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A typical intrinsic semiconductor has a nearly full valence band ending at $E_v$ and a nearly empty conduction band starting at $E_c$, separated byAt finite temperature, the equilibrium Fermi level lies near the middle of the gap. In the nondegenerate parabolic-band approximation,where $N_c,N_v$ are the two effective density of states values. Equal densities of states put it exactly at midgap; unequal masses or valley multiplicities shift it slightly. The left panel gives the requested energy-level sketch, with doped cases for comparison.

Arsenic has five valence electrons, one more than silicon. As a substitutional donor impurity, it supplies a weakly bound extra electron which is largely thermally released into the conduction band at room temperature, leaving an ionized positive donor.

Consequently the mobile electron density increases far above the intrinsic value, so the conductivity increases substantially even at ppm dopant levels. The crystal remains macroscopically neutral: fixed donor charge compensates the added mobile charge.

Arsenic has five valence electrons, whereas silicon has four. A substitutional arsenic atom can therefore supply an extra electron after participating in the silicon bonding network. It acts as a shallow donor impurity, introducing a localized level $E_D$ just below the conduction band. Ionizing that donor costs much less energy than excitation across the full band gap.

At the lowest temperatures, donor electrons remain bound and the valence band is filled. Mobile-carrier populations are negligible: this is dopant freeze-out. Initial heating releases donor electrons into the conduction band, producing a rapid rise in conductivity governed by the shallow-donor energy scale. The positive ionized donor charge compensates the released electronic charge; the bulk need not acquire net charge.

At intermediate temperatures, nearly all donors are ionized but intrinsic electron–hole generation is still small. In an uncompensated sample,where $N_D$ is donor number density. The supply of donor electrons is finite, so further heating cannot keep creating them. This is the extrinsic plateau under the approximation of weakly varying mobility, and its carrier density is much larger than that of the undoped sample at the same temperature. In real material, scattering changes mobility, so donor exhaustion alone predicts a carrier-density plateau rather than guaranteeing perfectly flat conductivity.

At higher temperatures, intrinsic generation becomes comparable to and then exceeds the donor contribution. Once $n_i\gg N_D$, the populations approach $n\simeq p\simeq n_i$, and activation across the main gap causes the steeper rise. The relevant regimes are therefore freeze-out, donor exhaustion, and intrinsic excitation, not three successive types of dopant.

A direct band gap has its conduction band minimum and valence band maximum at the same crystal momentum. Near-edge absorption or emission can therefore conserve momentum with the electron and photon alone, since optical photon momentum is small on the Brillouin-zone scale. An indirect band gap has its extrema at different wavevectors and normally needs a phonon or another momentum-transfer process in addition to the photon.

Direct-gap materials are usually preferred for efficient light emission, because the radiative transition avoids the extra phonon-assisted process. Their stronger near-edge absorption also permits thinner photovoltaic absorbers. Indirect-gap materials can nevertheless make excellent solar cells: sufficiently thick optical paths, light trapping, long carrier lifetimes and suitable material quality can compensate for weaker absorption. The direct/indirect distinction alone does not rank every device.

This momentum distinction and the difference between bound exciton lines and the free-pair continuum are treated together in the primary effective-mass absorption theory.^[ref]

Three routes are donor doping, acceptor doping and above-gap illumination. All increase mobile carrier populations if the added states or excitations are effective rather than strongly trapped.

An appropriate donor impurity provides an occupied level just below the conduction band. Thermal ionization requires the smaller donor binding energy rather than the full host gap, supplying mobile electrons and n-type semiconductor behaviour. An appropriate acceptor impurity creates an available level just above the valence band; accepting a valence electron leaves a mobile hole, giving p-type semiconductor behaviour. Doping concentrations and compensation must permit useful carriers; an arbitrary impurity need not be an effective dopant.

For photoconductivity, absorbed photons with $h\nu \gtrsim E_g$ can excite valence electrons into the conduction band, producing electron–hole pairs. For the stated $2.7\mathrm{eV}$ gap, the nominal threshold is about $1240/2.7\simeq 459$ nm. Optical selection rules and absorption strength can shift the practical onset. The conductivity increase persists only while generation maintains carriers against recombination and trapping; it is not necessarily a permanent change to the material.

In each case, $\sigma =e(n{\mu }_n+p{\mu }_p)$ connects increased carrier density to conductivity. Defect-induced reductions in mobility can offset some of the gain, so the diagrams specify mechanisms rather than guarantee a numerical improvement for an unspecified preparation.

Sulfur substituting for phosphorus acts as a donor impurity, giving n-type semiconductor behaviour and usually increasing electrical conductivity. A P site normally supplies five valence electrons; S supplies six. After forming the surrounding bonds, the extra electron can occupy a donor level below the conduction band and be thermally ionized into that band. The donor impurity then carries positive charge.

This assignment is supported experimentally: infrared transmission measurements on S-doped gallium phosphide resolved bound-donor transitions and determined a sulfur donor optical ionization energy of about $0.104\mathrm{eV}$, far smaller than the host band gap.^[ref]

The effect depends on substitution on P sites, donor ionization, compensation by acceptors and charge carrier mobility. A dilute active donor population increases the electron term in $\sigma =e(n{\mu }_n+p{\mu }_p)$; not every dopant must donate a mobile electron at room temperature. Sulfur's larger Allen electronegativity does not overturn this electron-counting argument or make a P-site sulfur atom an acceptor.

A $2.24\mathrm{eV}$ band gap is consistent with a semiconductor. In pure gallium phosphide, the valence band is essentially full and the conduction band essentially empty at low temperature; thermal or optical excitation can produce mobile electrons and electron holes.

At room temperature, $k_{B}T\approx 0.026\mathrm{eV}$ is much smaller than the band gap, so intrinsic excitation is strongly suppressed. Pure material consequently has a low intrinsic carrier density, while semiconductor doping can markedly increase its electrical conductivity.

Assume thermal equilibrium, nondegenerate carriers, and isotropic parabolic band dispersions, with the supplied effective masses interpreted as density-of-states masses. Take one effective valley with two electron spin states for each band; any additional valley multiplicity must instead be absorbed into the masses or explicit degeneracy factors.

Let the conduction band edge be $E_c$, the valence band edge $E_v$, and $E_g=E_c-E_v$. The spin-summed conduction density of states per volume isUsing the nondegenerate limit of the Fermi-Dirac distribution and the substitution $x=(E-E_c)/(k_{B}T)$ givesThe factor $\sqrt{\pi }/2$ in the supplied integral produces the normalization. Counting missing valence band electrons similarly givesMultiplication cancels the chemical potential, establishing the semiconductor mass-action law:The result does not require an intrinsic semiconductor: semiconductor doping changes $n$ and $p$ separately while preserving this product under the stated equilibrium and nondegeneracy assumptions.

Take uncompensated donor impurities and initially approximate them as ionized. Charge neutrality then gives $n-p=N_d$, while the semiconductor mass-action law gives $np=n_i^2$. Solving the quadratic without prematurely discarding minority carriers,Since $N_d\gg n_i$, the hole density isUse the conduction population to locate the chemical potential:It moves toward the conduction band but remains about $10.4k_{B}T$ below it, so the mobile electrons are nondegenerate. Despite the donor binding energy being comparable with $k_{B}T$, the large number of available conduction states makes ionization almost complete at this dilute doping. For a singly occupied two-spin donor, $N_d^{+}/N_d=[1+2e^{(\mu -E_D)/(k_{B}T)}{]}^{-1}$, with $E_D=E_c-0.02584\mathrm{eV}$. The inferred value is $0.99984$, confirming the approximation to much better than one percent. Including this small correction raises the hole density only to about $4.5824\times 10^{10}{\mathrm{m}}^{-3}$.

The Boltzmann carrier density of a two-dimensional parabolic band can be derived directly. Use one isotropic parabolic conduction band, $E(\mathbf{k})=E_c+{\hslash }^2{k}^2/(2m_e^{\ast })$, with two spin states per wave vector. In the nondegenerate limit the Fermi-Dirac distribution reduces to its Boltzmann factor. The number per area is thereforeThe integral uses ${\int }_0^{\infty }k{e}^{-a{k}^2}dk=1/(2a)$. ThusEquivalently, the two-dimensional density of states is constant, $D_c=m_e^{\ast }/(\pi {\hslash }^2)$, and $n=N_c{e}^{-\beta (E_c-\mu )}$ with $N_c=D_c{k}_{B}T$. These expressions include twofold spin degeneracy and one valley. With a general total band degeneracy $g_c$, replace the factor $1/\pi$ by $g_c/(2\pi )$; extra valleys or occupied confinement subbands would change the coefficient.

Equal effective masses and equal spin and valley degeneracies give electron-hole symmetry. In the intrinsic semiconductor, charge neutrality requires equal numbers of conduction band electrons and valence band holes. It therefore pins the chemical potential to the band touching point, $\mu =0$. The Fermi-Dirac distribution givesMeasured relative to the filled valence band at zero temperature, the thermal internal energy includes both species. Removing a valence band electron at $-E$ costs the positive hole energy $E$, just as adding a conduction band electron at $E$ does. HenceDifferentiate at fixed area and fixed total electronic particle number, allowing creation of electron-hole pairs:There is no conflict with the equipartition theorem. The normalization $N_c$ counts only the excited conduction band electrons, while the heat capacity includes both electrons and holes; moreover, $N_c\propto T$ is not fixed. In fact $\Delta U/N_c={\pi }^2{k}_{B}T/(6\ln 2)$, and differentiating $N_c(T)$ supplies another equal contribution to the heat capacity. Finally, $\mu =0$ gives occupation $1/2$ at the touching point, so the excitation gas is not in the classical dilute limit even though $k_{B}T$ exceeds the vanishing band gap.

Undoped gallium arsenide is expected to be diamagnetic at zero temperature. Its valence band is filled and its conduction band empty, so there are no mobile conduction band electrons to give Pauli paramagnetism. The specified absence of localized magnetic moments removes a Curie law response and the usual mechanism for ferromagnetic ordering. The remaining induced orbital response is diamagnetism. This classification uses the ordinary weak-field band insulator picture; the band gap suppresses thermally generated carriers as $T\to 0$.

The number of formula units per volume follows from the mass density:Each neutral formula unit contains $31+33=64$ electrons, so ${\rho }_e=64n_f=1.5726\times 10^{30}{\mathrm{m}}^{-3}$. Let $R^2$ be the electron-weighted mean-square orbital radius, averaged over all these electrons. The supplied Larmor-Langevin susceptibility then gives the dimensionless SI volume magnetic susceptibilitywhere $a_0=0.0529177\mathrm{nm}$ is the Bohr radius. An explicit rough atomic-scale assumption $R\simeq a_0$ gives ${\chi }_0\simeq -2.6\times 10^{-5}$.

The quoted bulk data do not determine $R$: a nuclear charge does not specify an orbital radius in a screened many-electron chemical atom. Consequently the numerical estimate just made has atomic-model uncertainty; it is not a deduction with guaranteed 30 percent accuracy. For example, the alternative assumed $R=2a_0$ gives four times the magnitude while leaving every supplied bulk number unchanged. Atomic wavefunctions, an explicit screened-orbital model, or measured atomic magnetic susceptibilities of materials exhibiting diamagnetism would be needed to support that accuracy. Keeping $R$ explicit makes the later comparison independent of an unadvertised radius assumption. The bare electron mass belongs in this atomic formula; the conduction band effective mass is relevant instead to the large donor envelope in part (c).

Use the stipulated noninteracting impurity band approximation, with one donated electron per donor and two available spin states per donor orbital. The energy band is half filled. Approximating its spin-summed density of states per volume near the Fermi energy by $D_v(E_F)\simeq 2n/W$ gives the zero-temperature Pauli paramagnetismWith $\alpha =1$,A factor of order one depends on the density of states shape; using the conduction band $g$-factor rather than silently setting $g=2$ matters here.

To locate the comparison with the magnitude of the undoped Larmor-Langevin susceptibility, put $x=n^{-1/3}/a_b>1$. Numerically,For the stated atomic-scale estimate $R=a_0$ in part (b), equality with $|{\chi }_0|=2.599\times 10^{-5}$ requiresThus the dilute-branch crossover iswell below $a_b^{-3}=9.45\times 10^{23}{\mathrm{m}}^{-3}$. Within this model the positive spin response exceeds the negative background magnitude for lower dopant densities on this dilute branch: the exponential narrowing of $W$ wins over the loss of donors. Indeed $d\ln (e^x/x^3)/dx=1-3/x>0$ for $x>3$. This is not a claim of ferromagnetism.

The degenerate Pauli paramagnetism calculation requires $k_{B}T\ll W$, with an unimportant order-one filling factor. At the estimated crossing,The weak-field condition also requires $|g|{\mu }_{B}B\ll W$. For a different assumed atomic $R$, use $|{\chi }_0|=2.599\times 10^{-5}(R/a_0{)}^2$ in the displayed equation for $x$. An order-of-magnitude background $|{\chi }_0|\sim 10^{-5}$ gives the same density order and a crossover scale of order $10^{-5}\mathrm{K}$. At higher temperature than this extremely narrow bandwidth permits, the $n/W$ enhancement cannot be used; the independent-donor Curie law is the appropriate thermal scale if interactions remain negligible. At very small dopant density, disorder, localization and Coulomb interactions can also invalidate the assumed noninteracting impurity band. The formal divergence as $n\to 0$ is therefore a limit of the stated zero-temperature band model, not a prediction for isolated donors at a fixed nonzero temperature.

Let $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant, $T$ the absolute temperature, and $\mu$ the common equilibrium chemical potential. The conduction band minimum and valence band maximum are ${ϵ}_c$ and ${ϵ}_v$; their separation is the band gap $E_g={ϵ}_c-{ϵ}_v$. The positive parameters $m_e,m_h$ in the supplied density of states functions are the band effective masses of electrons and electron holes, not necessarily the bare electron mass.

The electron occupation is the Fermi-Dirac distribution $f(ϵ)=[e^{\beta (ϵ-\mu )}+1{]}^{-1}$. The appropriate statistical factors are $f_c=f$ for electrons and $f_v=1-f$ for electron holes. In a nondegenerate semiconductor, ${ϵ}_c-\mu \gg k_{B}T$ and $\mu -{ϵ}_v\gg k_{B}T$, so the Boltzmann approximation givesFor the conduction band, set $u=ϵ-{ϵ}_c$ and then $\xi =\beta u$. The supplied integral yieldsFor the valence band, set $u={ϵ}_v-ϵ$ to obtainThe effective density of states factors $N_c,N_v$ include the two spin states already present in the supplied density of states functions. Multiplication cancels the chemical potential and gives the semiconductor mass-action lawHere $n_i=\sqrt{N_c{N}_v}{e}^{-E_g/(2k_{B}T)}$ is the intrinsic carrier concentration. Equality $n=p=n_i$ additionally requires an electrically neutral intrinsic semiconductor. The product law also applies to an equilibrium doped semiconductor while both populations remain nondegenerate. The derivation assumes parabolic energy bands, the stated spin degeneracy, and a single common chemical potential.

Assume uncompensated donor impurities, negligible intrinsic electron holes, and a nondegenerate conduction band. Then charge neutrality equates the conduction electron number density to the ionized-donor number density:The low-temperature condition relative to the band gap supports neglect of intrinsic carriers; one also needs $n_i\ll n$ and the Boltzmann approximation for the conduction electrons. Combining part (b)(i) with the effective density of states result givesEliminate the chemical potential by writing $e^{\beta (\mu -{ϵ}_D)}=(n/N_c)e^{\beta ({ϵ}_c-{ϵ}_D)}$. With the donor binding energy $E_i={ϵ}_c-{ϵ}_D>0$,Writing $A(T)=N_c{e}^{-\beta E_i}$, the physical root of the resulting quadratic equation isThis lies between zero and $N_D$. At semiconductor carrier freeze-out, $A\ll N_D$, it becomes $n\simeq \sqrt{N_D{N}_c}{e}^{-E_i/(2k_{B}T)}$; at essentially complete ionization, $A\gg N_D$, it tends to $N_D$. Including donor charge-state degeneracy would replace $A$ by $A{g}_{\mathrm{empty}}/g_{\mathrm{full}}$.

There are three regimes. At the lowest temperatures, most donor impurities retain their electrons; thermal ionization gives semiconductor carrier freeze-out. The result from part (b) isThis produces the approximately straight descending low-temperature tail on the logarithmic scale plot against $1/T$. The activation energy in the exponential is $E_i/2$, because the chemical potential adjusts with donor ionization.

At intermediate temperatures, the donor impurities are essentially fully ionized while intrinsic excitation remains small. Charge neutrality gives $n\simeq N_D$, producing the extrinsic plateau, roughly from $20$ to $40\mathrm{K}$ in the figure.

At higher temperatures, excitation from the valence band to the conduction band creates electron-hole pairs. The intrinsic carrier concentration grows as $n_i\propto T^{3/2}{e}^{-E_g/(2k_{B}T)}$ for constant parabolic band effective masses and constant band gap. With ionized donor impurities, charge neutrality and the semiconductor mass-action law giveThus $n\simeq n_i+N_D/2$ when $n_i\gg N_D$. This explains the renewed rapid rise and the eventual intrinsic branch; that branch is governed by the band gap, rather than the donor binding energy.

The lower curve is a valence band and the upper curve is a conduction band, separated by an indirect band gap. The readable upper curve has a minimum just inside the X boundary and slightly below the plotted chemical potential. It is therefore partially occupied, while the visible valence states lie below the chemical potential and are filled at low temperature. The sample is electron-doped, with electron pockets around the six symmetry-related conduction minima of silicon.

The full high-symmetry path and part of the lower-left dispersion are absent from the held image. Source gap: those missing path labels and curve segments cannot be read or reconstructed from this copy. The identification of the conventional six silicon valleys and the valence maximum at $\Gamma$ uses the usual silicon band model, rather than purported readings of missing pixels.

Two methods for calculating such an electronic band structure are density functional theory, solving a self-consistent periodic electronic problem, and a tight-binding model, diagonalizing a hopping Hamiltonian built from localized atomic orbitals. Neither is a measurement of the plotted dispersion relation.

Assume equal excess electron and electron hole number densities $N_t$, with no additional doping imbalance. At zero temperature, both occupied carrier spheres have the same Fermi wavevector $k_F$ because their spin counting is the same:For $k<k_F$, the conduction band is occupied and the valence band has electron holes, so $f_c-f_v=1$. Beyond $k_F$ the occupations reverse and the medium absorbs. Thus the upper edge of the positive-optical gain interval is the supremumEquivalently $E_{\mathrm{Fc}}=E_g+{\hslash }^2{k}_F^2/(2m_c^{\ast })$ and $E_{\mathrm{Fv}}=-{\hslash }^2{k}_F^2/(2m_v^{\ast })$. With the supplied effective masses,Using the vacuum wavelength $\lambda =hc/(\hslash \omega )$ reverses the endpoints:These boundaries are idealized and open: the lower-energy edge has zero joint density of states, and the upper edge is the zero-temperature occupation discontinuity, not a finite-width gain peak. The calculation gives positive material optical gain, not the stronger requirement of exceeding the losses at laser threshold.

A double heterostructure laser sandwiches a narrow-band gap active layer between wider-band gap cladding layers. In a type-I band alignment, the conduction band edge is lower and the valence band edge higher in the active layer. Electrons therefore see an energy well, and electron holes also see a well because their excitation energy increases when the valence band edge decreases. Both carrier populations remain concentrated where stimulated emission occurs.

The active layer also normally has the higher refractive index, forming an optical waveguide. The optical normal mode overlaps the injected carriers strongly, and the wider-band gap cladding has less interband absorption at the emitted photon energy. Carrier confinement reduces leakage, while optical confinement reduces the required pump current and hence heating.

Writing $\Gamma$ for the fraction of the optical normal mode overlapping the active layer, the threshold condition is $\Gamma g={\alpha }_i+{\alpha }_m$, with internal and mirror losses ${\alpha }_i,{\alpha }_m$. A larger overlap and a higher carrier number density at a given current both lower the threshold current. The figure shows composition-induced offsets; a working junction also has band bending under bias, which is omitted to make the confinement mechanism visible.