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Define a signed offset $V_c$ so that the accelerating region adds kinetic energy $e(V-V_c)$. This convention concerns this gap; collector contact fields belong in the effective $V_r$. An Electron emitted with initial kinetic energy $E_0$ and undergoing $n$ identical excitation losses reaches the grid withEach loss excites a fresh chemical atom initially in its ground state. These are repeated payments of the same excitation cost. The index $n$ counts events along an Electron trajectory, not the ordinal number of an atomic excited state. The outgoing energy also does not locate the last collision: after an early collision an Electron can regain enough kinetic energy to cross the barrier.

At fixed $V$, let $g(E_0)$ be the normalized initial-energy probability density function within the ensemble of Electrons reaching the grid, and $w_n(E_0,V)$ the conditional probabilities for $n$ losses within that ensemble. Then the ideal collection law isTake ${\sum }_n{w}_n=1$ and exclude paths with insufficient kinetic energy for a loss at its actual collision position. The weights contain collision positions and rates, mean free paths and angular acceptance. This equation is a bookkeeping model, not a derivation of those weights. Multiplying all incident particles by one constant transmission is an additional approximation; real transmission and $I_0$ can vary with $V$.

One analytical extreme assumes $E_0=0$, one-dimensional motion and an immediate loss whenever the kinetic energy reaches $\Delta$. With $d=\Delta /e$, $x=V-V_c\ge 0$ and $0<V_r<d$, the number of losses and remaining kinetic energy areHere $\lfloor \cdot \rfloor$ is the floor function. Collection switches off for an interval of width $V_r$ after each reset, then recovers; the pattern repeats every $d$. For $d=4.9\mathrm{V}$, $V_c=1.2\mathrm{V}$ and $V_r=1.0\mathrm{V}$, an applied $V=11.4\mathrm{V}$ gives $n=2$, $r=0.4\mathrm{V}$ and rejection, whereas $V=12.2\mathrm{V}$ gives $n=2$, $r=1.2\mathrm{V}$ and collection. With $V_r=0$ the dips disappear in this idealization. With no collisions, there is only a single collection onset. Both limits expose the separate roles of loss and filtering.

To draw a smooth illustrative curve, extend this periodic rule mathematically to all $x$ and convolve it with a chosen normal distribution of voltage jitter, of standard deviation $\sigma$:Here $\Phi$ is the standard normal cumulative distribution function. Each summand is the smoothed indicator of the accepted interval $kd+V_r<x<(k+1)d$. The result satisfies $0\le C_{\sigma }\le 1$ and has period $d$. For the figure choose $\sigma =0.18\mathrm{V}$ and plot only well beyond the startup region. This is a chosen phenomenological curve, not a Boltzmann equation transport prediction or a fit to measured electric current. Its symmetric troughs center on $V_c+nd+V_r/2$, rather than the reset voltages $V_c+nd$. Finite excitation rates, several loss channels, space charge, initial angular and energy spreads, and real electrode geometry can change the trough shape and phase; a single convolution does not solve these effects.

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.

MIT's guide uses an illuminated vacuum photodiode, with collecting anode positive relative to the emitting cathode. A transimpedance amplifier gives $V_1=R_{F}I$; a multimeter reads its mean. The fluctuation path rejects below $100\mathrm{Hz}$, adds roughly tenfold voltage gain, then nominal hundredfold preamplification and a $1$–$50\mathrm{kHz}$ band-pass filter; the readout measures root mean square voltage. With diode bias off, a sinusoid at the test port calibrates $g(f)$. Restore bias, retain gain settings, vary lamp brightness, and repeat both readings. The guide warns about overload; it supplies a procedure, not measured results.^[ref]

The inference can be expressed without relying on nominal passband edges. Let $g(f)$ denote the calibrated voltage transfer after the $R_F$ conversion, so $K(f)=R_{F}g(f)$ and define $\mathcal{K}={\int }_0^{\infty }|K(f){|}^{2}df$. Under independent arrivals and an unchanged additive background,The slope of voltage variance against mean electric current can therefore estimate $q$. A free intercept accounts for a constant background; forcing the line through zero generally biases the slope. The root mean square readings must have their mean removed and then be squared. Subtracting two root mean square amplitudes does not subtract independent variances. Estimate the variance from successive records; averaging the waveforms first would suppress the fluctuation being measured.

Additional controls sharpen this interpretation. Block the light with detector bias, temperature, and gain unchanged; this preserves dark current and electronics while removing photocurrent. Repeat this dark comparison between illumination settings to check drift. If independent dark arrivals have mean magnitude $I_d$ and illuminated arrivals add $I_{\mathrm{ph}}$, their spectra add: $2q(I_d+I_{\mathrm{ph}})$. Subtracting the unchanged dark variance isolates $2q{I}_{\mathrm{ph}}\mathcal{K}$. This subtraction model assumes unchanged transfers and independence; switching off bias for gain calibration is a different control and need not preserve the dark background.

Vary measured $I_{\mathrm{ph}}$ rather than assuming lamp knob position is proportional to photocurrent. Check collection versus bias to identify a range where incomplete collection is negligible; a space charge limited tube can correlate transfers. Verify that the transimpedance amplifier remains linear and that its frequency response is unchanged over the illumination range. Changing signal bandwidth should change the excess variance through $\mathcal{K}$; compare inferred input spectra in several bands. Correlated fluctuations of the light source can add a contribution scaling approximately as $I_{\mathrm{ph}}^2$, while an electronics floor can hide the linear term at small electric currents. A fitted line over a short range alone does not establish independent transfers.

The gain calibration must include the same detector-to-readout path used for noise, or explicitly supply any omitted transfer. Detector capacitance, finite collection time, amplifier voltage and current noise, thermal electrical noise, pickup, clipping, and finite recording precision limit the inference. Current-dependent electrical impedance or heating can make a background current-dependent too. Since $\mathcal{K}$ contains squared gain, a small fractional error in gain produces approximately twice that fractional error in the predicted variance scale. Knowing a nominal gain and two cutoff settings is insufficient.

A space-charge-limited thermionic current occurs when the negative space charge of emitted Electrons creates an electric field that obstructs further transport across the gap. Raising the emitter temperature may then increase the surface supply without producing the corresponding collector electric current. The gap geometry and collecting electric potential difference matter even when the surface is stable.

To distinguish this from an emission-limited thermionic current, vary the collecting electric potential difference at fixed emitter temperature, then repeat at another temperature. A voltage-dependent transport region followed by a collection plateau separates a gap constraint from a supply constraint. A weak collector electric current alone cannot establish a high work function. Nor does simply choosing a very large collecting electric potential difference guarantee a valid zero-field measurement: the surface barrier may then change.

Release of Electrons from a heated material; one source of Electrons in an X-ray tube.

Heating a metal gives some Electrons enough energy to escape its surface. A heated diode cathode can therefore supply an electric current across a vacuum gap to a collecting diode anode. This makes temperature a control for the supply of Electrons. Conversely, measuring that supply at several temperatures lets us estimate the surface work function: the energy barrier measured from the Fermi level to the nearby vacuum level. A Richardson plot turns the strong exponential function dependence into a slope.

Three questions must be kept separate: how many Electrons cross the surface, how many traverse the gap, and how many the collecting diode anode records. The Richardson-Dushman law answers the first under specified surface assumptions. Space charge can obstruct the second, while incomplete collection can alter the third.