Just to confirm, in my application, I am using a 2.048V reference and bipolar input mode. Can I assume that my 16-bit input range is +/-1.06*2.048V=+/-2.17088V at the input? Further, does this represent +/-32768 or +/-65536 internally?
If I do not need to add gain between the input mux and the A/D (via the MUX OUT and ADC IN breakout pins) of the ADS1158, do I need to unity-gain buffer this path? Is there any advantage to doing so?
Thanks in advance,
Scott
Hi Scott,
You are correct, the full-scale input range is +/-1.06*Vref = +/-1.06 * 2.048V= +/- 2.17088 V. In digital codes, this gets represented in 2's compliment format as per table 7 in the data sheet. Therefore, the LSB size is 66.25uV per code and a full-scale input will have a decimal code of +/- 32767 codes. (There are +/- [(2^15) - 1] or +/- 32767 codes because the first bit is a sign bit. Also one code is never useded with 2's compliment format because it's compliment is itself).
The use of a buffer between MUXOUT and ADCIN is recommended if your input sources are not driven (or do not have a low output impedance). The ADC input is a switched capacitor circuit and the capacitors must charge to your input signal. Having a source with high output impedace creates an RC circuit with a long time constant. Therefore, when the input is sampled the ADC will not capture the true input signal voltage. See page 14 in the data sheet for a description of the ADC's input circuitry.
Notice that the Electrical Characteristics were tested with OPA227 buffers between the MUX and ADC input.
Regards,
Chris
