I apply a constant 4ma current I to a resistor R and I measure a voltage V across.
so the resistor \$ R = \dfrac{V}{I}\$.

But when I apply \$ I_2=20ma\$ to the same resistor R and I obtain the voltage \$V_2\$ then the resistor \$R= \dfrac{V_2}{I_2}\$ I expect the same.

but in first case for 4ma, I obtain 248.2 ohm and in 20ma case I obtain 248.7 ohm.

I can only measure the voltages with a daq box. If I have a device with a current loop output how can I then translate the voltages to currents, since I find different resistances for different currents? is there a way to obtain accuracy error? is there a standard for that?

  • \$\begingroup\$ The difference between the two resistance values is roughly 0.2%. What is the precision of your current and voltage measuring instruments? A measurement error of a few tens of µA or 10 mV is enough to make the difference you've observed. I wouldn't consider those two values “different” unless I was certain of my instrumentation. \$\endgroup\$ – Emmet Jul 29 '14 at 20:38
  • \$\begingroup\$ Using a 4-wire Kelvin force/sense connection to the shunt resistor, correct? Otherwise the voltage drop you are measuring would include the shunt resistor itself, plus its leads and wires. electronics.stackexchange.com/questions/77220/… \$\endgroup\$ – MarkU Jul 30 '14 at 1:54
  • \$\begingroup\$ applied currents are constant by a current supply. where will the voltage drop? daq has extremely high input impedance? all the current will pass through the resistor since it cant flow anywhere else. so i dont think u re right. \$\endgroup\$ – user16307 Jul 30 '14 at 3:03

Perhaps check the temperature coefficient of the resistor. With 20mA, the power dissipation in the resistor is 25x larger than with 4mA (power dissipation is proportional to I2). The resistor heats up as you increase your current. As it heats up, its resistance changes. Incorporating the temperature coefficient would improve your accuracy. Another option is to use a much smaller resistor so that the temperature change over the current range is small.

  • \$\begingroup\$ ok but even i know the temperature coefficient it wont work for me. i cannot measure the resistor temperature. i can only measure the voltage across. i there a relation between power dissipation ratio and resistance? \$\endgroup\$ – user16307 Jul 29 '14 at 18:22
  • \$\begingroup\$ There is a relationship between the temperature and the resistance and there is a relationship between the power dissipation and the temperature. The latter will depend on a number of environmental factors (any heat-sinking, airflow, environmental temperature, etc.). You could try to calibrate out this error by taking measurements, you could buy a resistor with a lower temperature coefficient, or you could try to heatsink the resistor effectively so that the temperature is tightly coupled to the environmental temperature (assuming that is well controlled...) \$\endgroup\$ – user49628 Jul 29 '14 at 18:27
  • \$\begingroup\$ A resistor used to measure the current in a loop is called a shunt resistor. You can reduce the shunt resistor value to reduce the effect of temperature variation. Also, you can use a resitor with a smaller temperature coefficient. \$\endgroup\$ – 2over0 Jul 29 '14 at 18:35
  • \$\begingroup\$ He can use low temperature coefficient resistors (low PPM) to reduce this error. \$\endgroup\$ – Kamil Jul 29 '14 at 18:35
  • \$\begingroup\$ @user16307 - You write "ok but even i know the temperature coefficient it wont work for me. i cannot measure the resistor temperature" but that's not true. You can put your finger on it. 20 mA through 248 ohms is about 100 mW. That will get a 1/8 watt resistor quite warm. Your resistance change is compatible with a 50 ppm / deg C resistor with a temperature rise of 40 C. Starting at ambient that will be ~ 65 C, and that's about as hot as you can stand to touch. So touch your resistor with 20 mA through it. How hot does it get? \$\endgroup\$ – WhatRoughBeast Jul 29 '14 at 19:24
  1. Use a lower resistance value. A smaller resistor will heat itself less for the same current. Values below 1 ohm are often used for current measurement applications.

  2. Use a resistor with a lower TCR. Certain types of resistors (for example, wirewound types) have much lower TCR than others.

Realize that there are errors in every measurement. You need to quantify exactly how accurate your measurement needs to be. Then you can choose a resistor (or other solution) that is adequate for your application.

The difference between 248.2 and 248.7 ohms is only 0.2%. In many applications it would be acceptable to just assume the resistor value is 248.5 ohms, (or even just assume the nearest standard value of 249 ohms) and say the errors are less than 1%.

Is there a way to obtain accuracy error?

Apply a known current using an exceptionally accurate current source (or use your existing source but measure the current with a highly-accurate meter in series with your device). Measure the current using your device. Compare the two.

  • \$\begingroup\$ isnt there a way to calibrate a resistor Voltage versus Current? i mean if i make a regression line for 2 points i.e. at 4ma and 20ma. cant i use the slope of it? \$\endgroup\$ – user16307 Jul 29 '14 at 18:35
  • \$\begingroup\$ Yes, you could do that. But the calibration is likely to be invalid if the ambient temperature changes, airflow over the resistor changes, etc. \$\endgroup\$ – The Photon Jul 29 '14 at 18:38
  • 2
    \$\begingroup\$ Do you know your 4 mA source was exactly 4 mA, +/- less than 0.008 mA? \$\endgroup\$ – The Photon Jul 29 '14 at 18:39
  • \$\begingroup\$ it is like 3.99ma for low and 19.95ma for high output. hand held constant current supply has 2 cc outputs around 4 and 20ma. this device has calibration certificate. \$\endgroup\$ – user16307 Jul 29 '14 at 18:49
  • 1
    \$\begingroup\$ Do you know the difference between precision and accuracy, the way these terms are used in metrology? \$\endgroup\$ – The Photon Jul 29 '14 at 18:54

Are you sure what you're seeing is because of a change in temperature? Temperature coefficients are typically tens of PARTS PER MILLION per degree. If you are using a carbon film resistor with a coefficient of 500ppm/degC, 100 degrees of change would cause a 5% change in resistance.

Let's start with the current you're producing. How do you KNOW that you're providing 4.000 milliamps and 20.000 milliamps?? Now, how are you measuring the voltage? What's the resolution of your DAC?

  • \$\begingroup\$ He only saw a 0.2% change in the resistance. A 4 degree change in the resistor temperature seems pretty reasonable. \$\endgroup\$ – The Photon Jul 29 '14 at 18:56
  • \$\begingroup\$ we have handheld constant current supply. it has certificate and it can output constant 3.99ma and 19.95ma according to their certificate. the resistor is around 250 ohm. and the daq is 16 bit. \$\endgroup\$ – user16307 Jul 29 '14 at 18:56
  • \$\begingroup\$ @user16307, does the certificate specify the error limits? If it just says 3.99 mA and not "3.990 mA" I wouldn't expect it to be any better than +/- 0.01 mA. \$\endgroup\$ – The Photon Jul 29 '14 at 18:58
  • \$\begingroup\$ so are u suspecting that the current supply has error thats why i find different results? ok it might be but how can i work around this issue for different voltages when translating to current. is calibration logical? and if it is, i need to obtain the calibration at different ambient temperatures. summer winter ect. \$\endgroup\$ – user16307 Jul 29 '14 at 19:01
  • \$\begingroup\$ @user16307, The first thing you need to do is quantify what amount of error is acceptable. \$\endgroup\$ – The Photon Jul 29 '14 at 19:06

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