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.
@Keno - "Yes, heart of the problem is 100% "the changing of voltage" Right. So stop beating around the bush and tell us what the voltages ARE. And if the 723 voltages are changing, then isolate the 723 section by removing the 3 base drive components (R9, D4 and D5), and find out why the 723 is acting up. Come on, this isn't hard. Isolate the problem section, determine the error, and fix it. Stop beating around the bush with generalities.
Best Answer
For those not familiar with 4 - 20 mA sensors and control loops, they are commonly used in industrial control systems. They have several advantages and these have been listed in Why is zero represented by 4mA in 4 - 20 mA industrial control systems?. Suffice it to say that the author is building a sensor which is powered by the loop. It has to work when less than 4 mA is available to power its internal electronics.
simulate this circuit – Schematic created using CircuitLab
Figure 1. This initial design concept won't work well as the current sensing resistor, R3, doesn't monitor the current passed by 0.4-2 V and U3-A.
A brief web image search threw up a Maxim application note High-Performance, High-Accuracy 4-20mA Current-Loop Transmitter Meets Toughest Industrial Requirements which shows a scheme very close to the OP's.
Figure 2. In this scheme we see that a GND has been created 10 Ω above the negative lead of the sensor. The feedback from R-sense is provided by R2. This scheme will include all the device's internal current in the output current control.
All the chip current in your design is running through R1 causing a significant voltage drop. Note in the Maxim design the addition of U4, an LDO (low drop-out) voltage regulator, to power the chips while U3, the precision voltage reference, only powers Vref. You should probably do the same although if you have enough supply headroom a regular linear regulator might do the trick if its quiescent current isn't too great.
The Maxim application note is worth a read.
The TL431 works as a shunt regulator the way you have it. That means as input voltage goes up it shunts (and wastes) current to ground. Put in a voltage regulator as Maxim suggest and just use the TL431 as a reference. According to the TL431 datasheet setion 9, "In order for this device to behave as a shunt regulator or error amplifier, >1mA (Imin (max)) must be supplied in to the cathode pin." I think you just need to size the resistor to give it that amount of current. That leaves you with 2 mA for the rest of your circuit.
simulate this circuit
Figure 3. Using the TL431 with a voltage follower.
The arrangement of Figure 3 may save you for now. Set the current through the TL431 as low as possible (ensuring that it will work properly) at the minimum supply voltage. Given that 20 mA will drop 10 V across your 500 Ω load resistor (R2) this leaves you with 14 V on the sensor less cable voltage drops and 24 V tolerance. Allow some current for the transistor base too. The transistor won't shunt any current. It will just open up enough to maintain the required voltage output.