5/27/2010 2:47 PM | |
Joined: 2/6/2009 Last visit: 3/26/2025 Posts: 939 Rating:
|
Dear Experts, I am using an Analog Input module (6ES7 331-7NF10-0AB0) having 16 bit resolution and I configured it for -10 to +10 voltage inputs. For the 16 bit resolution, accuracy of the voltage (Minimum change in voltage can be measured) is as shown as below, Input Range = -10 to +10 Span = 10-(-10) = 20 Bit Resolution = 16 = 2^ (16-1) //where, 1 bit for Sign. = 32768 Accuracy of Voltage = 32768 / 20 //Depend on the module Specs = 0.000610352 But my confusion is, The range of value for an analog input in program is 0 – 27648. So the accuracy of the Voltage corresponds to the Analog Input count is, Minimum Count = 1 Accuracy of Voltage= 20 / 27648 //Depend on the AI count = 0.00072338 From the above two answers which is correct. Could anybody please help to find out the actual resolution of the AI? Please don't mistake me. |
Last edited by: saravana kumar@VAES-Bangalore at: 5/27/2010 2:47 PM_______________________________________________________________ |
|
This contribution was helpful to
1 thankful Users |
5/27/2010 5:33 PM | |
Joined: 5/28/2008 Last visit: 3/24/2025 Posts: 4520 Rating:
|
Hello Kumar, Very good question. Thank you. First of all, please have a look on this threadand please down load the attachment in it and read it carefully. Now I want to correct your explanation. For the module you have, Input range: -10V to +10V which means: -27648 to +27648 units. NB: 1. the values: +27649 to +32511 are used for over-range. (which is equivalent to +10.004 to +11.758V) 2. the values: up to +32767 are used for over-flow. (which is equivalent to up to +11.759V) 3. the values: -27649 to -32511 are used for under-range. (which is equivalent to -10.004 to -11.758V) 4. the values: up to +32768 are used for under-flow. (which is equivalent to upto +11.759V) From the above explnation, we can conclude to: For the 16 bit resolution, the minimum increment / decrementunits = 1 (please revise the attachments) in the above thread link. +27648 = +10 V 0 = 0 V -27648 = -10 V Each 1 unit = 10 / 27648 = 0.00036169 V Now let us assume we have 8-bit resulion AI module. Then the minimum increment / decrement units = 128 Which means that each 128 units = 128 * 0.00036169 V = 0.046296. That means for the 16 bit AI module, if the changesin voltage of the AI was = 0.00036169 then the AI channel will response to this change by increment / decrement 1 count. And for the 8-bit resolution AI module, if the changes in voltage of AI was = 0.046296 then the AI channel will response to this change by increment / decrement 128 count. I hope this helps, With best regards, |
Ayman Elotaify |
|
This contribution was helpful to
28 thankful Users |
7/31/2011 10:43 PM | |
Joined: 10/10/2010 Last visit: 2/11/2025 Posts: 307 Rating:
|
by the way i was reading the manual carefully thank you again |
This contribution was helpful to
2 thankful Users |
6/26/2013 12:41 AM | |
Joined: 6/14/2013 Last visit: 3/18/2023 Posts: 218 Rating:
|
Thank You |
6/26/2013 5:23 AM | |
Joined: 4/17/2012 Last visit: 3/28/2025 Posts: 474 Rating:
|
Dear Ayman, Thanks a lot |
5/20/2015 4:03 AM | |
Posts: 1 Rating:
|
The range is from -10 to 10 v span is 20v each unit should be 20/27648=0.0007233796v is i am correct..? |
5/20/2015 12:40 PM | |
Joined: 3/5/2014 Last visit: 8/5/2022 Posts: 5521 Rating:
|
hi, no you are wrong. when you have -10 to +10V you have a range in SW by -27648 to +27648 So your formular is 20/55296 = 0,0003616898V or 0,3619898mV per Unit hope this helps, regards, |
FCK WAR! |
|
This contribution was helpful to
1 thankful Users |
5/24/2015 9:28 AM | |
Joined: 5/28/2008 Last visit: 3/24/2025 Posts: 4520 Rating:
|
Dear Jaan, No that's wrong. Please see the attached pdf file. For -10v to +10 V Analog Signals, equivalent units is -27648 to + 27648 AttachmentAnalog Value Processing.pdf (2802 Downloads) |
Ayman Elotaify |
|
This contribution was helpful to
1 thankful Users |
8/31/2016 5:47 PM | |
Joined: 9/3/2014 Last visit: 3/13/2019 Posts: 4766 Rating:
|
New question published by U_fc475158-26b9-4523-99e6-0100fd45bdc6 is split to a separate thread with the subject Analog module, lower resolution from 10bit to 8bit. Best regards |
2/6/2017 8:02 AM | |
Joined: 4/28/2015 Last visit: 6/3/2024 Posts: 112 Rating:
|
Dear Sir, Just to get a clear knowledge from your above explanation, I have calculated it for 4 -20 mA 13 Bit resolution cards, Please verify 4mA - 0, 20mA - 27648 so span = 20 - 4= 16 so for 1 unit change = 16/27648 = 0.00057870mA for a 13 bit module = 13*0.00057870 = 0.0075231mA thus for every 0.0075231mA change there will be 4 units increment or decrement. Am I doing it Wrong? Thanks |
Thanks |
|
We are working on a new user interface with better overview and more relevance.
Follow us on