I want to use this ADC, an LTC2344-18, a quad, 18-Bit, 400ksps ADC. On page 15 you can see that the analog voltage input range can be set to +-Vrefbuf. Vrefbuf can be set to 2.5V to 5V. This would mean the analog inputs could accept a voltage between -5V and +5V.
However on page 2, in the absolute maximum ratings section, it is stated that the analog voltage input range is from -0.3V to Vdd+0.3V.
Does it mean that the -5V to +5V range is maybe between a common mode voltage of let's say +5V? So it would mean that the full scale range will be between 0V and 10V?
-
\$\begingroup\$ See typical application, page 1, bottom picture. Do you see ... inputs which are noted "fully" differential ... but voltages must be for each inputs within +5V/0V ... \$\endgroup\$Antonio51– Antonio512021年09月14日 13:34:35 +00:00Commented Sep 14, 2021 at 13:34
1 Answer 1
This would mean the analog inputs could accept a voltage between -5V and +5V.
No.
it is stated that the analog voltaghe input range is from -0.3V to Vdd+0.3V.
Let's say that the input range is 0..Vdd, positive only, so it can't accept negative voltage at the inputs.
Does it mean that the -5V to +5V range is maybe between a common mode voltage of lets say +5V?
It's a differential voltage $$V=V_+-V_-$$ \$V_+=5V; V_-=0\rightarrow V=5V \$
\$V_+=0V; V_-=5\rightarrow V=-5V \$
It would make sense to use a common mode voltage of 1/2 input voltage range, for example 2.5V.