Using loop powered separator with two AI channels

[mikas_m]

Hello to all,

I have an application where I have one transmitter 4-20 mA (loop powered) and I need to connect its current output to two AI channels.
The transmitter is Siemens pressure transducer and its resistance is according to the manual around 650 ohm. AI card is SM331 and input resistance is about 250 ohm. Loop is 2-wire (2DMU) so the transmitter is powered by the AI channel. Right now, I need to use the same current signal to the other SM331 on another PLC.

In order to do so, I decided to use Siemens loop powered passive isolator with specification that its load at max output current is 400 ohm. Of course, other channel must be passive (4DMU). I'm worry that loop load will cause problems.
There is also an option to use this separator: http://www.loreme.fr/fichtech/CAL25ig_eng.pdf
I think higher resistance (load) of galvanic separator is better.

I wonder if this is correct way to solve this? How you solve similar problems (one transmitter in the field to two channels)?

Thank you

[OkiePC]

If I am reading your document correctly, it is not a signal "splitter" rather it is two separate channels in one housing, however it does claim a load capacity of 1300 OHm, so you might be able to wire its output in series through both of the PLC inputs.

If the PLC inputs are isolated type, that will probably be fine, but if they're not isolated, I think you should find a 4-20mA signal splitter which is sure to work:

https://www.acromag.com/catalog/219/...ered-splitters

[mikas_m]

Hello OkiePC,

thank you for your answer. Actually I want to use the following loop galvanic separator.

The actual wiring would be done according to the attached photo.
I'm worried by the actual voltage drop and accuracy.

[OkiePC]

That looks like it should work. I don't think your accuracy will be degraded, but there is only one way to know for sure. Do you have a calibrator? If not, get one. Wire your calibrator in place of the transmitter and set it for simulate mode. Send a range of signals through the system while checking the results in both PLCs...tweak the scaling in the PLCs if needed to obtain the accuracy you need.

[Lare]

It should work.
Keep in mind that this is phased out product.
Also (from memory) if your second loop have wire break or card failure, also your input side AI won't see mA loop.
This is because passive sirius needs load on output side for work.
If you use amplifier models, then there is no this problem

[Dirt]

Maybe I'm incorrect but:

Max current for just input and transmitter @24VDC:
24VDC / ( 650 Ohm + 250 Ohm ) = 26.67mA

Max current for input, transmitter and isolator @24VDC:
24VDC / ( 650 Ohm + 250 Ohm + 400 Ohm ) = 18.46mA

Will this configuration be able to reach 20mA @ 24VDC?

[danw]

Originally Posted by Dirt

Maybe I'm incorrect but:

Max current for just input and transmitter @24VDC:
24VDC / ( 650 Ohm + 250 Ohm ) = 26.67mA

Max current for input, transmitter and isolator @24VDC:
24VDC / ( 650 Ohm + 250 Ohm + 400 Ohm ) = 18.46mA

Will this configuration be able to reach 20mA @ 24VDC?

No. The issue is not the maximum current, the issue is how much resistance 24V can drive in the circuit.

The Siemens transmitter drops some voltage for operational purposes and can drive current through a resistance of 650 ohms in the remaining circuit. So if the powered/active analog input is 250 ohms, there's 400 ohms of loop resistance left over for the isolator input and wiring (wiring is typically only a couple ohms).

The maximum current in the loop will be 20.0mA when things are functional (the transmitter regulates the loop current). The maximum current might go up to 21.5 or some such value if the transmitter failed and its output went to fault-mode, something around 21.5mA.

A transmitter will fails to achieve 20mA when the loop resistance is greater than what it and the DC power supply can drive. In this case, if the isolator input resistance were 500 ohms and the PLC input resistance 250 ohms, the total resistance of 750 ohms is greater than the transmitter's 650 ohm max load. The loop current would max out at about 85% (650/750) or at about 17.6mA even though the loop current should be higher.

[Dirt]

Originally Posted by danw

No. The issue is not the maximum current, the issue is how much resistance 24V can drive in the circuit.

The Siemens transmitter drops some voltage for operational purposes and can drive current through a resistance of 650 ohms in the remaining circuit. So if the powered/active analog input is 250 ohms, there's 400 ohms of loop resistance left over for the isolator input and wiring (wiring is typically only a couple ohms).

The maximum current in the loop will be 20.0mA when things are functional (the transmitter regulates the loop current). The maximum current might go up to 21.5 or some such value if the transmitter failed and its output went to fault-mode, something around 21.5mA.

A transmitter will fails to achieve 20mA when the loop resistance is greater than what it and the DC power supply can drive. In this case, if the isolator input resistance were 500 ohms and the PLC input resistance 250 ohms, the total resistance of 750 ohms is greater than the transmitter's 650 ohm max load. The loop current would max out at about 85% (650/750) or at about 17.6mA even though the loop current should be higher.

Thanks, then I get it. Thought the transmitter had 650 Ohms and upwards (to regulate current) instead of it's ability to drive 650 Ohms in the circuit other than itself.

Misunderstood the wording "pressure transducer and its resistance is according to the manual around 650 ohm" and didn't know what spec sheet to look for.

[mikas_m]

Hello guys,
sorry I wasn't specific enough regarding the transmitter's resistance. In the attachment, you can see what manual says.
I have measured AI card output voltage and it is about 24V.
So the transmitter's load is (23-10.5)/23 =590 ohm. With line's resistance it is about 600 ohm.
I have tested and at first I thought it worked but then I noticed that mA cannot go above 18 mA. In the documentation of AI card I have found out that input resistance is less than 300 ohm.
So, max current is 24/(600+300+400) = 18.5 mA.
At 4 mA and 12 mA (simulated from the transmitter itself) it is a perfect match, but current loop cannot go above 18 mA.

The problem is solved by using signal splitter with external power supply. Other option was to use passive inputs on the AI cards and to connect 26V in series with the transmitter.


The conclusion is that if power supply is 24V or slightly less, then passive galvanic separators should not be used if actual measurements are above 16 mA.

[Dirt]

So I was right after all?