Ripple voltage

Ripple voltage $V_r$ is the peak-to-peak variation left on the output of a Peak rectifier — the small sawtooth wobble that remains after the smoothing capacitor has averaged the rectified waveform. It is the residual AC riding on the DC, and minimising it is the main design objective of the smoothing stage.

Where the formula comes from

Between conduction bursts the diode is off and the load draws its current $I_L$ entirely from the capacitor. A capacitor discharged by a roughly constant current droops linearly: $i=Cdv/dt$, so $dv=(i/C)dt$. Over a discharge interval $\Delta t$ the voltage falls by

$V_r\approx \frac{I_L\Delta t}{C}$

$I_L$ is the DC load current, $C$ the smoothing capacitance, and $\Delta t$ the time the capacitor is left to discharge — essentially one full period for a half-wave rectifier, or half a period for a full-wave (because the next peak comes twice as often). With line frequency $f$, the period is $1/f$, so:

$V_r\approx \frac{I_L}{fC}\text{(half-wave, one cycle between peaks)}$

$V_r\approx \frac{I_L}{2fC}\text{(full-wave, both half-cycles refill the cap)}$

The factor of 2 for the full-wave case is its key advantage: peaks arrive twice as often, the capacitor discharges for only half as long, so the ripple is half (or, for a given ripple target, you need only half the capacitance).

Reading the formula and typical values

Ripple shrinks with larger $C$, higher $f$, or smaller load current $I_L$. You cannot change the line frequency (50/60 Hz), and $I_L$ is set by the load, so the design knob is the capacitor. A typical small linear DC supply uses a few thousand microfarads to keep the ripple down to a few hundred millivolts.

Quick example. A full-wave supply delivering $I_L=0.5$ A at 60 Hz with $C=4700\mu$F:

$V_r\approx \frac{0.5}{2\times 60\times 4700\times 10^{-6}}=\frac{0.5}{0.564}\approx 0.89\text{V peak-to-peak}$

Halve the capacitor and the ripple doubles; double the load current and it doubles too. The trade-off does not come free — a smaller ripple means a shorter, sharper diode conduction burst and therefore a larger peak diode current, which is why Conduction angle and ripple are designed together. The remaining ripple is what a downstream Zener voltage regulator or active regulator must reject; see DC value of a rectified waveform for why the smoothed output sits near $V_p$, not the raw average.

Conduction angle $\omega \Delta t\approx \sqrt{2V_r/V_p}$; average/peak diode currents far exceed $I_L$ because charge is replenished in a short burst.