I am trying to feed a 5V signal to AD7705 which is a 16 bit ADC and expects differential inputs. Do I need to divide my 5V signal using bridge and then feed it to the ADC or can I use it as a single ended ADC.
What confuses me is the need of Vref as 2.5V. I understood that the differential inputs don't need and Vref as the inputs are compared in respect to each other.
If I do need to split up my signal, then I assume a four resistor bridge of same values should give me center tapped output to feed into the differential inputs. What standard grade "R"values should be selected? The ADC has a buffer in built, thus input impedance is not much issue, I guess.
Datasheet link:http://www.analog.com/media/en/technical-documentation/data-sheets/AD7705_7706.pdf
Best Answer
Do not use a bridge. The result would be measuring how the bridge resistors vary with temperature, not measuring variations of the signal you want to measure.
Probably you can simply connect a fixed voltage to one of the inputs of the ADC and your signal to the other one. But watch out for whether the resulting common mode input could affect the conversion result.
Even better would be to simply switch to a single-ended ADC.
You still need to tell the ADC what voltage range you want to measure.
For example, the conversion result of a differential ADC might be given by
$$ D = 2^n \frac{V_+ - V_-}{V_ref}$$
But check the datasheet for your IC because there's other possibilities (for example to allow for both positive and negative differential inputs).
If you want 16-bit accuracy (or even 16-bit precision) you should probably not be using standard grade resistors anywhere in your signal path. But again, a bridge circuit won't convert your single-ended signal to differential.