# How does an ADC with an unequal split supply work?

I'm having some trouble verifying my thought process for how a differential-input ADC works, especially when done in a case when run off an unequal split supply. I'm hoping to get some clarification here. For example, if I have a differential-input ADC running off a -5V/+10V supply, with an internal reference of 2V, how is the analog data converted and what is it referenced to?

The way I'm thinking of it is this. With a -5V/+10V supply, the 'center' is at +2.5V. I'm feeding a differential output voltage from a sensor (also powered from -5/+10) into the ADC, with the CM at +2.5V. Let's say that this sensor momentarily has the differential outputs equal to each other, meaning both are exactly +2.5V.

Would the ADC read this as 'zero'? I believe it would, based on the fact that the difference between AIN+ and AIN- is exactly zero. However, since the internal voltage reference is 2V, and both my inputs are above 2V, would this cause an issue? Or, would the 2V reference be referenced to the 'center' of the supply voltages (+2.5V), meaning that the full-scale range of the ADC is then from 2.5-4.5V (referenced from the supply voltages -5/+10V)?

Lastly, how does an ADC compare a differential signal to a voltage reference? Is it correct to say that the differential signal is converted to the single-ended signal by taking the difference between V+ and V-, and then doing the comparison then?

Thanks for the help everybody!

• We will need to know the specific ADC you are using; the many flavours and implementations have a wide variety of operating parameters. Jan 8, 2016 at 8:37
• It is common for ADCs to have selectable output formats : the same ADC may read 0 to 5V as -n to +n with 0 = 2.5V (signed output) or 0 to 2n with 0 = 0V (unsigned output). So for specific answers you need to ask about a specific ADC.
– user16324
Jan 8, 2016 at 12:01