# How to acquire current values in data acquisiton accurately when using a shunt resistor in a current loop?

Here is the setup I use.

Since the DAQ system is only able to read voltages first I record the voltages by a shunt resistor. But I need the "current values" at the end. Transducer will make the current vary between 4 to 20 mA.

If the resistance is constant with respect to current one could just divide the voltage readings to resistance and obtain the current in the loop.

But I observed that the resistance is not the same for different currents. For example for a constant applied 20mA current the resistance is 248.74ohm and for 4 mA 248.27ohm. Is it better in that case to obtain a calibration expression (current to voltage) and calculate the currents in that way instead of taking the resistance a constant?

• How sensitive is your measurement to these errors? What are your other system errors? You're talking about a 0.2% error between 4 and 20mA - that is already extremely good for an uncalibrated system. Commented Nov 11, 2013 at 10:50
• i obtain a current voltage line. my question is should i use this line to calculate the currents from voltage readings or should i choose a constant resistance value? which is more accurate? Commented Nov 11, 2013 at 10:57

At 20 mA the resistor will be dissipating more power than at 4 mA. Say the resistor is 250 ohm (as per your previous question How to achieve common ground for a single ended input?), the power will be 100 mW at 20mA whereas at 4 mA the power will only be 4 mW.

If your resistor warms up 10 ºC and is a 25ppm/ºC type the resistance change will be from 250 ohm to 250.06 ohms or down to 249.94 ohms.

If your resistor warms up 20 ºC and it is a 50ppm/ºC type the resistance will change from 250 ohm to 250.25 ohms or down to 249.75 ohms.

This seems around the same order as what you are measuring. I calculate a change in resistance of 0.47 ohms - could this be due to self-heating?

• I obtain that line(current input - voltage output) with 2 measurements i.e. for 4mA and 20mA constant current. I obtain a line because I dont have another measurement point. My constant current source outputs only 4 and 20mA. I obtain the calibration expression (current-voltage) as linear but Im not sure it should be linear in real since I can only measure at 2 steps. Commented Nov 11, 2013 at 11:29
• Precisely what is the make and model of your resistor - let's get to the heart of the problem. Did yesterday's advice on the previous question help? Commented Nov 11, 2013 at 11:32
• the point is the resistor has no color code. it is totally black. i have no data sheet. thats why i try this method. some engineer before be measured it as 249.5 something 2 years ago. but they want me to use this resistance. so thats why im measuring again Commented Nov 11, 2013 at 11:37
• The error you get is likely to be this - without a spec you are in the dark. Get a resistor like this one: uk.farnell.com/vishay-foil-resistors-vpg/1625250r000t9r/… it is 0.01% and has a temperature coefficient of 0.2ppm per ºC or maybe something a bit less expensive. The error you are getting can easily be down to the resistor or maybe the offset voltage drift of your amplifier. Did the info I gave yesterday help? If so please consider accepting the answer or upvoting it. This site favours those who show some form of appreciation for useful advice. Commented Nov 11, 2013 at 11:47
• it didnt help much because im not getting clear answers to my questions. i have to use this resistor. thanks for ideas anyway. my question was is that better to calibrate versus current or take resistance as a constant. im asking which is more accurate. i cant change the setup unforunately not in my control Commented Nov 11, 2013 at 11:55