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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.

A half-wave rectifier with one diode and a resistor load passes one half of each input cycle.^[ref]

For an original ideal example, let an ideal voltage source supply the sinusoid $v_S(t)=V_p\sin (\omega t)$, with $V_p=5\mathrm{V}$ and angular frequency $\omega >0$. Connect its positive reference terminal to the diode anode, the diode cathode to the upper terminal of a load resistor with electrical resistance $R_L=1\mathrm{k\Omega }$, and the lower terminal to the source return. Let $v_L$ be the upper-minus-lower load electric potential difference, and $i_L$ the downward load electric current. Assume the ideal diode model, zero source electrical resistance, and no filter capacitor, inductor or switching delay.

The Kirchhoff cycle law gives $v_S=V_D+v_L$, and Ohm's law gives $v_L=R_L{i}_L$. If $v_S>0$, the blocking assumption would give $V_D=v_S>0$, contradicting the ideal diode model, so the diode conducts with $V_D=0$. If $v_S<0$, conduction would give $i_L=v_S/R_L<0$, also forbidden, so the diode blocks. At the crossings both values vanish. ThusThe peak load electric current is $5\mathrm{mA}$. The output is unidirectional but still varies in time, reaching zero throughout every negative half-cycle. Over a period of a function $2\pi /\omega$, write the wave phase as $\theta =\omega t$. Its average value of a function and root mean square follow from two integrals:Consequently the average load power is $v_{L,\mathrm{rms}}^2/R_L=6.25\mathrm{mW}$, not ${\overline{v}}_L^2/R_L$. The ideal diode has zero loss; the resistor converts supplied energy into Joule heating. The same switching argument predicts $v_L=\min (0,v_S)$ when the diode is reversed. Directional selection changes the waveform and its average value of a function; it does not by itself produce a constant output.

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.

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.