Conductors, insulators and semiconductors

The categories describe how electronic charge can respond

A metal conductor has electronic states whose occupation can change slightly under a weak electric field, allowing a substantial net current. Partly occupied or overlapping bands provide nearby available states. Scattering still limits ordinary metal transport, so a good conductor need not have zero resistance. The earlier Charge carriers and electronic energy bands lesson develops occupation, holes and mobility; here those mechanisms explain the categories rather than supply a resistance by themselves.

In a simple gapped semiconductor, a filled valence band and an empty conduction band at the ideal zero-temperature starting point supply no ordinary drift current. Thermal excitation or absorbed light can create mobile conduction electrons and valence-band holes. Deliberately added species can also alter populations. At a given temperature, both the available mobile population and its mobility matter. Semiconductor transport is especially useful because populations can be controlled, not because every semiconductor specimen has a resistance midway between every metal and every insulator.

An insulating state has very little steady conduction under the stated conditions. In a common electronic band picture, occupied states are separated from accessible conducting states by a gap and few carriers become mobile. Bound charges can still respond through polarisation. Real insulation can have finite leakage through its interior or along other paths; its measured current need not be exactly zero.

These are useful physical descriptions, not three bins separated by one universal resistivity or band-gap cutoff. Wide-bandgap semiconductors are a real materials class. A gap alone does not state carrier occupation, mobility, temperature, field or contact behaviour. Nor does the word semiconductor mean that every uniform piece rectifies current like a diode: a suitably contacted homogeneous specimen can have an ohmic low-field range.

Current counts charge per time; voltage measures energy per charge

Current I is the net charge passing a chosen cross-section per time. For a steady current, I = ΔQ/Δt; one ampere is one coulomb per second. A current of 0.200 A maintained for 5.00 s transports 1.00 C through each section of a single steady series path. Charge is not used up in the resistor: without accumulating charge there, the incoming and outgoing currents agree.

Voltage is a potential difference between two specified points, not a quantity located at one point without a reference. One volt is one joule per coulomb. In the passive resistor comparisons here, V denotes the positive voltage drop in the direction of positive conventional current I. A 0.200 V drop corresponds to a decrease of 0.200 J of electrical potential energy per coulomb of positive charge moving through that drop. The sign convention does not require the actual mobile carriers to be positive.

The electrical power transferred to that passive segment is P = VI: energy per charge multiplied by charge per time. Thus 0.200 V × 0.200 A = 0.0400 W, or 0.0400 J/s. In an ordinary resistor this energy is dissipated and can heat the specimen. A power calculation does not determine its temperature rise without thermal information.

At a nonzero steady current, the ratio R = V/I has units of ohms, Ω = V/A. A constant R over a stated range gives V = IR. Merely calculating a ratio at one operating point does not establish that the same ratio applies elsewhere. At V = I = 0, the ratio 0/0 is undefined; a nearby linear response, rather than that quotient, defines the ohmic resistance.

Material response meets geometry

For a uniform ohmic conductor, resistance R = ρₑL/A, where ρₑ is electrical resistivity. Resistivity and mass density use similar symbols in some texts but are different quantities. A long thin wire can have greater resistance without a different material resistivity.

Use ρₑ for electrical resistivity, in Ω m, and σ for conductivity, in S/m. A siemens is an ampere per volt. In the scalar linear transport model, σ = 1/ρₑ and J = σE, or E = ρₑJ. Here J is current density in A/m² and E is the electric-field component along the tested direction in V/m. The material state, temperature and direction belong with a reported ρₑ. The subscript avoids confusing it with mass density in kg/m³.

Now stipulate a straight specimen with uniform cross-sectional area A, uniform resistivity and a steady, uniform axial current. End regions are excluded from the gauge length L. The interior field has magnitude E = V/L and J = I/A. Substitution gives V/L = ρₑI/A, hence V = IρₑL/A and R_bulk = ρₑL/A. The voltage must span that same gauge region. Using a voltage that includes extra contact drops changes the quantity.

The units check the result: (Ω m) × m/m² = Ω. Conversely, ρₑ = R_bulkA/L has units Ω m. A square millimetre is (10⁻³ m)² = 10⁻⁶ m², not 10⁻³ m². Keeping area and length conversions distinct prevents a millionfold error when an area number in mm² is treated as m². Units, scales and orders of magnitude provides the wider unit-scaling preparation.

Longer material at the same current density needs more voltage drop because the same field acts over more length. A larger area carries more total current at the same J. These two facts explain the L/A factor. Doubling area halves R when other conditions remain fixed; doubling a circular wire’s diameter multiplies area by four and quarters R, since A = πD²/4.

This is a bulk, ohmic, uniform-path model. Nonuniform current spreading, anisotropy, changing temperature or a barrier contact can require a different interpretation. At dimensions where a local bulk transport description fails, extrapolating the same formula is not justified. No particular material is certified as satisfying these assumptions by the calculations below.

A lower component resistance can accompany a higher material resistivity

Use the following invented M and S states at a common stated test temperature within separately stipulated ohmic ranges. M is stipulated to be a metal conductor; S is stipulated to be a semiconducting material in a conducting state. Contacts and end effects are excluded. The letters and numerical values are teaching inputs, not property data for named commercial materials.

M has ρₑ = 2.00 × 10⁻⁸ Ω m, L = 0.500 m and A = 0.0100 mm² = 1.00 × 10⁻⁸ m². Therefore R_M = 1.00 Ω. At I = 0.200 A, its gauge voltage is 0.200 V. Its interior J is 2.00 × 10⁷ A/m² and E is 0.400 V/m, giving the same voltage by EL. The first figure connects this uniform field to the voltage accumulated along the length.

S has a million times M’s resistivity, 0.0200 Ω m, but its gauge length is only 100 µm = 1.00 × 10⁻⁴ m and its area is 4.00 mm² = 4.00 × 10⁻⁶ m². Its bulk resistance is only 0.500 Ω. This reversal is possible because the two objects have very different L/A. Equal externally applied voltage would not create equal internal fields in their different lengths.

For a material comparison, normalize by the stated geometry or test matched geometry under matched conditions. At M’s dimensions, the supplied S resistivity would give 1.00 × 10⁶ Ω, rather than 0.500 Ω. That is a conditional calculation, not permission to drive a specimen outside its measured range. A component’s low R alone cannot classify its material.

Read a range of current–voltage evidence before extending it

An ohmic test checks proportionality, not only whether current increases with voltage. In the second figure, invented response A passes through the origin with constant I/V = 0.500 A/V over the shown positive range; its resistance is 2.00 Ω. On an I-versus-V plot, the straight-line slope is conductance 1/R. On a V-versus-I plot it is R. Interchanging the axes without changing the units reverses the interpretation.

Invented response B also has increasing current, but I/V is not constant: at 2.00 V its current is 0.800 A, while at 6.00 V it is about 1.714 A. The corresponding V/I ratios are 2.50 Ω and 3.50 Ω. A constant 2.50 Ω inferred at the first point would predict 2.40 A at 6.00 V, exceeding the supplied response. B is not ohmic across that range.

The smooth B curve is an original guide generated from I = V/[R₀ + kV], with R₀ = 2.00 Ω and k = 0.250 Ω/V for 0 to 6 V. This relation merely makes the chosen teaching curve reproducible; it is not a microscopic transport or thermal model and is not fitted to a specimen. A curved relation alone cannot identify whether heating, a changing carrier response, contacts or another cause is responsible. Measurements at negative bias and outside the shown interval remain separate questions.

Control or measure specimen temperature and allow the stated steady conditions before comparing ranges. Increased lattice vibrations commonly raise resistivity in ordinary metals over a suitable range. In a semiconductor, increasing temperature can increase the mobile population while also changing mobility; the balance and dopant state matter. It is wrong to assign every semiconductor one fixed negative temperature coefficient.

A limited linear approximation ρₑ(T) ≈ ρ₀[1 + α(T − T₀)] can be used when supplied and supported over the intended interval. Here α has units K⁻¹, and temperature differences have the same numerical value in kelvin and Celsius degrees. The approximation need not survive a large temperature change, phase change or different carrier regime. Self-heating also depends on the drive: at fixed V, P = V²/R, while at fixed I, P = I²R. Neither condition alone gives a temperature prediction.

Determine which path and voltage the measurement includes

In a simple linear series model, a two-terminal reading includes R_2 = R_bulk + R_extra, where R_extra collects the current leads and contacts outside the chosen gauge. Each drop is I times its own resistance, so the same current gives a total voltage equal to their sum. A contact improvement can lower R_2 without changing the interior material. A higher contact resistance therefore does not prove a higher bulk resistivity. A real interface need not behave as a fixed linear resistor.

The final figure stipulates R_bulk = 1.00 Ω and two extra contributions of 0.200 Ω each. At I = 0.100 A, the outer voltage is 0.140 V, but the defined interior gauge drop is 0.100 V. If that gauge has L = 0.100 m and A = 0.500 mm², the actual model resistivity is 5.00 × 10⁻⁶ Ω m. Using the outer 1.40 Ω with that interior geometry instead reports an apparent 7.00 × 10⁻⁶ Ω m, 40% too high.

Separate potential sensing can address this particular problem: drive the current through outer connections and sense voltage between defined inner gauge points. In the ideal limit of negligible sense current, the sense-lead voltage drops are negligible. This is the four-terminal principle: separate current and potential connections. Correct gauge placement, current distribution, leakage, offsets and temperature still need assessment. Calibration, uncertainty and repeatability explains why a repeatable number alone does not validate the measurement model.

At high resistance, an unintended leakage route can instead carry current in parallel with the intended specimen. For the same applied V, the branch currents add. Attributing all of them to the specimen overestimates its conductance and underestimates its resistance. A series-contact correction is not a correction for every parallel path.

A current below the stated detection limit is a bounded observation, not proof of zero conductivity. The time after applying voltage matters too: charging and polarisation transients can differ from steady leakage. Check the path, voltage, temperature and observation time before calling a material an ideal insulator. The wider connection to interfaces, current distribution and actual components continues in From measured transport to device behaviour.

Categories have conditions

Conductors, semiconductors and insulators describe transport behaviour within a relevant range. Temperature, fields, defects and dimensions can change observed response. At sufficiently strong fields, an insulating material may break down; the ordinary low-field classification does not promise unlimited resistance.

A material can be useful for insulation because its leakage is sufficiently small under a specified condition, while still having nonzero electronic or other transport. The band description above focuses on electronic carriers and is not a complete model of ionic conduction or a disordered material. Deliberate dopants, traps and boundary states require the further mechanisms in Doping, defects and interfaces.

Conduction and polarisation also answer different questions. An insulator need not be electrically unresponsive, and a small low-field leakage does not establish its behaviour under a large or rapidly changing field. Polarisation, dielectric response and breakdown develops those responses and the evidence needed for breakdown. The present resistance arithmetic is not an insulation rating.

A useful report states the material state and direction; gauge dimensions and voltage points; current/voltage range; temperature and relevant time conditions; and which contact, leakage and uncertainty effects were assessed. Distinguish the observed range from any extrapolation. A conditional material classification and a measured component resistance can then be related without being confused.

Worked examples

Try the reasoning

1. Is resistivity measured in ohms alone? A uniform gauge has L = 30.0 mm, A = 0.120 mm², V = 0.100 V and I = 0.400 mA. Find R, ρₑ and σ with SI units.

   Show the answer and reasoning

   No. R = 250 Ω, ρₑ = 1.00 × 10⁻³ Ω m and σ = 1.00 × 10³ S/m.

   I = 4.00 × 10⁻⁴ A, L = 0.0300 m and A = 1.20 × 10⁻⁷ m². Then R = V/I = 250 Ω and ρₑ = RA/L = 0.00100 Ω m. Conductivity is its reciprocal. Entering 0.120 as an area in m² instead would inflate the inferred resistivity by 10⁶; the raw resistance would still be 250 Ω.

2. Does a larger contact resistance prove larger bulk resistivity? In a new series model, the bulk resistance is 2.00 Ω and the total extra contact/lead contribution rises from 0.100 to 0.500 Ω. Compare outer R before and after. What independent observation would test the bulk-change claim?

   Show the answer and reasoning

   No. The outer reading rises from 2.10 to 2.50 Ω while the stipulated bulk resistance remains 2.00 Ω.

   Series drops add at the same current. A suitable separate potential measurement over the same interior gauge, with negligible sense current and checked geometry/temperature, can test whether its V/I changed. The outer increase is not by itself evidence of a new bulk resistivity.

3. An unchanged uniform circular wire has its diameter halved at the same length in a stipulated geometric comparison. What is the resistance ratio? Separately, another model has ρ₂/ρ₁ = 1.50, L₂/L₁ = 0.800 and A₂/A₁ = 0.500. Find R₂/R₁.

   Show the answer and reasoning

   Halving diameter quarters area and multiplies R by 4. The separate combined-change ratio is 2.40.

   Use area, not diameter, in R = ρₑL/A. The second ratio is 1.50 × 0.800/0.500 = 2.40. A real shaping process can also alter the material state; the first answer expressly assumes unchanged resistivity rather than predicting it from the shaping step.

4. An ideal reshaping model keeps material volume and resistivity fixed while increasing a uniform wire’s length by a factor of 1.50. What happens to area and resistance? Can unchanged resistivity be assumed for a real plastically worked wire?

   Show the answer and reasoning

   Area becomes 2/3 of its old value and resistance becomes 2.25 times the old value. The unchanged-resistivity assumption needs separate justification in a real wire.

   Constant AL gives A₂/A₁ = 1/1.50. Thus R₂/R₁ = 1.50/(1/1.50) = 1.50² = 2.25. Plastic processing, defects or temperature could change the real resistivity, so the result belongs to the stipulated geometry model.

5. A supplied local model has R₀ = 10.0 Ω at 20 °C and α = 0.00400 K⁻¹ through 70 °C. Ignore dimensional change. Find R at 70 °C. Compare current and power at fixed 2.00 V; then find power at fixed 0.200 A instead. Does either power give the temperature rise by itself?

   Show the answer and reasoning

   R becomes 12.0 Ω. At fixed 2.00 V, current falls from 0.200 to about 0.1667 A and power from 0.400 to about 0.3333 W. At fixed 0.200 A, power instead becomes 0.480 W. No temperature rise follows without thermal information.

   The supplied temperature increment is 50 K, giving R/R₀ = 1 + 0.00400 × 50 = 1.20. Use P = V²/R for fixed voltage or P = I²R for fixed current. The temperatures are supplied endpoints, not predicted by those electrical equations, and α is not a universal metal or semiconductor constant.

6. At steady 10.0 V, an intended insulating specimen is stipulated to have R_s = 2.00 × 10¹¹ Ω. A separate fixture leakage branch across the same voltage has R_f = 1.00 × 10¹⁰ Ω. Find each current, total current and the apparent resistance if all current is assigned to the specimen.

   Show the answer and reasoning

   The specimen current is 50.0 pA and fixture current 1.00 nA. Total current is 1.05 nA, giving apparent R ≈ 9.52 × 10⁹ Ω rather than 2.00 × 10¹¹ Ω.

   Each branch current is V/R. Convert 50.0 pA to 0.0500 nA before adding it to 1.00 nA. Dividing 10.0 V by 1.05 × 10⁻⁹ A gives about 9.52 GΩ. This error underestimates specimen resistance, unlike a positive series-contact contribution. No leakage path or measurement uncertainty was actually characterised by these invented values.

7. After a stated settling time, a checked steady-current measurement at 5.00 V establishes only that a passive specimen’s current magnitude is below 1.00 nA. Neglect other paths under the supplied model. What resistance statement follows? Does it prove infinite resistance or performance at a higher voltage?

   Show the answer and reasoning

   It gives a lower bound R > 5.00 × 10⁹ Ω for a nonzero current under these conditions; zero current would be consistent with an ideal infinite-resistance limit. Neither is proof of infinite resistance or higher-voltage performance.

   For a positive current magnitude, dividing the fixed voltage by a smaller-than-limit current gives a larger-than-limit resistance. The result is a detection-limit bound, not a precise value or a confidence interval. Transients, unintended leakage and measurement uncertainty would need their actual assessments before making a real bound; the stated test does not establish breakdown behaviour.

8. A short broad specimen has a lower measured R than a long thin metal wire. Its ideal band sketch has a gap. A student therefore calls it the better bulk conductor, then calls it an insulator because the gap exceeds an assumed fixed cutoff. What evidence and interpretation are missing?

   Show the answer and reasoning

   Normalize the measured interior resistance by geometry and state the operating conditions; the raw R ranking and one arbitrary gap cutoff establish neither claim.

   A gap says which ideal energy states are unavailable, not the occupied mobile populations or mobilities at the test condition. Semiconductors include wide-bandgap materials. Transport/structure evidence under stated conditions is needed for a material description; contacts and current paths must be checked before using ρₑ = RA/L. The supplied M/S comparison already shows how geometry can reverse the R ranking.

What to carry forward

• Connect each electronic category to available states and mobile response; no universal one-number cutoff classifies every material.
• Current is charge per time and voltage is energy per charge. Resistance belongs to a defined path and range; resistivity has units Ω m.
• The uniform ohmic relation R_bulk = ρₑL/A follows from J = I/A and E = V/L. Normalize geometry before comparing materials.
• A current–voltage ratio at one point does not establish constant resistance over another range, and temperature depends on more than electrical power alone.
• Series contacts, parallel leakage and detection limits change what a measurement supports. Preserve the defined gauge and conditions before making a material or device claim.