2/6/2017 10:40 AM  
Joined: 9/23/2005 Last visit: 8/21/2017 Posts: 1252 Rating: (146) 
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: 8/21/2017 Posts: 51 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 (25 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: 8/21/2017 Posts: 1252 Rating: (146) 
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, 

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