Measuring a small resistance, ~0.001 ohm











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2
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I have a low resistance resistor that I would like to use to measure the current through another circuit, but first I need to accurately determine its resistance.



Its data sheet says it is 0.001 ohm +/- 1%





schematic





simulate this circuit – Schematic created using CircuitLab



A basic setup is to apply a voltage to the resistor, measure the current going through it and the voltage across it, the set up in the diagram above.



I am having some trouble understanding my current and voltage measurements and how the probes may effect the circuit.



The first thing is measuring the voltage. If I touch my probes together I get a reading of 0.02 mV. Is this something I need to offset in my measurement? For instance, if I measure 0.16 mV, should I subtract the 0.02 record the measurement as 0.14 mV?



Secondly the current measurement, when my power supply is OFF, my multimeter reads 3.7 mA, as I vary the current limit it is 3.7 mA less, than what the power supply indicates. For example, if set to a 200 mA current limit, the multimeter reads 197.3 mA.



The idea is to use one multimeter to verify the current to the resistor, as the power supply reading may not necessarily be accurate. However, as I write this I realise my power supply has an OFF button for the voltage output and for the device itself. When the device is OFF, the current reading is 0 mA, when it is ON, but the output is OFF, the reading it 3.7 mA, and when the output is turned ON (with the current limit 200 mA) the current reading is 197.3 mA.



So I reason that the current reading on the multimeter is probably fine as is. It is just when the supply is ON, even if the output is off, there is a small current leaking through.



But I am still not sure about the voltage.





Edit
The diagram I meant was this, with 2 separate meters one for current one for the voltage across the resistor.





schematic





simulate this circuit










share|improve this question




















  • 3




    It is unclear to me if you're doing a proper 4-point measurement, if not measuring like in your schematic can introduce HUGE measurement errors as the resistance you want to measure is MUCH smaller than for example the series resistance of the Current meter. Let alone the resistance of the wires. What you need to do is a 4 point measurement where you create separated connections/wires for the current and the voltage, go read: en.wikipedia.org/wiki/Four-terminal_sensing and: allaboutcircuits.com/textbook/direct-current/chpt-8/…
    – Bimpelrekkie
    18 hours ago








  • 1




    If you can get a low voltage supply or battery to supply a 50 to 100mV with 50 to 100A drop then your offset errors are reduced. But DMM calbration is important.
    – Tony EE rocketscientist
    18 hours ago










  • Check the power rating of the resistor is suitable before putting a large current through it, if that is how you end up measuring the resistance.
    – Andrew Morton
    14 hours ago










  • What's the part number?
    – Nick T
    12 hours ago










  • @AndrewMorton 100 A is probably a bit excessive, but (15 A)² × 1 mΩ = 0.225 W, so I wouldn't worry about the resistor so much as the supply.
    – Nick T
    12 hours ago

















up vote
2
down vote

favorite












I have a low resistance resistor that I would like to use to measure the current through another circuit, but first I need to accurately determine its resistance.



Its data sheet says it is 0.001 ohm +/- 1%





schematic





simulate this circuit – Schematic created using CircuitLab



A basic setup is to apply a voltage to the resistor, measure the current going through it and the voltage across it, the set up in the diagram above.



I am having some trouble understanding my current and voltage measurements and how the probes may effect the circuit.



The first thing is measuring the voltage. If I touch my probes together I get a reading of 0.02 mV. Is this something I need to offset in my measurement? For instance, if I measure 0.16 mV, should I subtract the 0.02 record the measurement as 0.14 mV?



Secondly the current measurement, when my power supply is OFF, my multimeter reads 3.7 mA, as I vary the current limit it is 3.7 mA less, than what the power supply indicates. For example, if set to a 200 mA current limit, the multimeter reads 197.3 mA.



The idea is to use one multimeter to verify the current to the resistor, as the power supply reading may not necessarily be accurate. However, as I write this I realise my power supply has an OFF button for the voltage output and for the device itself. When the device is OFF, the current reading is 0 mA, when it is ON, but the output is OFF, the reading it 3.7 mA, and when the output is turned ON (with the current limit 200 mA) the current reading is 197.3 mA.



So I reason that the current reading on the multimeter is probably fine as is. It is just when the supply is ON, even if the output is off, there is a small current leaking through.



But I am still not sure about the voltage.





Edit
The diagram I meant was this, with 2 separate meters one for current one for the voltage across the resistor.





schematic





simulate this circuit










share|improve this question




















  • 3




    It is unclear to me if you're doing a proper 4-point measurement, if not measuring like in your schematic can introduce HUGE measurement errors as the resistance you want to measure is MUCH smaller than for example the series resistance of the Current meter. Let alone the resistance of the wires. What you need to do is a 4 point measurement where you create separated connections/wires for the current and the voltage, go read: en.wikipedia.org/wiki/Four-terminal_sensing and: allaboutcircuits.com/textbook/direct-current/chpt-8/…
    – Bimpelrekkie
    18 hours ago








  • 1




    If you can get a low voltage supply or battery to supply a 50 to 100mV with 50 to 100A drop then your offset errors are reduced. But DMM calbration is important.
    – Tony EE rocketscientist
    18 hours ago










  • Check the power rating of the resistor is suitable before putting a large current through it, if that is how you end up measuring the resistance.
    – Andrew Morton
    14 hours ago










  • What's the part number?
    – Nick T
    12 hours ago










  • @AndrewMorton 100 A is probably a bit excessive, but (15 A)² × 1 mΩ = 0.225 W, so I wouldn't worry about the resistor so much as the supply.
    – Nick T
    12 hours ago















up vote
2
down vote

favorite









up vote
2
down vote

favorite











I have a low resistance resistor that I would like to use to measure the current through another circuit, but first I need to accurately determine its resistance.



Its data sheet says it is 0.001 ohm +/- 1%





schematic





simulate this circuit – Schematic created using CircuitLab



A basic setup is to apply a voltage to the resistor, measure the current going through it and the voltage across it, the set up in the diagram above.



I am having some trouble understanding my current and voltage measurements and how the probes may effect the circuit.



The first thing is measuring the voltage. If I touch my probes together I get a reading of 0.02 mV. Is this something I need to offset in my measurement? For instance, if I measure 0.16 mV, should I subtract the 0.02 record the measurement as 0.14 mV?



Secondly the current measurement, when my power supply is OFF, my multimeter reads 3.7 mA, as I vary the current limit it is 3.7 mA less, than what the power supply indicates. For example, if set to a 200 mA current limit, the multimeter reads 197.3 mA.



The idea is to use one multimeter to verify the current to the resistor, as the power supply reading may not necessarily be accurate. However, as I write this I realise my power supply has an OFF button for the voltage output and for the device itself. When the device is OFF, the current reading is 0 mA, when it is ON, but the output is OFF, the reading it 3.7 mA, and when the output is turned ON (with the current limit 200 mA) the current reading is 197.3 mA.



So I reason that the current reading on the multimeter is probably fine as is. It is just when the supply is ON, even if the output is off, there is a small current leaking through.



But I am still not sure about the voltage.





Edit
The diagram I meant was this, with 2 separate meters one for current one for the voltage across the resistor.





schematic





simulate this circuit










share|improve this question















I have a low resistance resistor that I would like to use to measure the current through another circuit, but first I need to accurately determine its resistance.



Its data sheet says it is 0.001 ohm +/- 1%





schematic





simulate this circuit – Schematic created using CircuitLab



A basic setup is to apply a voltage to the resistor, measure the current going through it and the voltage across it, the set up in the diagram above.



I am having some trouble understanding my current and voltage measurements and how the probes may effect the circuit.



The first thing is measuring the voltage. If I touch my probes together I get a reading of 0.02 mV. Is this something I need to offset in my measurement? For instance, if I measure 0.16 mV, should I subtract the 0.02 record the measurement as 0.14 mV?



Secondly the current measurement, when my power supply is OFF, my multimeter reads 3.7 mA, as I vary the current limit it is 3.7 mA less, than what the power supply indicates. For example, if set to a 200 mA current limit, the multimeter reads 197.3 mA.



The idea is to use one multimeter to verify the current to the resistor, as the power supply reading may not necessarily be accurate. However, as I write this I realise my power supply has an OFF button for the voltage output and for the device itself. When the device is OFF, the current reading is 0 mA, when it is ON, but the output is OFF, the reading it 3.7 mA, and when the output is turned ON (with the current limit 200 mA) the current reading is 197.3 mA.



So I reason that the current reading on the multimeter is probably fine as is. It is just when the supply is ON, even if the output is off, there is a small current leaking through.



But I am still not sure about the voltage.





Edit
The diagram I meant was this, with 2 separate meters one for current one for the voltage across the resistor.





schematic





simulate this circuit







resistance current-measurement multimeter voltage-measurement






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago









Peter Mortensen

1,58031422




1,58031422










asked 18 hours ago









Dave

737




737








  • 3




    It is unclear to me if you're doing a proper 4-point measurement, if not measuring like in your schematic can introduce HUGE measurement errors as the resistance you want to measure is MUCH smaller than for example the series resistance of the Current meter. Let alone the resistance of the wires. What you need to do is a 4 point measurement where you create separated connections/wires for the current and the voltage, go read: en.wikipedia.org/wiki/Four-terminal_sensing and: allaboutcircuits.com/textbook/direct-current/chpt-8/…
    – Bimpelrekkie
    18 hours ago








  • 1




    If you can get a low voltage supply or battery to supply a 50 to 100mV with 50 to 100A drop then your offset errors are reduced. But DMM calbration is important.
    – Tony EE rocketscientist
    18 hours ago










  • Check the power rating of the resistor is suitable before putting a large current through it, if that is how you end up measuring the resistance.
    – Andrew Morton
    14 hours ago










  • What's the part number?
    – Nick T
    12 hours ago










  • @AndrewMorton 100 A is probably a bit excessive, but (15 A)² × 1 mΩ = 0.225 W, so I wouldn't worry about the resistor so much as the supply.
    – Nick T
    12 hours ago
















  • 3




    It is unclear to me if you're doing a proper 4-point measurement, if not measuring like in your schematic can introduce HUGE measurement errors as the resistance you want to measure is MUCH smaller than for example the series resistance of the Current meter. Let alone the resistance of the wires. What you need to do is a 4 point measurement where you create separated connections/wires for the current and the voltage, go read: en.wikipedia.org/wiki/Four-terminal_sensing and: allaboutcircuits.com/textbook/direct-current/chpt-8/…
    – Bimpelrekkie
    18 hours ago








  • 1




    If you can get a low voltage supply or battery to supply a 50 to 100mV with 50 to 100A drop then your offset errors are reduced. But DMM calbration is important.
    – Tony EE rocketscientist
    18 hours ago










  • Check the power rating of the resistor is suitable before putting a large current through it, if that is how you end up measuring the resistance.
    – Andrew Morton
    14 hours ago










  • What's the part number?
    – Nick T
    12 hours ago










  • @AndrewMorton 100 A is probably a bit excessive, but (15 A)² × 1 mΩ = 0.225 W, so I wouldn't worry about the resistor so much as the supply.
    – Nick T
    12 hours ago










3




3




It is unclear to me if you're doing a proper 4-point measurement, if not measuring like in your schematic can introduce HUGE measurement errors as the resistance you want to measure is MUCH smaller than for example the series resistance of the Current meter. Let alone the resistance of the wires. What you need to do is a 4 point measurement where you create separated connections/wires for the current and the voltage, go read: en.wikipedia.org/wiki/Four-terminal_sensing and: allaboutcircuits.com/textbook/direct-current/chpt-8/…
– Bimpelrekkie
18 hours ago






It is unclear to me if you're doing a proper 4-point measurement, if not measuring like in your schematic can introduce HUGE measurement errors as the resistance you want to measure is MUCH smaller than for example the series resistance of the Current meter. Let alone the resistance of the wires. What you need to do is a 4 point measurement where you create separated connections/wires for the current and the voltage, go read: en.wikipedia.org/wiki/Four-terminal_sensing and: allaboutcircuits.com/textbook/direct-current/chpt-8/…
– Bimpelrekkie
18 hours ago






1




1




If you can get a low voltage supply or battery to supply a 50 to 100mV with 50 to 100A drop then your offset errors are reduced. But DMM calbration is important.
– Tony EE rocketscientist
18 hours ago




If you can get a low voltage supply or battery to supply a 50 to 100mV with 50 to 100A drop then your offset errors are reduced. But DMM calbration is important.
– Tony EE rocketscientist
18 hours ago












Check the power rating of the resistor is suitable before putting a large current through it, if that is how you end up measuring the resistance.
– Andrew Morton
14 hours ago




Check the power rating of the resistor is suitable before putting a large current through it, if that is how you end up measuring the resistance.
– Andrew Morton
14 hours ago












What's the part number?
– Nick T
12 hours ago




What's the part number?
– Nick T
12 hours ago












@AndrewMorton 100 A is probably a bit excessive, but (15 A)² × 1 mΩ = 0.225 W, so I wouldn't worry about the resistor so much as the supply.
– Nick T
12 hours ago






@AndrewMorton 100 A is probably a bit excessive, but (15 A)² × 1 mΩ = 0.225 W, so I wouldn't worry about the resistor so much as the supply.
– Nick T
12 hours ago












2 Answers
2






active

oldest

votes

















up vote
8
down vote













You can't "accurately determine" such a low resistance using household DMMs, which are likely completely lack of any calibration whatsoever. Accuracy class of your household DMMs is way worse than 1%, especially on the sensitive low-voltage range, and all your results will be suspect and useless. Plus you need to control temperature of contacts very carefully and avoid thermal gradients, otherwise thermal EMF will screw all your results.



Instead, you should trust the datasheets (if the manufacturer is reputable), and follow ALL recommendations on how to mount/connect/use this device, including PCB pad shapes/solder technique and configuration of voltage sense traces. A resistor is much simpler device than digital multimeter, so it is easier to believe that the class accuracy (1%) is better sustained in resistor than in multimeter. If anything, you should consider calibrating your DMMs using this resistor, not the other way around.






share|improve this answer




























    up vote
    6
    down vote













    I agree with both of your statements, subtract the 20 µV (or add it, depending on polarity), and use the meter measurement of current directly. You have no way of correcting for a span error on either meter.



    When measuring such a low resistance, it's vital to use a 4-wire technique. You will also want to use a relatively high current to minimize errors.



    To verify 1 mΩ to within, say, ±0.1% (since the resistor is allegedly ±1%) requires a measurement to within 1 µΩ, meaning you will have to be very careful with the connections. Thermal EMFs can also be a problem, which you can partially mitigate by reversing the polarity and making another measurement.






    share|improve this answer























    • Is the Wheatstone bridge method also applicable? Just out of interest.
      – Solar Mike
      17 hours ago






    • 2




      @SolarMike It's possible to incorporate a 4 terminal measurement into a wheatstone bridge arrangement, but not very useful. A Wheatstone bridge is a solution to a different problem.
      – Neil_UK
      17 hours ago













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    8
    down vote













    You can't "accurately determine" such a low resistance using household DMMs, which are likely completely lack of any calibration whatsoever. Accuracy class of your household DMMs is way worse than 1%, especially on the sensitive low-voltage range, and all your results will be suspect and useless. Plus you need to control temperature of contacts very carefully and avoid thermal gradients, otherwise thermal EMF will screw all your results.



    Instead, you should trust the datasheets (if the manufacturer is reputable), and follow ALL recommendations on how to mount/connect/use this device, including PCB pad shapes/solder technique and configuration of voltage sense traces. A resistor is much simpler device than digital multimeter, so it is easier to believe that the class accuracy (1%) is better sustained in resistor than in multimeter. If anything, you should consider calibrating your DMMs using this resistor, not the other way around.






    share|improve this answer

























      up vote
      8
      down vote













      You can't "accurately determine" such a low resistance using household DMMs, which are likely completely lack of any calibration whatsoever. Accuracy class of your household DMMs is way worse than 1%, especially on the sensitive low-voltage range, and all your results will be suspect and useless. Plus you need to control temperature of contacts very carefully and avoid thermal gradients, otherwise thermal EMF will screw all your results.



      Instead, you should trust the datasheets (if the manufacturer is reputable), and follow ALL recommendations on how to mount/connect/use this device, including PCB pad shapes/solder technique and configuration of voltage sense traces. A resistor is much simpler device than digital multimeter, so it is easier to believe that the class accuracy (1%) is better sustained in resistor than in multimeter. If anything, you should consider calibrating your DMMs using this resistor, not the other way around.






      share|improve this answer























        up vote
        8
        down vote










        up vote
        8
        down vote









        You can't "accurately determine" such a low resistance using household DMMs, which are likely completely lack of any calibration whatsoever. Accuracy class of your household DMMs is way worse than 1%, especially on the sensitive low-voltage range, and all your results will be suspect and useless. Plus you need to control temperature of contacts very carefully and avoid thermal gradients, otherwise thermal EMF will screw all your results.



        Instead, you should trust the datasheets (if the manufacturer is reputable), and follow ALL recommendations on how to mount/connect/use this device, including PCB pad shapes/solder technique and configuration of voltage sense traces. A resistor is much simpler device than digital multimeter, so it is easier to believe that the class accuracy (1%) is better sustained in resistor than in multimeter. If anything, you should consider calibrating your DMMs using this resistor, not the other way around.






        share|improve this answer












        You can't "accurately determine" such a low resistance using household DMMs, which are likely completely lack of any calibration whatsoever. Accuracy class of your household DMMs is way worse than 1%, especially on the sensitive low-voltage range, and all your results will be suspect and useless. Plus you need to control temperature of contacts very carefully and avoid thermal gradients, otherwise thermal EMF will screw all your results.



        Instead, you should trust the datasheets (if the manufacturer is reputable), and follow ALL recommendations on how to mount/connect/use this device, including PCB pad shapes/solder technique and configuration of voltage sense traces. A resistor is much simpler device than digital multimeter, so it is easier to believe that the class accuracy (1%) is better sustained in resistor than in multimeter. If anything, you should consider calibrating your DMMs using this resistor, not the other way around.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 15 hours ago









        Ale..chenski

        26.1k11861




        26.1k11861
























            up vote
            6
            down vote













            I agree with both of your statements, subtract the 20 µV (or add it, depending on polarity), and use the meter measurement of current directly. You have no way of correcting for a span error on either meter.



            When measuring such a low resistance, it's vital to use a 4-wire technique. You will also want to use a relatively high current to minimize errors.



            To verify 1 mΩ to within, say, ±0.1% (since the resistor is allegedly ±1%) requires a measurement to within 1 µΩ, meaning you will have to be very careful with the connections. Thermal EMFs can also be a problem, which you can partially mitigate by reversing the polarity and making another measurement.






            share|improve this answer























            • Is the Wheatstone bridge method also applicable? Just out of interest.
              – Solar Mike
              17 hours ago






            • 2




              @SolarMike It's possible to incorporate a 4 terminal measurement into a wheatstone bridge arrangement, but not very useful. A Wheatstone bridge is a solution to a different problem.
              – Neil_UK
              17 hours ago

















            up vote
            6
            down vote













            I agree with both of your statements, subtract the 20 µV (or add it, depending on polarity), and use the meter measurement of current directly. You have no way of correcting for a span error on either meter.



            When measuring such a low resistance, it's vital to use a 4-wire technique. You will also want to use a relatively high current to minimize errors.



            To verify 1 mΩ to within, say, ±0.1% (since the resistor is allegedly ±1%) requires a measurement to within 1 µΩ, meaning you will have to be very careful with the connections. Thermal EMFs can also be a problem, which you can partially mitigate by reversing the polarity and making another measurement.






            share|improve this answer























            • Is the Wheatstone bridge method also applicable? Just out of interest.
              – Solar Mike
              17 hours ago






            • 2




              @SolarMike It's possible to incorporate a 4 terminal measurement into a wheatstone bridge arrangement, but not very useful. A Wheatstone bridge is a solution to a different problem.
              – Neil_UK
              17 hours ago















            up vote
            6
            down vote










            up vote
            6
            down vote









            I agree with both of your statements, subtract the 20 µV (or add it, depending on polarity), and use the meter measurement of current directly. You have no way of correcting for a span error on either meter.



            When measuring such a low resistance, it's vital to use a 4-wire technique. You will also want to use a relatively high current to minimize errors.



            To verify 1 mΩ to within, say, ±0.1% (since the resistor is allegedly ±1%) requires a measurement to within 1 µΩ, meaning you will have to be very careful with the connections. Thermal EMFs can also be a problem, which you can partially mitigate by reversing the polarity and making another measurement.






            share|improve this answer














            I agree with both of your statements, subtract the 20 µV (or add it, depending on polarity), and use the meter measurement of current directly. You have no way of correcting for a span error on either meter.



            When measuring such a low resistance, it's vital to use a 4-wire technique. You will also want to use a relatively high current to minimize errors.



            To verify 1 mΩ to within, say, ±0.1% (since the resistor is allegedly ±1%) requires a measurement to within 1 µΩ, meaning you will have to be very careful with the connections. Thermal EMFs can also be a problem, which you can partially mitigate by reversing the polarity and making another measurement.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 12 hours ago









            Nick T

            11.4k23968




            11.4k23968










            answered 18 hours ago









            Spehro Pefhany

            201k4146400




            201k4146400












            • Is the Wheatstone bridge method also applicable? Just out of interest.
              – Solar Mike
              17 hours ago






            • 2




              @SolarMike It's possible to incorporate a 4 terminal measurement into a wheatstone bridge arrangement, but not very useful. A Wheatstone bridge is a solution to a different problem.
              – Neil_UK
              17 hours ago




















            • Is the Wheatstone bridge method also applicable? Just out of interest.
              – Solar Mike
              17 hours ago






            • 2




              @SolarMike It's possible to incorporate a 4 terminal measurement into a wheatstone bridge arrangement, but not very useful. A Wheatstone bridge is a solution to a different problem.
              – Neil_UK
              17 hours ago


















            Is the Wheatstone bridge method also applicable? Just out of interest.
            – Solar Mike
            17 hours ago




            Is the Wheatstone bridge method also applicable? Just out of interest.
            – Solar Mike
            17 hours ago




            2




            2




            @SolarMike It's possible to incorporate a 4 terminal measurement into a wheatstone bridge arrangement, but not very useful. A Wheatstone bridge is a solution to a different problem.
            – Neil_UK
            17 hours ago






            @SolarMike It's possible to incorporate a 4 terminal measurement into a wheatstone bridge arrangement, but not very useful. A Wheatstone bridge is a solution to a different problem.
            – Neil_UK
            17 hours ago




















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