# How can you measure low resistance accurately?

I got resistors having 1Ω and 0.1Ω. They will be used for measuring current incident to MCU. But when I tested the resistance by using my DMM, the value was fluctuated a lot and it was never what it should be. How can you measure resistance around 0.1Ω accurately?

I attached + - probe of the multimeter together and then it showed 0.5Ω on the screen. Also resistance of my alligator cable varied with the position I attached the probes. Does it mean that multimeter is not a good choice to measure such low resistance?

I'm using Fluke 177 digital multimeter.

• Commented Sep 4, 2015 at 6:01

Best way to do this is the Volt and Ampere meter method:
then you just need to apply a voltage V.
And then you can easily calculate the resistance, from the measured voltage and current.

• That's how I do that indeed :-) I would like to add that if Rl is very low you would need a very small value for V also because I = V/R so the current would become very large. So what I do is add a (power) resistor in series with the supply so that the maximum current is limited, for example to 1A. In case you have only one multimeter that would also help as you could measure the voltage across the power resistor to determine the current, then you don't need the current meter (which needs to remain in circuit). Commented Sep 4, 2015 at 6:49

The Fluke 177 has a maximum resolution of 0.1$\Omega$, so it's not a meter suitable for a direct measurement of these resistors. In the lowest resistance range it has an accuracy of 0.9% + 2 Digits. So it will have an accuracy which would result in 200% error for the 0.1$\Omega$ and 20% error for 1$\Omega$.

You can use the approach given by Gregory Kornblum or Bruce. Just be sure not to use too much current as self heating might cause the value to drift (or you could kill the resistor if you go over the top).

There are special low resistance meters - so called Milliohm-Meter, there are some which offer a resolution of 0.01µ$\Omega$ (which would be way overkill here).

They internally work on the same principle. They use different constant currents based on the resistance range. For example the Hioki RM3543 would use a 100mA or 1A current to measure a 0.1$\Omega$.

They also use 4 wire resistance measurement to cancel out the effect of the measurement leads. Just like in the approach given, 2 wires are connected to the current source and 2 wires are used to sense the voltage directly on the resistor.

Use separately current source and volt meter. Apply 0.1A and measure voltage on your resistor.

• You might want to extend that answer to make the idea a bit more clear. Commented Sep 4, 2015 at 8:36
• Why would not that be clear? The man uses DVM, he probably knows what I and V are.
– user76844
Commented Sep 4, 2015 at 8:41
• @GregoryKornblum Yeah but one line answers often get automatic criticism even if they are pretty clear. Commented Sep 4, 2015 at 19:24
• I would use the f word for the criticism, but it's forbidden. Anyway, i guess historian wouldn't ask DVM questions, as well as a writer or a priest. So one who asks needs short and clear answer rather than a long one, other he would read a book.
– user76844
Commented Sep 4, 2015 at 19:29

Typical approach would be to use differential amplifier with the gain of 10/100 across the resistor while putting current through it.

Or you can get a low ohm meter. If you do a lot of measurements. (I have had the same fluke, for over 30 years, its a bench meter). If you calculate via V and I, a HIGH precision resistor is required (or measure it with a VERY good meter).

• Welcome to EE.SE. To make this a good answer I suggest that you include the model of the Fluke meter, state its measurement capability at low R values and explain the principle of operation. Then explain what the problem is with measuring via V and I. There is a schematic tool on the editor toolbar if you wish to add a schematic of a test setup. You also need to improve your capitalisation and word and punctuation spacing to gain some credibility with your answer. Commented Oct 2, 2017 at 20:03