I have designed this simple analog temperature transmitter. Specifically, this is a 2-wires loop-powered transmitter. The circuit is intended to measure temperatures from -50 to 400 °C. The first stage (Blue square)is a conditioning circuit which convert the input range voltage from thermocouple (-2.431 mV to 21.848 mV) to 0-5V. The second stage convert the 0-5V to 4-20 mA.

The first stage by itlself works properly, but when connecting the second stage, that is, the one to convert voltage to current, the regulated voltage from the TL431 goes down. The second stage also works properly by itself. The regulated voltage is 6V, but it goes to 2.4V when joining the two stages. The regulator is supposed to deliver 3.6 mA max.

I am using TLV2772 Op amps. I also have LT1013. The LM35 is used for cold juction compensation.

Any suggestions about handling this problem? enter image description here

  • \$\begingroup\$ So to be clear V1 is the power for whole circuit? Which is current limited by R2? Do you know what the total current draw is for the circuit without R2? \$\endgroup\$
    – crowie
    Jul 17, 2016 at 5:58
  • \$\begingroup\$ Oh btw it doesnt seem like V1 is referenced to ground either. \$\endgroup\$
    – crowie
    Jul 17, 2016 at 6:09
  • \$\begingroup\$ Actually it is through R2 and R3. Try removing R2 and putting a ground reference at the bottom of R3 and see what happens. \$\endgroup\$
    – crowie
    Jul 17, 2016 at 6:17
  • \$\begingroup\$ Resistor between U3 output and Q1 comes to mind. \$\endgroup\$
    – winny
    Jul 17, 2016 at 7:15
  • \$\begingroup\$ How much base current does Q1 need at 20 mA? \$\endgroup\$
    – winny
    Jul 17, 2016 at 7:16

2 Answers 2


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.

enter image description here

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.

  • \$\begingroup\$ So you mean I should not use the TL431 as the main supplier? \$\endgroup\$
    – user115094
    Jul 17, 2016 at 14:32
  • \$\begingroup\$ Good find, I shall delete my crap answer! \$\endgroup\$
    – Andy aka
    Jul 17, 2016 at 14:34
  • \$\begingroup\$ @XavierPachecoPaulino: Yes. The way you have it set up it will have to cope with large variations in voltage. With the 500 Ω load voltage varying from 2 to 10 V the available voltage for the sensor will vary from 22 V down to 14 V. The TL431 should be kept running at a nice even current to minimise drift. In any case, I suspect it's not intended as a power Zener - only for precision reference into a high impedance load. \$\endgroup\$
    – Transistor
    Jul 17, 2016 at 14:47
  • \$\begingroup\$ @transistor: 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. \$\endgroup\$
    – user115094
    Jul 17, 2016 at 14:52
  • \$\begingroup\$ See the update. A 78L05 seems to have a quiescent current of 1 mA. That seems a bit too much for your application. You'll have to do some searching for a "low quiescent current voltage regulator". ;^) \$\endgroup\$
    – Transistor
    Jul 17, 2016 at 15:12

The TL431 requires about 0.3mA to maintain regulation. In your circuit it is set for a shunt voltage of 6V and is feed by a 4.8kΩ resistor, so to deliver 3.6mA you need a supply voltage of at least 4.8kΩ * (3.6mA+0.3mA) + 6V = 24.72V. But you only have 24V!

Even worse, voltage dropped across R2 and R3 reduces the voltage across R1, so it feeds even less current into U2 as the loop current increases. Also many receiving devices have significant voltage drop even at low current (is this what R2 is supposed to represent in your circuit?).

Feeding the shunt regulator with a resistor is incompatible with 4-20mA current loop because it is intolerant of voltage drop caused by wiring and receivers. You should feed it with a constant current (perhaps via a constant current diode or current regulator IC) so that line voltage drop doesn't starve the regulator of shunt current.

R15 also reduces the maximum voltage output of your transmitter because it drops about 9V when the circuit is producing 20mA. Perhaps you don't need maximum voltage to drive a particular instrument, and you might only be using 24V because that's what you have (in which case R15 could help to reduce dissipation in Q1). However this is limiting the application of your device.


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