14 bit resolution it should give me 250mV per LSB, more than I need - but that's okay.

Now I'm unsure if I can feed my -1V signal into this ADC, as in all the diagrams it shows VSS tied to GND, not a negative source (like an Op Amp). I expect I can't tie this to a negative source because of the I2C output which would swing negative? <- Just a guess.

Can anyone help me understand if this will just work as described?

Edit: I realize I can bias the pH probe with 1.024V which will put it precisely in the middle of the reference voltage for the Op Amp, but I was hoping to make use of the Gain function of the ADC in certain situations, but if this solve this problem I'm happy to do that.

Many thanks.

You can't use the part you link with a negative input voltage (wrt ground)
It is a differential input, but you need to keep the signals between Vss-0.3V and Vdd+0.3V - see the common mode input range in the datasheet (picture below)

You would either need to level shift the input, or use a dual supply ADC. Some examples here.

• Thanks Oli, I think level shifting the input would be the simplest solution. Thank you. – James Oct 15 '11 at 11:24

The ADC operates on a single supply, meaning $V_{SS}$ is ground, and $V_{DD}$ the positive supply. The datasheet lists under AMR (Absolute Maximum Ratings) that no voltage should ever be lower than $V_{SS} - 0.3V$. Note that AMR are indeed Absolute Maximum, and that you're not supposed to operate the device at the given limits.

edit
The $\pm$2.048V mentioned in the datasheet is differential, meaning that $V_{IN+} - V_{IN-}$ can range between -2.048V (like $V_{IN+}$ = 0V and $V_{IN-}$ = +2.048V) and +2.048V (like $V_{IN+}$ = +2.048V and $V_{IN-}$ = 0V). So the differential voltage can be negative while both input are positive.

• Thanks Steven, do you know the device advertises +/- 2V operation in this case? – James Oct 15 '11 at 11:23
• @James - added to my answer. – stevenvh Oct 15 '11 at 12:18