Electronic – Voltage dividers and ADC’s

Input Leakage Current

To determine your resistors voltage drop from the gate you need to use the leakage current from the datasheet. Microchip specifies an "Input Leakage Current" on their datasheets. The [datasheet that I have looked up][1] specifies an input leakage current of 1uA. This could cause a .1V or 100mV, which is only double what Robert calculated, probably not a problem on your signal.

Now remember, if you are dividing a 30V signal down to 30/11 (2.7v) volts full read then the 100mV is added to this, causing up to 3% error on your 30V signal.

If you need a resolution of 1V, divide that by 11 and then add the 100mV. This 100mV could be larger than the 1V signal.

Input Capacitance

Robert is correct, there will be a capacitance, but this really specifies an amount of time that is needed to take the ADC measurement. This also, combined with the input resistance you chose, creates a low pass filter; if you were wanting to measure signals with a higher frequency, you are not going to be able to capture them.

Reducing the error

The easiest way is to either reduce your resistance on your divider, or to buffer your signal. When you buffer the signal you will replace the PIC's leakage current with your op-amps leakage current which you can get quite low.

This 1uA is a worst case, unless it costs you a large amount to make minor changes to the design, fab your design and test how bad it is for you.

Please let me know if there is anything I can do to make this easier to read.

Lots of things in your question. So lets take them one by one.

Suppose I have a pair of 12-bit ADC's, I can imagine that they can be cascaded to get <= 24-bit output. I can think of simply using one for the positive range and the other for the negative range, though there probably will be some distortion in the cross-over region. (suppose we can ignore are a few error bits or, perhaps, place a 3rd ADC to measure the value around 0 volts).

Not really - you would get 13-bit resolution. One can describe operation of 12 bit converter as deciding in which of the 4096 bins (2^12) inputs voltage is. Two 12bits ADCs would give you 8192 bins or 13-bit resolution.

Another option I had been thinking of is using a single hi-speed ADC and switching the reference voltages to get a higher resolution at lower speed.

Actually this is how Successive Approximation Converter work. Basically one-bit converter (aka comparator) is used with digital to analog converter that is producing varying reference voltage according to successive approximation algorithm to obtain digitized sample of the voltage. Note that SAR converters are very popular and most of ADCs in uC are SAR type.

Also there should be a way of getting a real-valued result with using one fixed-ref ADC and then switching the aref's of the secondary converter to get more precise value in between.

Actually it is awfully similar to how pipeline ADCs are working. However instead of changing reference of the secondary ADC the residue error left after first stage is amplified and processed by next stage ADC.

Any comments and suggestions are welcome. I am presuming that a quad 8-bit (or dual 12-bit) chip is less expensive then a single 24-bit chip.

Actually there is a reason for that as having 24bit converter is not as simple as arranging in some configuration four 8-bit converters. There is much more to that. I think that key misunderstanding here is thinking that one can just "add" number of bits. To see why this is wrong it is better to think of ADC as circuit that is deciding to which "bin" input voltage belong. Number of bins is equal to 2^(number of bits). So 8 bit converter will have 256 bins (2^8). The 24 bit converter will have over 16 millions of bins (2^24). So in order to have the same number of bins as in 24-bit converter one would need over 65 thousands 8-bit converters (actually 2^16).

To continue with the bin analogy - suppose that your ADC have full scale of 1V. Then the 8-bit converter "bin" is 1V/256 = ~3.9mV. In case of the 24-bit converter it would be 1V/(2^24) = ~59.6nV. Intuitively it is clear that "deciding" if the voltage belongs to smaller bin is harder. Indeed this is the case due to noise and various circuit nonidealities. So not only one would need over 65 thousands of 8 bit converters to get 24 bit resolution but also those 8bit converters would have to be able to resolve to 24-bit sized bin (your regular 8-bit converter would not be good enough as it is able to resolve to ~3.9mV bin not 59.6nV bin)