From measured transport to device behaviour

A device adds contacts and geometry

Measured bulk conductivity is one input to a device. Contacts, interfaces, film thickness, current crowding and heating can determine the observed voltage–current relation. A two-terminal reading may include contributions beyond the specimen interior.

A fictional film is expected to have an interior resistance of 8 Ω, yet the voltage across its outer terminals divided by current is 12 Ω. A second pair of voltage probes gives 4 Ω. Those three numbers can all be correct: they can describe different parts of one current path. Before changing a material property, identify the two points defining each voltage.

A contact is the junction through which current enters or leaves the material. The adjoining conductor, an interfacial layer and the material beneath it can each contribute. A metal–semiconductor interface can impede carrier injection through an electronic barrier; an oxide, residue or changed interfacial structure can also alter the response. Some contacts have an approximately linear low-field range, while others rectify or depend strongly on bias. A constant contact resistance is therefore a model assumption, not a property guaranteed by the metal name.

Here positive conventional current I flows from the higher to the lower electrical potential. A positive voltage drop V is measured in that direction. The earlier Conductors, insulators and semiconductors lesson defines current, voltage, resistivity and a uniform ohmic resistance. This lesson follows those quantities through contacts, an actual sensing span and an operating thermal state. It does not require a transistor or diode model.

Keep the film dimensions with the measured voltage

Use a separate original isothermal teaching model at a specified temperature. A homogeneous film has length L = 4.00 mm, width W = 1.00 mm, thickness t = 1.00 µm and supplied resistivity ρ = 2.00 × 10⁻⁶ Ω m. Ideal end contacts inject current uniformly across the full width; the interior is ohmic and the sensing taps do not disturb it. There is no parallel leakage path. All values are invented, not a named material or a fabrication specification.

The current cross-section is A = Wt = 10⁻³ × 10⁻⁶ = 10⁻⁹ m². Thus R_film = ρL/(Wt) = 8.00 Ω. For a uniform film it is convenient to define sheet resistance R_sheet = ρ/t = 2.00 Ω per square. The number of squares along the current path is L/W = 4, a dimensionless aspect ratio, so R_film = R_sheet(L/W). Per square describes geometry cancellation; it is not a new unit of resistivity.

Put noninvasive voltage taps at x₁ = 1.00 mm and x₂ = 3.00 mm, measured from the film’s left end. Their spacing ℓ = x₂ − x₁ is 2.00 mm, not the whole 4.00 mm. At I = 10.0 mA the film drop is 80.0 mV, but the tap-to-tap drop is 40.0 mV. The sensed ratio is R_span = V_sense/I = ρℓ/(Wt) = 4.00 Ω.

Inferring ρ from that inner reading requires its own span: ρ = (V_sense/I)Wt/ℓ. Inserting the full L instead would give 1.00 × 10⁻⁶ Ω m, half the supplied value. Scaling R_span by L/ℓ recovers 8.00 Ω only because the stated uniform current, thickness and material response apply between and beyond the taps. A local constriction or nonuniform film can invalidate that extrapolation.

A real film can also have a different intrinsic resistivity from bulk stock because its defect structure, interfaces or scattering differ. Measuring a thin-film sheet resistance does not establish the bulk resistivity until conducting thickness and current distribution are known. Charge carriers and electronic energy bands and Doping, defects and interfaces explain mechanisms that can change the material response itself.

Separate the force path from the sense path

Add a 0.500 Ω lead and a 1.50 Ω current contact at each end of the same isothermal film. The force leads carry the imposed current through all five contributions. Their total is R_2 = R_lead,left + R_contact,left + R_film + R_contact,right + R_lead,right = 12.0 Ω. The outer-terminal drop is therefore 120 mV at 10.0 mA. Charge is not consumed in each resistor: the same steady current crosses every element of this unbranched path.

The separate sense leads observe the potential difference between the two interior taps. An ideal voltmeter draws zero current, so its sense leads and sense-contact resistances have no I_sense R drop. Current-contact and force-lead voltage drops lie outside this inner span. This explains the benefit of a four-terminal arrangement: the voltage and current are assigned different paths. It does not turn every four-terminal number into the full film resistance.

Here R_2 − R_span = 12 − 4 = 8 Ω. That difference contains 4 Ω of unsensed film as well as 3 Ω of current contacts and 1 Ω of force leads. Calling all 8 Ω contact resistance would double the actual fixed series contribution. First compare the same film length: R_2 − R_film = 4 Ω is the supplied contact-plus-lead total. Even this difference does not separate the two contacts from the two leads by itself.

A real voltmeter has finite input resistance and can draw a small sense current. Its input resistance must be large relative to the resistance presented by the sensed port if loading is to be negligible. In a separate supplied port model, an unloaded 40.0 mV source behind 1.00 kΩ connected to a 1.00 MΩ meter gives V_meter = 40.0 × 1000/1001 = 39.9600 mV. The small reduction follows from voltage division; it is not a new property of the specimen. This port example is distinct from the 4 Ω film span.

Separate sensing also does not remove the physical influence of current injection, invasive taps, offsets or heating. If current crowds near an electrode, a tap placed there can sample a different potential field from a one-dimensional interior model. Compare positions, geometry, temperature and the measurement quantity before assigning the result to ρ. Calibration, uncertainty and repeatability explains why repeatable indications can still share a bias.

Use a controlled length change to test the decomposition

One terminal measurement cannot determine both a film contribution and an unknown fixed series contribution. Use a fictional matched family instead: three strips share the same width, conducting thickness, material state, temperature and end-contact/lead contribution R_fixed. Only their uniform interior lengths differ. Stipulate that changing length does not change the contacts or current distribution. This is an ideal comparison, not an automatic property of a real patterned test structure.

The series model is R_2(L) = R_fixed + [ρ/(Wt)]L. The table gives 6, 8 and 12 Ω for lengths 1, 2 and 4 mm. The increase from 1 to 4 mm is 6 Ω over 3 mm, giving a slope of 2.00 Ω/mm. Subtracting the 2 Ω film contribution at 1 mm leaves an intercept R_fixed = 4.00 Ω. All three records lie on that same line.

With W = 1.00 mm and t = 1.00 µm, convert the slope to 2000 Ω/m before multiplying by Wt: ρ = 2000 × 10⁻⁹ = 2.00 × 10⁻⁶ Ω m. The slope identifies the assumed uniform interior response. The intercept is everything assumed independent of length; it is not automatically twice an interface resistance. Only after independently establishing the 1.00 Ω total force-lead contribution and equal contacts can the remaining 3.00 Ω be assigned as 1.50 Ω per contact.

A fit can be misleading if thickness, temperature, contact preparation or lateral spreading changes with length. An intercept does not by itself measure a specific contact resistivity in Ω m², nor a microscopic barrier height. Inspect whether the changed dimension actually isolated the intended contribution. Comparable results and independent reproduction develops matched comparisons and what fresh evidence adds.

A total current does not state the local loading

In a uniform straight interior, current density j = I/(Wt) is current per area. The original 10.0 mA strip gives j = 10⁷ A/m². For a locally ohmic isotropic region, E = ρj and the local conversion of electrical energy to heat per volume is p_vol = jE = ρj², in W/m³. Electrical power for the full uniform resistor is P = IV = I²R. These are different scales: local power density, total power and temperature are not interchangeable.

The original film interior dissipates (0.0100)² × 8 = 0.800 mW. Its volume is LWt = 4.00 × 10⁻¹² m³, so the uniform p_vol is 2.00 × 10⁸ W/m³. This large-looking per-volume number describes a very small stipulated volume; it is not a temperature or a damage threshold.

For a separate uniform interior with half the width, unchanged ρ, length and thickness, the same current doubles j. Local p_vol is four times larger, but the volume is half as large, so total power only doubles to 1.60 mW. At a contact edge or abrupt constriction the field can instead be nonuniform. Current entering through a small patch must spread into the adjoining material; some regions carry more current per area than others. This is current crowding, and j is not given everywhere by one nominal cross-section. The uniform-width comparison illustrates the geometry sensitivity; it does not solve that contact field.

Contact heat can also be concentrated in a different place from interior heat. In the ideal 12 Ω isothermal model, the film receives 0.800 mW, the two contacts together 0.300 mW and the two force leads together 0.100 mW. Their sum is 1.20 mW = I²R_2. Where that energy goes, and how readily each region loses it, determines the resulting thermal state. A four-terminal reading can exclude a contact voltage drop while the contact still physically affects the device.

Operating current can change the resistance being measured

Use a separate fictional resistive element to examine thermal feedback. It has independently supplied reference resistance R₀ = 100 Ω at bath temperature T_b = 300 K and a local temperature relation R(T) = R₀[1 + αΔT], where ΔT = T − T_b and α = 0.00400 K⁻¹. This relation is supplied only for 0 ≤ ΔT ≤ 30 K. Geometry is held fixed, and the element is approximately uniform in temperature. These constants are not a named material’s data or a recommended operating range.

The mounting has constant thermal resistance Θ = 50.0 K/W in this model: at steady state a temperature rise ΔT loses heat to the bath at P_out = ΔT/Θ. Θ represents the whole heat path, not electrical resistance or material thermal conductivity. Heat capacity would matter while warming; it is not needed to find this stated steady balance. Heat capacity and thermal conductivity develops why thermal energy, heat-transfer rate and geometry need separate descriptions.

With an imposed constant current I, the element generates P_in = I²R₀(1 + αΔT). Only this element’s electrical dissipation enters the thermal balance; no neighbouring contact heat, radiation change or additional source is included. Equating P_in and P_out gives ΔT = ΘI²R₀(1 + αΔT), hence ΔT = [ΘI²R₀]/[1 − ΘαI²R₀]. The numerator has units K and the denominator is dimensionless. Use a positive denominator and check that the solution lies inside the supplied temperature window.

At 25.0 mA, the balance gives ΔT = 3.16456 K, R = 101.266 Ω and V = IR = 2.53165 V. At 50.0 mA it gives ΔT = 13.1579 K, R = 105.263 Ω and V = 5.26316 V. Both are within the stipulated relation. The four-terminal element voltage can therefore be nonlinear as a function of imposed current even though the element is ohmic when its temperature is held fixed. Excluding contact drops has not kept the element at 300 K.

The second figure plots generated and removed power against temperature rise. Their intersections solve the coupled electrical/thermal account. Replacing R by its cold value would give rises 3.125 and 12.5 K, overlooking the temperature dependence. This construction is a steady model, not a transient or complete stability analysis. A formal solution outside 0–30 K cannot extend the supplied resistance relation or establish a real failure temperature.

The electrical constraint matters too. If voltage across the element is fixed, power is V²/R(T), not I² times an independently fixed current. A positive α increases heating at fixed current but reduces it at fixed voltage for the same trial temperature. Neither drive condition alone supplies a safe operating limit. In an actual device, temperature gradients, changing contacts, radiation and irreversible structural changes can require a different model.

Test the representation

Four-terminal arrangements can reduce some contact contributions, but geometry and current paths still matter. Compare measured and modelled quantities under the same thermal and electrical conditions. A good fit does not uniquely identify the microscopic transport mechanism.

An increased terminal ratio can follow a changed interior, a changed contact, a smaller conducting section or a hotter operating state. A nonlinear curve can likewise follow an interfacial barrier or thermal feedback. One curve does not select a unique cause. Separate voltage spans, dimensional evidence, independently observed temperature and matched changes of contact or mounting can discriminate between these explanations.

Keep the intended quantity explicit. An isothermal material resistivity, a tap-to-tap resistance, an outer-terminal voltage and the power delivered to a particular region answer different questions. Record the actual geometry, electrical drive, contact state, observation points and temperature represented by each. A fit to the complete device cannot silently become a measurement of one microscopic mechanism or a device qualification.

Worked examples

Try the reasoning

1. Does a four-terminal method solve every geometry problem?

   Show the answer and reasoning

   No.

   The current and voltage field still require suitable interpretation.

2. Why measure heating during operation?

   Show the answer and reasoning

   Transport can depend on temperature.

   The device may leave the conditions of the original material test.

3. The original uniform 4.00 mm strip is instead sensed across only 1.00 mm at the same 10.0 mA. Find R_span and V_sense. May 12 Ω minus that ratio be called the fixed contact/lead contribution?

   Show the answer and reasoning

   R_span = 2.00 Ω and V_sense = 20.0 mV. The 10.0 Ω direct difference includes 6.00 Ω of unsensed film.

   A quarter of the uniform length has a quarter of the 8 Ω film resistance. Compare equal lengths first; the fixed total remains 4 Ω under the unchanged supplied model.

4. A separate matched family has L = 1, 2 and 4 mm and R_2 = 8, 11 and 17 Ω. Width is 0.500 mm and thickness 2.00 µm. Find the slope, fixed total and conditional ρ. Does this identify each contact?

   Show the answer and reasoning

   Slope = 3.00 Ω/mm, fixed total = 5.00 Ω and ρ = 3.00 × 10⁻⁶ Ω m. Individual contacts remain unidentified.

   The slope is (17 − 8)/(4 − 1). The intercept is 8 − 3 × 1. The cross-section is 0.500 × 10⁻³ × 2.00 × 10⁻⁶ = 10⁻⁹ m², so 3000 Ω/m times that area gives ρ. Leads and unequal contacts could share the same intercept.

5. Two uniform films at the same temperature both give R_sheet = 2.00 Ω per square. Their independently known conducting thicknesses are 1.00 and 0.500 µm. Does equal sheet resistance establish equal resistivity?

   Show the answer and reasoning

   No. The inferred resistivities are 2.00 × 10⁻⁶ and 1.00 × 10⁻⁶ Ω m under the uniform model.

   Use ρ = R_sheet t. If resistivity had instead been unchanged while thickness halved, sheet resistance would have doubled. The equal observed ratio alone does not identify why the material states differ.

6. For two uniform interiors of equal length, thickness and ρ, one has half the width. Keep current fixed. Compare current density, local p_vol and total power; is four times the local density four times the total heating?

   Show the answer and reasoning

   j doubles, p_vol quadruples and total power doubles. The heated volume also halves.

   j = I/(Wt), p_vol = ρj² and volume = LWt. The original 0.800 mW interior therefore becomes 1.60 mW. This is an interior geometry model, not a solved contact-crowding field or a temperature prediction.

7. Use the separate 40.0 mV unloaded port behind 1.00 kΩ, now with a 10.0 kΩ meter. Find the reading and relative loading reduction. Can an ideal infinite-input assumption be kept?

   Show the answer and reasoning

   V_meter = 36.3636 mV; the reduction is about 9.09%. The ideal no-loading interpretation is inaccurate for this supplied circuit.

   Voltage division gives 40.0 × 10/(10 + 1) mV. Relative reduction is (40.0 − 36.3636)/40.0. This distinct port model does not assign 1 kΩ to the earlier 4 Ω span.

8. The same separate thermal element is mounted with Θ = 100 K/W while R₀, α and imposed 50.0 mA remain unchanged. Is its operating resistance unchanged? Check the supplied 0–30 K range.

   Show the answer and reasoning

   ΔT = 27.7778 K and R = 111.111 Ω, still within the supplied range. The changed mounting alters the operating reading without changing R₀.

   The numerator becomes 25.0 K and the denominator 1 − 0.100 = 0.900. The heat loss is 27.7778/100 = 0.277778 W, equal to I²R. A larger thermal resistance removes less heat at a given rise.

9. At imposed 0.100 A in the original Θ = 50 K/W thermal model, algebra gives ΔT = 62.5 K. Has the lesson established that actual operating temperature or a failure threshold?

   Show the answer and reasoning

   No. That value is outside the supplied 0–30 K resistance relation; no steady temperature or failure threshold is established there.

   Within the supplied range, generated power is 1.00(1 + 0.00400ΔT) W while heat loss is 0.0200ΔT W. Their difference is 1.00 − 0.0160ΔT W, still positive at 30 K, so there is no balance in this range. Behaviour beyond it requires new information, not unlimited extrapolation.

10. A device gives a rising V/I ratio as current increases. A separate-contact reading shows the same trend. Has this isolated a non-ohmic bulk transport mechanism?

    Show the answer and reasoning

    No. Separate sensing can remove specified series drops while the sensed interior still changes temperature or current distribution.

    Measure the relevant thermal state and geometry and compare under matched conditions. An injection-induced field change or invasive probe can also remain. A single nonlinear curve and a good fit do not distinguish all these causes.

What to carry forward

• Follow the complete current path.
• Keep contact, geometry and thermal effects distinguishable.
• A sensing span has its own length; terminal-minus-sense is not automatically contact resistance.
• A matched length series can identify a uniform slope and fixed sum only within its geometry and state assumptions.
• Local current density and power density need a current-field model; total electrical power alone does not give temperature.
• Separate voltage sensing does not prevent self-heating or validate a cold material property at an operating state.