This should be answered formally by someone who you are paying and who signs his opinion and can be held accountable for it - or ata least who specifically states their qualifications, competence and degree of culpability in the event of subsequent "issues".
There is more to this than can be answered without a deeper level of understanding than your abstraction will allow and than can be even partially answered well, without many to and fro q&As, in this sort of forum. We'll try regardless :-).
" ... the more reliable the output (ie closer to the analog input). ...
Your use of the term "reliable" is non standard and demonstrates a lack of understanding of the process you are attempting to monitor. This is not meant to be rude - just factual. And important.
If you are making this instrument then somebody with a good degree of technical understanding needs to go over the proposal and design in detail. If you are buying it then the name on the nameplate is more important than the specs of the ADC. eg you can be about certain that if it say "Agilent" (and is genuine) then it will do the job well and be 'reliable enough' [tm]. If it says "eg "Golden Sparrow" you may wish to look elsewhere.
ADC resolution affects the potential accuracy of the result. As the required resolution increases the other factors which affect achieved accuracy and resolution become increasingly important.
- Example: I have a well built but low cost digital scale with a full scale reading of 500 grams and a resolution of 0.01 gram (10 mg). I can obtain about 300 mg of reading variation by holding my wristwatch about 30mm above it in one orientation and about half that when the watch is turned at 90 degrees. An electric fan heater with coiled heating element of about 150mm dia, when operated within a 100 mm of the scales turns them into a random number generator).
ADC technology influences susceptibility to external interference. Successive approximation, Flash, dual slope, ... all have pros and cons. A dual slope system will reject 50 Hz superbly if designed to do so but may fail miserably in a 60 Hz mains environment. RF interference (cellphones, mobile phones, pagers, wristwatches !!!, other instruments, ...) MAY cause issues. The ADC proper is part of this but the overall design needs to address the actual requirement.
Potentially more may be said if better information on actual requirement becomes available.
What country?
Here is what I was thinking before:
This is a basic inverting opamp circuit. L1 and C2 are only there to filter the power supply to the opamp a little. They are not strictly required, but might reduce noise depending on what else is going on around this circuit and how clean the 3.3 V supply is.
R3 and R4 form a voltage divider that holds the + input at a fixed 500 mV. C1 removes noise that might be coming from the 3.3 V supply.
The real meat of this circuit is R2, the feedback resistor, and R1, the thermistor. The way the feedback is arranged, the opamp will do whatever it takes to keep its - input the same as the 500 mV on its + input. That means there will be a constant 500 mV accross the thermistor. The current thru the thermistor is then a function of its temperature. This current only comes from R2, so the voltage accross R2 is inversely proportional to the resistance of the thermistor.
This circuit won't utilize the full A/D input range, but will do significantly better than just a bare resistive divider as was discussed in your previous question. At 140 Ω, the current will be 3.57 mA, which produces 1.68 V accross R2. This voltage on R2 is added to the 500 mV bias, so the minimum OUT value is 2.18 V. At 98 Ω, OUT is 2.90 V, for a total range of 719 mV. That would be 223 counts of your A/D, which is more than typical thermistor accuracy can support.
You can get a wider output range by using a lower bias voltage and making R2 bigger accordingly. The value of R2 is directly proportional to the gain of this circuit. I showed 500 mV as a example because it seemed like the maximum sufficient value, but 250 mV would give you more than twice the A/D range. I wouldn't go much lower than that since other errors and sources of noise would start to get significant.
One advantage of this circuit is that it keeps a low voltage on the thermistor, which makes self-heating negligeable. At the worst case, the 500 mV is applied to 98 Ω, which causes a dissipation of only 2.6 mW. If you use 250 mV bias it goes down by a factor of 4 to 640 µW. Unless you have a very unusual situation, that amount of self-heating should be irrelevant.
One issue to keep in mind is that the output is dependent on the 3.3 V supply level. However, since you specifically mentioned 3.3 V, it sounds like it is produced by a regulator, so should be fine.
If you only have 3.3 V power available, you need a rail-to-rail opamp as I show. A TL081, for example, would not work here without both higher and lower supply voltages.
Best Answer
The simplest way is to use a resistor pullup (or pulldown) matched to the thermistor range to achieve the maximum voltage range output. 140Ω / 98Ω is a ratio of 1.43. To get the maximum response with this being one of the resistors of a voltage divider, we want to divide that range in half, which means taking the square root of the ratio. Sqrt(1.43) = 1.20. This means the center value of the voltage divider should be when the thermistor is 1.20 times its minimum, which is also its maximum divided by 1.20, which is 117 Ω. The nearest common value of 120 Ω will be close enough to still give you basically the maximum possible output.
So now we have:
The R2-R1 voltage divider divide ratio will change as a function of temperature as R2 changes. C1 is there only to reduce noise. You know a thermistor just can't change that fast, so it will reduce some of the high frequency content that you know can't be real signal. In this case, it will start attenuating above around 250 Hz, which is well above what any ordinary thermistor can do.
The next step is to figure out what voltage range you will get. This is just solving the divider for the two extreme cases, which are 120/(120 + 98) and 120/(120 + 140). Multiplying these by the 3.3V input, we get 1.82 V and 1.52 V, for a total range of 293 mV.
If you just run the voltage divider output straight into the A/D input, then you will be using 8.9% of the range, or about 91 counts. If 1 part in 91 is good enough, then you don't need to do anything further.
To get better resolution, you can amplify this signal about the midpoint of half the supply voltage. To bring it to a full scale signal, you'd need a gain of 3.3V / 293mV = 11. It's good to leave some headroom and not force the opamp to go completely rail to rail, so a gain of 8 or so would be good. That would give you lots more A/D counts of the temperature range than the accuracy of the parts can support.