Weighing system, load cell transmitter scaling for PLC

coco243

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Join Date
Oct 2012
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timisoara
Posts
11
Hi,

I have to make an weighing system with load cells. Here is what I had understood until now:

Load cells:
Sensibility often 2mv/V
Recommended excitation voltage: often 10V
Rated capacity: let's say 5kg=5000g

At maximum load(5kg) we will have 2mV/V*10V=20mV to output

So at 5000g we have 20mV, then for 1g we will have:

5000g.......20mV
1g.......X mV

X=20/5000=0,004mV=4uV
So, my load cell, for 1g will indicate 4uV, for 2g 8uV and so on. It's seems
that the division per gram is 4uV that beeing the sensibility of the load cell

I want to evaluate the weight with a PLC, and so I will need a transmitter/amplifier whatever so my PLC can be capable of reading the load cell.I want that the transmitter to take out an 4-20mA analog signal.

On my PLC I have 27648 value at 20mA, so I have 20mA/27648=723nA per division.

Now my question is, and please give me concrete, not abstract answers:

On the 4-20mA range of the transmitter how do I know what value represents 1g.

I don't know to make the correlation between load cell sensor, load cell transmitter and PLC. I don't know how to chose and parameterize the transmitter. What I know is that my system has to measure with a tolerance of +/-3 grams but I don't understand correlation between transmitter resolution with the one of the PLC.

If you have experience with load cells, and transmitter and you want to help me please guide me to understand this process.

Thank you.
 
I think "sensitivity" or "resolution" is meant; sensibility is half of a Jane Austen book title.


Let me re-read the OP. ...
 
E.g. see here


https://www.sensorsone.com/0-20mv-to-4-20ma-signal-conversion/



Assuming linear behavior for all conversions:


Code:
        Load        mV to mA         PLC ana-
        cell        converter        log input
   0g    =>    0mV    =>        4mA     =>          0count
5000g    =>   20mV    =>       20mA     =>      27468count
(20mA - 4mA) = (27648count - 0count) * C

C = 16mA / 27648count ~ 0.000579 mA/count ~ 579nA/count (not 723)



but really you want to go back to the mV, or grams, from the load cell:


(5000g - 0g) = (27648count - 0count) * K


K = ?g / ?count ~ ?g/count (nominal resolution)


1/K ~ ?count/g (nominal scaling)



Of course there will be noise and other factors, and all of this assumes linear behavior in all components, so the actual performance (precision and/or accuracy) will be worse than that.
 
Hi!

I'm not expert in the subject.... But my understanding is that accuracy depends on which class loadcell is.

I think that for your application, a full load of 5000 gr and a maximum permissible error of +/-3 gr (3/5000*100 = +/- 0.06%), and also as a good practice you should use loadcell to 75% of its maximum load..... a good option may to use a loadcell C6 (6000 divisions) with a full load of 7kg.

As you must use a loadcell in accordance with your application requirements, your measuring instruments must to be able to have the capability to work with the class of your loadcell.

And now days there are load cell with communication protocols instead of analog signals.... For example, ethernet or Can open.

Jus my opinion.

Saludos.
 
I agree with the communication option for weights of that resolution & range.
There are many other factors accuracy of the analogue card, noise etc.
The best idea it to purchase both the loadcell(s) and amplifier from the same supplier, A better alternative is something like a complete weighing solution i.e. Mettler Toledo that is capable of communication. calibration is essential do not just rely on the given specs.
One thing you have to remember is that any system that uses a load cell will have an offset so the mechanical components of the scale you put the product on or in will have an offset, so if your equipment that sits on the load cell weighs 1000 grams then the load cell will already be at a given mv/uv, the converter/ amplifier will have to be scaled so that 0g will = 4ma. With that in mind, the loadcell will have to have a span greater than the max weight to be weighed + the weight of the weigh pan equipment. allowing for possible overload that may happen (dropping load onto weigh pan)
You do not need to know what value is at any given weight. The loadcell/amplifier if calibrated for example 4ma = 0g and 20ma = 500g then the raw analogue should be scaled in the plc to that range for example Raw := 0g = 0, 500g = 27468 so scale it to 0.0-500.0g i.e. that would give you a count of 54.936 (the value will be raw integer so assume 55 per gram)
So assuming it is calibrated, excluding accuracy of the combined PLC card, Load cell & amplifier, noise etc. gives you 0.0181818181818182‬ grams per raw count.
There are a number of posts on here regarding scaling.
The one thing I suggest is you convert the raw to a real number (floating point(s), do the maths to scale it (most PLC's have functions for scale and will accept an integer value and scale it into a real number).
 
Hi,

Thank you all for your responses, I studied a little bit the subject and I had understood what drbitboy said and I came to the following conclusions and please correct me if I am wrong:

If I have a load cell with the following characteristics:
sensitivity: 2mV/V
excitation voltage: 10V
nominal load 5kg

Afferent to 1g I will have
Code:
(5000-0)=(20-0) * rate => rate=5000/20=250g/mV

So 1g=1/250mV=0,004mV=4uV
[B]1g=4uV[/B]

Further if I connect my load cell to a transmitter with a 4-20mA output I will have the following:
Code:
(20mA-4mA)=(20mV-0V)* rate => rate=16/20=0,8

Now I want to calculate 4uV how many mA will have:

0,8*4*10-3+4=3,2uA + 4mA, and so, if 4mA represents 0V, the rest, meaning 3,2uA reprezents the 4uV

For now we have 1g=4uV=4mA+3,2uA

From post no. 5 we know that 1 count =578nA and I have to read 3,2uA on the PLC input, and 579nA being less than 3,2uA it means that I can theoretical read the corresponding of 1 g in Amperes.

And now finally I have to see 1g how many counts it have:

Code:
((4mA +3,2uA) - 4mA)/578nA/count=5,5 count

Finally I will have 1g=4uV=4mA+3,2uA=5,5 count

Please correct me If my logic is wrong to see where the mistake is.


@kallileo, a configuration from Laumas I want to use too, can I ask you some questions about the transmitter?

Thank you all for your respones and I wait for your confirmation or not to see
if I am on the right path.

Thank you,
 
All that looks very impressive
But isn’t he just going to wire it up, set the zero (tare) on the amplifier , hang a weight 5kgs , set the span on the amplifier
Then drop a scale block in his Siemens plc

Done
Drink coffee
 
I think you are missing the point, the load cell will already be weighing the pan/ vessel it is on so lets assume a pan of 500g so in theory the converter will already be giving a reading above 4ma, most of these amplifiers will have a setup procedure so for example the pan/vessel is cleared of any product, the zero is set to give 4ma at the current weight of the pan so the load cell is already giving out the microvolts of the pan loading. You then add the full load and calibrate it to the 20ma (lets assume on a 5kg loadcell this will be 3 or 4kg).
I have never heard of trying to calculate the microamps after all your plc is converting it to raw integer. Surely you do not need to know this, the load cell (s) you use plus the load cell amplifier must be chosen for the weighing criteria the accuracy will depend on load cell accuracy, converter accuracy, plc analogue conversion accuracy. You have already determined that the accuracy per gram is 5.296 (27468 / 5000 )counts so as the count is an integer it would read 5 per gram, however, at 10g this would theoretically be 52.96 counts or 53 so 10/ 53 = 5.3 so you can see that there is an error that gets lower as the weight increases. If you are going to choose analogue as your preferred method then you may struggle to get the accuracy.
On the main point "you want to know the raw value (count) per gram" you already have the answer, given the conversion factors of 0-5000g and 0-27468 you just scale it accordingly. I suggest you read the articles on Metler Toledo or Applied Weighing websites, there are many other factors to be considered.

OUT = scaled output value 0-5000.0 g
In = analogue raw input value 0-27468
Out_High = high limit of the scale for the scaled output value 5000.0
Out_Low = low limit of the scale for the scaled output value 0.0
In_High = high limit of the scale for the analogue input value 27468
In_Low = low limit of the scale for the analogue input value 0
The following formula for calculating the scaled value :
Out = [(Out_High – Out_Low) * (In -In_Low) / (In_High – In_Low)] + Out_Low
 
Thank you drbitboy for confirmation.

Thank you parky for your concern that I am going in a wrong direction.

I had insisted with calculations because I wanted to understand the whole thing, I wanted to understand how 1g is represented on the output of the load cell, and then how the transmitter represents this 1g to it's output to the PLC, and all those because I didn't know how to measure and represent the smallest part that can be represented in PLC.

It was a lot of information and I didn't know to manage it, I will start to read transmitter's datasheet and I will be attentive to calibration.
If you can give me some hints about to what characteristics pay attention I will be glad.

Thank you,
 
I didn't read through everything, so maybe I missed it, but what you are weighing matters. Depending on the process, you may need a trade legal indicator. Make sure you do your homework upfront.
 

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