Resolution and range analog I/O module ac500
I have some trouble understanding range and resolutin on this module.
Example; the AX521 analog i/o module has 12bit resolution on the 4-20mA signal.
How does this correspond to the fact that it scales to the range 4.000-20.000 (digital 1-27648), and below the table it says "The represented resolution corresponds to 16 bits"?
What am I missing here? Trying to understand this...the measuring itself is no problem.
Voted best answer
Yes, it is. To put it simply if You change a analog signal to discret (digital) signal with x bit resolution, it means You are using x bits to describe this anagol value. This means it can differentiate between 2^x states of this signal. In other words it can only show change of 1/(2^x) part of signals range. For a 4-20 mA signal in 16 bits resolution a change of:
(20mA - 4mA)/2^16 = 0,144140625 μA
is the smallest recorded change (only if all 16 bits are used entirely for this range, which normally is not the case, as most of AI modules can take input signal smaller and bigger then this or work on entirely different range, like -20...20mA, which means they must be able to convert these other values to bits too, using the same notation).
Hope this clears things a bit (no pun intended),
"Resolution of the analog input is related directly to the precission of the reading of this analog input.
Generally, each AI is read in a Word (16 bits)
If we have an 8-bit resolution AI, by the physical changing of process value, only 8-bits will change accordingly.
The same for 12, 13, 14 & 15 bit resolution.
So, for 8-bit resolution,
We have 8-bits changes its status according to the physical changes in the process (Bits 8 to 15)
We have one bit for sign.
The remaining 7 bits is not used.
So when changes happen in the physical value, it affects the analog value.
And as we have 7 bits not used, so the physical should change by at least a value equivalent to count value (27= 128) to be monitored by the analog value.
By increasing the resolution, this value became lower, so the less changes in the physical value can be monitored in this case, and of course the precission will be better."
Makes sense - is it correct?