Ayman Elotaify

 

mhmss

Ayman Elotaify

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.

Thank you a lot Mr. Ahmad
i read that word start from right with high byte and the low byte will be in the left : the structure for example MW0= MB0-MB1>>> it means the first digit in the word bit0 will be M1.0 not M0.0 ...this is right? and how it is agreed with analog value which looks reversed what i know

Ayman Elotaify

Attachment : Analog Input Representation.pdf

 

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Split from How 27648 Analaog Maximum Value ?.

 Hello,

Let me explain my point of view in my post you reffered to.

Analog value is reppresented in a word (two bytes). Example (IW 128) is represented in bytes 128 & 129.

To understand resolution issue, you should know or remember that changes of physical quantaties you are measuring using transducer (i.e level, flow, pressure ,..etc) is being represented and changes in an electrial value in some standard ranges like ( (0-10V), 4-20mA) ,...etc)

The mission of analog Input modules is to take these electrical value and tell the CPU the analog input is now equal to : x for instance. It done using ADC (Analog to digital Converter).

But not each changes (small or big) happen in the physical value the transmitter measures can be known by the CPU. it depends on the AI module and the change itself

Examples

• 16-bits resolution, the minimum change in analog value which can be reported to the CPU: is 1 count = 0.00072338 mA.
• 8-bits resolution, the minimum change in analog value which can be reported to the CPU: is 128 count = 128 * 0.00072338 mA = 0.09259 mA

Now see the attached picture. you will found that each analog value is represented in two bytes (high byte & low byte). Always, low byte is written in the right hand side and high byte is written in left hand side.

The misunderstanding is comming from the meanining of low / high bytes.

Low byte contains bit No. 0, 1, 2 ,. ...7 which is MB1 in your example "MW0"
High byte contains bit No. 8, 9, 10,. ...15 which is MB0 in your example "MW0"

Please see the next post attachment.
 

 Dear Sir, 

             Please describe me once. We have AI input which is coming in 16bit word format and after that it is converting high byte to low byte and again 16bit word format is generated. Why it is done like this? 

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Split from word start from right with high byte and the low byte will be in the left - how it is agreed with analog value which looks reversed .

Hello,

from your question as is, is hard to answer exactly why that procedure is prepared. My personal opinion and experience says that procedure is prepared for some other device usage (via communication channel f.ex.) which use different Endian system as S7 use. This is some best logical answer in my opinion. I recommend check Endian (little/big) system for data usage. Typical situation can be Modbus with some other Endian type devices where swap of bytes on WORD register need to be set because then is impossible correctly read data due swapped bytes in WORD data type.

Closed cause question posted twice.

The second thread with the subject Word to analog conversion in TDC(For pressure transducer) is published within the conference SIMATIC TDC, FM458, T400.