research32 min studyLesson 981 in Physics

Four-probe resistivity and contact effects

Extract a material resistivity while testing geometry, voltage loading, offsets, and current heating.

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What you’ll understand

  • Distinguish measured resistance, sheet resistance, and bulk resistivity.
  • Use current reversal to separate a linear voltage from a stable offset.
  • Identify contact and geometry errors that survive a four-terminal measurement.

A four-terminal measurement separates the leads that inject current from the leads that sense voltage. This is a powerful experimental design, but it does not make every measured voltage an intrinsic material property. The current still needs a path through the sample, voltage contacts still occupy space, and a cold sample can be heated by the measurement itself. The first inference is therefore a relation between a specified terminal configuration and a transport model. Only after checking that model should one quote a bulk resistivity.

From a voltage ratio to a material coefficient

Consider a homogeneous diffusive bar with uniform cross-sectional area AA and voltage-contact separation LL. Assume the current density is nearly uniform between the voltage contacts, transport is linear, and the relevant resistivity tensor component lies along the bar. Resistance scales with length and inverse area; resistivity removes those geometric factors. The thickness of a thin film is especially important because an uncertain conducting thickness can dominate the uncertainty even when the voltage ratio is measured precisely.

A thin homogeneous isotropic sheet of arbitrary simply connected shape can instead use the van der Pauw construction. Four small ohmic contacts lie on its boundary, and two positive characteristic resistances determine the sheet resistance through an implicit equation. The familiar factor of pi divided by log two applies when those characteristic resistances are equal. Holes, anisotropy, finite contacts, or inhomogeneous parallel layers require additional treatment; a convenient sample shape does not remove these assumptions.

R4=VI,ρ=R4AL,e−πRA/Rs+e−πRB/Rs=1R_4=\frac{V}{I},\qquad \rho=R_4\frac{A}{L},\qquad e^{-\pi R_A/R_s}+e^{-\pi R_B/R_s}=1
R4, RA, RB, and sheet resistance Rs are in ohms; bulk resistivity rho is in ohm metres. For a uniform film of thickness t, rho=Rs t. The bar and sheet formulas describe different specified geometries.↔ Scroll wide equations horizontally.

What the extra terminals remove

An ideal voltmeter draws no current. Consequently the series resistance of its sense leads and their contacts causes no voltage drop. Current-contact resistance still contributes to the compliance voltage and dissipated power, but is excluded from the sensed voltage if the probes sample the intended interior region. A real voltage input has finite impedance, so excessive source impedance, leakage, or poor contacts can load the circuit. Nonohmic injection can also distort the current distribution and invalidate a simple bar analysis.

Stable thermoelectric and amplifier offsets can be removed by reversing the applied current while preserving a fixed voltage-polarity convention. The odd part in current isolates the linear resistive response, while the even part estimates the offset. This subtraction works only if the offset remains stable over the reversal sequence. Drift, current-dependent heating, and nonlinear transport can survive it. Recording several positive and negative currents makes these failures visible instead of hiding them inside one ratio.

V(+I)=IR+V0,V(−I)=−IR+V0,R=V(+I)−V(−I)2IV(+I)=IR+V_0,\qquad V(-I)=-IR+V_0,\qquad R=\frac{V(+I)-V(-I)}{2I}
The same positive magnitude I is used in both runs, with fixed voltage leads. V0 is a stable current-independent voltage. This equation is an offset-removal model, not a proof that the sample is ohmic.↔ Scroll wide equations horizontally.

A small voltage does not guarantee a gentle measurement

Joule heating scales quadratically with current in the linear regime. A low measured sample voltage can coexist with substantial heating at a resistive current contact. Compare current levels and sweep directions, allow thermal settling, and monitor whether the inferred resistance changes with excitation. Reciprocity comparisons can diagnose inconsistent wiring or sample inhomogeneity, subject to reversing magnetic field when the appropriate Onsager relation requires it. No universal current threshold applies to samples with different cooling and contact conditions.

Uncertainty in a geometric conversion is often shared across an entire temperature sweep. A thickness error rescales every inferred resistivity rather than creating independent noise at each temperature. Report the raw resistance and sample dimensions as well as the converted quantity. A resistance drop can suggest a transition, but a connected low-resistance filament may short the voltage probes without occupying most of the sample. Bulk thermodynamic or magnetic evidence is needed before assigning a bulk phase fraction.

P=I2R,(uρρ)2≃(uRR)2+(uAA)2+(uLL)2P=I^2R,\qquad \left(\frac{u_\rho}{\rho}\right)^2\simeq\left(\frac{u_R}{R}\right)^2+\left(\frac{u_A}{A}\right)^2+\left(\frac{u_L}{L}\right)^2
P is power in watts for the specified resistor. The uncertainty expression is first-order propagation for independent small errors; correlated dimensions require covariance terms.↔ Scroll wide equations horizontally.

Current reversal and a bar geometry

A hypothetical uniform bar has L=2 mmL=2\,\mathrm{mm} and A=0.10 mm2A=0.10\,\mathrm{mm}^2. With fixed voltage polarity, currents ±1 mA\pm1\,\mathrm{mA} give +12 μV+12\,\mu\mathrm V and −8 μV-8\,\mu\mathrm V. Find the resistance, offset, and resistivity.

  1. The odd voltage is (12−(−8))/2=10 μV(12-(-8))/2=10\,\mu\mathrm V. Dividing by the positive current magnitude gives R=0.010 ΩR=0.010\,\Omega.

  2. The even voltage is (12+(−8))/2=2 μV(12+(-8))/2=2\,\mu\mathrm V. This estimates the stable offset; using only the positive-current reading would overestimate resistance by twenty percent.

  3. Convert area to 10−7 m210^{-7}\,\mathrm m^2 and length to 2×10−3 m2\times10^{-3}\,\mathrm m. Their ratio is 5×10−5 m5\times10^{-5}\,\mathrm m, so ρ=5×10−7 Ω m\rho=5\times10^{-7}\,\Omega\,\mathrm m.

The result

The inferred resistance is 0.010 ohm, the offset is two microvolts, and the bar resistivity is five times ten to the minus seven ohm metres.

Explore the model

Interactive lab

Interactive lab · function

Quadratic excitation heating

For each fresh paused comparison Reset and set coefficient a and position x to (0.5,1)(0.5,1), then (0.5,2)(0.5,2). Predict y values 0.5 and 2. Interpret x as dimensionless current only in this power-scaling analogy. The exact quadratic lab does not solve current spreading, contacts, thermal anchoring, or van der Pauw inversion.

Quadratic excitation heatingThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forDoubling an ohmic current quadruples its dissipated power.

Check your understanding

Which error can persist in an otherwise ideal four-terminal resistance measurement?

Try it yourself

A hypothetical isotropic van der Pauw sheet has RA=RB=10 ΩR_A=R_B=10\,\Omega and thickness 100 nm100\,\mathrm{nm}. Find its sheet resistance and bulk resistivity, using π/ln⁡2≃4.532\pi/\ln2\simeq4.532.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Set the two exponentials equal, then multiply sheet resistance by thickness in metres.

Compare with the solution

Each exponential equals one half, giving Rs=10π/ln⁡2≃45.32 ΩR_s=10\pi/\ln2\simeq45.32\,\Omega. The thickness is 10−7 m10^{-7}\,\mathrm m, so ρ=4.532×10−6 Ω m\rho=4.532\times10^{-6}\,\Omega\,\mathrm m. The result assumes a uniform conducting sheet with small boundary contacts, not an arbitrary multilayer film.

Take it with you

  • A terminal resistance needs a geometry and transport model before it becomes a bulk resistivity.
  • Current reversal removes a stable offset but does not remove nonlinear or thermal artifacts.
  • Four-terminal sensing cannot establish a bulk phase fraction.

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