When measuring very low resistances the problem quickly arises that every multimeter sucks in measuring very low resistances.

Most multimeters 'give up' below 10 ohms, taking very long to get a result and the accuracy is horrible.

There are meters that can measure low resistances, which are 4-wire resistance meters.
But they are expensive, a crap one is more than 50 euro's, a half-decent one is more than 200.

In this instructable I will explain how to make one for about 2 dollar/euro in parts.

Not included in this cost is:

• A half-decent multimeter, although you can get one for \$15 dollars on ebay now.
• Basic soldering skills and soldering tools.

## Step 1: When Do You Need 4-wire Measurement

4-wire resistor measurement is if you need an accurate measurement in <1 Ohm.

If you are working with high-amps (for example, model cars, car batteries, etc) or want to measure current you often need to know what the resistance of your wires/steel strip/connector/etc is.

This is very valuable information to calculate temperature rise in cables and prevent overheating.

Checking how much voltage drop there would be at higher currents.

A normal meter just says 0.0 Ohm (often whilst beeping annoyingly), so how to do this without breaking your bank account?

## Step 2: How Does 4-wire Measurement Work

First a bit of theory, how do resistor measurements work.

To measure a resistor you need Ohm's law.

V = I * R or put in a different way: R = U / I.
So Resistance = Voltage divided by Current.

If you want to measure resistance, you basically need a known current running trough your resistor, and measure the voltage over the resistor.

This is how all resistance measurement work.

So why does you meter fail too work?
2 reasons:

1. The current is too low to measure the voltage.
2. It is also measuring the resistance of the test leads.

This is where 4-wire resistor measurement come in.

By running 4 wires to your resistor you can separate the current running trough the resistor and the voltage over the resistor eliminating the test leads.

## Step 3: Making Your Own

You need 2 parts:

1. A accurate voltage measurement.
2. A consistent current source and current meter to check the source.

An average multimeter is usually a quite good low voltage (mV) meter, so you can use the one you already have.

To make a consistent current source we are going to use a voltage regulator and a resistor.

The voltage regulator should be a positive one (makes calculation easier), the lower the voltage the better (less heat).

LM317 is 1.25V, an 7805 is 5V, etc. I'll be doing the calculations for a LM317.

We should be aiming for a current in the 1A range (low enough not to damage anything, but high enough to be easy to measure, higher is better).

In theory an exact 1A would make for very easy reading, but getting that exact without using high precision, expensive components is actually quite hard and would mean lots and lots of tuning.

The resistor would be a value rougly the same as the voltage (R = U/I, with I = 1) for 1 A.
1.25 V / 1 A = 1.25 Ohm

Please note that this resistor will get hot, the load on the resistor will be P = U * I, 1.25 V * 1 A = 1,25W.

The LM317 also needs cooling, because if we run it at 4V, it will be generating (4*1 = 4W - 1,25 W) 2,75W of heat.

## Step 4: Putting It Together.

I would suggest to directly solder the parts, breadboard can be a bit troublesome with contact resistance.

I'm using 8 0.25W 10Ohm standard resistors in parallel giving me a 1.25 Ohm resistor of 2W.

I mounted the LM317 on a heatsink I had lying around, old equipment such as power supplies and amplifiers are a great source for these.

When mounting any component on a heatsink, cleaning off the old thermal grease an putting a dab of new thermal grease is always a good idea. For this project it is not critical, but it is a good best practise.

## Step 5: Calibrating Your Current Source.

Before you can measure resistors you want to accurately know the current trough your current source.

You need a power supply, your multimeter in A DC mode, and some wire.

As a power supply I'm using a lab power supply, but you can use any power supply as long as it is something linear (like an old heavy wall-wart).
Just keep the voltage low, you don't want to go more than above 2 volts above your regulator specification because else the regulator will get very hot very fast. (for an LM317 this is about 4V, an 7805 would need 7V).
This specification is in the datasheet of you regulator under dropout voltage or 'input to output voltage'.
The LM317's dropout voltage is 3V, plus the 1.25V drop over your resistor, plus cable losses you will probably want about 5-6V for this circuit, any more is just wasted heat.

After turning it on you want to wait a few minutes for the device to warm up, 5-10 minutes should do it, you will see that your multimeter will no measure a change in current. This is called 'thermal equilibrium'.
In professional gear this is taken more seriously, often never turning equipment off to make sure it is stable.

Yes it will get hot, mine gets up to roughly 70 degrees Celsius, this is not a problem.

After the measurement is stable you can read the value from your multimeter.

This is now your current, you can label your device with this value, because it is not going to change anymore.

Mine stabilized at 1.015 A, which is pretty good for non-precision parts.

## Step 6: How It All Works in the End.

Now to get it all working and measure a known shunt resistor.

Your power supply and resistor/regulator combo make the current source shown above.

After leaving it on again for a few minutes (to reach the same stable temperature as before) you can read the voltage on your multimeter.

After that it is a simple calculation to get a resistance value.

Ohms law: R = U / I.
U = the measured voltage.
I = the current you labelled your device with.
R = the resistance in Ohms

For the shunt I have here I found a voltage of 3.8mV.
The current from the last step is 1.038A.

R = U / I = 3.8 mV / 1.015 A = 3.74 mOhms, yes that is 1/1000's of Ohms.

How accurate is this?
The shunt is labeled 75mV 20A, calculating that gives a resistance of:
R = U / I = 75 mV / 20 A = 3.75 mOhms.

So it seems it is very accurate, as long as your multimeter is.

Success! We have measured resistors in the low milliOhms where a multimeter would give up at < 1 Ohm.

Now we are ready to take on the world! or at least its low value resistors.

<p>Thanks for sharing! Exactly what I was looking for! This is my version of it.</p><p>As you can see in the pictures I have used a LM317 and 2.7&Omega; resistor and get 0.48A which still seems to yield nice results (I measured some connections to ~1m&Omega; and others to ~20m&Omega; which seems around right).</p><p>Please note that you don't use the LM317 as voltage regulator (vr), but as constant-current-source; see 9.3.8 <a href="http://www.ti.com/lit/ds/symlink/lm317.pdf" rel="nofollow">http://www.ti.com/lit/ds/symlink/lm317.pdf</a> I'm not sure if all vr do support this mode, since even pinning is different with most vr (in-adjust-out vs in-gnd-out).<br><br>BTW: There are good and cheap (~2&euro;/\$) constant current circuits (with adjustable output) available (e.g. on ebay). You can adjust your current to whatever your probed &quot;resistor&quot; may withstand.</p>
Your setup looks nice. A bit much on the thermal paste, but that doesn't matter.<br><br>I am using the LM317 as current source indeed, all linear voltage regulators can do that, but the power losses get a lot higher, for 1A it is not 1.25W but 5W with a 7805.<br><br>If you grab a LM317 and just wire the ADJ pin to ground like a 7805, it will behave like a 1.25V voltage regulator.
Thanks :)<br>It's actually thermal &quot;silicone glue&quot;, which should make it okay to use a little too much (it's also there to keep it together). With only 480mA it doesn't get hot anyway - just a little warm.<br><br>I think I only now understand how this really works. It's not even that hard :D The IC always wants a specific Voltage between Ground/Adjust and Out. The only voltage drop between those is caused by the resistor with U=R*I . R is fixed, and the IC effectively stabilised I by keeping U constant.<br>Maybe you could add the figure from the datasheet and explain it with that. I guess that would make this instructable even more useful to add/elaborate on how this works. I'm not too bad with stuff like this, but it took me some time to understand (since you &quot;schematics&quot; don't even state which pin is which). If you knew how this works you would probably not need this 'able anyway.<br><br>I just trusted you and it worked :D I trust you with the cheap current sources as well :P
<p>On the cheap current sources from Ebay, most of these are switching and by design those can never supply a constant current.</p><p>For true constant current you really need the last step to be linear.</p><p>They provide a good average constant current, but it you would plot the current on a osciloscope you will notice the actual current is not constant at all.</p>
Thanks for this trick, however Using this technique can I measure the value of Relay contact resistance?
Yes, that is a very good application of this technique.<br><br>Be aware that you are also measuring the resistance of the internal wiring of the relay, not just the contacts.<br>And that relays contacts get worse over time, that might have an impact on your application.
Does it matter if the resistor is wire wound, or better to use standard quarter or half watt resistors in parallel? Thanks
<p>What matters is how thermally stable the resitor is, wire wound resitor are quite good at that.</p><p>Metal film is also fine, carbon resistors are not so stable.</p>
<p>Well done Instructable, knowing how to obtain this value accurately is indeed useful, thanks for the explanation.</p>