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4-20mA current loop

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eiaro

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I'm working on making a 4-20mA data logger, but since I'm not the only one on the loop I wonder how to proceed without opening the can of grounding worms.

If anyone had experience on how to do this, I'd appreciate the input.
 
eiaro said:
I'm working on making a 4-20mA data logger, but since I'm not the only one on the loop I wonder how to proceed without opening the can of grounding worms.

If anyone had experience on how to do this, I'd appreciate the input.

Generally you drop a resistor from your device and them measure the voltage across it via an A/D (differential voltage) to do what you need to do. What else do you need to know?
 
Assuming it's an existing current loop that is already powered and you just want to be able to passively obtain the value then a low value resistor wired in series with the existing loop, say 100 ohm or under should not cause any problem with the transmitter or other receiver of this current loop.

You should then use a instrumentation differential op amp input circuit so as not to require to have either end of this resistor connected to your new added circuit's common connection.

You can design the gain of this amp to scale the drop across this added resistor, .4 to 2 volt ( = 4ma to 20ma = 0 - 100%) to whatever voltage range your new application requires.

Lefty
 
Generally speaking, a 250 ohm resistor is the preferred value because at 4ma the voltage drop will be 1 volt and at 20ma, the drop will be 5v. From there, you use one or more op amp circuits to shift the reference voltage level and the range to suit your readout or logging equipment ranges. Other than not wanting to introduce any alternate signal paths (i.e., grounds), you will need to use a differential amp (instrumentation amp) as mentioned previously and you need to make sure that the addition of the 250 ohms will not be too much resistance for the source to be able to push the full 20ma. If it's already a long wire run or if the original load is well-over 250 ohms, there may be a problem. If you're doing this in cooperation with the original "load user", you can get things squared away. You might have to lower your resistance value substantially and kick up the gain of your op amp to keep the system working. On the other hand, if you're inserting your circuit as a clandestine monitor, shame on you.

Dean
 
I have to take into consideration that the signal might not be 4-20mA, but actually 0 to 25 mA. Since I'm also going to use this on more than one loop, I'll keep the voltage drop lower and adjust the gain on the opamp. I know that the original intended loop has no problem with another 250ohm load.

The shunt resistor is 20ohm, the gain on the opamp is close to 10x. This should give me 0-5V for 0-25mA.

The attached circuit is my design using LT1789. I'm still concerned for the input bias currents.
 

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eiaro said:
I have to take into consideration that the signal might not be 4-20mA, but actually 0 to 25 mA. Since I'm also going to use this on more than one loop, I'll keep the voltage drop lower and adjust the gain on the opamp. I know that the original intended loop has no problem with another 250ohm load.

The shunt resistor is 20ohm, the gain on the opamp is close to 10x. This should give me 0-5V for 0-25mA.

The attached circuit is my design using LT1789. I'm still concerned for the input bias currents.


Should work fine. You can simulate an external current loop with a couple of 9volt batteries in series and a pot to set current and see how your amp performs. Most current loop receivers I have worked on add some low pass filtering to help cut down on noise and EMI.

Lefty
 
There will be a low pass filter on the output side of the amp. I guess it was on this side you were referring?

Adding capacity and/or inductance to the loop isn't a good thing I suspect...
 
A current loop that goes to zero has problems, for if the loop opens, the monitoring equipment sees 0 ma as a legitimate current and interprets it as such. That's why the standard instrumentation loop goes no lower than 4ma -- if the loop opens, the zero current is interpreted as a fault.

Dean
 
That's one of the main advantage of a current loop over a voltage line, it's obvious when it's faulty.
 
Since I'm a parasite logging someone else's loop, I will prefer to see the errors as well. If I'm using a 12 bits ADC my resolution will drop from 3.9uA to 6.1uA, and that is good enough for my use.
 
eiaro said:
Since I'm a parasite logging someone else's loop, I will prefer to see the errors as well.
Do you have their permission to use it?

If not then I suggest you give this up as a bad job before it gets you into trouble.
 
I have permission. The plan is to hook into very old current loops and see if the old receiver is missing something since it only has a few samples/second.

Since the machines will be running while I'm logging, I must make sure I don't interfere with the original equipment while it's running.

Hope this won't get me in trouble... In any ways... :)
 
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