Charge carriers and electronic energy bands

Electrons have allowed energy states

In a solid, electronic states form bands separated in some cases by gaps. Electrical transport depends on available carriers, accessible states and scattering. A band picture explains more than counting valence electrons on isolated atoms.

An energy state is an allowed electronic arrangement described by quantum mechanics. In an isolated atom, electrons occupy discrete atomic states. When many atoms form a solid and their electronic states interact, the allowed energies can form groups of very closely spaced levels: bands. In an ordinary periodic crystal, electronic states can extend through many repeated atomic units. This is a collective solid-state description, not a set of isolated atoms each lending one electron in exactly the same way.

Allowed and occupied are different information. A band describes possible states; occupation describes which of them contain electrons. An empty allowed band contains no electrons to carry its electron current. Electrons also obey the exclusion principle: two electrons cannot occupy the same complete quantum state, including its spin. The available empty states near occupied ones therefore matter to how the distribution can respond to a weak applied field.

In the simple band picture used here, a partially occupied band permits a small redistribution among nearby states that can give net drift current. A completely filled band does not supply that ordinary direct-current contribution: its filled-state contributions cancel and there are no empty states within it for the same small redistribution. This does not mean its electrons are motionless or that bound charge cannot polarise. The present question is transport, not every possible response to a field.

Metals can have partially occupied bands or overlapping bands that leave accessible occupied and empty states. A gapped solid instead has an energy interval with no allowed bulk states in the ideal band model. Defects, disorder or interfaces can introduce additional states or localise carriers; the ideal diagram is a starting description, not a complete transport calculation for every solid.

An electron excitation can leave a mobile hole

For the gapped semiconductor picture, call the upper edge of the valence band E_v and the lower edge of the conduction band E_c. The band gap is E_g = E_c − E_v, an energy difference. The vertical axis in the figure is electron energy, not distance through the specimen. Drawing an electron higher on that axis does not mean it has moved physically to the top of the sample.

Begin with an ideal filled valence band and empty conduction band. An excitation that promotes an electron across the gap creates a conduction-band electron and an unoccupied valence-band state. Relative to the filled valence-band reference, that missing negative charge can be described as a positive carrier: a hole. Redistribution of the remaining valence electrons can move the hole. It is not an empty atomic site, a pore or a positive atomic nucleus travelling through the lattice.

One electron–hole pair generation event adds one conduction electron and one hole, so paired generation alone adds equal numbers of the two. Recombination can remove a mobile electron–hole pair from that population. Their concentrations need not be equal in every semiconductor: added species, contacts and prior excitation can establish different backgrounds. Those mobile counts are not the specimen’s total charge: atomic cores and fixed charged states also contribute. The next doping lesson examines those causes; here the carrier populations are supplied or measured inputs.

Thermal excitation, absorbed light and other processes can change occupation. The population depends on the available states and occupation statistics, not just a statement that the gap exists. There is no universal one-electron-volt boundary that separates all semiconductors from all insulators. A gap value alone also gives neither the mobility nor a component’s resistance. Energy-band diagrams do not count how many mobile carriers survive trapping or recombination.

Mobility describes a net response, not every carrier’s speed

Mobile carriers already undergo motion without an applied field. In uniform thermal equilibrium there is no preferred direction, so their average charge flow is zero even though they are not stationary. An applied electric field biases that motion and produces an average drift. A drift velocity is the average response of a population, not a straight unscattered trajectory followed by each carrier.

An electric field describes force per unit positive test charge; its direction is the force direction for a positive charge. Its units N/C are equivalent to V/m. Take x along a chosen positive direction and write the signed field component as ℰ_x, in volts per metre. At sufficiently low field in the chosen direction, define positive electron and hole mobilities μ_n and μ_p by v_dn = −μ_n ℰ_x and v_dp = +μ_p ℰ_x. The minus sign is because an electron has negative charge; hole drift follows the field. Mobility has units m²/(V s), so multiplying it by V/m gives m/s.

Lattice vibrations, charged impurities and other departures from an ideal periodic environment can scatter carriers and reduce the persistent directional response. Interfaces and defects can matter as well. The electronic band structure also affects that response. Scattering is not the same event as removing a carrier: a carrier can be deflected while remaining mobile. Trapping and recombination can instead change the mobile population.

Changing temperature or adding species can change population and mobility together. Heating can make more electronic carriers available in a gapped material while also changing scattering. Therefore neither “more thermal motion” nor “more carriers” alone specifies the resulting conductivity. The model below takes mobilities and concentrations at the actual stated material condition.

At high field the simple proportional relation can fail, for example when drift speed approaches a saturation regime. A value of μ measured at low field cannot simply be used to predict indefinitely increasing speed. In a spatially nonuniform population, diffusion can also carry charge even with no applied drift field. Here the worked current calculations use uniform populations and the low-field drift contribution only.

Population and mobility both matter

A simple carrier model writes conductivity as the sum of carrier concentration times charge magnitude times mobility for each carrier type. Increasing concentration does not guarantee a proportional conductivity increase if mobility changes. Temperature, impurities and interfaces can affect both.

Let e be the positive elementary-charge magnitude, n the mobile conduction-electron concentration and p the mobile hole concentration, both numbers per cubic metre. Current density J_x is signed charge crossing unit area per unit time, in A/m². A uniform population with signed average velocity v carries current density equal to charge per carrier times concentration times v.

For electrons, J_n = (−e)n v_dn = enμ_n ℰ_x. For holes, J_p = (+e)p v_dp = epμ_p ℰ_x. The two negative signs in the electron term cancel. Although electrons and holes drift in opposite directions, their conventional drift currents have the same direction for the same field. Conventional current follows positive charge flow, not electron motion.

Add the currents: J_x = J_n + J_p = e(nμ_n + pμ_p)ℰ_x. Defining conductivity σ by J_x = σℰ_x gives σ = e(nμ_n + pμ_p). This scalar equation is for the specified direction and regime. A strongly anisotropic material needs an appropriate directional or tensor description; a single number need not serve every orientation.

Check the units before calculating. For one contribution, C × m⁻³ × m²/(V s) = C/(V s m) = A/(V m) = S/m, where S is the siemens, A/V. Multiplication by field in V/m then gives A/m². Conductivity and current density are different quantities. A mobile-carrier concentration is also different from the number of atoms in the solid.

For the numerical examples use the explicitly rounded e = 1.60 × 10⁻¹⁹ C. The SI-defined magnitude is 1.602176634 × 10⁻¹⁹ C exactly; the rounding is for these chosen teaching calculations. The quantities nμ_n and pμ_p determine the separate contributions. The carrier type with more particles does not necessarily dominate if its mobility is much smaller.

More carriers can coexist with a lower conductivity

The table contains original invented bulk states at one stated temperature and low-field measurement condition. A and B are the before/after states of a fictional treatment; hole current is stipulated negligible in that comparison. C is a separate two-carrier example with equal supplied electron and hole populations. These are not measured properties of silicon or a prescribed treatment, and no generation, doping or mobility law is being fitted.

For A, enμ_n is 25.6 S/m. B has three times the mobile-electron concentration but one quarter the mobility. Their product changes by 3 × ¼ = ¾, giving 19.2 S/m, a 25% decrease. Comparing carrier numbers alone would predict the wrong direction. The result does not identify which scattering or trapping mechanism caused the supplied mobility change.

For C, the electron contribution is 19.2 S/m and the hole contribution 6.40 S/m, for a total of 25.6 S/m. C and A therefore share the same total conductivity in this model despite different carrier populations and contributions. Equality of one macroscopic response does not establish equality of microscopic state.

A measured conductivity does not determine every factor

A conductivity measurement constrains a sum of products. Even when one carrier type dominates, σ ≈ enμ_n contains two unknown factors if neither population nor mobility is independently known. Measuring a larger σ does not by itself prove that treatment added carriers rather than changed their mobility. A band gap does not supply those missing quantities.

Additional evidence can constrain the explanation. Carrier-sensitive measurements, field and temperature dependence, controlled illumination and structural/compositional information can test different hypotheses. Each method has its own model and what it samples. For example, a Hall-derived effective population or mobility is not automatically the separate drift parameter of every carrier in a two-carrier material.

Contact quality, sample uniformity, geometry and temperature affect how electrical measurements are interpreted. Those checks matter before attributing an apparent change to bulk electronic structure. Here no Hall inversion formula or measurement procedure is prescribed: the lesson’s task is to identify which independent evidence the conductivity alone lacks.

Finally distinguish transport in the material from a two-terminal component reading. Dimensions, contacts, junctions, nonuniform fields and heating can affect that reading. The next lessons develop resistance and device connections. The present band-and-carrier model supplies one part of that argument, under its stated spatial, field and material-state assumptions.

Worked examples

Try the reasoning

1. Does an energy gap alone specify measured resistance?

   Show the answer and reasoning

   No.

   A gap describes unavailable energies in an ideal band picture. Mobile populations, mobilities, sample dimensions and contacts still matter. Even equal gaps do not establish equal conductivity, and conductivity alone is not a component resistance.

2. What is mobility?

   Show the answer and reasoning

   A measure relating carrier drift response to electric field in a stated regime.

   In the low-field model the drift magnitude divided by field magnitude gives mobility. It is not the random thermal speed, and a zero drift average in uniform equilibrium does not mean all carriers are stationary.

3. A new electron-dominated case has n = 3.00 × 10¹⁵ cm⁻³ and μ_n = 400 cm²/(V s). Convert both to SI and calculate σ using the rounded e. What error occurs if the SI concentration is multiplied by the unconverted numerical mobility and labelled S/m?

   Show the answer and reasoning

   n = 3.00 × 10²¹ m⁻³, μ_n = 0.0400 m²/(V s), and σ = 19.2 S/m. The mixed-unit calculation overestimates this by a factor of 10,000.

   One cubic centimetre is 10⁻⁶ m³, so the concentration increases numerically by 10⁶ on conversion to m⁻³. One square centimetre is 10⁻⁴ m², so mobility becomes 400 × 10⁻⁴. The correctly converted product is (1.60 × 10⁻¹⁹)(3.00 × 10²¹)(0.0400).

4. For another electron-dominated treatment, n increases by a factor of 1.50 and mobility becomes 0.900 of its old value. Find the conductivity ratio. What mobility ratio would instead leave conductivity unchanged, with all other assumptions retained?

   Show the answer and reasoning

   Conductivity becomes 1.35 times the old value, a 35% increase. The unchanged-conductivity mobility ratio would be 1/1.50 = 2/3.

   With one carrier term, the ratio is the product of population and mobility ratios. This algebra assumes the same carrier model and negligible other contribution; it does not predict the mobility from a treatment name.

5. A fictional uniform two-carrier state has n = p = 5.00 × 10²⁰ m⁻³, μ_n = 0.0400 and μ_p = 0.0200 m²/(V s). The signed field is ℰ_x = −50.0 V/m in the linear regime. Find the signed drift velocities and total conventional drift-current density.

   Show the answer and reasoning

   v_dn = +2.00 m/s; v_dp = −1.00 m/s; J_x = −240 A/m².

   The charge-density magnitudes are 80.0 C/m³. Electron current is (−80.0)(+2.00) = −160 A/m² and hole current is (+80.0)(−1.00) = −80.0 A/m². Equivalently σ = e[nμ_n+pμ_p] = 4.80 S/m and J_x = 4.80(−50.0). Current follows the reversed field despite the electron drift pointing the other way.

6. A supplied state has n = 1.00 × 10²⁰ m⁻³, μ_n = 0.200, p = 1.00 × 10²¹ m⁻³ and μ_p = 0.00500 m²/(V s). Which carrier contributes more to σ, and what fraction of total conductivity is electronic?

   Show the answer and reasoning

   Electrons contribute 3.20 S/m, holes 0.800 S/m; electrons provide 80.0% of the 4.00 S/m total despite being less numerous.

   Compare nμ_n with pμ_p, rather than counts alone. The electron concentration is one tenth of the hole concentration but electron mobility is 40 times larger, giving four times the conductivity contribution. This is a stipulated teaching case, not a universal relation between electron and hole mobilities.

7. In a new exposure of the same 1.00 mm³ volume, 1.00 × 10¹⁰ pairs are initially generated, and only half remain mobile at the reading time. Both supplied mobilities stay unchanged. Find Δσ. Would the same answer follow if half the pairs survived but the mobilities also changed?

   Show the answer and reasoning

   Δσ = 0.128 S/m with unchanged mobilities. If mobilities also change, their new values are required.

   The surviving additional concentration of each type is 5.00 × 10¹⁸ m⁻³, halving 0.256 S/m. The total historical generation count cannot replace the current mobile population; nor can survival alone specify its drift response.

8. A learner calls the unoccupied valence-band state in the first figure a missing atom and takes the upward excitation arrow as an electron trajectory across the specimen. Correct both interpretations.

   Show the answer and reasoning

   The hole is missing electronic occupation relative to a filled valence band, not an atomic vacancy. The vertical arrow denotes a change of electron energy, not a spatial flight path.

   The diagram has an energy axis and no real-space coordinate. Its dot and ring are illustrative carrier symbols, not the complete density of states, a crystal image or a measured carrier count.

9. A nonuniform carrier population spreads from a high-concentration region towards a lower-concentration region without an externally applied field. Does J_drift = σℰ prove that its total current is zero?

   Show the answer and reasoning

   No. It sets that imposed-field drift contribution to zero; a concentration gradient can give diffusion current, and internal fields may also develop.

   The worked conductivity model assumed spatially uniform populations. Applying its drift-only conclusion to a nonuniform problem removes a relevant transport mechanism. A full account must include the actual gradients and fields rather than add an unsupported zero-current claim.

What to carry forward

• Distinguish allowed states from their occupation; a band sketch is not a carrier-count measurement.
• A hole describes an unoccupied valence-band state, and its drift current adds to electron drift current under the same field.
• Mobility relates net drift to field in a stated regime; it is not every carrier’s thermal speed.
• Conductivity combines carrier concentrations and mobilities, with units and both carrier contributions made explicit.
• A measured sum of products does not uniquely identify all factors, and a bulk transport model is only part of a device interpretation.