Wrong analog input readings

widelto

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• Aug 3, 2011

• #16

Your diagram is wrong, it should be like this:
from power supply to the transducer, then slc 500 then plc5 then return to 24 vdc common.

 

Lancie1

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• Aug 3, 2011

• #17

Guys, it is a current LOOP. As long as the current travels in the same direction through each device, electrically the order does not matter.

In real electron flow (not conventional flow), current flows out of the power supply on the negative terminal, enters most negative side of the next device, leaves + terminal, goes to most negaive side of second device, leaves plus terminal, goes to most negative side of third device, leaves plus terminal, and finally arrives back at the power supply + terminal.

It is possible that the added reistance of the extra plc input is changing the loop current some, but probably this is not a significant error source. To make the operator display readings of the two PLCs match, you must adjust the scaling factors in each PLC program, and maybe also in each operator panel display. Most likely they are not scaled with the same instructions, or use different scale factors.

 

Last edited: Aug 3, 2011

TConnolly

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• Aug 3, 2011

• #18

Agreed. Kirchoff's first law dictates that current is EXACTLY the same at all points in the loop. If loop impedence is too high for the transmitter then both inputs would read wrong. The problem is most likely a configuration problem somewhere. Scaling on the two PLCs will not be the same.

 

Last edited: Aug 3, 2011

widelto

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• Aug 3, 2011

• #19

May be I´m wrong but the transducer is a current source, not a passive element. So kirrchoff doen not apply here.

 

bernie_carlton

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• Aug 3, 2011

• #20

His diagram shows the transducer as a 2 wire device. So it does apply. If it were a 3 or 4 wire device then it would be supplying current. As a 2 wire device it can only modify its internal resistance to provide a 4-20 ma throughput.

 

widelto

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• Aug 3, 2011

• #21

ok, if you see the transducer like a variable resistor, yes kirchoff law apply here. Thanks.

 

showshocka

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• Aug 4, 2011

• #22

widelto said:

Your diagram is wrong,

Click to expand...

yes, i was thinking that it might be wrong because of the readings. i drew it out how my predecessors had it wired.

 

showshocka

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• Aug 4, 2011

• #23

Lancie1 said:

To make the operator display readings of the two PLCs match, you must adjust the scaling factors in each PLC program, .............Most likely they are not scaled with the same instructions, or use different scale factors.

Click to expand...

Lancie1, this was my exact analysis also. I'm just not a programmer...do you think i will be able to do something like that without have to take a course at MIT? and what did you mean by "[/B][/U]"?

 

showshocka

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• Aug 4, 2011

• #24

guys, i don't want this to go in the wrong direction, so just a quick reply will do for me. Am i missing a 250 ohm resistor in this HART loop? And other piece of info that might prove to be of some aid, the PS is some, 325ft+ away from the SLC500 and then the same signal comes back via the transducer to the PLC5 which is right next the the PS, so that's close to a 650 to 700ft current loop.

 

Last edited: Aug 4, 2011

Bob O

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• Aug 4, 2011

• #25

Have measured your signal with a meter and just the pressure transducer and a known level/volume in the tank?
If so, is the reading correct?
4mA = Empty
12mA = 50% full or X amount of weight
20mA = 100% Full or X amount of weight

I assume you know what your pressure transducer is calibrated for.

HTH

 

showshocka

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• Aug 4, 2011

• #26

Bob O said:

I assume you know what your pressure transducer is calibrated for.

HTH

Click to expand...

h20 , yes at empty 4ma at .001in water level

 

Bob O

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• Aug 4, 2011

• #27

Have measured your signal with a meter and just the pressure transducer and a known level/volume in the tank?
If so, is the reading correct?
4mA = Empty
12mA = 50% full or X amount of weight
20mA = 100% Full or X amount of weight

 

brucechase

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• Aug 4, 2011

• #28

showshocka said:

Lancie1, this was my exact analysis also. I'm just not a programmer...do you think i will be able to do something like that without have to take a course at MIT? and what did you mean by "[/b][/u]"?

Click to expand...

What Lancie1 means by the

and maybe also in each operator panel display.

Click to expand...

is that the operator display could have a scaling factor built in to the display itself. It is not uncommon for programmers to have a number displayed being off so instead of adjusting it in the plc, then the programmer goes into the display and adds some kind of scaling factor to make it appear correct. By the way - I try to avoid doing this because of the confusion it can cause (and the fact that you might have to adjust 2 different systems to get a number to display correctly.

To get a good (IMHO) instrument for reading 4-20ma without breaking the loop, I use the Fluke 771 (http://www.fluke.com/Fluke/usen/Cla...mp-Process-Clamp-Meter.htm?PID=69415&trck=771). These can be picked up for around $400 new and work great.

If you want to post the code, there are many here that can check the scaling factors and Input config.

 

showshocka

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• Aug 5, 2011

• #29

What i did do was rewired the + signal, now it's coming from the PS + to the transducer's +, from the transducer - to the SLC 500 + input analog, from the SLC analog - back to the + analog input of the PLC5. Now i have 0lbs from the PLC5' operators display an 8lbs at the SLC's operators readout while the tank is empty. that's good! i then put 2000lbs of water in the tank and their both reading the same but only reading 1,500lbs. i then put another 2000lbs in and now, there still reading the same, bout now 2500lbs. I'M BAFFLED! it's 4000lbs of water in the tank.

 

accidentalengineer

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• Aug 5, 2011

• #30

Is the tank symmetrical? Is it cylindrical or rectangular? In other words, if you double the volume of water, does the height of the water double as well? If the tank is symmetrical, then pressure should have a linear relationship with the volume of water.

For example: if your tank is cone shaped, then the relationship between the water pressure and volume will not be linear.