How do I take my large signal to into one of today's ADCs with low input voltage range?
Standard single-ended industrial signal levels of ±5 V, ±10 V, and 0 V to +10 V are not directly compatible with the differential input ranges of modern high performance 16-bit or 18-bit single-supply SAR ADCs. A suitable interface drive circuit is required to attenuate, level shift, and convert the industrial signal into a differential one with the correct amplitude and common-mode voltage that match the input requirements of the ADC.
Whereas suitable interface circuits can be designed using resistor networks and dual op amps, errors in the ratio matching of the resistors, and between the amplifiers, produce errors at the final output. Achieving the required output phase matching and settling time can be a challenge, especially at low power levels.
The circuit in CN-0180 uses the AD8475 differential funnel amplifier to perform attenuation, level shifting, and conversion to differential without the need for any external components. The ac and dc performances are compatible with those of the 18-bit, 1 MSPS AD7982 PulSAR ADC and other 16- and 18-bit members of the family, which have sampling rates up to 4 MSPS.
Is it practicable to use fully differential amplifier like AD8475 to drive ADC AD7794 (instead of AD7192) if only ADC input range is within +/-VREF? Considering 5V supply to ADC, VREF 2.5V and attenuation on FDA 0.4 with CM=VREF would expect bipolar 5V input range. Is there some major mistake here?