2/6/2017 10:40 AM  
Joined: 9/23/2005 Last visit: 11/23/2017 Posts: 1459 Rating: (174) 
That is because you calculate only for positive range, then you have 15 usable bits. Your card has 13 bit resolution, i.e. 12 bit + sign. 2^(1512) = 8 To be 100% sure, pls post the exact number of your cards. 
Regards, 

2/6/2017 11:59 AM  
Joined: 4/28/2015 Last visit: 11/23/2017 Posts: 57 Rating: (3) 
Thankyou for your advice, I have just assumed the card to be 13 bit just for an example. What I learnt from the Analog Manual is that the 13 bit resolution cards has 4 unused bits. Is the attachment below is correct? so is it 4*0.00057870 = 0.0023148 AttachmentAnalog Module.pdf (32 Downloads) 
Last edited by: Satheesh290692 at: 2/6/2017 12:12:13 PM 

2/6/2017 1:13 PM  
Joined: 9/23/2005 Last visit: 11/23/2017 Posts: 1459 Rating: (174) 
To clarify according to the latest Siemens naming convention  if it's stated that the card is i.e. 8 x 13 bits then its resolution is 12 bits + sign. Then my above calculations apply. If you mean resolution of 13 bits + sign (i.e. 14 bits according to the convention) then number of unused bits is 2. 
Regards, 

This contribution was helpful to1 thankful Users 
9/12/2017 5:03 PM  
Joined: 2/1/2017 Last visit: 9/13/2017 Posts: 6 Rating: (0) 
All, If i have been following this thread correctly, then for my input card "6ES7 3364GE000AB0"; which the card states it is a 15bit card, but the datasheet claims then it is 16 bit (15 bit +sign); then for a 420mA signal input I am calculating using a range of 16mA. Therefore: 16/27648 = 0.00057870mA. This is because I am only using the positive range so I exclude the 27648 and only use the +27648? So each count equates to 0.00057870mA? The problem then lies with the card resolution, the card says 15bit; but the manual says 16bit (15bit +sign), so if I do the calculation using the card description, meaning 15bit (14 bit +sign), the formula is then: 2^(1514) = 2 Thus: 2*0.00057870mA = 0.0011574mA So for every 0.011574mA change there will be a 2 unit (count) change respectively. So if my input counts when reading the input point is 4976 (the input count I saw today on a variable table), this would mean that my reading is only: Or would I use: 0.0011574*4976 = 5.759mA Using the manual description, this means 16bit (15bit + sign), then: 2^(1515) = 0 Does this mean then that the value above stays at 0.00057870mA per count, or it negates to zero because it would be: 0*0.00057870 = 0. If it is 0.00057870mA per count, then for 4976 counts the mA reading is only 2.8796mA? The descriptions are not clear when I apply my numbers, although I am quite sure the 4976 number is an actual counts input and not a scaled value, this could be part of the problem maybe? I am wondering what I am missing. Thanks in advance

9/13/2017 9:13 AM  
Joined: 9/23/2005 Last visit: 11/23/2017 Posts: 1459 Rating: (174) 
2^0 = 0 Come on. Sorry mate but you are missing something seriously.
And where is your 4mA offset ? When converted value is 0, then it means, that actual input current is 4mA. 
Last edited by: jacek d at: 9/13/2017 9:16:44 AMRegards, 

9/13/2017 12:02 PM  
Joined: 2/1/2017 Last visit: 9/13/2017 Posts: 6 Rating: (0) 
Jacek, Both very good points you raised there... I'll be honest, I have no idea how I missed them, must've got myself stuck in a loop and missed the obvious! Thinking 2^0 = 0, what was I doing !Everything to the power of zero is 1! I was starting to think about the physical switchable block as well being selected for 020 instead of 420mA, but then I realised my calculations were using just 16mA anyway; forgetting the already raised zero was there. Cheers! 
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