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Old November 25th, 2021, 10:00 AM   #1
Al-Naseri
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Micro820 Analog output 4-20mA with 2080-OF2

Gents

I'm working on a small project here and the customer requested to control the drive with 4-20mA by using Micro820 Analog output 4-20mA, Module 2080-OF2.
After getting all the logic up and running the only issue I have is I can't get 4-20mA to present on the output? even checked with Ammeter on the output of the module I can't see anything.

Typically after doing scaling the output is REAL data type but the module only accepts UINT, and by using UNIT going to lose some data precision.

Thanks in Advance!
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Old November 25th, 2021, 12:29 PM   #2
parky
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Never worked on an 820 but I assume it requires an integer of something like 0-32767 or perhaps 6553 - 32767 (4-20ma) so you must convert it to an integer. so you need to scale it back up & convert it to an integer.
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Old November 25th, 2021, 01:15 PM   #3
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The 2080-OF2 has 12-bit resolution, and the output range is 0 to 65535, so 4mA will be around 13107, and 20mA will be around 65535.

The 24 bits of resolution in the 32-bit float are irrelevant, because you are going to lose at least 12 bits of that. However, 12-bits of output resolution is one part in 4096, or about 0.025%; it will be the rare process that needs more resolution than that.

For getting the signal out, I suggest checking the obvious first:
  • Which output terminals are you testing with the ammeter?
  • What load (resistance) are put putting across those terminals?
  • I see that channel 0 of the 2080-OF2 is [Enabled] in CCW; did you download that configuration to the Micro820?
  • The -OF2 has Open- and Short-Circuit protection; is it possible that is causing the problem?
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Old November 25th, 2021, 04:10 PM   #4
Al-Naseri
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Thanks for the response.

So it sounds like I don't really understand how the 2080-OF2 module functions.
because on the simulator after the scaling I do see the number you mentioned above. However, after downloading the program into the PLC, I thought the output would show 4-20mA when I use a multimeter, " this is how I typically check 4-20mA out of any current transducer".
But based on your explanation it sounds like the output still "data" a number range between 0-65535. is this correct?
if yes, how the VFD would read it?
really sorry I'm kinda confused ������

if you have any link or PDF regarding this to share I would really appreciate that.


Thanks
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Old November 25th, 2021, 07:29 PM   #5
drbitboy
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https://lmgtfy.app/?q=2080-of2+user+manual
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Old November 26th, 2021, 04:39 AM   #6
parky
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@DR, 12 bit resolution is 4095 I think you have got confused with 15 or 16 bit that would be 32767 & 65535
If you could use a real (float) as the variable to move to the analogue channel it depends on what the value was for example 1000.0 would show up as 0 in an integer 1000.1 would show as 1638 & 1011.1 would show up as minus 14746
What I'm saying is that assuming it was possible to move a real into a variable (analogue card variable) it might only transfer 16 bits of the 32 bit float (not sure that is possible in an 820) but in some other PLC's as the internal registers are 16 bit by default with direct addressing of variables it is possible.
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Old November 26th, 2021, 07:00 AM   #7
drbitboy
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Quote:
Originally Posted by parky View Post
@DR, 12 bit resolution is 4095 I think you have got confused with 15 or 16 bit that would be 32767 & 65535

Really? You think the bitboy doesn't recognize in his bones the apparent inconsistency between "65535" and "12-bit resolution?" That it doesn't cause major shivers throughout his body as he types it?

Anyway, seriously ...

According to the user manual at this link, 2080-OF2 has 12-bit resolution, and the output count range is 0-65535:
xxx.png

I interpret that to mean that the -OF2 uses the high (i.e. Most Significant) 12 bits of the output, 16-bit UDINT count to determine an -OF2 channel's signal current. I suspect the low (Least Significant) 4 bits are truncated and ignored; they could of course be rounded and dropped, but at that point who cares? Either way, the smallest integral change, in the output UINT value sent to an -OF2 channel, that is guaranteed to change the output signal current, is 16.

Tim Wilborne has a YouTubes at this link incorporating the -OF2, and I think that supports my interpretation.
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Old November 26th, 2021, 07:24 AM   #8
drbitboy
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Quote:
Originally Posted by Al-Naseri View Post
But based on your explanation it sounds like the output still "data" a number range between 0-65535. is this correct?

No.

I apologize for not being clear.
  • If you write a UDINT value of 65535 to an -OF2 channel in the Micro820 program, the output current signal should be near 20mA.
  • If you send a UDINT value of 0, the output current should be near 0mA.
  • If you send a UDINT value of 13107 (=65535/5), the output current should be near 4mA (=20mA/5).
The non-linearity and repeatability are each 0.1% of full scale, and the Output error over the full temperature range is 2%, so I would not be bothered about dropping from 24 bits of resolution (or better) in a real to the high 12 bits of a UDINT i.e. a programmatic resolution limit that is less than 0.025%.

I don't know why we don't see an output current; I suspect it is something simple:
  • Which terminals are being used to sense and/or measure the output signal current (0-20mA)?
    • Are we sure those are the correct terminals for the -OF2 channel for current (0-20mA)?
  • Is it possible there was initially no circuit (i.e. no load) between the terminals, and the -OF2, which the manual states has open-circuit protection, faulted and needs a power cycle or some other action to reset?
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Old November 26th, 2021, 02:13 PM   #9
parky
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@DR: I thought it was unlikely you would make that mistake lol
However, doing that on an analogue output does not make sense, Siemens did that on analogue inputs but then again they are read only, the lower 3 bits were diagnostic bits so to get the true count you shifted the value 3 places.
Why would you do that on an output, if they were diagnostic bits then your move function would overwrite them, also a small change in your data would not reflect on the 12 bits WHY?.
Perhaps they shift them internally still does not make sense perhaps someone has the answer, anybody?.
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Old November 26th, 2021, 06:12 PM   #10
Mickey
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See links ... about the 12 bit stuff

http://www.plctalk.net/qanda/showthr...bit+resolution

http://www.plctalk.net/qanda/showthr...bit+resolution
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Old December 3rd, 2021, 12:15 PM   #11
Al-Naseri
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Thank you guys for the Help. it finally worked with this setup
13107 = 4mA
65535 = 20mA

I typically use an Ammeter to see the 4-20mA, but here little different method
The output is a Varying voltage. after you connect it to the drive, it will "Covert" to Current because of the input resistance of the VFD.

Appreciate all help
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