How to get the resolution of Analog Input

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.

Ayman Elotaify

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,

The range is from -10 to 10 v span is 20v 

each unit  should be 20/27648=0.0007233796v

is i am correct..?

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,

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
Min_Moderator

Ayman Elotaify

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,

 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