B4 · Diodes

Bridge Rectifier and Smoothing: AC into DC

Link four diodes and you can fold even the negative half-cycles of AC upward. In this unit you enlarge the capacitor attached to them and see how a bumpy pulsation is smoothed into steady DC.

Enlarge the capacitor to fill the valleys

The faint curve is the peaks rectified by the bridge, and the bold gold curve is the output after the smoothing capacitor. Enlarge the capacitor and the valleys between peaks fill in, flattening the output more and more. Use the slider to grow it until the valleys almost vanish.

Smoothing capacitor CC = 6

Output ripple ΔVr

ΔVr ≈ Vpeak

There is almost no capacitor. The output follows the rectified peaks as they are, dropping all the way to 0 in the valleys.

Large ripple

The bridge gathers both half-cycles

Four diodes linked in a diamond form a bridge. When the input is positive, two opposite diodes turn on and drive current through the load in one direction; when the input is negative, the other two turn on. Either way the current through the load flows in the same direction, so the negative half-cycles are not thrown away but folded upward. This is full-wave rectification.

Full-wave: shallow, frequent valleys

Unlike half-wave rectification, which passes only one half-cycle, full-wave rectification turns both half-cycles into peaks. So the peaks line up densely at twice the input frequency, and with shorter valleys between them, the same capacitor can smooth the output much better. Whichever path the current takes, though, it passes through two diodes, so the peak height is lower than the input peak by two diode drops.

The capacitor fills the valleys

A capacitor is connected across the output. As each peak rises, the capacitor charges quickly to the peak; once the peak passes and the input falls, it discharges slowly into the load and holds the voltage up, lasting until the next peak refills it. The small swing left in between is the ripple: the larger the capacitor, the slower the discharge and the smaller the ripple, while a larger load current makes the ripple bigger. Combine full-wave rectification with a large capacitor and you get nearly flat DC.

ObserveVout = |Vin| - 2VD

The full-wave output is the input magnitude minus two diode drops.

Choosefripple = ? fin

Being full-wave, the ripple frequency is twice the input.

Fill inΔVr ≈ Iload / (2 fin ?)

Ripple is inversely proportional to the capacitor.

On your ownΔVr ≈ ?

Large load current and a small capacitor make the ripple grow.

Back to the first screen

With a small capacitor the output plunged all the way to 0 in the valleys, following the peaks. As you enlarged the capacitor, the charge filled in at each peak held up through the valleys and the output grew flatter and flatter. Beneath that flatness lie two things together: the bridge folding even the negative half-cycles into peaks, making the valleys short and frequent, and the capacitor filling those valleys. The rectification of B1 grew to use both half-cycles, and with smoothing added, the front end of a power supply was complete.

Bridge rectification uses four diodes to send both half-cycles of AC through the load in the same direction, folding the negative half-cycles upward (full-wave rectification). The output is a train of positive peaks at twice the input frequency. A smoothing capacitor charges at the peaks and discharges in the valleys, filling them in to give nearly flat DC with only a small ripple, whose size is ΔVr ≈ Iload / (2 fin C).

What comes next

Up to now the diode has set the direction of flow, cut it, or fixed it. In the next set you go one step further and meet a device that controls a much larger current with a small input current: the transistor. Its first unit starts from the principle by which a BJT governs a large collector current with a small base current.