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I am trying to use the 4-wire method to measure the resistance of small strip of carbon steel, which will have a resistance of about 0.3 mOhms. I am currently trying to test my circuit with a known 0.3 mOhms resistor. I am using the Arduino to make the measurements. I have a 5V supply and the circuit has a resistance of 150 Ohms, giving 33 mA current. The potential drop across the 0.3 mOhms resistor is expected to be 9.9 uV. This is amplified by the LT1006 by a gain of 1000. I am using oversampling (https://gumroad.com/l/eRCaGuy_NewAnalogRead) to achieve a resolution of 16-bits, so I a gain of 1000 should be sufficient.

A schematic of my circuit is shown below

enter image description here

I expect the resistor of the sensor to be given by V_sensor/V_100 * 100 Ohms.

I have used the circuit to measure 100 Ohms and 220 Ohms resistances successfully, using a gain of unity on the LT1006.

However, the issue is that the V_sensor I am measuring with the 0.3 mOhms is equal to my input offset voltage (0.002), i.e. its the same value as when the inputs of the LT1006 are shorted together. I have used a 10k pot to null the offset. But it only reduces it to 0.002 V.

Also, is it possible to achieve a higher gain on the voltage across the sensor, by replacing the 10 kOhms resistors on the LT1006 inputs by 100 Ohms. I read here (Kelvin "4 Wire" Resistance PCB Design Questions) that the input resistance for the 4-wire method should be 10 k.

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  • \$\begingroup\$ So what are your plans to combat the input offset voltage being of the same order as your signal? \$\endgroup\$ – Andy aka Jan 5 '17 at 11:57
  • \$\begingroup\$ According to the LT1006 datasheet (" For increased trim resolution and accuracy, two fixed resistors can be used in conjunction with a smaller potentiometer. For example, two 4.7k resistors tied to Pins 1 and 5, with a 500Ω pot in the middle, will have a null range of ±150µV.") The input offset can be reduced to 150 uV, which is still an order of magnitude greater than my input signal. \$\endgroup\$ – Afzal Jan 5 '17 at 12:00
  • \$\begingroup\$ So what plans do you have? \$\endgroup\$ – Andy aka Jan 5 '17 at 12:25
  • \$\begingroup\$ I'm not really sure. I'd expect I have to get an amplifier with a lower offset voltage? I haven't got much experience with electronics. My background is Mechanical. Only know the basics from my Mechatronics module. I want to measure the change in resistance of the sensor as it is eroded in a jet impingement rig. \$\endgroup\$ – Afzal Jan 5 '17 at 12:43
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    \$\begingroup\$ 40mA average can mean 4 A for 1% of the time - use a storage capacitor to generate the peak current and measure for a short period of time. \$\endgroup\$ – Andy aka Jan 5 '17 at 13:18
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Use an AC current source, and AC preamplifiers on the small developed voltage. There's no 'offset' if you use an AC signal instead of direct-coupled DC.AN-98 See Fig. 27

The use of FFT oscilloscope, or a phase-locked amplifier, might accomplish the desired result without having to wire up a dedicated measurement circuit.

A low-voltage measurement with DC sensitivity will always show artifacts, if not from offset voltage, then from thermocouple effects. AC measurements are the only good solution.

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I see the following ways (ordered by increasing impact on your circuit)

Don't increase the gain of the first stage beyond what you need to escape the input noise. Better put in a cheaper difference amp in the second stage.

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If you want to measure microohms, you need much more measuring current. A commercialy available microohmmeter with the range of 320 µΩ and a resolution of 1 nΩ uses 10 A current, that is 300 times more current than you tried. 10 mA is used for the range of 3.2 and 0.32 Ω, 100 mA for 320 and 32 mΩ and 1 A for 32 and 3.2 mΩ.

Several measurements are done for one single resistor value: zero current and current in both directions and both polarities of the voltage at the measured resistor. All these measurements are used to compensate offset error.

But a lot of know how is necessary to minimize errors due to thermoelectric voltages caused by different temperatures in the measurement circuit.

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