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A voltage multiplier made of rectifiers and capacitors that transfers electric charge from an alternating supply into a series stack of stored electric potential differences. The rectifiers allow charge transfer during selected portions of the cycle; the capacitors retain charge between transfers. The supply continually replaces charge removed by the load. Multiplication of voltage does not multiply energy: the input must supply the output power and losses.

In his 1951 lecture The Artificial Production of Fast Particles, Ernest Thomas Sinton Walton describes the fourfold multiplier's series storage capacitors and alternating transfer capacitor in Fig. 2, p. 190. This is a mechanism for producing a nearly steady accelerating potential from an alternating input.^[ref]

Define a stage as one storage increment of $2V_p$, with $V_p$ the alternating input's peak electric potential difference relative to its reference, not its peak-to-peak value. Assume ideal rectifiers, no load, no leakage or stray capacitance, a stiff alternating source and enough cycles to reach equilibrium. For $N$ such stages,The factor two follows from adding the input's next opposite excursion to an already stored peak. The subsequent series increments add. This ideal argument concerns equilibrium voltage; it supplies neither the charge-up time nor the useful load electric current.

To isolate the effect of drawing charge, consider the following original reservoir model. Each storage stage has equal capacitance $C$; a constant output electric current $I$ removes $\Delta q=I/f$ between replenishment events of frequency $f$. Every series reservoir loses this charge, hence its voltage falls by $a=I/(fC)$ per interval. Suppose replenishment removes a fixed fraction $1-\lambda$ of each reservoir's accumulated deficit, with $0\le \lambda <1$. This is an explicit phenomenological pump rule, not a solution of a particular multiplier's rectifier network. If $d_j$ is the deficit immediately after event $j$, thenHere $d_{+}$ and $d_{-}$ are the equilibrium deficits just after and just before replenishment. Linear discharge between events gives the average output and peak-to-peak variationFor chosen dimensionless settings $N=3$, $a/V_p=0.020$ and $\lambda =1/2$, the output varies between $5.88V_p$ and $5.94V_p$, averaging $5.91V_p$, instead of the unloaded $6V_p$. Starting with zero deficit, the recurrence approaches equilibrium geometrically with factor $\lambda$. Taking $I\to 0$ recovers the ideal ladder; increasing $C$ or $f$ reduces droop in this model. As $\lambda \to 1$, effective replenishment fails and the small-deficit assumption fails too.

Real multiplier stages have unequal charge-transfer burdens; their mean deficits need not be equal and depend on topology, input impedance and parasitic capacitance. Thus this reservoir model explains droop and voltage variation without claiming their exact dependence on stage count for a real Cockcroft-Walton generator. In particular, adding more stages cannot be judged solely by the unloaded $2N{V}_p$. For a beam sampling a slowly varying output, the electrostatic acceleration energy gain follows that instantaneous potential drop, so voltage variation also produces a spread of beam kinetic energies.

A diode is a two-terminal component with an asymmetric current-voltage characteristic: reversing its terminal electric potential difference need not reverse its electric current with the same magnitude. Name the terminals diode anode $A$ and diode cathode $K$. Define $V_D={\varphi }_A-{\varphi }_K$ using their electric potentials, and take positive electric current $I_D$ from $A$ to $K$. The bar in the diode symbol marks $K$. Positive $V_D$ is the forward direction; negative $V_D$ is the reverse direction.^[ref]

The asymmetry makes a rectifier possible: a sinusoid can drive electric current through a resistor on one half of each cycle. An ordinary resistor obeying Ohm's law instead reverses its electric current when the applied electric potential difference reverses. The useful idea is directional selection; the ideal diode model captures it, while the Shockley diode equation describes a smooth exponential function rather than an abrupt switch.

For the supplied linear meter model let ${\omega }_0^2=C/I$ and $2\gamma ={\Phi }_0^2/(RI)$. If the applied step is $i_s$, the settled deflection is ${\theta }_{\infty }={\Phi }_0{i}_s/C$. An underdamped response overshoots and rings about that value; a critically damped response reaches it monotonically without ringing; an overdamped response is also monotone from these initial conditions but has a slower final approach. The original plot uses the common final value and natural-frequency time scale.

The critical step response of a galvanometer has $\gamma ={\omega }_0$ and repeated homogeneous root $-{\omega }_0$. Apply $\theta (0)=\dot{\theta }(0)=0$ to the general constant-force solution to obtainIts derivative is ${\theta }_{\infty }{\omega }_0^{2}t{e}^{-{\omega }_{0}t}$, with the sign of the step. It is the fastest nonoscillatory response within this second-order family at fixed ${\omega }_0$, avoiding a needle that repeatedly crosses the reading or an unnecessarily slow overdamped settling.

For sinusoidal drive choose $e^{i\omega t}$ phasors. Writing $F_0={\Phi }_0{i}_0/I$, the response amplitude isFor positive $F_0$, choose the continuous lag branch from zero to $-\pi$ as frequency increases; an ordinary single-argument arctangent without the quadrant would give the wrong high-frequency phase.

For the critical galvanometer transfer function, the denominator factors as $({\omega }_0+i\omega {)}^2$. ThusFor $\omega \ll {\omega }_0$, the meter follows the quasi-static deflection ${\Phi }_0{i}_0/C$ with a small lag $\varphi \simeq -2\omega /{\omega }_0$. For $\omega \gg {\omega }_0$, its amplitude is $F_0/{\omega }^2$ and its lag approaches $-\pi$.

The stated components give $\gamma =12.5$ s⁻¹ and, if critical, ${\omega }_0=12.5$ rad/s, or $f_0=1.99$ Hz. At 50 Hz, $\omega =314.16$ rad/s andThe needle barely responds compared with its DC sensitivity and its mean deflection for a pure sinusoid is zero in this linear model. It is unsuitable for directly measuring 50-Hz amplitude or RMS current without an additional rectifier or another readout method. The calculation treats ${\Phi }_0$ as the stipulated torque/back-emf coefficient of the given equation; it does not infer an unprovided winding geometry.

In a p-n junction, diffusion and electron-hole recombination near the contact leave a depletion region of ionized dopants. Its built-in electric field opposes further majority-carrier diffusion. At thermodynamic equilibrium drift and diffusion currents cancel. Forward bias lowers the electrostatic barrier and injects electrons and electron holes across the junction; reverse bias raises the barrier and suppresses that majority-carrier injection. For an ideal diode below breakdown,Thus a modest forward voltage gives a rapidly increasing current, whereas reverse current remains approximately $-I_s$. This asymmetric current-voltage relation makes the junction a rectifier. Real leakage, electron-hole recombination and breakdown limit the ideal expression.

A rectifier converts an alternating input into a unidirectional output. Diode orientation selects which direction of electric current is transmitted.^[ref]