how to scale the 4 to 20mA signal

jcp

Member
Join Date
Feb 2010
Location
Multan
Posts
369
Hi friends'
i have a IC693ALG223 analog input module on the chasis with CPU 351 from GE funac system. I have the transmitter o to 100 PSIG which give 4 to 20 mA. I want at 3.75 mA and 20.25 mA the module tell me that transmitter out of range lower and high limit. For this i use the %I register of the module.
But the problem is the scalling. i use 4-20mA+(Enhanced) the module documentation tells me that the counts are for low limit is -8000 to +32759 and for high limit -7999 to +32760.
How apply and cofigure the module for proper scalling?
Please guide me if any one have a universal formula for scalling plz send me beacuase i have also some 0 to 25 inches of water col diffrential transmitter as well.
Thanking you i will remain
 
I'm not sure how you plan to read 20.25ma on a device that only goes up to 20ma.

However, using the 0 - 20ma range in the module, each count represents 0.625 microamps (abbreviated as "ua" for the purposes of this post)

Example: 32000 x .625ua = 20ma

So to get your low limit of 3.75ma, you would divide 3.75ma by .625ua (.00375 \ .000000625) which would give you 6000 counts for 3.75ma.

Use this as your low limit in your program (I would do it in the program rather than use the low limit in the analog module).

32000 would represent 20ma. If the module happens to go a bit overrange then you might reach your 20.25ma, but I wouldn't count on it.
 
Thanks for reply;
Sir now i explain plz guide me
0 to 100 PSIG transmitter 4 to 20 mA signal.
In analog module configration which limit i set for Low alarm and Hi alarm?
second how i calculate the ua/count?
Can i used the following formula to apply in PLC math function
[R(raw)-R(min)/R(max)-R(min)]*((Imax-Imin)+Imin)
Now the problem is that suppose i have set the module low limit =0 and high limt=32760 then
i calculate 32760/20=1638 now 1638*4=6552
so if i assume the transmitter give 6mA then
[9828-6552/32760-6552]*((20-4)+4)=2.5
I am stuck here? Please help me?
For 0 to 100 PSIG and 4 to 20 mA transmitter i think 100/16=6.25 Psi/mA am i right?
I need help on this issu i hope u favour me
 
Hi
Basically your card will give 0 counts for 4mA and 32000 counts for 20mA

So.. Scaling is

(Your Max Unit / 32000) * Actual counts

As for over and under range, I agree with rootboy ,suspect your not going to get that on this card but you can easily test by injecting mA and seeing what happens.
 
psig / whatever units you wish to use. I guess your not into metric units (bar) where you live
 
For any unit conversion on any PLC you can use four function math with the following, whichasumes integer math:

DataOffset = Data in input register at 4 mA (or zero analog signal of whatever range - for example 0 VDC on a 0-10 VDC transmitter)
DataSPan = Data at 20 mA - Data @ 4 mA
FACTOR = arbitrary factor (multiple of 10) needed to get proper resolution and accuracy
EU = Engineering Units x FACTOR
EUOffset = Engineering Units @ 4 mA x FACTOR
EUSpan = (Engineering Units @ 20 mA - Engineering
Units @ 4 mA) x FACTOR
DATA = Actual data reading in input register
EU = ((EUSpan x (DATA - DataOffset))/DataSpan) + EUOffset
ProcessValue = EU / FACTOR

For Example, 0-60 psig from 4-20 mA with resolution of 0.1 psig:
FACTOR = 10
Data @ 4 mA = 6240
Data @ 20 mA = 31208
DataOffset = 6240
DataSpan = 31208 - 6240 = 24968
EUOffset = 0 x 10 = 0
EUSpan = (60 - 0) x 10 = 600
DATA = 18975
EU = ((600 x (18975 - 6240)) / 24968) + 0 = 306
Process Value = 306 / 10 = 30.6 psig
 
Thanks for reply and guide. Its very helpful for me. But can u explain further what is the count values of low limit and for hi limit?
You write
Data @ 4 mA = 6240
Data @ 20 mA = 31208
where it comes from i set the analog module configration is
Low limit=0 and high=32000
Regards
 
Tom is providing an example of the calculations. Your actual results will probably be different.

Apply a 4 mA signal to the analog module and read the value in the PLC. That is your Data @ 4 mA.
Apply a 20 mA signal to the analog module and read the value in the PLC. That is your Data @ 20 mA.
 
Their is no effect of limits which set in analog module configration? I set 0 to 32000 or if i set 4-20mA+ and limits are -2000 to 32000 then what diffrence in the calculation?
 
The IC693ALG223 module can be configured in one of three ways:

When you select 4 - 20 mA range, the module reports a value of zero when it detects a 4 mA signal at the terminal and reports a value of 32000 when it detects a 20 mA signal.

When you select 0 - 20 mA range, the module reports a value of zero when it detects a 0 mA signal at the terminal and reports a value of 32000 when it detects a 20 mA signal.

When you select enhanced 4 - 20 mA range, the module reports a value of - 8000 when it detects a 0 mA signal at the terminal and reports a value of 32000 when it detects a 20 mA signal.

The 0 - 20 mA option gives you a non-zero value in the %AI register when the signal level from the field device falls between zero and 4 mA. It does this at the expense of resolution since a 4 mA signal will result in a value of approximately 6400. That results in a span of 25600 counts corresponding to the field device's 16 mA span compared to 32000 counts when you select 4 - 20 mA.

The enhanced 4 - 20 mA selection gives you the same resolution as the standard 4 - 20 mA setup. When you select enhanced 4 - 20 mA, signals from the field device less than 4 mA are reported as negative values in the %AI register.
 
Last edited:
Steve,
I am not a user of this product, but I did have a look at the manual and it looks to me that the raw data is a little different then what you have posted. The alarm values setting can be higher though. Correct me if I'm not reading this correctly. See pdf below for an exert from the manual.
 
Thanks for reply;
Sir now i explain plz guide me
0 to 100 PSIG transmitter 4 to 20 mA signal.

Yup, that should be what you get, if you have your analog module setup for 4 - 20ma


In analog module configration which limit i set for Low alarm and Hi alarm?

Why bother? Set your limits in the PLC where it is easier to modify them should it become necessary.


second how i calculate the ua/count?

If you have your module setup for the "normal" 4 - 20ma range, then you would have 32000 counts over a range of 16ma.

.016/32000 = .5ua per count

If you have your module setup for the 0 - 20ma range, then you would have 32000 counts over a range of 20ma.

.020/32000 = .625ua per count

If you have your module setup for the "enhanced" 4 - 20ma range, then you would have 40000 counts over a range of 20ma. But this also includes an 8000 count (negative) offset to be able to measure 0ma.

.020/40000 = .5 ma per count

You have to offset your count by -8000 for either 4 - 20ma range, it's just that the "enhanced" mode increases the range of the input to 40000 counts.

So in "enhanced" mode, 0ma would be equal to -8000 counts, 4ma would be equal to 0 counts, and 20ma would be equal to 32000 counts (btw, you would have to use signed integers with this range).

In each of these configurations you would take the appropriate I per count (either .5ua or .625ua) and multiply that by the counts that you see in your PLC (and add 4ma to the result if appropriate).

Conversely, you could calculate what to expect beforehand. For example (using a 4 - 20ma range, "enhanced" or not):

0 PSI would be 0 divided by 100 times 16ma + 4 ma = 4ma
25 PSI would be 25 divided by 100 times 16ma + 4 ma = 8ma
50 PSI would be 50 divided by 100 times 16ma + 4 ma = 12ma
75 PSI would be 75 divided by 100 times 16ma + 4 ma = 16ma
100 PSI would be 100 divided by 100 times 16ma + 4 ma = 20ma

You would then divide your result by the "I" per count (.625ua for normal 4 - 20ma, and .5ua for "enhanced" 4 - 20ma)

For the normal 4 - 20ma range it's straightforward:

50 PSI, would be 12ma divided by .5ua - 8000 (counts)

Or: 12ma / .5ua = 24000 - 8000 (which would give you 16000 counts in the PLC).

Should you have 2ma in your loop you would get:

2ma /5ua - 8000 = -4000 counts in your PLC.


For the 0 - 20ma range it would be:

0 PSI would be 0 divided by 100 times 20ma = 0ma
25 PSI would be 25 divided by 100 times 20ma = 5ma
50 PSI would be 50 divided by 100 times 20ma = 10ma
75 PSI would be 75 divided by 100 times 20ma = 15ma
100 PSI would be 100 divided by 100 times 20ma = 20ma

Then for 50 PSI, it would be 10ma divided by .625ua equaling 16000 counts.

I hope this helps. :)
 
Greetings JCP,

May I suggest you take a look at the scaling tutorial one of our forum members, Ron Beaufort, put together.

http://www.plctalk.net/qanda/showthread.php?t=15069

The tutorial covers the basic principles behind scaling analog values in a very easy to understand format. While the tutorial is written using an Allen Bradley PLC for the examples, the principles are the same not matter what PLC you are using.

I suggest you take the time to read the tutorial at the link I posted. It will help you to understand exactly what math steps to take to scale any analog signal from any kind of input to any kind of engineering unit.
 

Similar Topics

Good afternoon. 1794-ie12 module available, channel set to 4...20 mA. According to the table from the manual, we get from 0 at 4mA to 30840 at...
Replies
5
Views
2,201
Good day all, I am having bit of an issue getting my head around this RTD scaling issue. Be forewarned, my head has been known to be "THE" issue...
Replies
24
Views
8,698
i make a program ( plc siemens s7-300 ) with analog output 4-------20ma 0-------27648 -20-----80c temp. how to do this?
Replies
8
Views
2,207
i make a program ( plc siemens s7-300 ) with analog output 4-------20ma 0-------27648 -20-----80c temp. how to do this?
Replies
3
Views
1,927
Hi, I'm a noob after some help if possible. I have been using a Click PLC with a 0-20mA analog input module, quite happily, for some time now. I...
Replies
4
Views
1,701
Back
Top Bottom