In previous PART 2 the ratiometric question was discussed as follows:
THE RATIOMETRIC PROBLEM
Because of the ratiometric nature of the sensor, variations in the supply voltage of the sensor introduce a reading error as follows.
Supply...........Offset...........Error............Corrected & Linearised Error
V..................mV..............Litres............Litres
5,5................220..............7,74..............7,7
5,4................216..............6,02..............6,2
5,3................213..............4,73..............4,6
5,2................209..............3,01..............3,1
5,1................206..............1,72..............1,6
5..................202...............0,00..............0,1
4,9................199..............-1,29............-1,4
4,8................195..............-3,01............-3,0
4,7................192..............-4,3.............-4,5
4,6................188..............-6,02............-6,0
4,5................184..............-7,74............-7,5
Considering that the USB + 5V can vary in the range 4.75 to 5.25 V, We could have a random reading error in excess of +/- 3 litres, which might not be acceptable in some instances. The solution is to use a precision + 5 V supply to feed the sensor, which needs approx. 6 mA. To do this the jumper between pins 11 & 12 of M1 (DrDAQ Buffer) must be removed and the stable + 5 V applied to Pin 12.
Since the ratiometric error depends on supply voltage variations, IT IS INDEED POSSIBLE TO CORRECT OR ELIMINATE this error by measuring the supply voltage variation and generating a correction signal using the "Calculated Parameter" channels of Picolog. This is a GENERAL PROCEDURE, useful for any ratiometric device, made possible by the facilities offered by Picolog.
Considering the experimental data above, with supply voltage decreasing from 5 to 4.5V, offset (e.g. sensor output) decreases by 18 mV from 202 to 184 mV and going from 5 to 5.5 V offset increases by 18 mV from 202 to 220 mV.
Therefore the correction is -18 mV for increasing supply voltage and + 18 mV for decreasing supply voltage, hence:
Correction factor is:
Factor = +/-(18/500) = 0,0360
And correction voltage to be algebraically added to sensor output is: ((5*1000)-(Vcc*1000))*0,0360 [mV]
EXAMPLE: Vcc=5.23V Correction voltage = (5000-5230)*0,0360 = -8.28 mV
Here below three screens showing corrected tank level displays with varying sensor supply voltage. With the correction the ratiometric error becomes so low that the precison PSU is no longer necessary.
- High supply voltage
- Nominal 5 V supply voltage
- Low supply voltage
