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Measuring 24 V with an accuracy of .02 V


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#1 Joey2

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Posted 24 October 2013 - 02:18 PM

I am using a LabJack U12 to measure a DC voltage of approximately 24 V versus ground. I do this by reading ~1/3 of the voltage (~8 V, resistor divider) using single ended analog input. The absolute accuracy for this reading is specified as .2% of full scale, therefore .02 V so that the absolute accuracy of the 24 V measurement is ~.06V (= 24 V +- 0.06V). I would like to measure this 24 V value more accurately and I am imagining subtracting a fixed voltage of about 20 V and measuring the difference of about 4 V within an accuracy of .02 V which would then be the absolute accuracy of the 24 V measurement (= 24 V +- .02 V) which is satisfactory for me.

Can you suggest a  circuit would produce a very stable 20 V that I could subtract from my 24 V signal?

Is there a better way for me to get the accuracy I need?
Regards,
Joe



#2 LabJack Support

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Posted 24 October 2013 - 03:11 PM

Full-span might be a clearer term than full-scale.  If you are using the +/-10V range, full-span is 20V, so the accuracy spec is 0.002*20 => +/-0.04 V typical and 0.004*20 => +/-0.08 V max.

 
Nonetheless, since you are adding a resistive voltage divider out front, you are going to need to do a calibration anyway, so the absolute accuracy does not matter.  You just need a DMM or some other reference meter than can measure the 24 volts to your desired accuracy, and use that to take a couple points and come up with an equation relating voltage at the DUT (device under test, ~24 volts) to voltage reported in software by the U12.
 
On the +/-10V range, the resolution of the 12-bit U12 is 20/4096 => 5mV.  If your reference meter is accurate enough, your calibration should allow you to get an accuracy close to the resolution of 5mV.  That is referred to the U12 inputs, so on the other side of your /3 divider that would equate to 15mV on the 24V signal.  This will work best if you tailor your calibration to the range of interest, so perhaps get a cal point at 23V and another at 25V.
 
If you want to try something fancier, change your divider to something like /8, so that 20-28V is attenuated to 2.5 to 3.5V.  Connect that to AIN0, jumper CAL to AIN1, and do a differential reading of AIN0-AIN1 on the +/-1V range. CAL is a 2.5V reference, so the differential reading will swing from 0.0 to 1.0.  Calibrate the relationship of the 20-28V signal to the 0-1V signal.  The resolution on the +/-1V range is 0.5mV, so other the other side of the /8 divider you are looking at 4mV, which is an improvement over the 15mV above.


#3 Joey2

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Posted 25 October 2013 - 02:40 PM

Thanks for the reply. Do you have an estimate of how stable the calibration will be? That is, how often will I have to recalibrate?

Regards,

Joe



#4 LabJack Support

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Posted 25 October 2013 - 02:49 PM

I don't have a reason to particularly suspect instability over time.  You are only calibrating to a 12-bit level (244 ppm), which is not very extreme.

 

My guess is that the biggest source of instability would be temperature change affecting the resistors you are using in the voltage divider.  If you keep the divider and the U12 at room temperature, this should not be a noticeable problem.

 

One other suggestion would be to connect the U12 to a USB hub that has it's own power supply (self-powered hub).  This will provide the U12 a more stable power supply than if you connect directly to the host or to a bus-powered hub.



#5 Joey2

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Posted 26 October 2013 - 11:15 AM

Thanks again.

Joe




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