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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.

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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)

Input Range

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

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

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

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