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In the Lorentz oscillator model, each participating bound electron is displaced by an applied electric field against a restoring force. The model replaces complicated interband transitions by a driven damped harmonic oscillator. If $u$ is the electron displacement along the field and the electron charge is $-e$, its equation of motion isHere ${\omega }_0$ describes binding and $\gamma >0$ describes loss of coherent motion. For $E(t)=\mathrm{Re}(E_0{e}^{-i\omega t})$, the complex amplitude isThe induced electric dipole moment is $-eu$, so the electric polarization amplitude for number density $n$ isUsing $D={ϵ}_{0}E+P={ϵ}_0{ϵ}_{r}E$ givesThe paper's $ϵ$ is this dimensionless relative permittivity, rather than the absolute permittivity ${ϵ}_0{ϵ}_r$. Independent classes of bound electrons contribute additive susceptibilities, usually with different resonance frequencies and optical oscillator strengths.

Writing $D_{\omega }=({\omega }_0^2-{\omega }^2{)}^2+{\gamma }^2{\omega }^2$ separates the response intoAt zero frequency the real part is $1+{\omega }_p^2/{\omega }_0^2$ and the imaginary part vanishes. Near resonance the imaginary part has a positive photon absorption peak and the real part changes rapidly, passing through $1$ at ${\omega }_0$. Above resonance the real part can become negative when damping is sufficiently weak; at very high frequency it approaches $1$ from below while the imaginary part tends to zero. Finite damping removes the undamped divergence.

For a nonmagnetic material take the passive square-root branch $N=\eta +i\kappa =\sqrt{{ϵ}_r}$, with $\kappa \ge 0$. A plane wave then has an intensity attenuation coefficient $\alpha =2\omega \kappa /c$. Thus the imaginary permittivity expresses dissipation, whereas the actual photon absorption coefficient is determined by the imaginary refractive index. The normal-incidence power reflectance isA large refractive-index mismatch produces wave reflection even without loss. In the undamped model ${ϵ}_r<0$ between ${\omega }_0$ and $\sqrt{{\omega }_0^2+{\omega }_p^2}$, so the field is evanescent and wave reflection is total in a semi-infinite sample. With damping the response includes both wave reflection and photon absorption; a peak in the imaginary permittivity should not simply be identified with a reflectance peak.

To infer the short-wavelength background from $R=0.35$, assume weak photon absorption and take the physical semiconductor branch $\eta >1$. ThensoThis is an estimate of the valence-electron background in a frequency window above the doped-carrier response and below the effective bound-electron resonance. One reflectance value alone does not determine both parts of a complex dielectric constant. Even for real ${ϵ}_r$, a reciprocal-index branch gives the same $R$; the ordinary background index greater than unity selects the branch used here.

Using the supplied conventional cubic lattice parameter, the participating valence-electron number density isUsing $m=m_e$ for the stipulated free-electron comparison givesAn angular frequency “in eV” means its quantum energy $\hslash \omega$, not $h\omega$. No doped-carrier effective mass should be inserted into this valence-electron count.

The low-frequency limit of the bound-electron contribution obeysConsequentlyThis effective resonance is much larger than the quoted $0.2\mathrm{eV}$ band gap. There is no contradiction: the single oscillator has been assigned the strength of all sixteen participating valence electrons per conventional cell, whereas the band gap is only the lowest interband excitation threshold. A real spectrum distributes optical oscillator strength over many transitions. Equating the oscillator resonance to the lowest threshold would misrepresent the assumed background response. Here “low frequency” means low relative to ${\omega }_0$; the static response of the doped conducting sample also contains its free carriers.

The additional mobile carriers in doped indium antimonide contribute a Drude frequency response, obtained by setting their restoring frequency to zero. At infrared frequencies well below the effective valence resonance, the sum is approximatelyAt the shorter wavelengths the carrier term is small and $R$ approaches the background value. As wavelength increases, the negative free-carrier contribution lowers the real permittivity towards unity. For weak damping, refractive-index matching to vacuum gives the Drude-Lorentz reflectivity minimum atThis minimum is near ${ϵ}_r^{\prime }=1$, not its zero. At the slightly lower screened plasma frequency ${\omega }_{pc}/\sqrt{{ϵ}_b}$, the real permittivity crosses zero; at longer wavelengths it becomes negative and the material becomes strongly reflective. Finite damping prevents a perfectly vanishing minimum and rounds the edge. At still lower frequency the large conducting response also drives $R$ towards unity.

For fixed carrier effective mass and background permittivity, ${\omega }_{pc}\propto \sqrt{n_c}$ and ${\lambda }_{\min }\propto n_c^{-1/2}$. The observed shift of the minimum towards shorter wavelength as doping increases is therefore the expected free-carrier trend. The figure is an original model illustration, not a digitization or quantitative fit of the supplied curves; the carrier effective mass, damping and their density dependence have not been specified.

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.

For an electron wavepacket in one isolated band under a uniform electric field,Because $ϵ(k+2\pi /a)=ϵ(k)$, acceleration through successive Brillouin zones makes the velocity periodic rather than increasing indefinitely. For $ϵ={ϵ}_0-2J\cos (ka)$,for a positive field along the chosen axis. Integration givesThe excursion amplitude of a Bloch-oscillating wavepacket is $2|J|/(e|E|)$ about its mean and peak-to-peak extent $4|J|/(e|E|)$. The turning of velocity corresponds semiclassically to Bragg reflection near a zone boundary.

Coherent Bloch oscillations require scattering/dephasing times long compared with $T_B$ and a field weak enough to avoid substantial interband tunnelling. Ordinary metals usually scatter too rapidly; semiconductor superlattices and cold atoms in periodic optical potentials provide cleaner realizations. A completely filled band has canceling velocities, so its net current does not oscillate simply because individual wavepackets can.^[ref]

Assume three-dimensional parabolic band edges, with density-of-states masses $m_c,m_h$ and one valley per edge, each with spin degeneracy two. Their effective densities of states areFor the semiconductor mass-action law, integrating the square-root density of states against the Boltzmann tail gives $n=N_c{e}^{(\mu -E_c)/(k_{B}T)}$ and $p=N_v{e}^{(E_v-\mu )/(k_{B}T)}$. Multiplication cancels the chemical potential, givingIf there are additional valley degeneracies, multiply by their product; anisotropic bands use the appropriate density-of-states masses. In the later equal-mass model, $A=4m^3/(2\pi {\hslash }^2{)}^3$. No numerical mass or valley multiplicity has been invented.

When $\mu =E_c=0$, the conduction-edge occupation is exactly $1/2$, so the nondegenerate approximation for electrons fails at every positive temperature. Define the Normalized complete Fermi-Dirac integralFor the carrier densities with chemical potential pinned to a band edge, with the two specified equal masses, put $N(T)=2[m{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$. Direct Fermi-Dirac integration then givesThus $n\propto T^{3/2}$ exactly within the parabolic model. At low temperature, $p\simeq N(T)e^{-E_g/(k_{B}T)}$ is thermally activated; at $k_{B}T\gg E_g$, $p/n\to 1$ from below. Both vanish at zero temperature and increase at fixed chemical potential. The other states that pin $\mu$ can supply the compensating charge; one must not additionally impose $n=p$.

For the Pauli reduction of the pinned-edge mass-action product, the exact product relative to the classical law isIt approaches $f_0$ at low temperature and $f_0^2\simeq 0.58545$ in the formal high-temperature limit. The low-temperature activation factor is unchanged but its prefactor is reduced; the classical mass-action formula overestimates the product. The high-temperature limit here refers to the given unbounded parabolic bands, not a claim that real semiconductor band structures remain unchanged at arbitrarily high temperature.^[ref]

Finally, for a fixed positive $\mu$, the zero-temperature Fermi sphere has radius $k_{\mu }=\sqrt{2m\mu }/\hslash$, soThe low-temperature expansion at fixed $\mu$ gives the fixed-chemical-potential Sommerfeld density increase:The monotonic density at fixed chemical potential in a square-root density of states initially rises quadratically from $n(0)$. For $k_{B}T\gg \mu$, its leading behaviour crosses to $f_{0}N(T)\propto T^{3/2}$. The fixed chemical potential is what allows the particle number to change.

For nondegenerate electrons in a parabolic conduction band,when the density-of-states effective mass and valley multiplicity are fixed. The intrinsic carrier extrapolation across a crossover must therefore specify how the activation energy changes with temperature.

The usual intended estimate treats the quoted $0.9\mathrm{eV}$ as a constant intrinsic activation energy between the two temperatures, and identifies the crossover by $n_i(T_c)\simeq n_{\mathrm{ext}}$. ThenFor $T_1=350\mathrm{K}$, $T_c=280\mathrm{K}$ and $k_B=8.6173\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}$, the exponent is $-7.46$ and the prefactor is $0.8^{3/2}=0.7155$. HenceEquivalently, a fixed-gap intrinsic model with $E_g/2=0.9\mathrm{eV}$ gives this result. A crossover is gradual: if $n_{\mathrm{ext}}$ denotes the net ionized donor density and $n_i(T_c)=n_{\mathrm{ext}}$, charge neutrality gives the actual electron population $n(T_c)=(1+\sqrt{5})n_{\mathrm{ext}}/2$, rather than exactly $n_{\mathrm{ext}}$.

There are two limits to interpreting the numerical estimate literally. First, the printed data specify $E_c-\mu$ only at $350\mathrm{K}$, without specifying $E_g$ or the valence density of states. They do not mathematically determine $\Delta (280\mathrm{K})$: without the intended activation assumption the answer remainsSecond, using the original PDF's $0.9\mathrm{eV}$ and density implies $N_c(350)\simeq 2.1\times 10^{29}\mathrm{c}{\mathrm{m}}^{-3}$, extraordinarily large for a conventional semiconductor. This confirms that the numerical data should not be treated as a realistic material determination. No unprinted decimal-point correction has been assumed.

Band edges and optical transitions. A direct band gap has the conduction minimum and valence maximum at the same crystal momentum. A near-edge photon then connects the two almost vertically on an energy-versus-wavevector plot, because its momentum is small on the Brillouin zone scale. An indirect band gap has the extrema at different momenta, so an additional momentum transfer, commonly supplied by phonon absorption or emission, is required for a near-edge radiative transition. This reduces its probability relative to an allowed direct transition.

Near the zone centre, an isotropic direct-gap model has $E_c(k)=E_c+{\hslash }^2{k}^2/(2m_e^{\ast })$ and downward valence dispersions $E_{v,j}(k)=E_v-{\hslash }^2{k}^2/(2m_{h,j}^{\ast })$. The heavy-hole and light-hole bands are the upper branches with large and small positive hole masses: the heavy branch is flatter. In common cubic semiconductors these arise from the $j=3/2$ manifold, while spin-orbit coupling places the $j=1/2$ split-off valence band lower by ${\Delta }_{\mathrm{so}}$. Its thermally generated hole population is suppressed by approximately $e^{-{\Delta }_{\mathrm{so}}/(k_{B}T)}$ relative to upper-valence holes, apart from density-of-states factors. Neglect is justified when ${\Delta }_{\mathrm{so}}\gg k_{B}T$; it is not a universal exclusion of this band at arbitrary temperature.

For an indirect material such as Si, the conduction minimum need not be at $\Gamma$. In applying the carrier-density formulae, $E_c$ must mean the actual minimum of the included conduction valley. The supplied scalar effective masses and densities of states constitute a simplified band-edge model; they do not specify Si's full valley anisotropy or multiplicity.

Carrier populations. Let $f(E)$ be the Fermi-Dirac distribution. In the nondegenerate conduction tail, $f(E)\simeq e^{-(E-\mu )/(k_{B}T)}$. Set $x=E-E_c$ and integrate the given density of states:The integral is $\Gamma (3/2)(k_{B}T{)}^{3/2}=\sqrt{\pi }(k_{B}T{)}^{3/2}/2$. Defining the effective density of states givesThe factor two is already included in the supplied density of states; it must not be added a second time.

A hole is an unoccupied valence state. For $E$ sufficiently below $\mu$, $1-f(E)\simeq e^{(E-\mu )/(k_{B}T)}$. Substitution of $x=E_v-E$ gives in exactly the same wayTheir product eliminates the chemical potential:This is the semiconductor mass-action law at thermal equilibrium. In an intrinsic semiconductor, charge neutrality additionally givesIn a doped semiconductor the product still applies in this nondegenerate equilibrium model, but the two populations need not be equal.

Intrinsic chemical potential. Taking logarithms of $n=p$ yields the intrinsic-semiconductor chemical potentialFor unequal band degeneracies, use the first form with the full $N_c,N_v$. For the supplied mass ratio $m_h^{\ast }/m_e^{\ast }=1/2$, the shift from midgap at $300\mathrm{K}$ is approximately $-13.4\mathrm{meV}$, toward the valence edge.

Three temperature regimes. The original plot runs from high temperature on the left to low temperature on the right. At high temperature, thermal excitation across the gap dominates and $n\simeq p\simeq n_i$ rises rapidly with temperature. In the middle, donors and acceptors are effectively ionized, the material is extrinsic, and $n\simeq N_D-N_A$ makes the plateau. At low temperature, dopant freeze-out binds increasing numbers of electrons to donor levels, reducing the mobile-electron population as the sample cools. Its activation scale is a donor binding energy, not the full band gap. Compensation and incomplete ionization affect that regime.

Gap estimate from the original figure. On the dashed intrinsic branch, $n_i=C{T}^{3/2}{e}^{-E_g/(2k_{B}T)}$. Hence band-gap extraction from intrinsic carrier activation uses the slope of $\ln (n_i/T^{3/2})$ versus $1/T$. With $x=1000\mathrm{K}/T$ and $k_B$ expressed in eV/K, two points giveThe very compressed intrinsic branch is only approximately readable: $n_i\sim 10^{15}\mathrm{c}{\mathrm{m}}^{-3}$ near $x\sim 1.8$ and $n_i\sim 10^{13}\mathrm{c}{\mathrm{m}}^{-3}$ near $x\sim 2.5$. These give $E_g\sim 1.0\mathrm{eV}$, rather than a precision determination. Neglecting the thermal prefactor gives about $1.13\mathrm{eV}$ from those same rough points. Thus a gap of order $1$–$1.1\mathrm{eV}$ is justified by this schematic plot; its resolution does not support distinguishing a few hundredths of an electronvolt. The declining freeze-out branch must not be used to infer the full gap.

Room-temperature densities. At $300\mathrm{K}$, the supplied masses giveIn the fully ionized extrinsic plateau, put $N=N_D-N_A=10^{15}\mathrm{c}{\mathrm{m}}^{-3}$. Charge neutrality and the mass-action law implySince $n_i\ll N$, $n\simeq 10^{15}\mathrm{c}{\mathrm{m}}^{-3}$ is insensitive to the imprecise gap. The minority density is much more sensitive:Using $E_g=1.0\mathrm{eV}$ gives $n_i\simeq 5.8\times 10^{10}\mathrm{c}{\mathrm{m}}^{-3}$ and $p\simeq 3.3\times 10^6\mathrm{c}{\mathrm{m}}^{-3}$. Using $1.1\mathrm{eV}$ instead gives $n_i\simeq 8.3\times 10^9\mathrm{c}{\mathrm{m}}^{-3}$ and $p\simeq 7.0\times 10^4\mathrm{c}{\mathrm{m}}^{-3}$. These are explicitly conditional estimates within the stated mass model. The plot alone cannot fix a precise minority density because the uncertain gap occurs inside an exponential.

In the nondegenerate conduction band, the Fermi-Dirac distribution reduces to its Boltzmann distribution tail. ThusHere the substitution $x=(ϵ-{ϵ}_c)/(k_{B}T)$ leaves ${\int }_0^{\infty }{x}^{1/2}{e}^{-x}dx=\sqrt{\pi }/2$. For electron holes use the empty-state probability $1-f(ϵ)\simeq e^{(ϵ-\mu )/(k_{B}T)}$, not the conduction-band Boltzmann factor. This givesThe chemical potential cancels in the product. With $E_g={ϵ}_c-{ϵ}_v$,This semiconductor mass-action law holds in thermal equilibrium for nondegenerate carriers, including a doped semiconductor. Only in an intrinsic semiconductor must $n=p=n_i$. The factors $N_c,N_v$ are the effective density of states; the factor $1/2$ in the final expression follows from the two spin factors already included in (a).

A photodiode is a semiconductor diode producing photocurrent from photon absorption. Exciting an electron across the band gap leaves an electron hole. The depletion region's electric field separates the charge carriers; others can arrive by diffusion before electron-hole recombination. Contacts complete the electrical circuit.^[ref]

The useful bridge is from light to a measurable electric current: spectral responsivity converts incident radiant flux into photocurrent, while quantum efficiency counts collected signal charge carriers per incident photon. These answer different questions: how much electric current one watt produces, and how many incident photons contribute to the signal. The worked calculation below makes their relationship explicit.

Thorlabs' silicon FDS100 experiment sends a collimated, fibre-coupled laser beam through a beam splitter to the photodiode and a reference optical power meter. Varying reverse bias and electrical load tests linearity. The measured readout approaches saturation when its electric potential difference consumes the bias.^[ref] A separate low-signal setup uses a $637\mathrm{nm}$ laser, polarizers, an optical chopper, optical attenuators and a lock-in amplifier inside a light-tight enclosure. These are actual apparatus choices, not the parameters of the constructed example above.^[ref]

For an independent dark current check, Thorlabs enclosed a photodiode in nested metal boxes to exclude light and suppress interference, used a Keithley 6487 ammeter supplying reverse bias, and monitored temperature. Only one specimen per type was tested; the nearby temperature sensor did not directly contact the semiconductor. Such a dark control measures the baseline but cannot certify spectral responsivity.^[ref]

To infer spectral responsivity at a selected wavelength, measure the radiant flux actually reaching the test photodiode, rather than just the reference arm. A reference-arm measurement calibration must include the beam splitter ratio and path losses. With signed readouts in one fixed orientation, subtract a beam-blocked baseline and usefor direct electric current readout or a calibrated linear transimpedance amplifier, respectively. Repeating at several incident powers tests whether this ratio is constant. Blocking only the source estimates ambient light plus electronic baseline; covering the detector excludes ambient light too. Their difference distinguishes optical background from dark current and readout offset. For a broad optical spectrum, the ratio is a spectrum-weighted response, not generally $R$ at one wavelength.