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Siwarex CS

Created by: Nicky123 at: 5/7/2012 7:34 PM (20 Replies)

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5/7/2012 7:34 PM	Rate (0)
Nicky123 Posts: 59 Rating: (2)	Dear All, We are using Siwarex CS Module 7MH4 910-0AA01. And we are now trying to calibrate the module by Siwatool CS. Our load cell Characteristic as below: Load 0.5 ton. 2mv/V When we set 0 from Siwatool Cs when loadcell is totally empty then it accepts 0 & shows weight also 0. But when set adjustment weight as 80 kg on loadcell & we set from siwatool Cs as Adjustment Weight 1 valid then it shows error as “Adjustment weight too low”. We have tried to set all parameters for maximum weight range & Mimimum weight as well as numeral step but still we could not get success on this calibration. So please help us how can we calibrate this load cell? Our maximum weight requirement is only 80 Kg. So how much weight we have to use for calibration? Right now we are using 4 no,s of 20 Kg Loads for calibration. What should we set for Maximum weight range, Minimum weight & numeral step in our application? Please reply as soon as possible.
Nicky123 Posts: 59 Rating: (2)
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5/7/2012 8:17 PM	Rate (3)
SolutionFinder Posts: 559 Rating: (87)	Dear Nicky123, the message you get indicates that the signal difference from zero point to adjustment point is less than 5% of the max. signal difference. Normally you have to have an adjustment weight of more than 5% of the nominal load of ALL load cells together.This condition if if the load cells have the same characteristic value than the setting in the Siwarex CS. You have set the Siwarex CS to 2mV/V so it is like above if the load cells have also 2mV/V. You can also check in data record DR30 the "ADC digits after filtering" the actual value of the digits. The difference of the actual digit and the "Adj. digits0" must be >=3000. If you still have problems, we would need also the no. of load cells and some photos of the installation.
SolutionFinder Posts: 559 Rating: (87)
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5/7/2012 9:40 PM	Rate (0)
Nicky123 Posts: 59 Rating: (2)	Dear SolutionFinder, Thanks a lot for your reply. I understood the meaning of message now. So now i have to increase the adj weight 1 from 80 to some other vales to reach count difference more than 3000 & then i will get the result. Thanks a lot. I will check & revert back to you.
Nicky123 Posts: 59 Rating: (2)
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5/8/2012 2:20 PM	Rate (0)
Nicky123 Posts: 59 Rating: (2)	Dear SolutionFinder, Thanks a lot for your reply. The problem has been solved now. You were right. We have increased the Adj Weight 1 from 80 to 100 & the count value difference was reached up to 3200. So now we are able to do the calibration. We were struggling for this problem from last 2 days. Thanks a lot for your kind support.
Nicky123 Posts: 59 Rating: (2)
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5/9/2012 10:28 PM	Rate (0)
SolutionFinder Posts: 559 Rating: (87)	Thanks a lot Nicky123, For your valuable reply. This will help to others also.
SolutionFinder Posts: 559 Rating: (87)
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5/10/2012 8:12 AM	Rate (0)
AMAZINGAHMED Gold Member Joined: 8/27/2006 Last visit: 10/22/2024 Posts: 719 Rating: (88)	I have a question here, please. Nicky123said that he uses 4 load cells each of 20Kg. That makes his span of 80Kg. This span is equal to about 3000 digits for siwarex. Knowing that siwarex can transmit from 0 to 65536 digits, isn't it logic that from 0 to 80 equals (or nearly equals) 0 to 65536 digits ????? What is the difference if we have load cells of span 800Kg??? will this affect the accuracy???
	Last edited by: AMAZINGAHMED at: 5/10/2012 8:13 AM The more you know, The more you know that you don't know
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5/10/2012 2:13 PM	Rate (0)
Siwarexi Gold Expert Joined: 1/4/2007 Last visit: 11/13/2024 Posts: 1147 Rating: (175)	Hello AMAZINGAHMED, If you have 4 load cells each of 20 kg, then you have a total nominal load of 80 kg. 5% of the total nominal load (in our example 5% of 80 kg = 4 kg) representing 3004 digits. Best regards Siwarexi

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5/10/2012 2:20 PM	Rate (0)
AMAZINGAHMED Gold Member Joined: 8/27/2006 Last visit: 10/22/2024 Posts: 719 Rating: (88)	Siwarexi Hello AMAZINGAHMED, If you have 4 load cells each of 20 kg, then you have a total nominal load of 80 kg. 5% of the total nominal load (in our example 5% of 80 kg = 4 kg) representing 3004 digits. Best regards Siwarexi Thanks for your reply, I got what you said, but in this case,Nicky123has calibrated using 80Kg as adjustment weight (which is full load) and the digits difference between 0 and full load was less than 3000 !!!! how is that possible??? or do I miss something??
	The more you know, The more you know that you don't know
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5/10/2012 2:55 PM	Rate (0)
Siwarexi Gold Expert Joined: 1/4/2007 Last visit: 11/13/2024 Posts: 1147 Rating: (175)	Hello AMAZINGAHMED, I see, you are right. Well I think Nicky123 has mechanical problems, e.g. the transport lock, … . He has to control allmeasuremt signals from all load cells separately. Best regards Siwarexi

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5/10/2012 3:42 PM	Rate (0)
SolutionFinder Posts: 559 Rating: (87)	AMAZINGAHMED Siwarexi Hello AMAZINGAHMED, If you have 4 load cells each of 20 kg, then you have a total nominal load of 80 kg. 5% of the total nominal load (in our example 5% of 80 kg = 4 kg) representing 3004 digits. Best regards Siwarexi Thanks for your reply, I got what you said, but in this case,Nicky123has calibrated using 80Kg as adjustment weight (which is full load) and the digits difference between 0 and full load was less than 3000 !!!! how is that possible??? or do I miss something?? My dear Friend AMAZINGAHMED, I think you got confused about what Nicky123 has written. Please read it carefully what it said is "Right now we are using 4 no,s of 20 Kg Loads for calibration." Means they are using 4 loads of 20 kg means total 80 Kg for calibration purpose only. Means for setting Adj weight 1 they are using 80 Kg load with 4 no.s of 20 Kg load. Am i right Nicky? Please correct me if I am wrong. Max load for the laodcell wold be different.
SolutionFinder Posts: 559 Rating: (87)
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