Questions on analog input output signals for 1200 CPU

Hello every one ... please i have questions on analog input output signals for 1200 CPU

Assume the voltage range for the module and the analog sensor is [0 to 10v]:

Q-1- : if used analog input /output module with resolution 14 bit the range of norm-x will be [0 to 16384] or [0 to 27648] ?????

Q-2- : i mean the range   [0 to 27648] is constant for all resolutions bits or not. what about 10 ,11,12,13,14,15,bit resolution????

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Split from Where to post my question?.

 Hello.

mhmss

Hello every one ... please i have questions on analog input output signals for 1200 CPU

Assume the voltage range for the module and the analog sensor is [0 to 10v]:

Q-1- : if used analog input /output module with resolution 14 bit the range of norm-x will be [0 to 16384] or [0 to 27648] ?????

Q-2- : i mean the range   [0 to 27648] is constant for all resolutions bits or not. what about 10 ,11,12,13,14,15,bit resolution????

The numeric (integer) value you see from the analog input stays within the same range, regardless of the resolution. The analog-to-digital is internally scaled, so that your program is blind to the actual resolution used. See the attached picture.

The "sneaky" part of the specification is that 14-bit resolution (example) specifies voltage range between -11.85V  to 11.85V (example). The normal usage range is between -10V and 10V. The reach of the analog is specified in the picture.

BR

But what if I want a 16 bit resolution? This can't be mapped to 0 - 27648 (0x6C0) without losing precision: log2(27648) ~ 14.8 bits

I understand the idea of under/overflows but:

Let's assume I'll want high precison and use a 15 or 16 bit analog card for 4 -20 mA (from 1200 or et200sp range). This will be mapped to 0 - 27648. Can't map it to the, better, -27648 .. 27648 ( which still isn't the full 16 bit resolution, near but not full). Only when I switch to bipolair voltage input...

Strange? Or is it me.

Hello.

enof

But what if I want a 16 bit resolution? This can't be mapped to 0 - 27648 (0x6C0) without losing precision: log2(27648) ~ 14.8 bits

I understand the idea of under/overflows but:

Let's assume I'll want high precison and use a 15 or 16 bit analog card for 4 -20 mA (from 1200 or et200sp range). This will be mapped to 0 - 27648. Can't map it to the, better, -27648 .. 27648 ( which still isn't the full 16 bit resolution, near but not full). Only when I switch to bipolair voltage input...

Strange? Or is it me.

Aha... that is where users are caught with the "resolution".

Assume the analog is configured for measuring {-10 .. +10}V.

The analog to digital conversion is in actual fact handling {-11.85 .. 11.85}V scaled over the full 16-bit integer. (see the picture of earlier) The {-10 .. +10}V is limited to the {-27648 .. 27648} numerical value. (ouch!!)

The 16-bit is not for the {-10 .. +10}V as one expects, it is for {-11.85 .. 11.85}V.

BR

Thanks guys, that's the piece I was missing. The resolution is over the full range including overrange. It's specified in the et200sp manuals. The 1200 manual simply uses 'resolution'.

That implies that a -10V .. 10V has a (max.) resolution of 16 bits while the 4 - 20 mA is limitied to 15 bits. That's half the resolution! (while the conversion is capable of doing 16 bits).

Well at least it's documented correctly (never 16 bits resolution at the uni polaire ranges).

So if I want full 16 bit resolution I should use the bipolair range.