I had a company design a custom terminal board for me. It does numerous things, but what I'm fighting with is the analog section. Standard 0-20ma inputs and , Thru a Precision 250 ohm resistor to convert to 0-5 volts for a ADC Card. ADC Card is a 16 channels single ended 12 bit analog.
Problem is, that no matter what the source is, the voltage always seems to be approx. 50mv lower than actual. I have an Omega MA Simulator in order to test this. If I send 4 ma then the ADC Card reads 0.950 volts and 20 ma =
4.950 volts. It appears to be pretty linear. Not sure what is causing it, because I have not seen this on my other boards designed with 250 ohm resistors.Anyway, since it's linear, and we built abunch of these boards, I need a correction factor in my scaling in order to make the engineering units correct. I would like to scale it on the voltage side since we know the offset is the same on all the channels and not offset the scaled value.
I have the following for each channel: Factor and Offset which works normally (When I don't need an offset)
Vin * Factor + Offset
I can't add the millivolt difference to Offset because offset is adding to the scaled value. But I could add it to Vin like this.... CrFactor = 0.50 ((Vin + CrFactor) * Factor) + Offset
The above works for Sensors 4-20ma, because we are actually scaling 1 - 5 volts or 0.950 to 4.950 and by adding the correction factor it brings it to 1.00 to 5.00
Follow?
Okay, But now, we actually have a sensor that is 0-20ma or 0-5volts, the above formula and correction factor does not work because Zero volts is already zero volts (no more voltage drop) so the actual voltage shown is
0 to 4.950Using the above correction factor, we end up with 0.050 to 5.000 which is incorrect and 0 to 4.950 is also incorrect. Would it be 0 - 4.900 ?
Not sure, Anyone have an idea how I can handle this voltage drop for all my channels? My software "could" look at MinVoltage and MaxVoltage and determine if the scalling starts at Zero or Negative and use a different formula for each case?
Richard
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