How 27648 Analaog Maximum Value ?

Petr Broza

 Hi Ayman,

Are you sure about this statement? If we have input 0-20mA and 8bit ADC, that mean 20/2^8=0.078mA is smallest current we can sense, that mean 27648/2^8=108 is smallest change in analog input. Am I right?

Ayman Elotaify

For 8-bit resolution; it would be able to measure (8/27648)*20mA. 

 Hello,

Yes you are right...  

Please see my answer from other thread about the resolution issue and also please see the attached file.

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,

 For 0-20mA AI,

For the 16 bit resolution, the minimum increment / decrementunits = 1 (please revise the attachments) in the above thread link.
+27648 = +20mA
0 = 0 mA

Each 1 unit = 20 / 27648 = 0.00072338 mA

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 * 0.00072338 mA = 0.09259 mA. 

Petr Broza

 Thanks for answer. Why the number is 128 (I see it's on first page of your pdf). I know it's 2^7, but if we use 128 as the minimal inc/dec then 27648/128=216 different values for 8bit ADC, so we missing 40 values(resolution is lower, than it could be, if we use all 256 values), is there any reason for this? Thanks

Ayman Elotaify

 Now let us assume we have 8-bit resulion AI module.
Then the minimum increment / decrement units = 128 

Hello,

If you look at table 5-2 of the attachment of my last post here, For 8-bit resolution, The minimum count value that would affect the channel is as follows:

For High Byte: 0000 0000;  For Low Byte: 1000 0000

Then the whole word became: 0000 0000  1000 0000 BIN = 128 DEC

I hope I could clear the resolution issue now.

Petr Broza

 Thanks for answer. Why the number is 128 (I see it's on first page of your pdf). I know it's 2^7, but if we use 128 as the minimal inc/dec then 27648/128=216 different values for 8bit ADC, so we missing 40 values(resolution is lower, than it could be, if we use all 256 values), is there any reason for this? Thanks

 Hello Again,

Why you consider that the module missed 40 values?!

if it really missed these values, then 27648 / 40 = 691.2 How could you write this value in BIN ?!

Again, for 8-bit resolution analog modules, the first bit which will be affected according to changes in the measured analog signal is Bit # 7 of the lower Byte. Which equal BIN 1000 0000 = DEC 128