I am evaluating some ADCs for my project and am trying to understand the operation of Adafruit's ADS1115 16 bit ADC (https://learn.adafruit.com/adafruit-4-channel-adc-breakouts/downloads) in single-ended mode. The ADS1115 ADC is from Texas Instruments and the datasheet can be found at https://cdn-shop.adafruit.com/datasheets/ads1115.pdf.
The datasheet says the fullscale input voltage is +/- 4.096 V - I believe that is the internal voltage reference - please correct me if that's wrong!
My confusion comes from the Adafruit documentation saying that "Single ended inputs can, by definition, only measure positive voltages. Without the sign bit, you only get an effective 15 bit resolution."
I thought that to convert my ADC bits back to Volts I need to take the output bits and multiply by the number of levels (e.g. 65536 for a 16 bit ADC) divided by Vref. So a Vref of 4.096 would give 6.25e-05 V/bit at 16 bit resolution but 1.25e-04 V/bit at 15 bit resolution.
I am guessing that what they are saying is that the 16 bit resolution is for differential input,which over the maximum voltage range of +/- 4.096 V is 65536/(2*4.096) = 1.25e-04 V/bit. That is equivalent to 15 bit resolution for single-ended measurement. This is different from the documentation of, say, the ADC in the Atmel SAMD-21 chip where you still get 16 bit resolution for single-ended measurement if I am reading that correctly.
Cutting through all of this, my question boils down to if it is correct to use 1.25e-04 V/bit for single-ended measurements from the ADS1115?
