5/27/2010 5:33 PM | |
Joined: 5/28/2008 Last visit: 10/7/2024 Posts: 4515 Rating: (839)
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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 |
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This contribution was helpful to28 thankful Users |
6/26/2013 12:41 AM | |
Joined: 6/14/2013 Last visit: 3/18/2023 Posts: 218 Rating: (9) |
Thank You |
6/26/2013 5:23 AM | |
Joined: 4/17/2012 Last visit: 8/30/2024 Posts: 474 Rating: (10) |
Dear Ayman, Thanks a lot |
5/20/2015 4:03 AM | |
Posts: 1 Rating: (0) |
The range is from -10 to 10 v span is 20v each unit should be 20/27648=0.0007233796v is i am correct..? |
5/24/2015 9:28 AM | |
Joined: 5/28/2008 Last visit: 10/7/2024 Posts: 4515 Rating: (839)
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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 (2612 Downloads) |
Ayman Elotaify |
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This contribution was helpful to1 thankful Users |
8/31/2016 5:47 PM | |
Joined: 9/3/2014 Last visit: 3/13/2019 Posts: 4767 Rating: (123) |
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: (3) |
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 |
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