s7 300 Analog Input Card 6ES7331-7KF02-0AB0

[shamusdb]

Hi All

I am using 2 x analog cards on my rack and they are described as 12bit cards in there title. The manual suggests that the resolution can be programmed per group but i can't seem to find out how to do this. I do select 4 wire (4-20mA) so would that be selecting the resolution behind the scenes, instead of me selecting 9bit,12bit or 14bit somewhere?

Also my understanding is that 12bit represents 4096 14bit is 16384 etc but with Siemens I always seem to put 27648 as upper limit (20mA) to get it to work which i have also done in this case.

My reason for requesting some feedback on this is that i'd like to know the smallest increment i can measure in 2 silos. One is 200000KG max reading and the other is 80000KG.

200000/27684 = 7.22KG (HMI display will be to the nearest 7.22kg)
80000/27684 = 2.89KG (HMI display will be to the nearest 2.89kg)

Is this the correct way to work this out?

thanks for your time

 

[AustralIan]

Regardless of the bit, siemens seem to pad the number so that it is 15 bits.

so for a twelve bit Analog to digital convertor, the last three bits are always 0.

This also means that your resolutions are:
200,000/4096 = about 50kg
80,000/4096 = about 20kg

The below digram is for positive values. negative values will ofcourse be 2's compliment

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
 0 [    12 bits of data       ] 0 0 0

 

[shamusdb]

Thanks very much for the reply. When it mentions in the hardware manual about selection of resolution ●
** "Programmable resolution at each channel group (9/12/14 bits + sign)" **
is this actually possible?

regards
Shaun

 

[Jeebs]

Properties > Inputs > Measuring

Integration time - Interference frequency - Resolution
-------------------------------------------------------
     2.5ms                  400Hz             9-bits
    16.6ms                   60Hz            12-bits
    20.0ms                   50Hz            12-bits
   100.0ms                   10Hz            14-bits

 

[shamusdb]

Thanks Jeebs, I think I get the picture now.

Regards
Shaun

 

[AustralIan]

My calculation was wrong. I forgot that the 9/12/14 bits are actually for the overflow range, which is 118.515% of your full range.

so the formulae are:
200,000 * 1.18515 / 2^(bit resolution)
80,000 * 1.18515 / 2^(bit resolution)

 

[shamusdb]

Thanks AustralIan