= Half-wave rectifier
{title2=$v_L(t)=\max(0,v_S(t))$}
= Half-wave rectification
{synonym}
A <half-wave rectifier> with one <diode> and a <resistor> load passes one half of each input cycle.<The Junction Diode by W. Marshall Leach, Jr.>[The Junction Diode, §1.12.1, printed p. 30: alternate-half-cycle conduction into a resistor load]{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.
\Image[/half-wave-rectifier-circuit-and-waveform.png]
{title=An ideal <half-wave rectifier> selects the positive half-cycle}
{description=Original schematic and calculated load <electric potential difference> for $v_S=5\sin(\omega t)\,\mathrm V$ and $R_L=1\,\mathrm{k\Omega}$. The <diode cathode> is the bar at $K$; positive load <electric current> flows downward through the <resistor>. Dashed orange: source <sinusoid>. Blue: load output, zero during the negative half-cycle. Green: its cycle-average <electric potential difference> $5/\pi\,\mathrm V$. No smoothing <capacitor> is present; these are ideal model predictions.}
The <Kirchhoff cycle law> gives $v_S=V_D+v_L$, and <Ohm's law> gives $v_L=R_Li_L$. 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. Thus
$$
v_L(t)=\max\{0,V_p\sin(\omega t)\},\qquad i_L(t)=\frac{v_L(t)}{R_L}.
$$
The 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>:
$$
\overline v_L=\frac1{2\pi}\int_0^\pi V_p\sin\theta\,d\theta=\frac{V_p}{\pi}=1.59155\,\mathrm V,\qquad
v_{L,\mathrm{rms}}=\left(\frac1{2\pi}\int_0^\pi V_p^2\sin^2\theta\,d\theta\right)^{1/2}=\frac{V_p}{2}=2.500\,\mathrm V.
$$
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.
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