BACKGROUND:
This question is about the leakage current into the ADC input channel(s) of SAR ADC's such as the ADS7844, ADS7841, ADS7828 or many other TI ADCs in this class.
The datasheets provide only a "Typical" leakage current spec (usually +/- 1uA ), but do not provide "Min" and "Max" values for leakage current, or temperature drift specs. We are designing an improvement for a legacy product that was designed without considering this leakage current. This ADC (the ADS7828) measures several DC voltages, but all have very high source resistances (~1 Mohm), so clearly the leakage current MUST be considered for accurate measurement. If we knew the Min and Max values of leakage we could determine what the maximum error will be and proceed accordingly. Having only the "Typical" value doesn't provide enough information for our application. We would prefer not to have to perform our own "device characterization" by testing, for example, a batch of 100 or more units to empirically predict the Min and Max leakage currents.
SPECIFIC QUESTIONS:
1) Does TI have additional device characterization data for the range of leakage currents we can expect to see in production quantities, and over the device operating temperature range?
2) We have measured the leakage current on a few devices and found much lower levels than the typical spec (0.300uA to 0.500uA on 5 devices). Can TI indicate whether the +/- 1uA typical spec is a conservative number and the vast majority of devices will have much lower leakage currents?
3) Is there anybody who has designed-in this class of capacitive SAR ADC's who can share more information on the leakage current range?
4) Since it appears that this leakage current is essentially the input offset current of the Track and Hold (Sample and Hold) op amp, it makes sense that the leakage current can be either positive or negative, as indicated by the "+/- 1 uA" specification. Could someone please confirm that for production quantities, the leakage current on a given device can be either positive or negative?
Thanks very much!
Leo