Some sensors act like current sources, and I have seen it several times, especially for very long wires even at outdoors like wind vanes. 4-20 mA current loops are used instead of 0-10 V voltage for instance.

What can be the physical explanation for this? How is current more advantageous?

(I'm also wondering in terms of EMI interference whether a current loop signal is more immune and why.)

Please explain this concept by using circuit diagrams, voltage current sources with some components. How common mode interference is coupled in both cases, etc. and why a current loop is immune to noise.


After reading the answers, here is what I understand(click to see the simulation diagrams and corresponding plots):

enter image description here

I apply common mode Vcm interference in all scenarios.

In the first top figure a current source with 1Giga Ohm impedance is transmitted via an unbalanced/inbalanced cable and even the receiver is single ended the output is immune to noise. (1G Ohm makes the noise small, the lesser this Rcur the more the noise at receiver)

In the middle figure a voltage source is transmitted via an unbalanced cable and the receiver is single ended, the output is very noisy.

In the bottom figure a voltage source is transmitted via a balanced cable and the receiver is differential-ended, and common mode noise is eliminated.

Is my conclusion/simulation correct to represent this question?

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    \$\begingroup\$ Mostly noise immunity and wire voltage drop tolerance. \$\endgroup\$ – KalleMP Apr 4 '18 at 8:34
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    \$\begingroup\$ @KalleMP "Use comments to ask for more information or suggest improvements. Avoid answering questions in comments." also apply to short answers. \$\endgroup\$ – pipe Apr 4 '18 at 9:41
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    \$\begingroup\$ Another key point is that 4-20mA always has 4mA to power the sensor. It is a 2 wire, remote powered, isolated sensor. This gives it noise advantages quite aside from the current loop aspect. \$\endgroup\$ – Henry Crun Apr 4 '18 at 10:30
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    \$\begingroup\$ Another note is that you can detect when you have a wire break with a 4-20ma signal. \$\endgroup\$ – MadHatter Apr 4 '18 at 13:25
  • \$\begingroup\$ "Is my conclusion/ simulation correct to represent this question?" I think the 3rd example looks so good because the CMR of your receiving amplifier is (almost) perfect. If the CMR was not that good (more realistic) its ouput would be worse than in 1st example. \$\endgroup\$ – Curd Apr 4 '18 at 14:07

Actually what matters for immunity against noise is the power that is needed to disturb the singal.

I.e. a current signal at an input with nearly zero impedance is just as bad as a voltage signal at an input with nearly infinite impedance.

What is needed is a receiver with non-zero as well as non-infinite impedance so that the signal involves some power.

  • if the information is coded as voltage, there still should be some current flowing into the receiver and
  • if the information is coded as current, there still should be some voltage across the receiver.

So both cases are similar, but you just have decide whether is is better to code the signal as voltage or as current (another alternative would be coded as power). For measurement purposes voltage or current signals are most appropriate.

A good wire for a current signal just needs to ensure that no current is lost (or inserted), i.e. ideally no leakage, i.e. perfect isolation. This can be accomplished in practice quite well.

A good wire for a voltage signal needs to ensure that no voltage is lost, i.e. ideally no voltage drop, perfect conductance along the wire. Unless you are using a superconductor this is almost impossible to accomplish in practice.


simulate this circuit – Schematic created using CircuitLab

In any case receiver resistance should be well above 0 and well below infinity.
It's easy to have the isolation resistance practically infinite.
It's practically impossible to have the series resistance 0.

Therefore if the signal has to be sent down some distance along a wire it is better to use a current signal than a voltage signal.

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    \$\begingroup\$ You say the only reason to use current loop is because if we use voltage then voltage drop will be an issue when it comes to long cables. How about in terms of EMI or common mode noise? Is one superior to the other if the same twisted shielded pair used? \$\endgroup\$ – cm64 Apr 4 '18 at 9:59
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    \$\begingroup\$ If EMI can induce a voltage it will be turned into a current by the receivers impedance and it will affect the current signal just as it would affect a voltage singal. All that matters is the power needed to make a difference. \$\endgroup\$ – Curd Apr 4 '18 at 10:08
  • \$\begingroup\$ Can you explain what you say by using a voltage source and current source and some resistors for two systems? hard to picture the sentences for me. \$\endgroup\$ – cm64 Apr 4 '18 at 10:35
  • \$\begingroup\$ I think drawing both cases (current/voltage source) doesn't help much because it doesn't matter if the signal source is a current or voltage source (Thevenin source or a Norton source) because both are equivalent. What matters, however, is if the signal is coded as current or voltage if the wire has not 0 resistance. \$\endgroup\$ – Curd Apr 4 '18 at 10:47
  • \$\begingroup\$ Is that a voltage source? What are those infinite resistances? Why too implicit? \$\endgroup\$ – cm64 Apr 4 '18 at 11:03

Current is great in that it is equal at all parts of a conductor. I.e. if you are pushing in 15 mA from one side, the other side is seeing 15 mA even if it is 200 m away. This is very easy to sense and makes data transmission reliable.

The same is not true for voltage. If your conductor has a high impedance and has electrical interference, then your input voltage signal will degrade and a valid voltage may not reach the other side.

The noise immunity comes from the fact that current loops are a low impedance system. See here why this matters: Why are high impedance circuits more sensitive to noise?

  • \$\begingroup\$ Will it be more immune to all sort of EMI or common-mode interference better than voltage signal? \$\endgroup\$ – cm64 Apr 4 '18 at 8:39
  • \$\begingroup\$ @cm64 see my edit \$\endgroup\$ – Makoto Apr 4 '18 at 8:44
  • \$\begingroup\$ Leakage current through the isolation of the conductor should be so small to be neglectable. \$\endgroup\$ – Uwe Apr 4 '18 at 9:22
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    \$\begingroup\$ i still dont get how current loop is less immune to noise \$\endgroup\$ – cm64 Apr 4 '18 at 10:03
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    \$\begingroup\$ @Makato constant current sources are high-impedance. This is what makes the loop immune to cable R. The receiver is low R, and this helps capacitively coupled noise \$\endgroup\$ – Henry Crun Apr 4 '18 at 10:28

Current signalling has different advantages in different situations, so there are several different answers.

In the case of low frequency signalling.

A constant current source (sender) has a very high impedance (and a CV one has a very low impedance). So when you put fairly high series resistance in, it has no effect: the CC source is already super high, what effect is a few hundred/thousand extra ohms going to make? Likewise when you couple noise into the cable (C1,2) the high source R means that both wires go up and down together - it is common mode noise and has no effect on the current. Meanwhile the receive end has a low R. This damps down any capacitively coupled noise, and is robust.

A voltage system is the opposite. The source should have very low impedance. Series R is going to matter. The rx needs to be very high input impedance or you get a voltage divider. It will capacitively pickup noise, and will be prone to damage. Capacitively injected noise flows through RSource, and you get differential voltages at the receiver.


simulate this circuit – Schematic created using CircuitLab

In the case of high frequency signalling (e.g video)

The current loop has basically constant voltage on both sides of the cable. Therefore capacitance across the cable does not pass any current, and does not have any effect. The signal is immune to cable C, and is immune to extra C added to protect from noise and emi. Much less power is used as C does not have to be driven.

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    \$\begingroup\$ This much easier to follow. But shouldn't Rsource in the current loop be in parallel to the current source? qph.fs.quoracdn.net/main-qimg-3fdb4e6f9d02023a7235d50600f91031 \$\endgroup\$ – cm64 Apr 4 '18 at 11:37
  • \$\begingroup\$ Im trying to understand how common mode noise is almost eliminated in current loop configuration. Basically focusing on your first diagram. I just wanted to be sure if Rsource is correct first. \$\endgroup\$ – cm64 Apr 4 '18 at 11:47
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    \$\begingroup\$ Also, an impedance-bridged configuration like you describe for voltage signalling ... probably falls afoul of what was said in another answer about using POWER to get good SNR. \$\endgroup\$ – rackandboneman Apr 4 '18 at 14:15
  • \$\begingroup\$ It's only an idea: Rsource is a property of I1 itself. It is what you calculate if you plot the curve of I vs Rload, and calculate Rsource from the slope of the line. As I is always exactly the same, you are left calculating that the Rsource is infinity \$\endgroup\$ – Henry Crun Apr 4 '18 at 19:32

As far as I'm concerned, these are the two main reasons for choosing current loops in several cases:

  • You don't care about the length/resistance of your wires. You can change a 3m wire to a 50m one, changing its resistance, the signal will be the same (as long as the source can deliver enough voltage/power, of course).
  • You can detect damage and failure. If you get 0mA, either you sensor or your wire is broken. With voltage loops is not that easy to figure out.

About EMI, it won't affect most of the times. EMI usually comes at (very) high frequencies, way faster than your signal changes, so you can filter it.

Also, it seems this is related to the old pneumatic controls systems, where 3-15psi range was used.


Something else to remember regarding analogue signals is the ability to then integrate the HART communication protocol. HART (Highway Addressable Remote Transmitter) is a digital signal which is overlaid on top of the analogue signal allowing for additional information to be sent via the same wiring. Most smart industrial instruments nowadays operate with HART capability. So the benefits are far greater than just voltage drop and EMI.


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