We know that in a 4/20 mA loop current, 4 mA represent 0% and 20 mA represent 100%. For loop current we use an IC like XTR115, and my circuit is fed from this IC.

If my circuit needs higher current than 4 mA to run, how can I obtain it?


You can increase the voltage drop and use a (very efficient) DC-DC converter to convert the higher voltage drop to a higher current. For example, if you need 3V at 20mA (60mW) you could drop 17V or so and if your DC-DC is efficient enough you will stay under the 4mA. Chances are you will have to design this, most chips and modules use too much quiescent current to make this work.

Optionally, and this is common practice if you are not even close on the current, is to use an external supply and DC-DC isolation to provide the necessary current. There are many kinds of devices that cannot operate from the small amount of power in a loop-powered instrument. Of course it increases the wiring cost and may make intrinsic safety more complex.

Another option (probably less popular these days) would be to use a different analog current loop standard such as 10-50mA.



simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) 2-wire 4 - 20 mA sensor and (b) 3-wire sensor configurations.

One way out of the problem - but perhaps not easily achievable with the XTR115 chip - is to add a third wire. You now have a greatly increased power budget.

enter image description here

Figure 2. Different arrangements for 2-wire and 3-wire transmitters. Source: e2e.ti.com.

2-wire 4-20mA Transmitter:

  • Accepts loop-supply from 4-20mA receiver
  • Sinks a return current between 4-20mA

3-wire 4-20mA Transmitter:

  • Accepts power and GND from a local supply
  • Sources an output current between 4-20mA

For the 3-wire transmitter you may be better using the XTR111.

enter image description here

Figure 3: XTR111 application note extract.

  • \$\begingroup\$ You generally need to add a separate isolated power source for your device. (Or the 4->20 part of the circuit needs to be isolated with respect to the added power suppy domain. \$\endgroup\$ – Michael Karas Jun 21 '17 at 18:28
  • \$\begingroup\$ Thanks, Michael. I've never designed one - only used both 2 and 3-wire. I've updated to make the 3-wire a source device. \$\endgroup\$ – Transistor Jun 21 '17 at 19:11

When designing a transmitter with the XTR115/6 your options are limited by the chip specifications. In particular the 5 volt regulator output current is limited to < 12 mA. There is a knee in the voltage output at 1 mA of current where it drops to about 4.5 volts where it remains up to Imax.

You will find that if you are designing industrial 4-20mA devices, this is about the maximum power budget you have available since the loop specifications are made to limit power for intrinsically safe environments.

Note that if instead of using the internal regulator, you add a buck converter to your circuit, for example, you are adding "energy storage" devices to the design which will force you to go through certification if you desire an IS rating. If you already have other "non-simple" devices in your design, this may not be a deciding issue.


You may find this app note from TI regarding harvesting energy from a 4-20 mA loop applicable to your situation. They walk through a design using a TPS62125 to provide a 10 mA supply with some higher peak current capabilities.

  • 1
    \$\begingroup\$ Note that you don't have to use the built-in regulator in the TI XTR-series chips. We use those chips in several products but use an external regulator. \$\endgroup\$ – Dwayne Reid Jun 21 '17 at 18:23
  • \$\begingroup\$ if I use an external regulator, is not change my current of the loop? \$\endgroup\$ – Rashid Jun 21 '17 at 18:28
  • \$\begingroup\$ The loop current is not affected so long as the loop (-) terminal goes through the XTR chip before hitting your ground connection. In other words, don't connect your external regulator to the Loop (-) terminal. Instead, connect your regulator ground to the XTR ground terminal. \$\endgroup\$ – Dwayne Reid Jun 21 '17 at 18:40
  • \$\begingroup\$ @Dwayne - it may help the OP if you could specify how much power you are able to generate with an outboard regulator. \$\endgroup\$ – Glenn W9IQ Jun 21 '17 at 18:43
  • \$\begingroup\$ I haven't yet designed anything that required more than the 4mA limit. We use an external regulator in our products to generate a 3.3 Vdc supply rail that doesn't require using a low-dropout regulator. If our 3.3V rail was derived from the 5V output of the XTR chip, we would need to use a low-dropout regulator with its attendant large-value output bypass capacitor. By powering the external regulator from the incoming loop (+) line, we can use an emitter-follower type regulator which requires tiny capacitors. Intrinsically-Safe, Class 1, Div 1, Groups A, B, C, D. \$\endgroup\$ – Dwayne Reid Jun 21 '17 at 18:48

You have a few options but it depends on how much current you require and how much burden voltage you can tolerate.

The obvious solution is to use a small DC-DC converter to step the incoming voltage down to a lower value. This will give you more current at the expense of increasing the minimum burden voltage.

Please modify your question stating exactly what your requirements are.

  • \$\begingroup\$ the output of xtr115 is +5 volt, and I need this voltage and can not use smaller voltage and maybe it is low for my plan,except my micro , i have some opamp and switch like adg436 that need more current than 4ma \$\endgroup\$ – Rashid Jun 21 '17 at 18:20
  • \$\begingroup\$ Can you not change your design to use different parts? Much of what we do requires significantly-less than 3mA from the loop power supply. \$\endgroup\$ – Dwayne Reid Jun 21 '17 at 18:42
  • \$\begingroup\$ ADG436 draws less than 0.5 mA. Op-amps with quiescent current in this range are also readily available. OP, please be more clear what your problem is. \$\endgroup\$ – The Photon Jun 21 '17 at 18:54

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