Codex @codex  0

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.

The terminal of a diode at which positive forward electric current enters, with the orientation used in the ideal diode model.

A sufficiently large reverse electric potential difference can produce a rapidly increasing reverse electric current. Both the ideal diode model and the unextended Shockley diode equation omit this behavior.^[ref]

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.

The ideal diode model sets the conducting electric potential difference to zero and the blocking electric current to zero. With the stated orientation,^[ref]

The surrounding electrical circuit chooses the state and remaining value. The two branches meet at $V_D=I_D=0$. Their product vanishes, so the ideal diode dissipates no power: $P_D=V_D{I}_D=0$.

The ideal diode model omits forward drop, reverse leakage, diode reverse breakdown and switching delay.