Why does the regulated voltage go down?
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 &ndash; 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.


  [OP comment:] So, do you recommend any basic circuit to supply 6V (may vary) 3.5 mA max? As I want a current loop of 4-20 mA, my circuit should consume less than 4mA. TL431 seems not to be the best option.


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.


  [OP's comment:] I will see what I can do since I only have the TL431 for now. 








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.