Measuring a small resistance, ~0.001 ohm
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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%
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.
simulate this circuit
resistance current-measurement multimeter voltage-measurement
|
show 2 more comments
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%
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.
simulate this circuit
resistance current-measurement multimeter voltage-measurement
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
|
show 2 more comments
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%
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.
simulate this circuit
resistance current-measurement multimeter voltage-measurement
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%
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.
simulate this circuit
resistance current-measurement multimeter voltage-measurement
resistance current-measurement multimeter voltage-measurement
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
|
show 2 more comments
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
|
show 2 more comments
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.
add a comment |
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.
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
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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.
add a comment |
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.
add a comment |
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.
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.
answered 15 hours ago
Ale..chenski
26.1k11861
26.1k11861
add a comment |
add a comment |
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.
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
add a comment |
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.
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
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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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