four wire(Kelvin) measurement with very thin lead wire

Discussion in 'General Electronics Chat' started by tangentspace, Nov 10, 2016.

  1. tangentspace

    Thread Starter New Member

    Nov 8, 2016
    18
    0
    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?
     
  2. MrAl

    AAC Fanatic!

    Jun 17, 2014
    6,383
    1,382
    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.
     
  3. crutschow

    Expert

    Mar 14, 2008
    23,095
    6,850
    Then you have a faulty setup.
    Post a diagram of what you are doing.
     
  4. tangentspace

    Thread Starter New Member

    Nov 8, 2016
    18
    0
    Here is the diagram. The only thing special is I am using very thin lead wire(AWG44), and the DUT has low resistance(about 10mohm)
    upload_2016-11-10_12-33-46.png
     
  5. crutschow

    Expert

    Mar 14, 2008
    23,095
    6,850
    How is I being generated?
    What type of voltmeter is Vm?
     
  6. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,989
    1,848
    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?
     
  7. kubeek

    Expert

    Sep 20, 2005
    5,584
    1,093
    Unless the resistance of the force wiring is so high that the meter is unable to output the measuring current it expects.
     
  8. tangentspace

    Thread Starter New Member

    Nov 8, 2016
    18
    0
    The source is from a voltage square wave generator. The voltage is measured on an oscilloscope.
     
  9. crutschow

    Expert

    Mar 14, 2008
    23,095
    6,850
    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.
     
  10. tangentspace

    Thread Starter New Member

    Nov 8, 2016
    18
    0
    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
     
  11. crutschow

    Expert

    Mar 14, 2008
    23,095
    6,850
    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.
     
  12. tangentspace

    Thread Starter New Member

    Nov 8, 2016
    18
    0
    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.
     
  13. crutschow

    Expert

    Mar 14, 2008
    23,095
    6,850
    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?
     
  14. tangentspace

    Thread Starter New Member

    Nov 8, 2016
    18
    0
    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.
     
Loading...