How to get the resolution of Analog Input

Satheesh290692

 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  = 8 * 0.00057870mA

thus for every 0.0075231mA 0.0046296mA change there will be 8 units increment or decrement.

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^(15-12) = 8

To be 100% sure, pls post the exact number of your cards.

jacek d

 

Satheesh290692

 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  = 8 * 0.00057870mA

thus for every 0.0075231mA 0.0046296mA change there will be 8 units increment or decrement.

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^(15-12) = 8

To be 100% sure, pls post the exact number of your cards.

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

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.

jacek d

 

Satheesh290692

 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  = 8 * 0.00057870mA

thus for every 0.0075231mA 0.0046296mA change there will be 8 units increment or decrement.

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^(15-12) = 8

To be 100% sure, pls post the exact number of your cards.

All,

If i have been following this thread correctly, then for my input card "6ES7 336-4GE00-0AB0"; which the card states it is a 15-bit card, but the datasheet claims then it is 16 bit (15 bit +sign); then for a 4-20mA 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 15-bit; but the manual says 16-bit (15-bit +sign), so if I do the calculation using the card description, meaning 15-bit (14 bit +sign), the formula is then:

2^(15-14) = 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:

0.0011574*(4976/2) = 2.8796mA? But this is below the 4-20mA range?

Or would I use:

0.0011574*4976 = 5.759mA

Using the manual description, this means 16-bit (15-bit + sign), then:

2^(15-15) = 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

JakeGraves

Using the manual description, this means 16-bit (15-bit + sign), then:

2^(15-15) = 0

2^0 = 0 Come on. Sorry mate but you are missing something seriously.

JakeGraves

If it is 0.00057870mA per count, then for 4976 counts the mA reading is only 2.8796mA?

jacek d

 

JakeGraves

Using the manual description, this means 16-bit (15-bit + sign), then:

2^(15-15) = 0

2^0 = 0 Come on. Sorry mate but you are missing something seriously.

JakeGraves

If it is 0.00057870mA per count, then for 4976 counts the mA reading is only 2.8796mA?

And where is your 4mA offset ? When converted value is 0, then it means, that actual input current is 4mA.

 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 0-20 instead of 4-20mA, but then I realised my calculations were using just 16mA anyway; forgetting the already raised zero was there.

Cheers!