four wire(Kelvin) measurement with very thin lead wire

Thread Starter

tangentspace

Joined Nov 8, 2016
18
If you do a Four-wire(Kelvin) resistance measurements, the 4-wire configuration is supposed to remove the effect of the resistance of the lead wires, but what happens when the lead wires get very thin? I see the measured resistance start to go up. Why does this happen?
 

MrAl

Joined Jun 17, 2014
7,761
Hi,

Why does the resistance of the wires go up when they get thinner?
That's because the total resistance is based partly on the cross section of the wire.
I say "partly" because if the wires are 1000 feet long there is a lot of resistance but if they are only 0.1 millimeters long then the resistance is low even if they are thin wires.
So the cross section (thinness) AND the length determine the resistance.

The extra two wires are for sensing the voltage and they do not have to pass much current so there is no voltage drop. If you use the primary set of wires though you can see a large voltage drop so the reading gleaned from that measurement could be very incorrect. Using a second set of leads eliminates that error.

Secondary effects can set in if the primary set is too long because then the required current for the measurement may no longer be achievable. If the current has to be say 1 amp and the voltage supply supplying that current can not overcome the total primary resistance (which includes the wires) then the measurement can not be made in the same manner as usual. That can be compensated for but only if that is part of the test setup procedure.
 

crutschow

Joined Mar 14, 2008
25,269
If you do a Four-wire(Kelvin) resistance measurements, the 4-wire configuration is supposed to remove the effect of the resistance of the lead wires, but what happens when the lead wires get very thin? I see the measured resistance start to go up. Why does this happen?
Then you have a faulty setup.
Post a diagram of what you are doing.
 

ErnieM

Joined Apr 24, 2011
8,053
If you do a Four-wire(Kelvin) resistance measurements, the 4-wire configuration is supposed to remove the effect of the resistance of the lead wires, but what happens when the lead wires get very thin? I see the measured resistance start to go up. Why does this happen?
The size of the wire does not matter one bit.

How do you "see" this? Is this the observation of an experiment or just something you thought up?
 

crutschow

Joined Mar 14, 2008
25,269
The source is from a voltage square wave generator.
There's your problem.
You can't use a voltage source, it needs to be a current source (that's a current source icon in your diagram, not a voltage source).

The voltage source's current is affected by the lead resistance, which you don't want.
You need a constant current through the DUT for the 4-wire measurement to work properly.
That's why you need a current source, which is not affected by lead resistance.
 

Thread Starter

tangentspace

Joined Nov 8, 2016
18
There's your problem.
You can't use a voltage source, it needs to be a current source (that's a current source icon in your diagram, not a voltage source).

The voltage source's current is affected by the lead resistance, which you don't want.
You need a constant current through the DUT for the 4-wire measurement to work properly.
That's why you need a current source, which is not affected by lead resistance.
It's hard to find a current source signal generator, so I am using a voltage pulser to 50ohm, in series with my device(milliohms) to act as the current source.
upload_2016-11-11_15-54-22.png
 

crutschow

Joined Mar 14, 2008
25,269
It's hard to find a current source signal generator, so I am using a voltage pulser to 50ohm, in series with my device(milliohms) to act as the current source.
View attachment 115180
That only works if the wires connecting the generator and the DUT are much less than 100Ω.
Otherwise it doesn't act as a current source.
To be a proper current source it must have an effective output impedance much greater than the wire plus DUT resistance (ideally infinite).

Why are you applying an AC signal instead of DC.
It's much easier to make a DC current source than an AC one.
 

Thread Starter

tangentspace

Joined Nov 8, 2016
18
That only works if the wires connecting the generator and the DUT are much less than 100Ω.
Otherwise it doesn't act as a current source.
To be a proper current source it must have an effective output impedance much greater than the wire plus DUT resistance (ideally infinite).

Why are you applying an AC signal instead of DC.
It's much easier to make a DC current source than an AC one.
I am doing a thermoelectric measurement that requires AC signal. The wires need to be very thin to avoid heat leakage to the DUT. I am using AWG42 for the lead wires. They resistance of the lead wires is less than 0.1ohm.
 

crutschow

Joined Mar 14, 2008
25,269
I am using AWG42 for the lead wires. They resistance of the lead wires is less than 0.1ohm.
That doesn't sound right.
AWG42 copper wire has a resistance of about 1.66Ω/ft as stated here.
Thus the total length for both leads would be only 0.7 inches for 0.1Ω total lead resistance. o_O
I'm sure they're longer than that.

I think you need to come up with a better current generator.
What frequency do you need?
What current?
 

Thread Starter

tangentspace

Joined Nov 8, 2016
18
That doesn't sound right.
AWG42 copper wire has a resistance of about 1.66Ω/ft as stated here.
Thus the total length for both leads would be only 0.7 inches for 0.1Ω total lead resistance. o_O
I'm sure they're longer than that.

I think you need to come up with a better current generator.
What frequency do you need?
What current?
I am only using about 5mm to 15mm of lead wire.
I am using frequency in the range of 0.01-10000Hz. The current I need is 0.1Amp.
Thanks for your help.
 
Top