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In a periodic crystal potential, atomic levels broaden into bands of allowed bloch states. Coupling between states of the same symmetry opens avoided crossings, including Bragg-coupled states in the nearly free-electron picture. Bonding and antibonding combinations provide a complementary tight-binding picture. A semiconductor has a filled valence band and an empty conduction band at zero temperature separated by a forbidden energy interval; thermal excitation, light and doping provide mobile electrons and holes. The gap depends on the crystal's periodic potential, bonding and symmetry, not on a classical collision barrier. In a direct gap, the relevant extrema coincide in wavevector.

For the stated parabolic edges, take $E_v(k)=-{\hslash }^2{k}^2/(2m_h^{\ast })$ and $E_c(k)=E_g+{\hslash }^2{k}^2/(2m_e^{\ast })$. Negligible photon momentum and no phonon make the transition nearly vertical in $E$ versus $k$. parabolic-band optical energy sharing divides the excess photon energy $0.32\mathrm{eV}$ in inverse proportion to the effective masses:Thus the initial and final electron energies, relative to the valence-band maximum, areTheir difference is $0.5000\mathrm{eV}$. The selected $|k|$ is about $9.33\times 10^8{\mathrm{m}}^{-1}$; either sign gives the same energy in this symmetric approximation.

A shallow donor supplies an electron in a screened impurity potential, so its localized level lies just below the conduction edge. A shallow acceptor binds a hole and has an electron-occupation level just above the valence edge; thermal ionization supplies mobile holes. The level positions are measured relative to the relevant band edge, and deep impurities need not obey this hydrogenic approximation.

For the supplied bound-carrier mass and unit effective ionic charge, dielectric screening and effective mass change the hydrogen binding energy toThis corresponds to $E_b/k_B\simeq 110\mathrm{K}$ and an effective Bohr radius about $6.35\mathrm{nm}$. The binding is shallow compared with the band gap and typical atomic binding. Trivalent denotes chemical valence, not an unscreened hydrogenic charge of three. An acceptor physically binds a hole; the question's stated $0.1m_e$ is used as its stipulated bound-carrier mass rather than silently replacing it by the separate band-edge $0.4m_e$.

A completely filled valence band carries no net current: its occupied crystal-momentum states have velocities that cancel. Near a valence-band maximum, an almost full band is conveniently described by the few missing electrons, or electron holes. Removing an electron of charge $-e_0$, with $e_0>0$, changes the charge relative to the filled band by $+e_0$. This is a quasiparticle description of electronic motion, not a mobile positive ion or a Positron.

For an isotropic parabolic maximum, ${ϵ}_v(\mathbf{k})=E_v-{\hslash }^2{k}^2/(2m_h^{\ast })$. Assign the missing-electron state momentum ${\mathbf{k}}_h=-{\mathbf{k}}_e$ and hole energyIt has positive effective mass $m_h^{\ast }$. Its velocity equals the velocity of the missing electron state, and its positive-charge current replaces the negative of that state's electronic current. Semiclassical motion is thus $\hslash {\dot{\mathbf{k}}}_h=+e_0(\mathbf{E}+{\mathbf{v}}_h\times \mathbf{B})$. With momentum relaxation time ${\tau }_h$, the low-field mobility is ${\mu }_h=e_0{\tau }_h/m_h^{\ast }$ andIf electrons also contribute, $\sigma =e_0(p{\mu }_h+n{\mu }_e)$: the two species' longitudinal conductivities add.

A Hall bar separates carrier concentration from mobility. Apply current density $j_x$, magnetic field $B_z$, and measure transverse electric field $E_y$. For one isotropic hole population, transverse force balance gives $E_y=v_x{B}_z$, soFor thickness $t$, the Hall-voltage magnitude is $|V_H|=IB/(e_{0}pt)$. A longitudinal four-terminal resistance measurement supplies $\sigma$ independently. Use ohmic contacts, known conducting thickness, controlled temperature and low enough current to avoid self-heating. Reversing magnetic field removes even-in-field longitudinal offsets from contact misalignment; reversing current also helps reject thermoelectric offsets. These are central controls, rather than merely taking one transverse voltage reading.^[ref]

The simple inference assumes one carrier type and a known Hall scattering factor. More generally $R_H=r_H/(e_{0}p)$, so $\sigma R_H$ measures Hall mobility rather than automatically the drift mobility. With two carrier populations at weak field,Its sign alone need not identify the more numerous carrier. A Hall experiment also cannot determine $m_h^{\ast }$ and ${\tau }_h$ separately: their ratio sets mobility. cyclotron resonance supplies an independent mass measurement by sweeping frequency or magnetic field through $\omega =e_{0}B/m_c$ and detecting resonant absorption. The measured cyclotron effective mass can depend on band anisotropy and need not equal every curvature mass; the isotropic parabolic model identifies them. Its linewidth provides a scattering-time estimate only with a specified damping model.

semiconductor doping generates holes by adding acceptor impurities. In silicon, a substitutional group-III atom such as boron provides one fewer valence electron than silicon. A shallow acceptor level lies slightly above the valence-band edge: accepting a valence electron makes the dopant negatively ionized and leaves a mobile hole in the valence band. The impurity stays fixed in the lattice. charge neutrality and the nondegenerate equilibrium mass-action law giveFor fully ionized, uncompensated acceptors in the extrinsic regime, $p\simeq N_A$ and $n\simeq n_i^2/N_A$; with fully ionized donor compensation, $p\simeq N_A-N_D$ while holes remain dominant. The Fermi level moves towards the valence band. At low temperature, semiconductor carrier freeze-out suppresses mobile electron holes; at high temperature intrinsic electron-hole generation can dominate. Increasing dopant density also changes scattering, so a conductivity increase alone cannot distinguish increased carrier density from changed mobility. Temperature-dependent Hall and conductivity measurements separate these effects, subject to the carrier and scattering qualifications above.

Semiconductor doping introduces impurities with levels close to the band edges. A donor impurity can release an electron to the conduction band, leaving a fixed positive ion; an n-type semiconductor has electrons as majority carriers. An acceptor impurity can take an electron from the valence band, leaving a mobile hole and fixed negative ion; a p-type semiconductor has holes as majority carriers. The opposite species remains as a minority carrier population, and a neutral bulk sample balances the ionised dopants with mobile carriers.

At a p-n junction, concentration gradients initially drive electrons and holes across the join. Their removal near the join exposes negative acceptor ions on the p side and positive donor ions on the n side, forming a depletion region. Its electric field opposes further majority-carrier diffusion. Equilibrium has a flat Fermi level, no net current and spatially bent band edges. With p on the left and n on the right, the electron band edges fall towards the n side; the difference is the equilibrium barrier in electron energy. A flat Fermi level does not mean flat conduction and valence edges.

A semiconductor has a filled valence band and an empty conduction band at zero temperature in the intrinsic idealisation, separated by a modest band gap. Thermal excitation or deliberately introduced impurities can create mobile electrons and electron holes. Its conductivity is consequently very sensitive to temperature and doping. A metal has available electronic states at the occupied energy boundary, through a partly filled band or overlapping bands, and need not thermally excite carriers across a gap.

For a direct band gap, the conduction-band minimum and valence-band maximum occur at the same crystal momentum. An electron can recombine with a hole near these extrema and emit a photon while conserving crystal momentum, since optical photon momentum is small on the lattice scale. For an indirect band gap, the extrema lie at different crystal momenta. A band-edge radiative transition then ordinarily needs a phonon as well, making it less probable and allowing nonradiative processes to compete more effectively.

gallium arsenide has a direct fundamental gap, while silicon has an indirect one. This makes efficient band-edge emission and hence a light-emitting diode easier to obtain from gallium arsenide. It does not mean silicon has no optical transitions or that every direct-gap sample emits efficiently: defects, surfaces and carrier confinement also matter.

Doping changes the mobile carrier population by introducing controlled impurity levels. In an intrinsic semiconductor, the filled valence band is separated from the nearly empty conduction band by a band gap $E_g$. Thermal excitation produces an electron-hole pair, so the electron and hole number densities obey $n=p=n_i$. A electron hole represents a missing valence-band electron and behaves as a mobile positive charge carrier. For nondegenerate bands in thermal equilibrium,Here $N_C,N_V$ are the effective density of states, $E_C,E_V$ the band edges, and $E_F$ denotes the chemical potential, conventionally called the Fermi level. The semiconductor mass-action law holds in this nondegenerate equilibrium regime.

In a tetrahedrally bonded elemental semiconductor such as silicon, a substitutional group-V atom has an extra outer electron relative to the host. It is a donor impurity: a shallow level just below $E_C$ can ionise, releasing an electron to the conduction band and leaving a fixed positively charged donor. The resulting n-type semiconductor has electrons as majority carriers and holes as minority carriers. Conversely a group-III substitution supplies one fewer bonding electron. An acceptor impurity can take an electron from the valence band, leaving a mobile electron hole and a fixed negative acceptor; the resulting p-type semiconductor has holes as majority carriers.

The impurity state is shallow because the electron or hole responds to a screened Coulomb potential and an effective mass, rather than the bare atomic potential. A simple hydrogenic donor estimate iswhere ${ϵ}_r$ is the relative permittivity. This is a useful scaling model, with real band anisotropy and central-cell effects omitted. Dopants are fixed in the lattice; their ionisation changes mobile carrier numbers without making a uniformly doped bulk sample carry a large net electric charge.

The charge neutrality condition isWhen shallow dopants are almost fully ionised and $N_D-N_A\gg n_i$, an n-type region has $n\simeq N_D-N_A$ and $p\simeq n_i^2/n$. A p-type region similarly has $p\simeq N_A-N_D$ and $n\simeq n_i^2/p$. Thus compensation by the opposite impurity type reduces the majority-carrier density; the total dopant concentration is not generally the mobile carrier density. At sufficiently high impurity density the degenerate semiconductor regime requires the full Fermi-Dirac distribution and these exponential approximations fail.

There are three useful temperature regimes. At low temperature, dopant freeze-out leaves many shallow levels un-ionised. In the extrinsic regime, the dopants are largely ionised and set the majority-carrier density. At high enough temperature, intrinsic pair creation dominates and $n,p$ approach $n_i$. The Fermi level moves towards the conduction band with donor doping and towards the valence band with acceptor doping. The electrical conductivity isDoping therefore controls conductivity and its carrier type, but ionised impurities also scatter carriers and can reduce charge carrier mobility. Carrier density and mobility must both be considered.

A p-n junction forms when p-type and n-type regions are joined within a semiconductor. Initially the concentration gradients drive diffusion current: electrons diffuse from the n region to the p region and holes in the opposite direction. Recombination removes mobile majority carriers near the join, exposing fixed positive donors on the n side and fixed negative acceptors on the p side. This forms a depletion region and an electric field directed from n to p. Its drift current opposes diffusion. At equilibrium the net current vanishes and the Fermi level is uniform, although the band edges vary spatially through band bending.

For an abrupt nondegenerate junction with fully ionised dopants, its built-in potential isUnder the depletion approximation, the space-charge densities are $-e{N}_A$ on a p-side width $w_p$ and $+e{N}_D$ on an n-side width $w_n$. Integrating Poisson equation and imposing zero field outside gives $N_A{w}_p=N_D{w}_n$ andMost of the depletion width therefore extends into the more lightly doped side. The schematic shows the impurity levels before contact and the resulting space charge and potential after contact; there is no externally extractable equilibrium battery voltage from the built-in field alone.

A forward bias, with p positive relative to n, reduces the barrier from $V_{\mathrm{bi}}$ towards $V_{\mathrm{bi}}-V$ and injects minority carriers across the junction. In the ideal low-injection, diffusion-controlled regime this produces the Shockley diode equation, $I=I_s[e^{eV/(k_{B}T)}-1]$. Reverse bias increases the barrier and the depletion region; only a small thermally generated minority-carrier current flows before breakdown. Reverse-biased depletion regions also give a voltage-dependent junction capacitance. Real devices have recombination, leakage, series resistance and breakdown mechanisms beyond the ideal law. Controlled doping creates majority and minority carriers, and their junction produces a depletion barrier that makes rectification possible.

Semiconductor doping introduces controlled impurities to alter equilibrium carrier populations and electrical conductivity. In an intrinsic semiconductor, thermally excited electrons in the conduction band and electron holes in the valence band have equal densities $n=p=n_i$. A shallow donor impurity, such as phosphorus in silicon, supplies an electron from an impurity level just below the conduction-band edge. Once ionized it is positively charged, and the material is n type. An acceptor impurity, such as boron in silicon, accepts a valence electron into a level just above the valence-band edge, leaving a mobile hole and a negatively charged ionized impurity; the material is p type.

With dopant densities $N_D,N_A$, charge neutrality requiresFor fully ionized, nondegenerate dopants at equilibrium, the semiconductor mass-action law gives $np=n_i^2$. In a substantially n-type sample, $N_D-N_A\gg n_i$, soThe corresponding p-type result is $p\simeq N_A-N_D$ and $n\simeq n_i^2/(N_A-N_D)$. The majority population grows while the minority population decreases; adding donors does not simply add equal numbers of free electrons and holes.

The Fermi energy shifts toward the conduction band for n-type material and toward the valence band for p-type material. In the nondegenerate approximation,where $N_C,N_V$ are effective band-edge state densities. At very low temperature, dopants may not be ionized and carrier freeze-out occurs. At intermediate temperatures, ionized dopants dominate the carrier density; at sufficiently high temperature, intrinsic excitation can dominate. Heavy doping can produce degenerate carrier statistics, invalidating these Boltzmann expressions.

The electrical conductivity is $\boxed{\sigma =e(n{\mu }_e+p{\mu }_h)}$. Doping usually raises the majority carrier density, but ionized impurity scattering can lower mobility, so conductivity need not scale indefinitely with dopant concentration. Joining p-type and n-type regions creates a p-n junction, with a depletion region and built-in electric field established by carrier diffusion and compensating drift. The fixed ionized dopants, not an absence of all charge, account for the depleted region's net charge density.

Take a p-n junction made from one semiconductor, with the p-type semiconductor on the left and the n-type semiconductor on the right. At thermal equilibrium the chemical potential $\mu =E_F$ is horizontal throughout. Before contact the chemical potential lies nearer the valence band on the p side and nearer the conduction band on the n side; after contact the band edges bend to permit one common value.

In this orientation both the conduction band edge $E_C$ and the valence band edge $E_V$ decrease from p to n by $e{V}_{\mathrm{bi}}$. For a homojunction their separation $E_g=E_C-E_V$ is constant. The acceptor impurity level $E_A$ lies just above $E_V$ where acceptors are present, and the donor impurity level $E_D$ lies just below $E_C$ where donors are present. These impurity levels follow the local band bending; they do not form one additional horizontal band across the device.

The shaded depletion region has few mobile electrons or electron holes and contains fixed negative ionized acceptor impurities on the p side and positive ionized donor impurities on the n side. Thus its electric field points from n to p. Its neutral outer regions have negligible electric field, so the band edges become flat there. In the depletion approximation, charge neutrality requires $N_A{x}_p=N_D{x}_n$; the symmetric drawing illustrates $N_A=N_D$ rather than assuming that equality is universal. Diffusion and field-driven drift current cancel at equilibrium: no applied voltage does not mean zero internal electric field or zero built-in electric potential difference.

In a direct optical transition, a photon promotes an electron between the valence band and conduction band at essentially the same crystal wavevector: optical photon momentum is small on the scale of the Brillouin zone. An indirect optical transition also needs a phonon or another momentum source. Near an indirect gap, phonon emission requires photon energy approximately $E_g+\hslash \Omega$, while phonon absorption permits $E_g-\hslash \Omega$ with a thermally occupied phonon.

Strong direct interband absorption needs both many allowed transitions and a nonzero optical matrix element. The joint density of states counts pairs with $E_c(\mathbf{k})-E_v(\mathbf{k})=\hslash \omega$; nearly parallel bands and critical points of this energy difference can enhance it. Dipole selection rules and polarization weight that count, so a large joint density of states alone does not guarantee a strong absorption peak.

The supplied infrared spectrum lies below the interband threshold and instead probes free-carrier response. In the Drude model, $ϵ(\omega )={ϵ}_{\infty }-{\omega }_p^2/[\omega (\omega +i/\tau )]$, with ${\omega }_p^2=n{e}^2/({ϵ}_0{m}^{\ast })$. At long wavelengths the mobile electrons screen the field and the reflectivity is high. As wavelength decreases through the screened plasma edge, the real dielectric function becomes positive and propagating light can enter. The reflectivity reaches its minimum near index matching, $ϵ\simeq 1$, then rises toward the bound-electron background reflectance.

For background ${ϵ}_{\infty }=16$, its refractive index is four and the normal-incidence Fresnel equations give $R_{\infty }=[(4-1)/(4+1){]}^2=9/25=0.36$. The short-wavelength side of the figure is approaching a reflectance of this order, consistent with that background; its shortest plotted wavelength still has a carrier correction. Here “infinity” means high frequency relative to the free-carrier scale while remaining below the remote interband resonances.

The Drude-Lorentz reflectivity minimum is near $ϵ=1$, not the plasma-edge zero $ϵ=0$. Reading the deep minimum as ${\lambda }_{\min }\simeq 22\mu \mathrm{m}$ and neglecting damping there gives ${\omega }_p^2=({ϵ}_{\infty }-1)(2\pi c/{\lambda }_{\min }{)}^2$. The carrier effective mass from a reflectivity minimum is thereforeThe graph supports only a rough read-off: shifting the minimum by $0.5\mu \mathrm{m}$ changes this mass by about five percent. The deep narrow minimum supports the weak-loss assumption $\omega \tau \gg 1$ in this region, although finite scattering explains why the long-wavelength reflectivity is below unity. The estimate also assumes one dominant carrier species, a nonmagnetic bulk sample and a nearly constant background dielectric function. If the plasma-edge zero is used instead, its denominator is ${ϵ}_{\infty }$ and its wavelength is slightly longer; the two wavelengths must not be interchanged.

A positive electric potential difference between gate and substrate attracts electrons towards the insulating interface and repels the mobile electron holes of the p-type semiconductor substrate. The exposed, negatively charged acceptor impurities form a depletion region; the thin mobile electron layer at its surface is the semiconductor inversion layer. The conduction band and valence band bend downwards towards the positively biased gate, because the electron potential energy is $-e\psi$.

To specify the conventional MOS inversion threshold, let $N_A$ be the ionized acceptor density, $n_i$ the intrinsic carrier concentration, andIn the neutral bulk the intrinsic level lies $e{\varphi }_F$ above the semiconductor chemical potential. A positive surface potential ${\psi }_s$ lowers both band edges by $e{\psi }_s$, giving, in the nondegenerate approximation,At ${\psi }_s={\varphi }_F$, the surface is intrinsic: $n_s=p_s=n_i$. The usual onset of strong inversion instead meansThus the surface electron density has grown to the bulk majority-carrier density. The chemical potential need not actually enter the conduction band. The phrase “just forms” can also refer to the earlier weak-inversion crossover; the diagram uses the conventional strong-inversion threshold and labels it explicitly.

Within the depletion approximation the negative space charge is approximately $-e{N}_A$ over a widthwhere ${ϵ}_s$ is the absolute semiconductor permittivity. Its outer edge is where the bands become flat and mobile electron holes recover their bulk density. The electrical insulator blocks steady carrier transport while sustaining an electric field; the metal gate supplies positive surface charge balancing depletion and inversion charge. Under gate bias, the gate and substrate electrochemical potentials are different, so a single equilibrium Fermi level should not be drawn through the whole biased stack.

A solar cell uses the photovoltaic effect to turn photon energy into electrical work. In a semiconductor, above-gap photon absorption generates an electron in the conduction band and an electron hole in the valence band. A p-n junction, together with selective electrical contacts, collects the two carrier species at opposite terminals. Carrier diffusion and the built-in electric field help separate them before electron-hole recombination. Illumination maintains different quasi-Fermi levels for electrons and electron holes; their splitting, divided by $e$, sets the available terminal voltage in an ideal device. The equilibrium built-in potential alone cannot continuously deliver work.

For an ideal junction, the illuminated output current density isHere $J_{\mathrm{sc}}$ is the photogenerated short-circuit current density and $J_0$ the diode saturation current. The useful operating point maximizes $VJ$, rather than either current or voltage separately. With solar-cell fill factor $\mathrm{FF}=P_{\max }/(V_{\mathrm{oc}}{I}_{\mathrm{sc}})$,Series resistance, shunts, incomplete photon absorption and nonradiative recombination reduce the attainable efficiency.

The Shockley-Queisser limit removes these avoidable losses but retains the following physical constraints. There is one semiconductor junction with one band gap; each absorbed above-gap photon yields at most one collected electron-hole pair; carriers thermalize to the cell temperature; and the inverse optical process produces unavoidable radiative recombination. If ${\Phi }_{\odot }(E)$ is the incident photon flux per unit energy, ideal photon absorption givesThe same ability to absorb light requires the junction to emit it. At a forward voltage $V$, the emitted photon population has a chemical potential $eV$; in the ordinary nondegenerate limit the resulting recombination current is the exponential diode term above. Maximizing $VJ$ with this radiative $J_0$, and then choosing $E_g$, gives the limiting efficiency.

A small band gap collects many photons but gives a low voltage and large thermalization losses. A large band gap permits a larger voltage but transmits too much of the incident spectrum. Below-gap photons are lost to transmission, excess photon energy above the gap becomes heat, and radiative emission and its associated entropy loss keep $e{V}_{\mathrm{oc}}$ below $E_g$. The optimum single-junction conversion is roughly thirty percent under ordinary unconcentrated sunlight; the precise ceiling depends on spectrum, cell temperature and optical geometry. The original detailed-balance calculation gave about $30\%$ at $E_g\simeq 1.1\mathrm{eV}$ under its specified assumptions.^[ref] Multiple junctions and optical concentration change the assumptions and consequently change the bound.

Dopant atoms and carrier type. Substitutional semiconductor doping places a small fraction of foreign atoms on host lattice sites. In a tetrahedrally bonded group-IV semiconductor, a group-V substitution such as phosphorus in silicon has an extra valence electron. A shallow donor level just below the conduction edge can release that electron, leaving a positively charged donor. In an n-type semiconductor, mobile electrons are the majority carriers and the chemical potential shifts toward the conduction band.

A group-III substitution such as boron in silicon has one fewer valence electron. An acceptor level just above the valence edge captures an electron from the valence states, leaving a mobile electron hole and a negatively charged acceptor. In a p-type semiconductor, electron holes are the majority carriers and the chemical potential shifts toward the valence band. The material can remain electrically neutral in both cases: mobile carrier electric charge is compensated by fixed ionized impurity electric charge. Dopant freeze-out, compensation and degeneracy qualify the simplest one-carrier-per-dopant picture.

Making the junction. Choose p material at $x<0$ and n material at $x>0$. Electrons initially diffuse from n to p, and electron holes from p to n, recombining near the interface. Negative ionized acceptors are exposed on the p side and positive ionized donors on the n side. They produce an electric field directed from n to p. The field opposes further majority-carrier diffusion until drift and diffusion balance and the equilibrium electrochemical potential is uniform.

The depletion regions are depleted of mobile majority carriers, not of the fixed dopant charges that create the space charge. At equilibrium the conduction and valence edges bend together, while the chemical potential is flat. The specified $\Delta {\varphi }_0$ is a electric potential difference: the corresponding initial chemical-potential energy separation is $e\Delta {\varphi }_0$. This convention is required by the quoted units and width formulas. Write ${\epsilon }_s=ϵ{ϵ}_0$ for the absolute permittivity; the paper's $ϵ$ is relative permittivity.

Poisson equation and widths. Let the depleted intervals be $-w_p<x<0$ and $0<x<w_n$. Fully ionized number densities $N_A,N_D$ correspond to electric charge densities $-e{N}_A,+e{N}_D$, respectively. ThusThe factor $e$ converts a dopant concentration to a space-charge density. Requiring zero field in both neutral regions gives charge neutrality,With $\mathcal{E}=-{\varphi }^{\prime }$, direct integration gives the triangular fieldContinuity at the interface is the same neutrality condition. The field is negative throughout the depleted interval. Set $\varphi (-w_p)=0$. Its integral givesPotential continuity therefore requiresCombining it with neutrality yields the abrupt-junction depletion width and capacitance resultsThe more lightly doped side depletes farther, because equal and opposite space charges must be exposed.

Applied bias. Define positive $V$ as forward bias, raising the p-side terminal potential relative to the n-side terminal. It lowers the remaining barrier from $\Delta {\varphi }_0$ to $\Delta {\varphi }_0-V$. Reverse bias has $V<0$ and raises that barrier. The same Poisson integration applies, with this replacement:Forward bias narrows the depleted region; reverse bias widens it. Under bias there need not be one equilibrium chemical potential across the device, so the equilibrium flat-level argument must not be reapplied unchanged. The barrier/width change is illustrated in the next figure.

Differential capacitance and electric charge sign. Put $B=\Delta {\varphi }_0-V>0$ and $N_{\mathrm{eff}}=N_A{N}_D/(N_A+N_D)$. The signed p-side fixed electric charge per unit area isDifferentiating with respect to forward electric potential difference gives the positive capacitance per unit areaThe n-side electric charge is $Q_n=-Q_p$, so $d{Q}_n/dV=-C_A$ instead. If $Q$ denotes a positive space-charge magnitude, the positive capacitance is $-dQ/dV$ for this forward-voltage convention. This resolves the sign implicit in the paper's positive $dQ/dV$. The total device capacitance is $C=\mathcal{A}{C}_A$, where $\mathcal{A}$ is its cross-sectional area.

Finite device and numerical sketch. For the symmetric device, $N_A=N_D=N$ and each side has thickness $d$. Before full depletion,Using ${\epsilon }_s=12{ϵ}_0=1.0625\times 10^{-10}\mathrm{F}{\mathrm{m}}^{-1}$, the zero-bias width on each side is $4.07\mu \mathrm{m}$, less than $d=10\mu \mathrm{m}$. The total zero-bias capacitance is $\boxed{C(0)\simeq 1.30\mathrm{pF}}$.

Full depletion occurs when $w=d$, soAt larger reverse bias the depleted thickness cannot exceed $2d$. In the ideal fully depleted slab, incremental terminal electric charge changes the field uniformly through that fixed thickness, giving the full-depletion capacitance of a symmetric junction slabThus the idealized piecewise sketch over the specified range isIt is continuous at full depletion, rises through $1.30\mathrm{pF}$ at zero bias, and has a formal inverse-square-root divergence as $V\to 0.5\mathrm{V}$ from below.

There are two physical limits to this extension. Once all dopants are exposed, their fixed space charge no longer increases, but the terminal electrodes can still acquire incremental electric charge; differentiating only the fixed ionized electric charge would incorrectly predict zero capacitance. Near $V=\Delta {\varphi }_0$, the depletion approximation itself fails as mobile carriers and forward injection become important, so the formal infinite endpoint is not a prediction of an actual infinite measured capacitance. The fully depleted plateau assumes ideal contacts and no breakdown, injection storage or extra series capacitance.

Nondegenerate carrier product. Assume three-dimensional parabolic energy bands near their extrema, spin degeneracy two, and no additional uncounted valley multiplicity. Near the conduction edge, a Boltzmann approximation to the Fermi-Dirac distribution givesAngular integration and the Gaussian radial integral yieldFor valence electron holes, the empty-state probability is $1-f(E)\simeq e^{(E-\mu )/(k_{B}T)}$. Measuring electron hole kinetic energy downwards from $E_v$ similarly givesThe factors two count spin once in each band. Multiplying cancels the chemical potential and proves the semiconductor mass-action law:In an intrinsic neutral semiconductor, $n=p=n_i$. In a doped equilibrium semiconductor the same product holds under the stated nondegenerate assumptions, even when $n\ne p$. Extra valleys or several valence branches must be included in the effective density-of-states factors rather than counted inconsistently.

Equilibrium junction. Put the p region on the left and the n region on the right. Before contact their majority carriers are electron holes and electrons, respectively. Diffusion initially sends electrons from n to p and electron holes from p to n. This leaves negative ionized acceptors on the p side and positive ionized donors on the n side, creating a depletion region and an electric field directed from n to p. At equilibrium this field balances diffusion, so the net electric current vanishes.

For a homojunction the gap remains constant while the conduction band and valence band edges bend downwards from p to n by $e{V}_{\mathrm{bi}}$. The equilibrium chemical potential is flat. Since $E_{c,v}(x)=E_{c,v}^{\mathrm{ref}}-e\varphi (x)$, the electrostatic potential rises from p to n. The carrier profiles satisfyWith fully ionized nondegenerate dopants, the neutral p side has $p_p\simeq N_A$, $n_p\simeq n_i^2/N_A$, while the neutral n side has $n_n\simeq N_D$, $p_n\simeq n_i^2/N_D$. Thus the electron concentration rises across the junction and the electron hole concentration falls, with both small relative to the far-side majority densities in much of the depleted region. The ideal depletion approximation neglects these mobile populations when calculating the fixed space electric charge, but the actual equilibrium profiles are smooth.

The built-in potential follows from these Boltzmann factors:There is no applied terminal bias; the internal electrostatic potential variation does not contradict a constant electrochemical potential.

Opposing carrier fluxes under reverse bias. Use $V>0$ for the magnitude of reverse bias and $I>0$ for reverse electric current, as in the PDF. The junction barrier increases by $eV$. The Shockley diode equation in that orientation can be writtenThe reverse-directed contribution is supplied by minority electrons on the p side and minority electron holes on the n side. They reach the depletion edge by diffusion and are swept across by the built-in field. The opposing contribution is associated with majority carriers thermally crossing the barrier in the forward direction; reverse bias suppresses it by the Boltzmann factor. At zero bias detailed balance makes the two equal, while at large enough reverse bias the opposing flux becomes negligible.

In the ideal neutral-region diffusion model,Here $D_n,D_p$ are minority-carrier diffusion coefficients, $L_n,L_p$ their diffusion lengths, and $\mathcal{A}$ is junction area. The two terms inside this positive prefactor are the electron and electron hole contributions; both flow in the same reverse-current direction. This separates the two carrier-species contributions from the two opposing fluxes in the Shockley diode equation. Calling the saturation term a drift current refers to its passage across the depleted region, and does not eliminate the minority-carrier diffusion needed to supply it.

Reading the measured curves. At $eV\ll k_{B}T$, the ideal equation gives $I\simeq I_{\mathrm{sat}}eV/(k_{B}T)$, initially proportional to reverse electric potential difference. At $eV\gg k_{B}T$, it approaches a plateau $I\simeq I_{\mathrm{sat}}$. The thermal electric potential differences at the three temperatures are about $21$–$26\mathrm{mV}$, consistent with a rise at small reverse biases and near saturation at larger ones. Higher temperature produces a much larger plateau because thermal electron-hole excitation raises $n_i^2$ exponentially.

The lower-temperature curve has a noticeable residual electric potential difference dependence rather than a perfectly flat plateau. Depletion-region generation, surface leakage or other departures from the ideal diffusion model can contribute. The sharp upturn at the largest reverse electric potential differences is reverse breakdown, outside the unextended Shockley diode equation; the plot alone does not identify a unique microscopic breakdown mechanism. The original horizontal axis duplicates the final decade label, so a precise breakdown electric potential difference cannot be read reliably. Neither issue prevents estimating the gap from the intermediate-bias saturation region.

Band-gap estimate. Use plateau electric currents at a common intermediate reverse bias, away from both the small-bias rise and the breakdown upturn. Approximate readings are $I_{\mathrm{sat}}(298\mathrm{K})\sim 2.5\times 10^{-6}\mathrm{A}$, $I_{\mathrm{sat}}(273\mathrm{K})\sim 1.5$–$2\times 10^{-7}\mathrm{A}$, and $I_{\mathrm{sat}}(248\mathrm{K})\sim 10^{-8}\mathrm{A}$. Kelvin temperatures are required in the activation law.

For fixed effective masses, $n_i^2\propto T^3{e}^{-E_g/(k_{B}T)}$. If the remaining transport prefactor is approximately constant, extracting a semiconductor band gap from diode saturation current therefore usesUsing the widest temperature separation and the rounded readings givesNeglecting all algebraic prefactors and fitting $\ln I_{\mathrm{sat}}$ instead gives about $0.70\mathrm{eV}$. These differ by roughly $0.07\mathrm{eV}$, a material correction on the graph's precision scale. Thus an honest graph-based conclusion isThe nonflat low-temperature plateau, rough electric current readings, unprovided temperature dependence of $D/L$, and possible gap variation prevent a more precise extraction. If depletion-region generation dominated rather than the specified $I_{\mathrm{sat}}\propto n_i^2$ mechanism, its activation slope could involve a different energy scale; the displayed estimate explicitly uses the diffusion-dominated saturation model requested here.

A positive gate voltage attracts electrons towards the oxide interface of the p-type semiconductor and repels mobile electron holes. The uncovered ionized acceptors form a negatively charged depletion region. Increasing the voltage further creates a thin electron-rich semiconductor inversion layer at the surface. The conduction band and valence band both bend downwards towards the positively biased gate, because an electron's electrostatic energy is $-e\psi$.

In an ideal nondegenerate substrate with acceptor density $N_A$, write ${\varphi }_F=(k_{B}T/e)\ln (N_A/n_i)>0$ and surface potential ${\psi }_s$ relative to the neutral bulk. The electron and hole surface densities areThe first change of majority-carrier type occurs at ${\psi }_s={\varphi }_F$, where $n_s=p_s=n_i$. The conventional MOS inversion threshold for appreciable strong inversion isAt threshold the inversion sheet is just developing, while a much thicker depletion region lies between it and the neutral p-type bulk. Inversion does not require the conduction band to cross the Fermi level: the surface can remain nondegenerate.

The metal and semiconductor Fermi levels need not be equal. The ideal oxide blocks particle exchange, so the gate and substrate can maintain different electrochemical potentials. In a biased MOSFET, the gate-to-substrate Fermi-level difference is ${\tilde{\mu }}_g-{\tilde{\mu }}_s=-e{V}_g$, when the applied voltage is defined in the electron-reservoir convention. Each region can separately be in internal equilibrium. If an external zero-voltage connection instead equilibrates the reservoirs, their Fermi levels coincide, but different work functions can still cause band bending. Here the chemical potential shown in a band diagram is the electron electrochemical potential; the purely chemical contribution alone is not the equilibrium criterion for charged particles.

Equilibrium junction. Put the p-type side on the left and n-type side on the right. When they are joined, electrons diffuse from n to p and electron holes diffuse from p to n. This leaves negative ionized acceptors on the p-side and positive ionized donors on the n-side of the depletion region. The resulting built-in electric field points from n to p, opposing further diffusion. At equilibrium there is one constant Fermi level, and drift and diffusion currents cancel for each carrier species.

In the nondegenerate, fully ionized limit, the neutral-region carrier densities areThe conduction band and valence band lie higher on the p-side and bend down towards the n-side. The gap between them stays fixed for a homojunction. If $\psi$ increases towards the n-side, both edges shift by $-e\psi$, whileThus $n$ rises and $p$ falls continuously across the junction; neither mobile population is an exact step function. Their depletion exposes fixed dopant charge density, rather than an absence of every kind of charge. The built-in voltage is

Light emission. Forward bias lowers the barrier and injects minority carriers across the p-n junction. In a light-emitting diode, electron-hole recombination converts some of their energy into photons. A direct band gap permits efficient radiative transitions with nearly conserved crystal momentum; nonradiative channels compete. Near-edge emission has $\hslash \omega \simeq E_g$ and $\lambda \simeq hc/E_g$. Material composition and confinement select the emission colour. Reverse-bias saturation is not the ordinary LED operating mode.

Reverse current. Take $V=V_R>0$ and $I_R>0$ to mean reverse voltage and reverse current magnitudes. The Shockley diode equation in this convention isThe bias-independent $I_S$ term is the supply of thermally excited minority carriers from the neutral regions, swept across the junction by its electric field. The opposing injection current is reduced by the extra barrier energy $e{V}_R$ and is $I_S{e}^{-e{V}_R/(k_{B}T)}$. Their equality gives zero net current at zero bias; for $e{V}_R\gg k_{B}T$, the reverse current approaches $I_S$ until leakage or breakdown invalidates this ideal law. The collecting field need not create the minority carriers; their thermal supply limits the diffusion-dominated current.

For an ideal long junction of area $\mathcal{A}$, steady diffusion with recombination gives the diode minority-carrier diffusion currentIt depends on doping, geometry, carrier diffusion coefficients and lifetimes, and especially the intrinsic carrier concentration. Depletion-region generation current and surface leakage can add nonideal contributions to the measured reverse current.

Band-gap estimate from the data. Read a nearly flat region well above the small-bias knee and below the high-bias upturn, rather than estimating $I_S$ in the breakdown region. For a diffusion-dominated junction with nearly temperature-independent gap and effective masses,so $I_S=C(T)T^3{e}^{-E_g/(k_{B}T)}$, where $C(T)$ includes transport and doping factors. Extracting a semiconductor band gap from diode saturation current amounts to measuring the slope of $\ln [I_S/(C(T)T^3)]$ against $1/T$: it is $-E_g/k_B$. If the prefactor is taken approximately constant over a short interval,Approximate readings from the original PDF plateaus areUsing the two warmer curves and $k_B=8.6173\times 10^{-5}eV/K$ givesKeeping the $T^3$ factor but taking the residual transport factor constant changes this estimate to approximately $0.57\mathrm{eV}$. The justified graphical estimate is thereforeThe colder curve is a genuine limitation of the supplied data: the $0$ to $-45{}^{\circ }\mathrm{C}$ pair gives only about $0.32\mathrm{eV}$ with a constant prefactor, and the endpoint pair gives about $0.42\mathrm{eV}$. Thus all three cannot support one precise diffusion-only activation slope. A depletion-region generation current, approximately proportional to $n_i$ rather than $n_i^2$, can have a roughly half-gap activation energy; other leakage and temperature-dependent transport can also alter the slope. The PDF alone does not identify their relative contributions. A reliable precision gap measurement would use more temperatures and verify the transport regime instead of forcing one fit through these three readings.

At equilibrium, majority electrons undergo diffusion from n to p, while majority electron holes diffuse from p to n. Their departure exposes positive donors on the n side and negative acceptors on the p side. The resulting electric field points from n to p and opposes further diffusion. Equilibrium is reached when the chemical potential is constant across the junction. The built-in potential is this electrostatic adjustment, not a discontinuity of the material's band gap.