Energy Bands and Carrier Transport: From Bonds to Drift, Diffusion and the Continuity Equations

In an intrinsic semiconductor every conduction electron came from a broken bond, so it left a hole behind and n = p = n_i. The Fermi level E_F — the energy at which the occupation probability is one half — then sits close to the middle of the gap, and it is worth knowing exactly how close, because the exam sometimes asks. The intrinsic level is E_i = (E_c + E_v)/2 + (kT/2) ln(N_v/N_c), and for silicon with N_c = 2.8 × 10¹⁹ and N_v = 1.04 × 10¹⁹ cm⁻³ the correction is (0.0259/2) × ln(0.371) = −0.0128 eV. So E_i lies 12.8 meV below midgap, which is half a kT. Calling the intrinsic Fermi level 'midgap' is therefore an approximation good to half a thermal voltage, and for every purpose in this syllabus that is good enough — but the reason it is not exactly midgap is worth having, and the reason is that electrons and holes have different effective masses, so the two bands do not offer the same number of states.

Doping makes the material extrinsic. Substitute a group-V atom — phosphorus, arsenic, antimony — into a silicon site and it brings five valence electrons where four bonds are needed, so the fifth sits in a donor level only about 45 meV below E_c for phosphorus. At 300 K that is 45/25.9 = 1.74 kT, shallow enough that essentially every donor is ionised, which is why we write n ≈ N_D without apology. A group-III atom — boron, aluminium, gallium — brings three electrons where four are needed, leaving an acceptor level about 45 meV above E_v that captures an electron from the valence band and produces a hole. The band edges do not move, the gap does not change, and no new states appear inside the bands. What moves is E_F. In n-type material E_F climbs toward E_c; in p-type it sinks toward E_v. That is the entire band-diagram content of doping, and it is what makes a junction possible: bring an n-region and a p-region together and the two Fermi levels must line up at equilibrium, which is only achievable by bending the bands — and the amount of bending is the built-in potential of the next chapter.

Push the doping far enough and the description above breaks. As N_D approaches the conduction-band effective density of states — around 2.8 × 10¹⁹ cm⁻³ for silicon — E_F climbs to within a kT of E_c and then enters the band. The material is now degenerate: the Boltzmann approximation n = N_c exp(−(E_c − E_F)/kT) fails and full Fermi–Dirac statistics are needed, the gap narrows slightly because the impurity wavefunctions overlap into a band of their own, and the semiconductor starts behaving metallically. This is not a curiosity to be avoided; it is deliberately exploited. Every ohmic contact on an integrated circuit is made by doping the semiconductor degenerately (written n+ or p+) right under the metal, because a degenerate surface makes the barrier at the metal so thin that carriers tunnel straight through it instead of having to be thermally lifted over it. The rule of thumb to carry is the validity condition for everything else in this chapter: the Boltzmann form is safe while E_c − E_F exceeds about 3kT, that is while N_D stays a couple of orders of magnitude below N_c.

Everything in this section follows from two statements. The first is charge neutrality: a lump of semiconductor sitting on the bench has no net charge, so p + N_D⁺ = n + N_A⁻, and with complete ionisation that is p + N_D = n + N_A. The second is the mass-action law np = n_i², which comes from writing n = N_c exp(−(E_c − E_F)/kT) and p = N_v exp(−(E_F − E_v)/kT) and multiplying: the E_F cancels, leaving np = N_c N_v exp(−E_g/kT) = n_i², a quantity that depends on the material and the temperature and not on where the Fermi level sits. Solving the two together for n-type material gives the exact result n = N_D/2 + √((N_D/2)² + n_i²), and it is instructive to see how little the second term matters: at N_D = 10¹⁷ and n_i = 1.5 × 10¹⁰, the correction is of relative size (n_i/N_D)² ≈ 2 × 10⁻¹⁴. So n ≈ N_D is not laziness, it is accurate to fourteen figures — but the exact form is the one to reach for when the doping is comparable to n_i, which happens either in very lightly doped material or at high temperature.

Once n is known the Fermi level follows from n = n_i exp((E_F − E_i)/kT), so E_F − E_i = kT ln(n/n_i). For our N_D = 10¹⁷ cm⁻³ sample, E_F − E_i = 0.0259 × ln(10¹⁷/1.5 × 10¹⁰) = 0.0259 × 15.713 = 0.407 eV, so the Fermi level has climbed 0.407 eV above the intrinsic level and now sits about 0.15 eV below the conduction-band edge. The p-type mirror image is E_i − E_F = kT ln(p/n_i), which for N_A = 10¹⁷ gives the same 0.407 eV downward. Compensation is where the two equations earn their keep. Dope a sample with both N_D = 5 × 10¹⁶ and N_A = 2 × 10¹⁶ cm⁻³. The donors and acceptors do not annihilate; each is still there, still ionised, still scattering carriers. But charge neutrality only sees the net, so n ≈ N_D − N_A = 3 × 10¹⁶ cm⁻³ and p = 2.25 × 10²⁰/3 × 10¹⁶ = 7500 cm⁻³. The material behaves electrically as though it were doped at 3 × 10¹⁶, and this is exactly how a p-well is made inside an n-substrate, and how a base region is formed by diffusing acceptors into a donor-doped collector.

Band edges are not flat lines; they are the extremes of curves plotted against crystal momentum k. In gallium arsenide the minimum of the conduction band and the maximum of the valence band both sit at k = 0, the centre of the Brillouin zone, so an electron can fall from one to the other without changing momentum: a direct gap. In silicon the valence maximum is at k = 0 but the conduction minimum lies out along the ⟨100⟩ direction, and in germanium it lies along ⟨111⟩; both are indirect gaps. Now put a photon into the picture. A photon of 1.12 eV has a wavelength of 1107 nm and therefore a momentum smaller than the width of the Brillouin zone by roughly three orders of magnitude — it carries plenty of energy and essentially no momentum. So a direct transition needs one photon and nothing else, while an indirect transition must find a phonon to supply the missing momentum at the same instant. Requiring two particles to arrive together instead of one is what makes indirect transitions orders of magnitude less probable, and every consequence in this section is that single fact propagating.

Apply a field E and a carrier accelerates, scatters off a phonon or an ionised impurity, and starts again. Averaged over many scattering events the result is a steady drift velocity proportional to the field, v_d = µE, where the mobility µ has units of cm²/V·s and is nothing more than the proportionality constant. The drift current density is then J_drift = q(n µ_n + p µ_p)E = σE, which is Ohm's law with σ = q(n µ_n + p µ_p) and ρ = 1/σ. Work the reference sample: n-type silicon at N_D = 10¹⁷ cm⁻³, where the mobilities at that doping are µ_n = 800 and µ_p = 350 cm²/V·s. The electron term is σ_n = 1.6 × 10⁻¹⁹ × 10¹⁷ × 800 = 12.8 S/cm, so ρ = 1/12.8 = 0.0781 Ω·cm. Now compute the hole term and see what happens to it: σ_p = 1.6 × 10⁻¹⁹ × 2250 × 350 = 1.26 × 10⁻¹³ S/cm, which is fourteen orders of magnitude smaller. In extrinsic material the minority carriers contribute nothing whatever to conductivity, which is why every resistivity formula you will use in this paper has one term in it — and also why minority carriers, which are irrelevant here, become the entire story the moment a junction injects them.

Mobility itself is not a constant of the material but a sum of scattering channels, combined by Matthiessen's rule 1/µ = 1/µ_lattice + 1/µ_impurity. Lattice (phonon) scattering worsens as the crystal vibrates harder, so µ_lattice falls roughly as T^(−3/2). Ionised-impurity scattering does the opposite: a faster carrier is deflected less by a fixed charge, so µ_impurity rises roughly as T^(3/2) and falls as the impurity count rises. The two together give a mobility that peaks at some intermediate temperature and that falls with doping — which is exactly what the numbers in this chapter show, µ_n dropping from 1350 in lightly doped silicon to 800 at 10¹⁷ cm⁻³. Two examinable consequences. First, in a compensated sample the scattering counts the total ionised impurity concentration, not the net: our N_D = 5 × 10¹⁶, N_A = 2 × 10¹⁶ sample has n = 3 × 10¹⁶ but is scattered by all 7 × 10¹⁶ ionised centres, so its mobility is lower than that of a uniformly 3 × 10¹⁶ n-type sample. Second, in the extrinsic regime n is pinned at N_D while µ falls with temperature, so the resistivity of a doped semiconductor rises with temperature — a positive temperature coefficient, the opposite of the intrinsic behaviour everyone remembers, and the basis of the diffused silicon resistor's well-known positive drift.

The second transport mechanism needs no field at all. If carriers are more concentrated in one place than another, random thermal motion carries more of them out of the crowded region than into it, and the net flux is proportional to the concentration gradient. Fick's law gives the particle flux, and multiplying by charge gives the current: J_n,diff = +q D_n (dn/dx) and J_p,diff = −q D_p (dp/dx). The signs are worth pausing on, because they are a standing source of lost marks. Holes diffuse down their gradient and carry positive charge with them, so a positive gradient gives a negative current — hence the minus sign. Electrons also diffuse down their gradient, but they carry negative charge, so the two sign reversals cancel and the electron diffusion current comes out with a plus sign. Write both down once, in that form, and check them against the physics rather than trying to remember which one has the minus.

Now compare the two mechanisms on real numbers, because the comparison is what tells you which one to keep in a given region of a device. Suppose an excess electron concentration falls linearly from Δn = 10¹⁵ cm⁻³ at x = 0 to zero at x = 1 µm. The gradient is 10¹⁵/10⁻⁴ = 10¹⁹ cm⁻⁴, so J_diff = q D_n (dn/dx) = 1.6 × 10⁻¹⁹ × 20.7 × 10¹⁹ = 33.2 A/cm² — a very large current density from a gentle-sounding gradient, and the whole reason diffusion cannot be neglected. Now let the same 10¹⁵ cm⁻³ of electrons sit in a modest field of 10 V/cm: J_drift = q n µ_n E = 1.6 × 10⁻¹⁹ × 10¹⁵ × 800 × 10 = 1.28 A/cm², which is 26 times smaller. That ratio is the structural fact behind every diode and bipolar transistor calculation you will do. In the quasi-neutral regions the majority carriers screen out any field so effectively that only a few volts per centimetre survive, and minority-carrier transport there is pure diffusion. In the depletion region the field is tens of kilovolts per centimetre and drift dominates completely. Devices are analysed region by region not for convenience but because in each region one of these two terms is genuinely negligible.

At equilibrium generation and recombination are both happening constantly and at exactly equal rates, which is what np = n_i² records. Disturb the balance — shine light on the sample, or forward-bias a junction into it — and an excess population Δn = Δp appears, decaying back by recombination at a rate Δn/τ where τ is the carrier lifetime. Three recombination paths compete. Band-to-band recombination is the direct fall from E_c to E_v and, as the last section explained, is feeble in an indirect-gap material. Shockley–Read–Hall recombination goes through a trap level near midgap contributed by a defect or an unwanted impurity, and in silicon it is the dominant path by a wide margin. Auger recombination hands the energy to a third carrier instead of a photon and only matters at very high carrier density, which is why it limits heavily doped emitters and high-injection solar cells. Under low-level injection — Δn much smaller than the majority concentration — the decay is a clean single exponential Δn(t) = Δn(0) exp(−t/τ), and that assumption is what makes every analytical device model in this syllabus tractable.

Work an illumination example on the reference sample, N_D = 10¹⁷ cm⁻³ with τ = 1 µs. Illuminate it uniformly at an optical generation rate G_L = 10²⁰ cm⁻³s⁻¹. In steady state generation equals recombination, so G_L = Δn/τ, giving Δn = Δp = 10²⁰ × 10⁻⁶ = 10¹⁴ cm⁻³. Check the low-injection assumption first, because it is the assumption that could invalidate everything after it: Δn/n_0 = 10¹⁴/10¹⁷ = 0.001, so the majority population changed by one part in a thousand and low-level injection holds comfortably. The photoconductivity is Δσ = q(Δn µ_n + Δp µ_p) = 1.6 × 10⁻¹⁹ × 10¹⁴ × (800 + 350) = 0.0184 S/cm, against a dark conductivity of 12.8 S/cm — a change of just 0.14 per cent. Switch the light off and the excess decays with a 1 µs time constant, reaching a tenth of its value in τ ln 10 = 2.30 µs.

One warning about τ itself, because it is the parameter students most often treat as a material constant and it is nothing of the sort. Lifetime is inversely proportional to the density of recombination centres, so it is set by how clean the crystal is and by what has been deliberately put into it. Float-zone silicon can reach milliseconds; the same wafer contaminated with gold or platinum drops to nanoseconds, and that contamination is done on purpose in fast recovery diodes, where a long lifetime would mean a long reverse-recovery time and switching losses. Surfaces are recombination centres too — a free surface has dangling bonds and is described by a surface recombination velocity — so a small or thin device can be limited entirely by its surfaces rather than its bulk, which is why passivation layers exist and why a solar cell's rear surface field is worth the extra process step. When a question quotes a lifetime, it is quoting a fabrication choice, not a property of silicon.

Poisson's equation relates the field to the charge that produces it: dE/dx = ρ/ε_s, or equivalently d²φ/dx² = −ρ/ε_s, with ρ = q(p − n + N_D − N_A). For silicon ε_s = 11.7 × 8.854 × 10⁻¹⁴ = 1.036 × 10⁻¹² F/cm. Inside a depletion region the free carriers have been swept out, so ρ is just the ionised dopant charge and the field profile is a straight line whose slope you can compute in one step. For N_D = 10¹⁶ cm⁻³, dE/dx = qN_D/ε_s = 1.6 × 10⁻¹⁹ × 10¹⁶/1.036 × 10⁻¹² = 1.54 × 10⁹ V/cm², which is 154 kV/cm for every micron of depletion. That single number tells you why depletion regions are fractions of a micron wide rather than tens of microns: a volt or two of band bending is used up almost immediately. The same equation, applied across a gate oxide instead, is what gives the MOS capacitor of the next chapter its Q = C_ox V behaviour.

The continuity equations are conservation of carriers written per unit volume: the rate at which electrons pile up in a slice equals what flows in minus what flows out, plus what is generated minus what recombines. In one dimension, ∂n/∂t = (1/q)(∂J_n/∂x) + (G_n − R_n) and ∂p/∂t = −(1/q)(∂J_p/∂x) + (G_p − R_p) — again the sign difference is only the charge of the carrier. Substitute the drift-diffusion expressions for J, keep only diffusion because we are in a quasi-neutral region, use R − G = Δp/τ_p for low-level injection, and look for a steady state. Everything collapses to the minority-carrier diffusion equation D_p (d²Δp/dx²) = Δp/τ_p, whose solution is Δp(x) = Δp(0) exp(−x/L_p) with L_p = √(D_p τ_p). That exponential, and that length, are the answer to almost every diode question there is. For our sample, L_p = √(9.07 × 10⁻⁶) = 3.01 × 10⁻³ cm = 30.1 µm and L_n = √(20.7 × 10⁻⁶) = 45.5 µm, both with τ = 1 µs. Notice that L_n > L_p purely because D_n > D_p, which traces all the way back to electrons being more mobile than holes.