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Overcurrent protection for 4-20mA inputs

Hi all,


Is there any known an simple way to protect 4-20mA inputs for overcurrent ?

I know it is not that difficult to do at the source side of the Loop, but I am looking for to the sensing side, the way it is show at the picture below.



Any suggestion will be appreciated!

Thank you! Very Happy
 
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Boncuk

New Member
A current loop is supposed to provide accurate voltage, in that particular case 1 - 5V at 4 - 20mA and a burden resistor of 250Ω.

Anything connected parallel with the burden resistor results in inaccurate current to voltage conversion.

My advise: Don't do anything
 
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RadioRon

Well-Known Member
I'm a bit puzzled also. Typically the regulation of current, and any protection against overcurrent, is done within the source. In the unlikely event that the source fails to do this, the sensor side can be protected from overcurrent by insuring that the load resistance can handle the overcurrent without burning up, and that the sensing circuit placed across that resistor is tolerant to overvoltage to whatever degree you think appropriate. Overvoltage protection is pretty straightforward with the zener idea. Other overvoltage protection devices exist for higher voltage transients.
 
What are you expecting to happen?, and why is your input designed so as to make it a problem?.
I already had a problem before with a sensor that started sink too much current (about 3A instead of the 4-20mA), so I plan to protect my sensing side since I will use the sensing circuit in union to third-party supplies to the 4-20mA loop.


Do you think it is better protect the loop power supply instead of protecting the sensing side?

I was thinking of applying a current limiter circuit to the sensing side, so it would be protected against some problems in the loop source and the sensor (load).


Thank you
 
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I'm a bit puzzled also. Typically the regulation of current, and any protection against overcurrent, is done within the source. In the unlikely event that the source fails to do this, the sensor side can be protected from overcurrent by insuring that the load resistance can handle the overcurrent without burning up, and that the sensing circuit placed across that resistor is tolerant to overvoltage to whatever degree you think appropriate. Overvoltage protection is pretty straightforward with the zener idea. Other overvoltage protection devices exist for higher voltage transients.

You might be right. Protecting the loop source seems to be more straightforward ! Thank you!
 

ericgibbs

Well-Known Member
Most Helpful Member
I already had a problem before with a sensor that started sink too much current (about 3A instead of the 4-20mA), so I plan to protect my sensing side since I will use the sensing circuit in union to third-party supplies to the 4-20mA loop.


Do you think it is better protect the loop power supply instead of protecting the sensing side?

I was thinking of applying a current limiter circuit to the sensing side, so it would be protected against some problems in the loop source and the sensor (load).


Thank you
hi,
If your sensing resistor circuit greatly exceeds 20mA then there is most likely a failure in the 4-20mA converter module.

Adding a current limit in the 250R will not protect the converter.

As you maybe aware, even with a short circuited load resistor [the250R] the current should remain within 4 to 20mA.

Most 4-20mA converters require a overhead voltage of approx 5V, they are designed to operate from 12Vdc thru to 24Vdc

The simplest way would be to fit a 250mA fuse in the +V supply line.
 
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SPDCHK

Member
I had the same problem with an ABB PLC. We solved the problem by replacing all the 250Ω 1/4Watt resistors with 1/2Watt resistors and replacing the original supplied 100mA fuses with 50mA Fast blow fuses.

Our original problem was that the 1/4W resistor burnt out long before the fuse could protect the circuit. A 1/4W resistor required only 31.6mA to equal 250mW and burn. At least with a 500mW resistor, the current limit could go up to 44.7mA before the resistor burns out. We used 50mA fuses because that was the lowest fuse rating available at the time.
 
I had the same problem with an ABB PLC. We solved the problem by replacing all the 250Ω 1/4Watt resistors with 1/2Watt resistors and replacing the original supplied 100mA fuses with 50mA Fast blow fuses.

Our original problem was that the 1/4W resistor burnt out long before the fuse could protect the circuit. A 1/4W resistor required only 31.6mA to equal 250mW and burn. At least with a 500mW resistor, the current limit could go up to 44.7mA before the resistor burns out. We used 50mA fuses because that was the lowest fuse rating available at the time.
Hi SPDCHK,

Thank you for your information.
It is possible that I implement a solution like you did.

Thank you,
Rodrigo
 
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hi,
If your sensing resistor circuit greatly exceeds 20mA then there is most likely a failure in the 4-20mA converter module.

Adding a current limit in the 250R will not protect the converter.

As you maybe aware, even with a short circuited load resistor [the250R] the current should remain within 4 to 20mA.

Most 4-20mA converters require a overhead voltage of approx 5V, they are designed to operate from 12Vdc thru to 24Vdc

The simplest way would be to fit a 250mA fuse in the +V supply line.


Eric,

It seems I am making the things more complicated than they really are.

Instead of protecting the 4-20mA input with an overcurrent limiter circuit, I will use a simple fast acting of of 50mA and a rated 250 resistor (1W or 1/2W), as it was said by SPDCHK. Also, I will implemente my own supply to the 4-20mA loop so I can implement overcurrent and shortcircuit protection on it.

What could be a good short-circuit protection for the source?

Here is my circuit for overcurrent protection:



Thank you,
Rodrigo
 
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ericgibbs

Well-Known Member
Most Helpful Member
Eric,

It seems I am making the things more complicated than they really are.

Instead of protecting the 4-20mA input with an overcurrent limiter circuit, I will use a simple fast acting of of 50mA and a rated 250 resistor (1W or 1/2W), as it was said by SPDCHK. Also, I will implemente my own supply to the 4-20mA loop so I can implement overcurrent and shortcircuit protection on it.

What could be a good short-circuit protection for the source?

Here is my circuit for overcurrent protection:

Thank you,
Rodrigo
hi,
Have you considered using a LM317 as a current limiter in the +V supply input.?
 

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Boncuk

New Member
If you use an LM317L (TO92 package) you can limit the current between 1 and 100mA by selecting the appropriate reference resistor value. (The LM317 can be used down to 10mA)

The formula for reference resistor: VR(Ω)=VRef(V)/IRef(A)

For your application (4-20mA) use VR=1.25/0.02 - VR=62.5Ω.

Add the current used by the transmitter circuit to gain the maximum output current.

Boncuk
 

Mr RB

Well-Known Member
What about reverse connection? Or accidental connection to mains or some power source when/if the wires get munched? I would just use a 50mA fuse (maybe 2) and a tough bidirectional zener (transorb?) about 10v in parallel with the sense resistor. That should save you from all possible overload situations.
 
If you use an LM317L (TO92 package) you can limit the current between 1 and 100mA by selecting the appropriate reference resistor value. (The LM317 can be used down to 10mA)

The formula for reference resistor: VR(Ω)=VRef(V)/IRef(A)

For your application (4-20mA) use VR=1.25/0.02 - VR=62.5Ω.

Add the current used by the transmitter circuit to gain the maximum output current.

Boncuk


The LM317 may be a nice trick for me, but since I am going to get energy from a battery, it could be even better if I get an LM317 equivalent with high efficiency,

Do you know any equivalent with a higher efficiency ?
 

Danw

New Member
Yes, I know, it's a 10 year old thread . . . But the problem wasn't really covered, so here goes.

The problem is protecting the analog input (current measuring DAQ Device in the graphic below) in a 4-20mA, 2 wire, loop powered current loop. The transmitter (mislabeled transducer in the graphic below) is the loop current regulator, and there's the typical external 24Vdc power supply:

Current loop - short the transducer and the 24V is across the AI.jpg

Somehow there's a short across the terminals at the 4-20mA transmitter so it's current regulating function in the loop is nil. The short puts the 24Vdc directly across the dropping resistor at the analog input. Dropping resistors are not power resistors, they're light duty little precision resistors capable of handling up to 25mA (maximum fault current indication) because 4-20mA loops aren't supposed to go higher than that. Some AI's still use discrete, external dropping resistors wired across the AI terminals, and if an external dropping resistor burns out, it's replaceable. But more and more AI's are going to integrated dropping resistors that are not replaceable, so a short at the transmitter now creates an out-of-commission AI.

Some Distributed Control System (DCS) AI's incorporate 50mA fast acting fuses on each AI to protect the AI.
 

alec_t

Well-Known Member
Most Helpful Member

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