A 10-bit ADC at 5V has a resolution of 5mV. Your sensor has a sensitivity of 100mV per inch, which you reduce by your resistor divider to 50mV per inch. That means 10 ADC counts per inch. This isn't much, especially when you think that you'll always have some noise in your readings.
If you're interested in only a small range, like the 1.99V..2.09V, subtract the 1.99V from the input using a difference amplifier and amplify 50-fold to get a range from 0V to 5V.
If R1=R2 and R3=R4 then
\$V_{OUT} = \dfrac{R3}{R1} \times (V_2 - V_1) \$
I would still use a low-pass filter in software to filter out the noise.
There are a bunch of separate questions in the text here, so I'll talk about those too.
My first question is that if I use both op amp of the first AD712, then will it cause any issue? Can one opamp in an IC affect the other opamp(s) in the same package?
This won't do any harm - the IC manufacturers test their parts as if you'll be using both of the components.
I understand that voltage divider with these resistances will be fairly linear. Is this correct?
Hmm. The issue I see is that you're loading your low pass filter. This isn't necessarily a bad thing, but it might not do what you expect - you no longer have an RC filter, but an R(C || R) circuit, which has a different response.
Is it possible to change your circuit so that the low pass filter is built into the buffer? For example, you could use a generic 1st order active low pass filter:
Then, your voltage divider won't have an effect on the low pass filter (since the op-amp can supply the extra current to the R||C branch, leaving your voltage divider untouched).
I also understand that I lose the resolution, as the ADC can read 0.0049 volts as-is (10bit ADC with vref = 5v) which means that the ADC should be able to read 0.01volt difference when voltage is divided (.01 at input will be 0.0040 at divider and .02 will be 0.0090 at divider).
Right - that's the point. This way your voltage steps are bigger (bad) but your maximum measurement is higher (good).
Does the above sound logical or am I missing something fundamental?
This all makes sense to me.
All Op Amps are rail to rail with supply of +12 and -12. What if I supply +5..+5.5 to last buffer op amp so that its output never exceeds its rail which can allow me to skip the schottky diodes?
This could work. Be careful, though: most rail-to-rail op-amps can only get close to their rails when they're not supplying much current to their output. If your output is +4.95 V with a +5 V rail, when your ADC pulls in a spike of current, the output will momentarily drop. I don't know how far - this depends on how much current your ADC takes and how good your op-amp is near the rail - but it's something to think about.
How much gain can I get from DC perspective without any problems?(I tried to understand the gain-vs-frequency charts but I cant get it. DC is 0 Hz or 1 Hz)
I don't really know how to answer this - it might be a stability question, and I don't have much experience there. DC is 0 Hz.
Should I put some bypass capacitors on signal line? e.g. after trans-impedance amp , put a small cap in parallel to let the high frequency signal (noise) to ground out...
how do I decide the value of Capacitor such that it does not cause delay in my signal?
You're already doing this! Half of your low pass filter is a capacitor to ground.
You can choose a capacitor based on the amount of resistance around it. As you know, a resistor and capacitor in series make a low pass filter, which has some amount of phase shift associated with it. The decision is essentially a tradeoff between the cutoff frequency and how much phase shift you can tolerate at your signal's frequencies.
side note: this is an enormous question. If you can narrow down your question into something more specific, you'll get much better + faster answers. A lot of people will stop reading as soon as they see a wall of text, so the shorter your question is, the better.
Best Answer
What you have will work to attenutate 110 V to 4.0 V. However, there will be considerable power dissipation. R11 will dissipate about 1 W at 110 V in, so you definitely can't use a ordinary 0805 resistor. If you really need the low output impedance of 400 Ω, then get a resistor that can handle the power.
If you can do with a higher output impedance, then you could use larger resistors. For example, 100 kΩ at top and 3.77 kΩ at bottom will dissipate less than 120 mW total with 110 V in. The output impedance in that case will be 3.63 kΩ, which is still low enough to go directly into many microcontrollers.
The capacitor as you show will filter short term spikes. No, it won't hurt your accuracy at all. It may actually enhance it due to reducing high frequency noise. Make the low pass filter rolloff frequency as low as possible without cutting into the signal you actually want. For example, with 100 kΩ and 3.77 kΩ resistors, a 1 µF cap will make a LPF with a rolloff of about 44 Hz. If your valid signal is only up to 20 Hz or so, that would work fine.
Added:
Clabacchio makes a good point in a comment. At up to 220 V you not only have to think about the power rating of the resistor, but also its voltage rating. This is the value it can safely withstand and stay within spec, not arc, or vanish into a puff of greasy black smoke. However, the resistor itself isn't the only issue voltage-wise. You also have to consider the closest distance between pads on the PC board. There are various regulations and guidelines for spacing and creapage distance depending on the intended use, the nature of the high voltage, etc.
All this means that a single 0805 resistor is probably not good enough in your case. You can use a bigger package, or in some cases you can string multiple smaller packages together. Some regulations require minimum spacing that has to be met somewhere, and it's not valid to accumulate it in pieces like with multiple resistors. High voltage standoff is one place where thru hole parts still make sense sometimes, although that's for a lot more than 220 V.