4-wire ohm resistance measurement

[mdanh2002]

Hi,

I got a HP3468A bench multimeter for a cheap price from eBay. This units comes with 4-wire ohm measurement capability and I was surprised at the accuracy, compared with the traditional 2-wire method, despite the fact that the unit is second hand and probably have not been calibrated for years.

I read through the basics of 4-wire resistance measurement method here and achieved some basic understanding of it. However, as soon as I finished reading, a stupid question came across my mind - as the 4 probe leads for the 4 wire method will still need to be connected to the resistor/device under test (DUT) anyway, then why would we need 4 leads at all? Why don't we simply just have 2 leads like the traditional 2-wire method, and use an internal voltmeter and ammeter to calculate the 4-wire resistance?

Apologize if this question sounds obvious, but I can't think of any answers. Can anyone help?

 

[crutschow]

The purpose of the 4-wire measurement is to cancel the effects of the measurement wire and contact resistances. The measurement current is sent down one set of wires and the voltage drop across the resistor is measured with the other set of wires. Thus any voltage drops in the current leads are ignored and, since there is negligible current in the voltage measurement leads, the resistance of those leads contributes no significant error. The 4-wire measurement is generally only needed when trying to accurately measure low value resistances where lead resistance can be a significant percentage of the test resistance.

 

[mdanh2002]

The purpose of the 4-wire measurement is to cancel the effects of the measurement wire and contact resistances. The measurement current is sent down one set of wires and the voltage drop across the resistor is measured with the other set of wires. Thus any voltage drops in the current leads are ignored and, since there is negligible current in the voltage measurement leads, the resistance of those leads contributes no significant error. The 4-wire measurement is generally only needed when trying to accurately measure low value resistances where lead resistance can be a significant percentage of the test resistance.

Yes, I understand it's for measuring low resistance while compensating for the leads resistance. But still wondering why would we need 4 leads (2 pairs) if both pairs are going to be connected to the same thing? Why can't we have a single pair of test leads and the logic for 4-wire measurements can still be applied?

 

[crutschow]

Yes, I understand it's for measuring low resistance while compensating for the leads resistance. But still wondering why would we need 4 leads (2 pairs) if both pairs are going to be connected to the same thing? Why can't we have a single pair of test leads and the logic for 4-wire measurements can still be applied?

You may understand that it's used for low resistance measurement but you don't understand why. Please re-read my explanation. If you used only two leads then the voltage drop in the test leads due to the test current would be unknown and would not be compensated, so the measurement would be in error. The compensation is not in any "logic", it's intrinsic to the 4-wire connection.

 

[dr pepper]

If your measuring resistances higher than 100r or so then you can link the measuring and supply terminals together as the erros will be small, some meters came with metal links for the job.

To understand cruts's explanation better connect 4 leads to a low resistance about 0.1 ohm, connect all 4 leads to the resistance, then connect the 2 measurement leads to the meters output terminals rather than the resistance, you'll be surprised at the difference.

 

[mdanh2002]

Hi,

Thanks for the fast replies.

I am watching this video:

,

which shows me how to achieve 4-wire resistance measurements with just just a normal multimeter and attempt it, to better understand the concept. From the video the steps are as follows:

1. Use a current limiting power supply, set to a low voltage, connect it to a multimeter in current range and adjust the current to a desired value. This is to get a constant current source.
2. With the setup in (1), connect a resistor, e.g. 1 kOhm, to be tested across the power supply output.
3. Measure the voltage across the resistor using another multimeter
4. The resistance is R = V / I

However I tried it and observe some very strange results. With R = 1kOhm and the power supply current 0.7A, as indicated on the power supply current indicator and on the multimeter when in ampere range, I measure 70mV across the resistor, which means the resistance is 70mV/0.7 A = 0.1 Ohm! However, when I changed the multimeter to measure in mA range, I get 7mA of current, which result in a resistance of 70mV / 7 mA = 10Ohm. Neither is correct. I am using a Fluke 17B to measure current, a Lodestar 8103 power supply and a cheap Victor VC921 to measure voltage.

The same behaviour is achieved with different set of multimeters. The power supply current indicator shows 0.7A in both cases.

Can anyone advice what I am doing wrong here? Why are the current measurements different in different ranges, and why is the resistor value incorrectly calculated?

 

[JimB]

With R = 1kOhm and the power supply current 0.7A

Using that well known expression which is usually attributed to Mrs Ohms little boy...

V = I x R

In your case V = 0.7 x 1000 = 700 volts. Which again is probably not correct.
So, sort out your units and be aware that taking an ammeter in/out of circuit can often change the resistamce of a circuit quite considerably.

JimB

 

[crutschow]

I suspect an error in your current reading. If it were 70mA instead of 0.7A then your voltage reading of 70mV would correspond to a resistance of 1kΩ, which is the resistor value you stated was being tested. Did you measure the current with the 1kΩ resistor in series with the ammeter?

 

[mdanh2002]

Using that well known expression which is usually attributed to Mrs Ohms little boy...
V = I x R

Of course I known Ohm's law. It's not that I don't know what I am doing. Just some observations that I am posting to find out the reason why.

Did you watch the video in the link? I followed the instructions there exactly to maintain a constant current from the power supply and measure the voltage across the resistor. My question is why would the multimeter read differently in different ranges and why isn't the value of the resistor calculated correctly.

crutschow, yes, I checked the unit and they are correct.

 

[nsaspook]

Current source resistance measurement methods.
Your classic 'Fluke type' two terminal meter uses the constant current method to measure resistance.

2 terminal:
**broken link removed**
The four terminal method can be used when lead resistance values become significant because of higher sense currents for better accuracy measurements of the DUT.
4 terminal:
**broken link removed**

 

[KeepItSimpleStupid]

4-wire ohms works to a point, The Multimeter has a 10 M-ohm input Z, so at some point the input Z of the meter will dominate. The 4 wire technique is BEST USED for low resistances where the wire resistance is appreciable.

Your measuring the current through the device and the voltage across the device. The point of contact of the voltage measurement determines the actual resistor your measuring.

The method totally breaks down when trying to measure 1 G-Ohm resistances usually because of the input Z of the multimeter.

I've set up system to measure about 2E-12 amps and up to 100 V on the sample and I've designed systems from scratch that worked at low voltages (a few mV) and 100 mA.

There is 5 wire ohms technique too. It's more dependent on the ratios of resistors in the measuring equipment. Keithley.com has some meters that use the 5-wire technique. There is also a 6 wire technique as well.

With extremely high values of resistance a Zero Resistance Ammeter is used which has a few mV of a voltage drop. This is combined with a high voltage source where a few mV doesn't make a great deal of an error. So, in this case, 4-terminal is NOT a good technique.

With the 4-wire technique, you can measure milli-ohms of resistance when your current carrying leads have a resistance of say 2 ohms. For your voltmeter 10 M-ohms+2 ohms is negligible and 10 M-ohms in parallel with a few milliohms is negligible.

 

[mdanh2002]

Got it working now. Followed the instructions in the video and managed to measure the resistance of a 60cm thin piece of wire to be 0.378 ohm using 2 multimeters, amazingly almost the same value reported by my HP 3468A multimeter in 4-ohm mode. Previously I forget to remove the multimeter which I use to measure current while adjusting the constant current from the power supply, hence the circuit is shorted causing the current limiting power supply to reduce the output voltage to a few hundreds mV and the measured voltage across the resistor is obviously wrong.

As for the change in the current measured by the Fluke 17B while switching between A and mA range, I figured it was due to a design flaw of this multimeter. If the range switch is in the mA range but the probes are connected to the A socket, then the current measured will be approximately 1/100 the actual value (showed 1mA instead of 100mA for example). I did not change the probe connections when changing the range switch and the measurement displayed was wrong.

 

[mdanh2002]

Also I would like to add that the method mentioned in the video to generate a constant current source using a current limiting power supply would only work if the resistor value is so small such that the amount of current flowing through it will be larger than the threshold set, causing the power supply to limit the current at the set value and becomes a constant power supply. If the resistor is large enough, e.g. 1kOhm like what I tried, the current flowing through it will be less than the threshold and hence the bench power supply is no longer a constant current supply, so the method cannot be used.

 

[chemelec]

Here are My Simple and Accurate, Low Ohm Meters


Contact Resistance, when doing High Resistance Tests is usually not an Issue.