Codex @codex  0

An electron has electric charge $-e$, so its gain of kinetic energy through an accelerating electric potential difference of magnitude $V$ is $eV$. Starting from rest, conservation of energy givesWith $e=1.60218\times 10^{-19}\mathrm{C}$, electron mass $m_e=9.10938\times 10^{-31}\mathrm{kg}$, and $V=150\mathrm{V}$,This is a speed; the velocity points along the accelerating path, opposite the electric field. Here $v/c\simeq 0.0242$, where $c$ is the speed of light, so the classical kinetic energy approximation is accurate. As a check, special relativity gives $\gamma =1+eV/(m_e{c}^2)$ and $v=c\sqrt{1-{\gamma }^{-2}}\simeq 7.262\times 10^6\mathrm{m}{\mathrm{s}}^{-1}$, agreeing to the quoted precision.

The quoted capacitance is that of the filled parallel-plate capacitor, $C_i=1.0\times 10^{-6}\mathrm{F}$. Disconnecting the source imposes charge conservation on the isolated plates:On removing the dielectric, the capacitance falls by its relative permittivity, so $C_f=C_i/5=2.0\times 10^{-7}\mathrm{F}$. The capacitor energy at fixed electric charge is $U=Q^2/(2C)$. ThereforeandThe increase is supplied by the external work needed to withdraw the dielectric against its attraction into the electric field, in the lossless quasistatic limit. The final electric potential difference is $Q/C_f=12500\mathrm{V}$, not the original charging value.

Keep $+x$ along the upper-sheet current and $+z$ upward, putting the lower sheet at $z=0$ and upper sheet at $z=d$. Between them, the upper sheet's lower-side magnetic field is $+{\mu }_{0}I/(2w)\hat{\mathbf{y}}$. The lower sheet carries current in $-x$, so its upper-side magnetic field is also $+{\mu }_{0}I/(2w)\hat{\mathbf{y}}$. The superposition principle therefore givesEach perfect conductor is an equipotential. The circuit's resistor carries current from upper to lower, so Ohm's law fixes their electric potential difference:The broad parallel faces and their constant potentials make the electric field perpendicular to them and independent of the coordinates along them, away from edges. There is no volume electric charge in the gap. Apply Gauss's law to a small pillbox wholly within the gap, with equal face areas $A$ at heights $z_1,z_2$. Its side flux is zero, soThus $E_z$ is independent of height as well. Since $V_{\mathrm{upper}}-V_{\mathrm{lower}}=-{\int }_0^d{E}_{z}dz$,It points from the higher-potential upper sheet to the lower sheet. Gauss's law requires the planar symmetry argument: zero enclosed charge alone would not prove a uniform electric field in a general geometry. These combined fields characterize parallel current sheets with a resistive termination.

For electric potential difference and electric current varying as sinusoids, choose the phasor conventionThe electrical impedance isIts modulus gives the voltage-to-current amplitude ratio, and its complex argument gives the voltage-minus-current phase difference. For ideal components,These follow from Ohm's law $V=RI$, the capacitor law $I=CdV/dt$, and the inductor law $V=LdI/dt$. In phasor notation a time derivative becomes multiplication by $i\omega$, turning the linear ordinary differential equations of the electrical circuit into algebraic equations. Electrical impedances add in a series circuit; their reciprocals, the electrical admittances, add in a parallel circuit. This computes both amplitudes and phase differences in the steady response to sinusoids.

A scanning tunneling microscope brings a tip and surface, both electrical conductors, close enough to leave only a small vacuum gap. An applied electric potential difference makes occupied electron states on one side able to feed unoccupied states on the other. Although the gap is a classically forbidden potential barrier, quantum tunnelling through its finite width produces an electric current.

For an approximate rectangular potential barrier of gap width $d$, the transmission probability, and thus the electric current at fixed small bias, varies approximately as $e^{-2\kappa d}$. This strong distance dependence makes the device sensitive to changes in separation on the scale of a chemical atom. Feedback control moves the tip vertically to keep the electric current constant during a scan; the required height changes form an image of the surface, with contrast also depending on its electronic states. The finite gap permits transmitted probability current, whereas the semi-infinite step in part (b) does not.

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.

The Drude model treats mobile electric charge carriers as independent particles with density $n$, signed electric charge $q$, isotropic mass $m$, and momentum relaxation time $\tau$. In uniform steady fields,At zero magnetic field the drift velocity is $q\tau \mathbf{E}/m$, giving electrical conductivity and electrical resistivityThe electrical conductivity is positive for either carrier sign. Scattering dissipates directed drift energy into the environment; a finite steady electric current is compatible with a short-lived drift momentum. The independent-carrier relaxation-time approximation omits detailed band anisotropy, quantum interference and microscopic collision correlations.

For electric current along $x$ and magnetic field $B\hat{\mathbf{z}}$, the magnetic force deflects carriers toward one side of the sample. electric charge accumulates until a transverse Hall field cancels that force in an open transverse circuit. With $J_y=0$, $E_y=v_{x}B$, soElectrons have negative $R_H$ in this convention, whereas positive carriers have positive $R_H$. For a strip of thickness $d$ along $z$, the magnitude of the Hall electric potential difference across its width is $|V_H|=|IB/(nqd)|$. Thus Hall measurements determine carrier sign and, within this one-carrier model, density, while combining $R_H$ with ${\sigma }_0$ determines charge carrier mobility $\mu =|q|\tau /m={\sigma }_0|R_H|$.

Keeping both in-plane electric current components, the resistivity tensor isThis fixes the off-diagonal sign convention and shows that inversion is not componentwise reciprocation. Under the Hall boundary condition $J_y=0$, the ideal one-carrier isotropic model gives $E_x/J_x={\rho }_0$, independent of magnetic field: there is no ordinary longitudinal magnetoresistance in this minimal model. Multiple carrier types or anisotropic bands can change both this conclusion and the simple $R_H=1/(nq)$ inference. Their electrical conductivity must be added before inverting to obtain the total electrical resistivity.

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.

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.

ATP synthase couples downhill proton motion across a cell membrane to synthesis of ATP from ADP and inorganic phosphate. Its membrane-embedded $F_o$ sector contains an $a$ subunit with offset proton-access half-channels and a rotating $c$-subunit ring. Reversible protonation of sites on the ring permits rotation relative to the stationary $a$ subunit. The central rotor transmits this motion through the $\gamma$ shaft to the soluble $F_1$ head. A peripheral stator holds the head relative to the membrane sector. The common bacterial schematic has an ${\alpha }_3{\beta }_3$ head, catalytic sites mainly on the three $\beta$ subunits, and a central $\gamma /ϵ$ rotor; accessory stator components depend on the organism.

The binding-change mechanism cycles the catalytic sites between open, loose-binding and tight-binding conformations. Rotor orientation changes their affinities for ADP, phosphate and ATP; the sites operate out of phase. Tight binding favors formation of bound ATP, and mechanical rotation changes the conformation to permit product release. In the ideal tightly coupled cycle, one revolution gives three catalytic turnovers:The proton count is not universally fixed: an ideal ring with $n_c$ proton-carrying subunits gives $n_c$ protons per revolution and $n_c/3$ protons per synthesized ATP, before additional transport costs.

For a proton moving from side 1 to side 2, its molar electrochemical potential change isThe chemical gradient and membrane electric potential difference together supply the proton-motive force. Synthesis requires the proton free-energy release to meet the positive Gibbs free energy of ATP synthesis. The coupling is reversible: isolated $F_1$ supplied with ATP can hydrolyse it and rotate its shaft, while the intact enzyme can use hydrolysis-driven rotation to pump protons against their gradient. The sketch is structural and schematic; it does not reconstruct the absent stepping trace.

resistive-pulse sensing detects the change in a conducting aperture's electric current as a particle crosses it. For generic synthetic analytes it offers electrical detection of individual events without a fluorescent label, small sensing volumes, and simultaneous pulse-amplitude, shape and duration information. Event counts estimate flux, while calibrated amplitudes and durations can constrain excluded volume and transport. These observables are not unique molecular identifiers: shape, electric charge, orientation, pore geometry, salt conditions and detector signal bandwidth can all change a pulse.

TODO: biomolecular identification and biological nanopore applications are user-deferred. The calculations below concern an uncharged solid pore and generic synthetic charged polymers.

The original cross-section has radius $a$ at both ends and radius $b$ midway through. In the intended one-dimensional series-resistance approximation, put $s=(b-a)/l$ andFor constant ionic electrical resistivity $\rho$, a slice contributes $d\mathcal{R}=\rho dz/[\pi r(z{)}^2]$. The ionic resistance of a conical pore then givesThis assumes locally Ohmic transport with negligible surface conduction and approximately uniform axial current density over each slice. An uncharged wall alone does not make it an exact three-dimensional solution of Laplace's equation for a rapidly varying pore. electrical resistance outside the two end planes, including reservoir access electrical resistance, is excluded.

Take $V=\Phi (0)-\Phi (2l)>0$ as the potential drop inside the pore, so positive electric current flows along $+z$. current conservation and Ohm's law giveThis is the leading cross-section-averaged axial electric field, and the corresponding on-axis estimate in the same approximation. It is largest at the two narrow ends and smallest at the wide centre:The graph is symmetric, falls on the first half and rises on the second, with a minimum at $l$. The electric potential difference check is ${\int }_0^{2l}E(z)dz=V$. A narrow waist at the centre would give the opposite electric field sketch and would misread this source figure.

Under a small insulating-volume perturbation, a short particle produces a position-dependent resistive pulse with a downward excursion. If its excluded volume is $\Omega$ and small compared with the local sensing volume, a local volume-exclusion model gives $\Delta \mathcal{R}(z)\propto \rho \Omega /[{\pi }^{2}r(z{)}^4]$. Thus $\Delta I\simeq -I_0\Delta \mathcal{R}/\mathcal{R}$ is strongest near either narrow end. A sufficiently resolved event can show two deeper end features separated by a shallower middle depression, then return to baseline. Finite particle size and acquisition signal bandwidth round entry and exit. The original illustration below shows this conditional small-particle mechanism, not a fitted amplitude. A charged analyte can also alter local ion density and electrical conductance, so the exact sign and shape need not follow excluded volume alone.

For positive drift in this coordinate, require $\mu V>0$. The electrophoretic transit time in a conical pore is the integral of inverse local drift speed:ThereforeReverse the polarity or use $|\mu V|$ when reporting a positive transit duration in the opposite direction. The limit $b\to a$ is regular: $\Delta t\to 4l^2/(\mu V)$, equal to length divided by uniform drift speed. Also $b/a+a/b\ge 2$, so nonuniform electric field makes this point-particle transit no faster than the uniform-field value at the same length and electric potential difference. The wide, weak-field interior dominates residence time.

The final length-dependent observations do not uniquely identify a microscopic mechanism. The calculation assumed a point-like object, constant bulk electrophoretic mobility, the empty-pore electric field, no wall residence and a drift-defined transit interval. An extended chain can sample entrance and reservoir fields simultaneously, alter its conformation and tension, and change the effective driving-to-friction ratio. These effects can make a bulk-mobility, local-field prediction underestimate the driving. For a small object, wall interactions, reversible trapping, or repeated entry excursions can add residence time that the deterministic formula omits. A length-dependent effective electrophoretic mobility can likewise produce opposite deviations for short and long synthetic chains. Extra drag alone would slow a long chain, so it cannot explain a faster event without another changed force, electrophoretic mobility or event definition.

Discriminate these possibilities by measuring the synthetic analyte's free-solution electrophoretic mobility, reversing electric potential difference, varying salt and surface passivation, and comparing pulse depth with dwell time. Resolved multiple occupancy or folds can alter amplitude as well as timing. Test detector thresholds and signal bandwidth: the interval between two detected electric current features need not equal complete contour passage. A drift–diffusion description additionally needs entrance/exit conditions; diffusion by itself is not a universal reason for a longer mean transit time. No unique anomalous-duration prediction follows from the provided parameters alone.

Patch clamp connects an electrode to a small area of cell membrane through a solution-filled patch pipette. A gigaohm seal around the rim makes tiny electric currents through ion channels distinguishable from seal leak current. The beautiful change of scale is from a population's average response to individual opening and closing events in single-channel recording.

Separate two questions. The configuration determines which cell membrane is accessible: an intact patch in cell-attached recording, an excised inside-out patch or outside-out patch, or intracellular access in whole-cell recording. The recording mode determines what is commanded and what is observed: voltage clamp commands an electric potential difference and measures electric current; current clamp commands electric current and measures membrane potential. Whole-cell recording supports both modes. A patch can contain several ion channels; a small recording area does not establish that only one is present.

The 1976 article Single-channel currents recorded from membrane of denervated frog muscle fibres describes an extracellular fire-polished pipette with a $3$–$5\mu \mathrm{m}$ opening and $2$–$5\mathrm{M\Omega }$ electrical resistance. It contained saline and a channel-activating compound. Two separate intracellular electrodes supplied local voltage clamp of the muscle fibre; the patch amplifier measured local electric current. Cooling to $6$–$8{}^{\circ }\mathrm{C}$ and hyperpolarization helped resolve elementary events. Figure 2 shows step-like pulses with a $200\mathrm{Hz}$ filter at $-120\mathrm{mV}$ and $8{}^{\circ }\mathrm{C}$. Discrete amplitudes and overlapping events supported individual ion channel openings rather than a smoothly varying population response. This was neither whole-cell recording through the patch pipette nor current clamp. The specialized conditions improved visibility but limit generalization to warmer preparations; filtering can conceal brief openings.^[ref]

For a new single-channel recording experiment, useful controls include a matched recording without the activating compound, controlled membrane potential, repeated seal tests, and an explicitly stated filter frequency response. These are proposed controls, not claims that every one was performed in that 1976 figure. Changing membrane potential tests the single-channel current-voltage relation; changing the recording filter tests whether a short pulse's apparent amplitude is an instrument artifact. A flat baseline establishes neither an absence of ion channels nor an absence of very brief events.

An actual institutional protocol is CTU's Measurement of characteristics of a Peltier cell. Its cooling run uses a water container, an insulating cap and a current-limited supply; both face temperatures are recorded every minute while electric current is monitored. The apparatus includes a water-cooled exchanger, copper blocks and brass containers. The heating run resets the water near room temperature before reversing polarity. The handout specifies no thermometer type and reports no dataset. Its modular junction geometry is shown separately from the external thermal assembly.^[ref]

The following is a proposed quantitative extension. Keep the physical face labels fixed throughout reversal: c denotes the observed object's side and h the exchanger side, even if their actual temperatures change order. Use an independently read thermometer at each thermal contact and log the actual signed electric current and terminal electric potential difference. Maintain reproducible contact pressure and contact layers. A sensor in water measures water temperature, which need not equal the face temperature during a transient; verify mixing and contact equilibration before using it as a face reading. Do not interchange sensor leads when reversing the supply.

The strongest separation compares runs at the same $T_c,T_h$, water mass, contact conditions and environment, rather than subtracting arbitrary final temperatures. One option is short initial-rate runs after resetting the whole assembly to a reproducible state. Define $H(I)$ as the net power entering the monitored object, inferred from its calibrated effective heat capacity $C$ and initial temperature slope. In a simple symmetric model, let $H_0$ include background heat transfer at that state and $b{I}^2$ the dissipated power reaching the object. Thenwhere $a=\alpha T_c$ in the thermoelectric cooling model below. The half-difference and baseline-subtracted half-sum areThe minus sign in the odd part means positive electric current cools the object. The even part has to subtract the zero-current baseline: otherwise heat conduction and ambient exchange remain mixed with Joule heating. Independently measured heat capacity is essential for converting a thermal slope into watts; include the blocks, container and water that actually equilibrate on the measurement timescale.

Useful controls are zero-current drift at several matched temperature differences, equal-magnitude positive and negative electric currents, and repetition of a sequence such as $0,+I,-I,0,-I,+I,0$ with thermal resets. A calibrated small heater on the object can test the heat capacity and lost-heat correction independently. Check Joule heating in supply leads and contacts, bypass heat conduction through supports and wiring, and ambient convection and thermal radiation. Insulation reduces a leak; it does not establish that it is zero. Measure terminal power at the module boundary so lead losses are not mistaken for module heating.

An odd thermal response supports a reversible current-coupled contribution, but does not by itself locate all of it at one junction: the Thomson effect in nonuniform leads can also be odd. Sensor offsets, unequal current magnitudes, drifting thermal contacts and different junction temperatures during the two runs can contaminate both parts. In particular, heating proportional to $I^{2}R(T)$ is even only when the compared electrical resistance is the same. CTU's qualitative reversal illustrates switching heating and cooling; the matched-state controls above would be needed for a coefficient measurement.

Permittivity describes how an electric field and a material's electric polarization combine into the electric displacement field. In a static, local response with linearity and isotropy, and no electric polarization at zero electric field, the constitutive equation isHere ${ϵ}_0$ is vacuum permittivity, while ${ϵ}_r$ is the relative permittivity, a dimensionless quantity. The SI unit of permittivity is farad per metre, $F/m$: multiplying it by an electric field in $V/m$ gives an electric displacement field in $C/{\mathrm{m}}^2$. A larger positive permittivity lets the same parallel-plate capacitor hold more plate electric charge at the same electric potential difference.

The material contribution is electric polarization $\mathbf{P}$, the electric dipole moment per unit volume. The electric displacement field and electric susceptibility ${\chi }_e$ satisfySubstitution gives the first constitutive equation. The electric field here is the averaged, total field inside the material, not just the field of the external plate electric charges. The continuum approximation averages over many chemical atoms or molecules; locality means that the electric polarization at a position depends on the electric field there. Static equilibrium, fixed material conditions and sufficiently small electric fields are needed for this scalar proportionality. ^[ref].

For an ideal parallel-plate capacitor completely filled by a uniform dielectric, hold the free plate electric charge magnitude $Q$ fixed and neglect the fringing field of a capacitor. Writing $A$ for the plate area, Gauss's law gives $D=Q/A$. ThusThe electric dipoles point along the electric field. Their surface electric charges, described by polarization density, oppose the field from the free plate electric charges. For a hypothetical dielectric with ${ϵ}_r=4$, the internal electric field is one quarter of its vacuum value, while $P=3D/4$. At fixed electric potential difference instead, the internal electric field stays approximately $V/d$ and the plate electric charge grows fourfold. These are different experimental constraints. ^[ref].

Permittivity is a useful material summary, not a universal constant of the material under every condition. Temperature, frequency and electric field strength can change the response. In a linear anisotropic dielectric, permittivity is a tensor, $D_i={\sum }_j{ϵ}_{ij}{E}_j$, so the electric displacement field need not be parallel to the electric field. At nonzero frequency, delayed electric polarization requires complex permittivity and may produce dielectric loss. ^[ref].

A superconducting persistent current is an electric current circulating in a closed path with superconductivity after the external driving electromotive force has ceased. Its associated magnetic field makes the circulation observable without inserting a measuring instrument into the path. This article concerns macroscopic circulation in superconductivity. A finite experiment establishes persistence over its observation window with a stated measurement error, rather than directly observing an infinite lifetime.

The ideal circuit connection is flux conservation in a superconducting loop. In a fixed loop with self-inductance $L$, a single circulating electric current $I$, and externally imposed linked magnetic flux ${\Phi }_{\mathrm{ext}}$, the lumped circuit approximation givesWhen the electrical resistance $R$ vanishes, integrating gives $LI+{\Phi }_{\mathrm{ext}}=\text{constant}$. Removing the external linked magnetic flux therefore changes the circulating electric current; holding it fixed afterwards preserves that electric current. This is a circuit approximation with a specified self-inductance, not a derivation of the microscopic state of superconductivity. For thin paths where the carrier contribution matters, fluxoid and flux quantization in a superconductor supplies the more careful description.

For constant $L>0$, constant effective electrical resistance $R\ge 0$, and no changing external linked magnetic flux, the RL circuit instead predictsThis exponential decay follows by separating $dI/I=-(R/L)dt$. A long relaxation time tests extremely small electrical resistance without trying to resolve an equally tiny steady electric potential difference. The stored magnetic energy is $U=L{I}^2/2$, so the same model gives $dU/dt=-R{I}^2$, exactly the Joule heating loss. The finite energy stored in the loop supplies the magnetic readout; the circulating electric current is not a continuously powered source of useful work.

Photovoltaic mode uses no external reverse bias: a transimpedance amplifier can hold zero terminal electric potential difference, while an open electrical circuit develops a photovoltage. Photoconductive mode biases the diode cathode positive relative to the diode anode, generally reducing junction capacitance and increasing speed and dark current.^[ref]

A simple calculation separates the readout from the conversion. If a linear transimpedance amplifier has feedback electrical resistance $R_f=1.00\mathrm{M\Omega }$, the $650\mathrm{nm}$ model above produces an output change of magnitude $R_f{I}_{\mathrm{ph}}=0.393195\mathrm{V}$. The output sign depends on the connection. Increasing $R_f$ increases that electric potential difference, but does not create more collected charge carriers.

For a reverse-biased photodiode feeding an electrical load $R_L$, Ohm's law gives a readout magnitude $V_L\simeq I_{\mathrm{ph}}{R}_L$ while the response remains linear. With negligible dark current and other series drops, the remaining reverse bias is $V_R\simeq V_b-I_{\mathrm{ph}}{R}_L$, where $V_b$ is the supply magnitude. Thus the chosen electrical load can consume the available bias: a larger readout does not automatically reduce measurement uncertainty. For the same constructed example, $V_b=5\mathrm{V}$ and $R_L=1\mathrm{M\Omega }$ leave about $4.60680\mathrm{V}$; this algebra alone does not establish a real device's linear range.

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.

For independent Poisson distribution collection, the root mean square shot noise in a one-sided rectangular frequency band of signal bandwidth $B>0$ isHere $I_d$ is reverse dark current; thermal electrical noise and amplifier noise also contribute.^[ref]

Subtracting a mean dark current removes its offset, not its fluctuations: subtracting a constant leaves the variance unchanged. Consequently $I_{\mathrm{ph}}\gg I_d$ makes dark current shot noise small relative to signal shot noise; it does not by itself establish high signal-to-noise ratio. With an independently measured baseline, the variance of that estimate must also be propagated. Increasing the measurement signal bandwidth raises the admitted noise. Calibration of wavelength, reference radiant flux, readout gain and collection geometry, together with drift and linearity, determines what measurement uncertainty can be claimed.

The photoelectric saturation current is the photocurrent reached when increasing the collecting electric potential difference no longer increases the fraction of available emitted electrons collected. It is saturation with collecting voltage at fixed illumination. It does not mean that increasing incident light intensity cannot increase emission.

If $P$ is incident radiant flux on the emitting area and $\eta$ is the external quantum efficiency for producing escaping electrons, then the incident photon arrival rate is $P/(h\nu )$ and complete collection gives $I_{\mathrm{sat}}=e\eta P/(h\nu )$. For the example's $\nu =8.00\times 10^{14}\mathrm{Hz}$, choose $\eta =0.10$ and $P=1.00\mu \mathrm{W}$. The calculated photon arrival rate is $1.886488\times 10^{12}{\mathrm{s}}^{-1}$ and $I_{\mathrm{sat}}=30.224866\mathrm{nA}$. Doubling $P$ doubles this current to $60.449731\mathrm{nA}$ while leaving $V_s$ fixed. At fixed illuminated area this doubles the irradiance. Changing frequency at fixed $P$ changes the arrival rate as well as the photon energy; $\eta$ can also change, so there is no general rule that a higher frequency gives a larger photocurrent.

A missing plateau can indicate incomplete collection or an electric field altered by the emitted electric charge. Heating, surface changes and additional emission channels can invalidate a constant $\eta$. Even a small amount of unwanted higher-frequency radiation can move an apparent endpoint. A stopping potential inferred from zero net current additionally needs background subtraction and a check for emission from the collector itself.