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For a zero-bias abrupt p-n junction, choose the p side at $x<0$ and the n side at $x>0$. Ionized acceptors and donors remain fixed while mobile Electrons and electron holes diffuse and establish a built-in electric field. In the depletion approximation, mobile carriers are neglected on $-x_p<x<x_n$ and the net charge density isCharge neutrality gives $N_A{x}_p=N_D{x}_n$. The quasi-neutral bulk can have a large mobile carrier density while its net charge density is nearly zero, because fixed dopant charge compensates it. With complete ionization and nondegenerate carriers, equilibrium also gives $np=n_i^2$; real carrier profiles vary smoothly rather than forming exact step functions at depletion edges.

The depletion approximation replaces the junction's transition region by a region without mobile carriers, leaving only fixed ionized dopants. For a uniformly doped abrupt p-n junction, $\rho =-e{N}_A$ on the p side, $\rho =+e{N}_D$ on the n side and $\rho \simeq 0$ outside. Charge neutrality gives $N_A{x}_p=N_D{x}_n$. Integrating Poisson equation gives the electric field and the electrostatic band bending.

A linearly graded p-n junction has net ionized doping $N_D-N_A=\alpha x$ near the metallurgical junction. In the depletion approximation, $\rho =e\alpha x$. With vanishing external field, neutrality places the depletion edges at $\pm w/2$ and integration of Poisson equation givesThe field points from n to p and has largest magnitude $e\alpha w^2/(8{\epsilon }_s)$ at $x=0$. The potential is cubic inside the layer.

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.

Take an abrupt equilibrium p-n junction with the p side at $x<0$ and the n side at $x>0$. In the quasi-neutral bulk, electron holes are the majority carriers on the p side and Electrons are the majority carriers on the n side. With ionized dopant densities $N_A,N_D$ and nondegenerate statistics,where $n_i$ is the intrinsic carrier concentration. The semiconductor mass-action law $np=n_i^2$ applies at equilibrium. Both mobile populations are small in the depletion region compared with the dopant densities.

The charge and carrier density of an abrupt p-n junction must distinguish mobile populations from net charge. Diffusion leaves negatively charged ionized acceptor impurities on the p side and positively charged ionized donor impurities on the n side. Under the depletion approximation,The neutral bulks have large carrier densities but nearly zero net charge density. The electric field points from the positive donor region toward the negative acceptor region, balancing diffusion at equilibrium.

The figure uses equal dopant densities only to show a simple symmetric sketch. Generally the more lightly doped side has the wider depleted interval. The carrier profiles are smooth, whereas the step-shaped net charge density is the ideal depletion approximation.

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.

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.

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 junction field-effect transistor has a conducting channel between source and drain, with gate regions forming reverse-biased p-n junctions to that channel. In an n-channel device, a more negative gate-source electric potential difference increases the reverse bias and widens the depletion region. The conducting cross-section shrinks, reducing the drain electric current. The gate ideally draws only a small leakage electric current, giving a high input resistance; the source-drain supply provides the controlled output power. Carrier motion in a p-channel device reverses the relevant polarities.

At small positive $V_{DS}$, the channel conducts approximately as a resistor. As the potential rises from source to drain, the local gate-channel reverse bias increases toward the drain. The depletion layers are therefore thicker there. For a symmetric n channel of half-thickness $a$, uniformly doped at density $N_D$, the one-sided depletion approximation giveswhere $w$ is the depletion width from each side and $V_{\mathrm{bi}}$ the built-in junction potential. The neutral channel closes locally when $w=a$. Define the positive geometrical depletion electric potential difference $U_P=e{N}_D{a}^2/(2{\epsilon }_s)$. The approximate drain-end pinch-off condition isMore negative gate bias makes this saturation occur at a smaller drain electric potential difference.

Pinch-off in a junction field-effect transistor does not mean that the drain electric current vanishes. Electrons arriving through the remaining neutral channel are swept across the high-field depleted region into the drain. Further increases in $V_{DS}$ mostly fall across that region, so the electric current changes only weakly: this is the saturation regime used for amplification. Channel shortening gives a finite output slope in real devices. Excessive reverse electric potential difference eventually causes breakdown, beyond the normal saturation model.

A sufficiently negative gate electric potential difference can instead deplete the entire channel from the source end and produce cutoff. In the simplified geometry its threshold is $V_{GS,\mathrm{off}}\simeq V_{\mathrm{bi}}-U_P<0$. A commonly used approximate transfer law in saturation isHere $I_{DSS}$ is the saturated drain electric current at zero gate-source bias. The square law is a useful device approximation, not an exact result for arbitrary channel profiles. It connects field control of depletion to a measurable transfer characteristic while keeping drain-end pinch-off distinct from complete gate cutoff.

Write ${\epsilon }_s={ϵ}_0ϵ$ and take $e>0$, $\alpha >0$, so the p side is $x<0$. In the depletion approximation, mobile Electrons and electron holes are negligible compared with the fixed ionized dopants. Their charge density isApplying Poisson equation givesOutside the depletion region, mobile carriers approximately neutralize the dopants, so $\rho \simeq 0$ and $V^{\prime \prime }\simeq 0$. Full dopant ionization alone does not remove the mobile-carrier charge: the result also requires the depletion approximation.

The electric potential difference is the negative integral of the field:ConsequentlyThe linearly graded junction depletion width has this cube-root dependence because of the linearly growing space charge: the charge on one side scales as $w^2$, the field as $w^2$, and its potential integral as $w^3$. The formula assumes $V_b-V_{\mathrm{ext}}>0$ and that the depletion approximation remains applicable.

The positive charge per unit interface area on the n side and negative charge on the p side areUse the signed p-side charge $Q=Q_p$ because $V_{\mathrm{ext}}$ is the p-side voltage relative to n. Differentiating the width relation givesUsing part (e), the positive junction capacitance per unit area isIt decreases with increasing reverse-bias magnitude as $(V_b+|V_{\mathrm{ext}}|{)}^{-1/3}$. Equivalently,so a linear $C^{-3}$ versus voltage plot diagnoses this ideal linear grading and can determine $\alpha$.

For this linearly graded junction capacitance, the charge convention matters: if “stored charge” means the positive magnitude $Q_n$, then $d{Q}_n/d{V}_{\mathrm{ext}}=-{\epsilon }_s/w$; the positive measured capacitance is $-d{Q}_n/d{V}_{\mathrm{ext}}=d{Q}_n/d{V}_R$ for $V_R=-V_{\mathrm{ext}}$. The total net depletion charge is zero and is not the stored terminal charge. At substantial forward bias, minority carrier storage and the failure of the sharp depletion approximation prevent extrapolation of the apparent divergence at $V_{\mathrm{ext}}=V_b$.

In an n-channel junction field-effect transistor, increasing drain voltage increases the local gate-channel reverse bias near the drain. The depletion region can meet there, ending the quasi-neutral channel before the drain. Carriers already in the channel are swept across the remaining high-field region, so drain-end pinch-off produces current saturation rather than zero current. A sufficiently negative gate voltage can instead deplete the whole channel and cut the device off. In the one-sided depletion approximation, a channel half-width $a$ and donor density $N_D$ give a depletion threshold voltage $U_P=e{N}_D{a}^2/(2{\epsilon }_s)$; with built-in junction voltage $V_{\mathrm{bi}}$, drain-end pinch-off occurs approximately at $V_{DS}=U_P-V_{\mathrm{bi}}+V_{GS}$.