Deciding Whether a Resistivity Run Is Trustworthy
A student measures the resistivity of a semiconductor strip using the four-terminal method: a constant current of 1.00 mA is passed through the outer contacts, and the voltage across the inner contacts is read with a digital voltmeter of 10 MΩ input impedance. The strip is 10.0 mm long, 2.0 mm wide and 0.50 mm thick. At room temperature the voltmeter reads 42.0 mV; at 80 °C it reads 18.0 mV. The lab thermometer is a mercury-in-glass type reading to ±0.5 °C, and the oven has a stated uniformity of ±3 °C. The student computes resistivity at each temperature and reports that the material is a semiconductor with a band gap derived from the two points.
1. Compute the resistivity at both temperatures and check whether the voltmeter loading is negligible.
Read the worked answer
Resistance and resistivity. R = V/I. At room temperature R = 42.0 mV / 1.00 mA = 42.0 Ω. Resistivity ρ = R A / L, with A = 2.0 mm × 0.50 mm = 1.0×10^-6 m^2 and L = 10.0 mm = 0.010 m. So ρ = 42.0 × 1.0×10^-6 / 0.010 = 4.2×10^-3 Ω·m.
At 80 °C: R = 18.0 mV / 1.00 mA = 18.0 Ω, giving ρ = 1.8×10^-3 Ω·m.
Voltage-probe loading. The voltmeter draws current through its 10 MΩ input in parallel with the sample between the inner probes. The sample resistance between probes is of order tens of ohms; the parallel combination is essentially the sample resistance. Fractional loading error ≈ R_sample/R_meter ≈ 42/10^7 ≈ 4×10^-6, utterly negligible. The four-terminal method plus a high-impedance meter is appropriate here.
Contact check. The current contacts carry 1 mA; contact resistance at the outer pads does not affect the inner voltage reading, which is the point of the four-terminal arrangement.
2. Assess whether a band gap can legitimately be extracted from just two temperature points, given the stated uncertainties.
Read the worked answer
The intended relation. For an intrinsic semiconductor, ρ ∝ exp(Eg/2kT), so ln ρ = const + Eg/(2kT). A plot of ln ρ against 1/T should be a straight line of slope Eg/2k.
Two-point extraction. Using the two values: ln(4.2×10^-3 / 1.8×10^-3) = ln(2.333) ≈ 0.847. And (1/(2k))(1/T2 − 1/T1) with T1 = 293 K, T2 = 353 K: 1/353 − 1/293 = −5.80×10^-4 K^-1. So Eg = 0.847 / (5.80×10^-4 / (2 × 8.617×10^-5)) ≈ 0.847 / 3.365 ≈ 0.25 eV. This is a plausible order for a semiconductor but rests on only two points.
Uncertainty problem. The temperature uncertainty dominates. With ±3 °C oven uniformity, T2 could be 350–356 K. Recomputing the 1/T difference over that range changes the slope by roughly 5–10%, i.e. Eg by tens of meV. The voltage readings themselves are precise, but the temperature is not.
Two points are insufficient. A straight line through two points cannot reveal curvature, contact effects, or a change of conduction mechanism. Extracting a band gap from two points assumes the intrinsic regime over the whole range, which is unverified. At least five or six temperatures spanning a wider range, plotted as ln ρ versus 1/T, are needed to confirm linearity and to attach a meaningful uncertainty to the slope.
3. Recommend improvements and state what would make the reported band gap credible.
Read the worked answer
Improvements.
- Take readings at six or more temperatures, ideally spanning 20–120 °C, and plot ln ρ against 1/T. Check for a straight line; curvature signals a change of regime or contact problems.
- Replace or calibrate the thermometer; use a thermocouple or platinum resistance sensor in contact with the sample, and record the sample temperature rather than oven air temperature, since the sample may lag.
- Allow thermal equilibration at each set point (wait until the voltage reading is stable) to avoid transient gradients.
- Reverse the current direction and average, to cancel any thermoelectric offset voltages at the probes.
- Repeat the thickness and width measurements with a micrometer, since ρ scales linearly with thickness and a 0.05 mm error is 10%.
Credibility criterion. The band gap is credible when the ln ρ versus 1/T plot is linear over the measured range, the slope's uncertainty (from a least-squares fit) gives Eg to within a stated tolerance, and the value agrees with independent measurements or literature for the identified material. A single two-point number, especially with ±3 °C temperature uncertainty, should be reported as an order-of-magnitude estimate at best.