0-20mA to 4-20mA

eisenpferdreiter

Hi Mishal11,

the answer to your question why there are no configurable settings with your module has been given detailed already by dchartier. Nothing more to say about it - its a cheap module and a compromise with low resolution.

Can you use your 4 to 20 mA Transmitter/Actuators with 0 to 20 mA inputs/outputs?
Yes, you can - but you have to be aware of the fact, that you will loose 20% accuracy. If this is ok with your application, you can do so as intended.

To help you for your decision:
the module has 8 bit resolution, that means 256 possible states. If you drop 20% you will reduce that to 200 states. As the module uses the Msb (most significant bit) as sign bit the 256 states are acatually +/- 128 and the 200 states are +/- 100 bit.

In your reply to William B. you state that you have inputs 0 to 10 bar.
If you use 0 to 20 mA you will have a resolution of 10bar/128 states, which means your resolution is0,078125 bar per bit, so roughly 0,08 bar resolution.
If you use 4 to 20 mA at 0 to 20 mA input, you would have a resolution of 10 bar/100 states, which is about 0,1 bar resolution. Actually this could still be good for many applications, but of course you will have to take into account for your decision.
You can check other inputs/outputs with same calculation.

Just as William B. stated already, you cannt use library functions FC105 and FC106, as these are designed for full range, but you only have 8 bit resolution.

For scaling inputs with 0 to 20 mA signal on 0 to 20 mA input, you will have to do following calculation:
Engineering Value= (((MAE - MAS) / effective states) * RAW_VALUE) + MAS

MAE...........................measuring area end - real
MAS...........................measuring area start - real
effective states..........with your module 128 - real
RAW_VALUE............digitized input value from module - byte

You see that you need to convert RAW_VALUE before you can do the calculation.
You can do so by using a MOVE block, where you move RAW_VALUE to a real variable.


If you use 4 to 20 mA signal on 0 to 20 mA input, calculation would be:
Engineering Value = (((MAE - MAS) / effective states) * (RAW_VALUE - 20%FS)) + MAS

MAE...........................measuring area end - real
MAS...........................measuring area start - real
effective states...........with your module 102.4 - real
RAW_VALUE.............digitized input value from module - byte
20%FS.......................20% full states, in your case 25.6 - real

So in your example with 0 to 10 bar on 4 to 20 mA with 0 to 20 mA input:
half of measuring range is 5 bar, which will be converted to 12 mA (middle between 4 mA and 20 mA).
Engineering value = (((10 - 0) / 102.4) * (76.8 - 25.6)) + 0 bar = 5 bar
the RAW_VALUE will be (128digit / 20 mA) * 12 mA = 76.8 digit

For calculating outputsyou take same equation, but calculate RAW_VALUE instead of Engineering Value, as Engineering Value would be given in that case.

I hope I could help you along.
Best regards

 Hello

i have the same probem with à température sensor (-10;150°C)

could you Please make the same calculation ? Tks 

Use Scalling Block. FC105

engr_Ahmed

Use Scalling Block. FC105

Hello engr_Ahmed,

have you read this?

"Just as William B. stated already, you cannt use library functions FC105 and FC106, as these are designed for full range, but you only have 8 bit resolution."

regards

AutoGen_46749

 

eisenpferdreiter

Hi Mishal11,

the answer to your question why there are no configurable settings with your module has been given detailed already by dchartier. Nothing more to say about it - its a cheap module and a compromise with low resolution.

Can you use your 4 to 20 mA Transmitter/Actuators with 0 to 20 mA inputs/outputs?
Yes, you can - but you have to be aware of the fact, that you will loose 20% accuracy. If this is ok with your application, you can do so as intended.

To help you for your decision:
the module has 8 bit resolution, that means 256 possible states. If you drop 20% you will reduce that to 200 states. As the module uses the Msb (most significant bit) as sign bit the 256 states are acatually +/- 128 and the 200 states are +/- 100 bit.

In your reply to William B. you state that you have inputs 0 to 10 bar.
If you use 0 to 20 mA you will have a resolution of 10bar/128 states, which means your resolution is0,078125 bar per bit, so roughly 0,08 bar resolution.
If you use 4 to 20 mA at 0 to 20 mA input, you would have a resolution of 10 bar/100 states, which is about 0,1 bar resolution. Actually this could still be good for many applications, but of course you will have to take into account for your decision.
You can check other inputs/outputs with same calculation.

Just as William B. stated already, you cannt use library functions FC105 and FC106, as these are designed for full range, but you only have 8 bit resolution.

For scaling inputs with 0 to 20 mA signal on 0 to 20 mA input, you will have to do following calculation:
Engineering Value= (((MAE - MAS) / effective states) * RAW_VALUE) + MAS

MAE...........................measuring area end - real
MAS...........................measuring area start - real
effective states..........with your module 128 - real
RAW_VALUE............digitized input value from module - byte

You see that you need to convert RAW_VALUE before you can do the calculation.
You can do so by using a MOVE block, where you move RAW_VALUE to a real variable.


If you use 4 to 20 mA signal on 0 to 20 mA input, calculation would be:
Engineering Value = (((MAE - MAS) / effective states) * (RAW_VALUE - 20%FS)) + MAS

MAE...........................measuring area end - real
MAS...........................measuring area start - real
effective states...........with your module 102.4 - real
RAW_VALUE.............digitized input value from module - byte
20%FS.......................20% full states, in your case 25.6 - real

So in your example with 0 to 10 bar on 4 to 20 mA with 0 to 20 mA input:
half of measuring range is 5 bar, which will be converted to 12 mA (middle between 4 mA and 20 mA).
Engineering value = (((10 - 0) / 102.4) * (76.8 - 25.6)) + 0 bar = 5 bar
the RAW_VALUE will be (128digit / 20 mA) * 12 mA = 76.8 digit

For calculating outputsyou take same equation, but calculate RAW_VALUE instead of Engineering Value, as Engineering Value would be given in that case.

I hope I could help you along.
Best regards

 Hello

i have the same probem with à température sensor (-10;150°C)

could you Please make the same calculation ? Tks 

Hello AutoGen_46749,

where is the problem? All necessary equations I have written already above...

However, I will do it detailled for you:

the module has 8 bit resolution, that means 256 possible states. If you drop 20% (as you drop 4 mA from 20 mA which is exactly 20%) you will reduce that to 256 + 0,8 = 204,8 states. As the module uses the Msb (most significant bit) as overrange bit the 256 states are acatually 128 and the 204,8 states are 102,4 states = 102 bit.

If you use 4 to 20 mA at 0 to 20 mA input, you would have a resolution of 150°C - (-10°C) /102,4 states = 160°C / 102,4 which is about 1,56 °C resolution. I am not sure if this is still good for your application. You will have to take into account for your decision.

If you use 4 to 20 mA signal on 0 to 20 mA input, calculation would be:
Engineering Value = (((MAE - MAS) / effective states) * (RAW_VALUE - 20%FS)) + MAS

MAE...........................measuring area end - real
MAS...........................measuring area start - real
effective states...........with your module 102.4 - real
RAW_VALUE.............digitized input value from module - byte
20%FS.......................20% full states, in your case 25.6 - real (20% of 128 states)

So in your example with -10 to 150°C on 4 to 20 mA with 0 to 20 mA input:
half of measuring range is 70°C, which will be converted to 12 mA (middle between 4 mA and 20 mA).
Engineering value = (((150°C - (-10°C)) / 102.4) * (76.8 - 25.6)) + (-10°C) = 70 °C
the RAW_VALUE will be (128digit / 20 mA) * 12 mA = 76.8 digit

So to convert RAW to engineering value in your case you would calculate:

Engineering Value = (((150°C - (-10°C)) / 102.4) * (RAW_VALUE - 25.6)) + (-10°C)

regards

Hello Eisenpferdreiter, 

I have similar problem. I'm using the same module as Mishal11 and I also have 4-20mA sensor. I wanted to use your conversion equation. I'm testing my control cabinet so in place of sensor I use current generator. I set it on 12mA.

My engineering values are 0-100%.

So if I correctly understood your answer my raw value should be something like 76,8.  I have 64.0 that's why the end value is 37% not 50%. Can you look at my function? I'm not sure if I did a good conversion of raw value.  

Hello drWall,

you can use also the follow:

• 20 mA are a value of 27648 inkrements at your input
• 4 mA are a value of 5530 inkrements at your input
• use the follow formula,
  the input value is the INT-value of your analog input

output value [%] = (input value - 5530) × 20 mA / 16 mA × 100

or

output value [%] = (input value - 5530) × 125

Hope this  is a easier way

regards
CaOd