krea123 Posted February 23, 2015 Report Share Posted February 23, 2015 Hi i m taking two voltage analog input and linearize them so i can read tempreature and purity value But the two value dont have enough sensitivity , input voltage is changing between the rate of 0-1 volt but the real engineering unit is 23.45 Celcius how can i fix this issue ? (i use the sample linearize program of the unitronics U90 software) Link to comment Share on other sites More sharing options...
krea123 Posted February 26, 2015 Author Report Share Posted February 26, 2015 i realized that my linearization code doesnt work for this issue i read 445 mA as an input(MI 1) but it couldnt be converted to MI 11 SI 84 value isnt changing (0) and SI 85 is always 20. how can i see the changing of the MI 11 value ? (i attached the screenshot of my linearization code ) Link to comment Share on other sites More sharing options...
krea123 Posted February 26, 2015 Author Report Share Posted February 26, 2015 i solved the two problem i want to take data from aquis 500 cr transmitter to the unitronics then i can export the data to excel now i am struggling to linearize the MI 1 value as seen in the code it seems that the sample linearization code doesnt work my read values from the MI 1(4-20mA) and desired values are listed (left MI 1- right desired): what function or mathematical calculation should i use for MI 1 in order to obtain the desired value ? 485(mA) - 6.81(mikrosiemens/cm) 484 - 6.80 624 - 10.2 615 - 10.0 612 - 9.90 606 - 9.76 600 - 9.61 550 - 8.40 483 - 6.76 480 - 6.70 (i connect the -/ + terminals of analog output aquis 500 cr(transmitter) to the COMMON/AN 1 terminal of unitronic jz 20 r 16 ) Link to comment Share on other sites More sharing options...
cantcliff Posted February 27, 2015 Report Share Posted February 27, 2015 Are you using the linearization code block? If not, it may simplify things for you. Referenced from page 120 of the Ladder Programming manual from Visilogic What you need to know for it: Bit rating of your analog in, 12 bit versus 10 bit, you MUST use the correct bit rating reference or your linearized value will come out wrong. 10 bit Use: X - Analog input MI X1 - 204 X2 - 1023 Y1 - zero point you expect at 4ma Y2 - Max value you expect at 20ma Y - Linearized Value Memory Integer or Long 12 bit Use: X - Analog input MI X1 - 819 X2 - 4095 Y1 - zero point you expect at 4ma Y2 - Max value you expect at 20ma Y - Linearized value Memory Integer or Long Example: 0-150 pressure sensor on a 12bit sensor with two decimal places X - MI1 X1 - 819 X2 - 4095 Y1 - 0 Y2 - 15000 (150.00 represented as an integer) Y - MI2 Link to comment Share on other sites More sharing options...
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