Forenote: I'm not an EE, I'm a Controls Engineer.

So if need to measure a resistance value of 240Ω ±2%. I need to figure out what voltage and current range I would need to measure to within 4.8Ω (= 240Ω * 2% I assume I did that right).

So for example (not actual numbers just examples), if my voltage on the line being measure is 9V and voltage input measures 0-10V ±1% of full scale and my current input measures 4 to 20mA ±1% of full scale. How would I calculate my resistance measurement accuracy?

ETA: Side Note I need to ultimately use a PLC to measure this.

ETA2: I added a diagram of the


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ It would help to know accuracy of the current source, in addition to the (example) 1% accuracy of the 0-10V voltage measurement. Is accuracy % full-scale? or is accuracy % of the reading? \$\endgroup\$ – glen_geek Feb 10 '17 at 16:15
  • \$\begingroup\$ Sorry added both. I forgot to add current accuracy. \$\endgroup\$ – TheColonel26 Feb 10 '17 at 16:20
  • \$\begingroup\$ What has "4 to 20mA" got to do with this question - are you trying to feed some signal on a current loop? Also 240 ohms on 9 volts is 37.5 mA = what is the significance of "9V at 10mA"? \$\endgroup\$ – Andy aka Feb 10 '17 at 16:22
  • \$\begingroup\$ It was proposed (not by me) to measure current directly with a 4-20mA analog input card on a PLC. \$\endgroup\$ – TheColonel26 Feb 10 '17 at 16:27
  • \$\begingroup\$ 10mA if we placed a second resistor in the circuit to bring load down down to between 4-20mA so it can be measured. I think that any way I slice it We can't use the 4-20mA PLC input. Regardless I'll remove it as it is irreverent to my question. I will change my example. \$\endgroup\$ – TheColonel26 Feb 10 '17 at 16:35

I need to figure out what voltage and current range I would need to measure to within 4.8Ω (= 240Ω * 2% I assume I did that right).

it will depends on how you measure it.

How would I calculate my resistance measurement accuracy?

Again, that's a function of how you measure it.

if you have access to high precision resistors, a ratio-metric approach would be good. it doesn't require a high precision current source.

The simplest ratio-metric approach would be an adc that takes differential reference inputs. Put your high precision resistor across the reference voltage inputs and you are ready to go.

if you have to adc the voltage across your resistor, you want its minimum resolution to be at most 2% of the voltage drop. For example, if your current is 1ma, and 4.8R -> 5mv. With a 5v Vref to a 10bit adc, that works, barely.

Typically the last 2-3 LSB of an adc are fairly noisy. So in that case, you want the resolution to be 15mv (= 3 * 5mv) on a 4.8ohm resistance. that least to a minimum current of 15v/4.8R = 3mv.

You will then do the same calculation to make sure that your adc doesn't max out at the top end of your adc. Or you potentially need to switch Vref dynamically or switch to a higher resolution adc.

but the basic thought is there.

the questions have nothing to do with 4-wire measurements.

  • \$\begingroup\$ I added a diagram to explain better what I am trying to measure. \$\endgroup\$ – TheColonel26 Feb 10 '17 at 16:57

I take it you are trying to measure the burden present in an active (but idle) 4-20 mA control loop. You need to measure the voltage across the resistor and the current through it. If you want the calculation to be to a specified accuracy then you need two calibrated meters, the current one is placed inline and the other measures the voltage across the resistor.

If the accuracy you need is 2% and you use professional level calibrated meters at around 0.5% then simply use the meter values (R=V/I of course). If you use uncalibrated meters, all bets are off.

ADD: Since you are using a PLC.

  1. Set PLC to 20 mA
  2. Press switch 2
  3. Measure voltage with a calibrated meter

R= V/0.020

Accuracy depends on the PLC current accuracy and your Voltmeter. This IS effectively a 4 wire Kelvin measurement since by placing the voltmeter directly at the load resistor you measure only the voltage across the burden resistor, thereby getting rid of any cable voltage drop.

If you have to measure all three resistors (940 Ohms) then you can adjust the current to 10 mA, then you only need 9.4 volts in the current loop.

enter image description here

  • \$\begingroup\$ added a diagram of what I am trying to measure. \$\endgroup\$ – TheColonel26 Feb 10 '17 at 16:57
  • \$\begingroup\$ This is not using the Kelvin method, we are required you use the kelvin method. Also with a 1540 ohm load the voltage drop would be 30V, and at 630 ohms voltage drop would be 12V my input is only 10V. \$\endgroup\$ – TheColonel26 Feb 13 '17 at 12:42
  • \$\begingroup\$ You completely miss the point.....the PLC METHOD IS EFFECTIVELY a 4 wire Kelvin method. You measure the current at the source (PLC sets the current) and measure the voltage at the load (resistor) which is what I said to do. This might help you: allaboutcircuits.com/textbook/direct-current/chpt-8/… \$\endgroup\$ – Jack Creasey Feb 13 '17 at 16:39
  • \$\begingroup\$ @TheColonel26 PLC can use loop voltages up to 48 V commonly, so there is little problem testing any of the resistors the OP suggested. Though the Op said he only wanted to measure the 240 Ohm resistor. \$\endgroup\$ – Jack Creasey Feb 13 '17 at 16:45

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