Half-wave rectifier

The half-wave rectifier is the simplest Rectifier: a single Diode in series with the load. It passes the half-cycles of one polarity straight through and blocks the half-cycles of the other polarity entirely, so the output is a series of one-sided humps with flat gaps between them.

Ideal-diode behaviour

With an Ideal diode model the analysis is two states. When the input $v_s$ has the polarity that forward-biases the diode, the diode is a short and the full input appears across the load: $v_o=v_s$. When $v_s$ has the opposite polarity, the diode is an open circuit, no current flows, and the output is zero: $v_o=0$. The transfer characteristic is two straight pieces: a unity-slope line through the origin for the conducting polarity, flat at zero for the other.

Positive cycles pass, negative blocked — positive humps with gaps.

A note on the slide’s orientation: it draws the diode with its cathode toward the source (anode toward the load), so this particular circuit passes the negative half-cycles and clips the positive ones — $v_o=v_s$ when $v_s<0$ and $v_o=0$ otherwise. Flipping the diode the other way (the more common textbook convention) passes the positive half instead. The physics is identical; only which half survives depends on which way the diode points.

Ideal half-wave rectifier — circuit, output waveform, transfer characteristic (slide passes the negative half-cycle).

CVD behaviour

The real diode does not turn on until it has about 0.7 V across it, so with the Constant-voltage-drop model the output is

$v_o=\{\begin{array}{ll}0& v_s<V_D\\ v_s-V_D& v_s\ge V_D\end{array}$

with $V_D\approx 0.7$ V. The conduction threshold is shifted away from zero by $V_D$, and the surviving peak is reduced to $V_p-V_D$. The transfer characteristic still has unity slope when conducting, just offset by $V_D$.

CVD model: $v_o=0$ for $v_s<V_D$, $v_o=v_s-V_D$ for $v_s\ge V_D$; the diode drop lowers the peak.

Two practical numbers

The Peak inverse voltage is the worst-case reverse voltage the diode must survive. When the input is at its blocked-polarity peak (magnitude $V_p$) the diode is off, no current flows so there is no drop across the load, and the entire input swings across the diode: $\text{PIV}=V_p$. The diode’s reverse voltage rating must exceed this.

The DC component is the time-average of the lumpy output. For an ideal half-wave-rectified sinusoid of peak $V_p$ — half-cycles of a sine averaged over a full period — the average is $V_p/\pi \approx 0.318V_p$. The derivation and the full-wave counterpart are in DC value of a rectified waveform. That meagre, lumpy average is why a half-wave rectifier alone is a poor DC source and almost always gets a smoothing capacitor.